https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Clapatron7568&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-25T15:21:15Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=136538 2015 AMC 8 Problems/Problem 25 2020-11-04T12:23:37Z <p>Clapatron7568: /* Solution 3 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown:<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the &lt;math&gt;4&lt;/math&gt; big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by &lt;math&gt;\mathrm{AA}&lt;/math&gt; similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be &lt;math&gt;x&lt;/math&gt;; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;. Note that the other solution we got, namely, &lt;math&gt;x=\dfrac{5+\sqrt 5} 2&lt;/math&gt;, is the length of the segment &lt;math&gt;5-x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;5-x&lt;/math&gt; together sum to &lt;math&gt;5&lt;/math&gt;, the side of the length of the large square, and similarly, the sum of the solutions is &lt;math&gt;5&lt;/math&gt;. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be &lt;math&gt;x&lt;/math&gt;, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> Thus the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> Don't use this in a contest scenario, only use when practicing math skills :)<br /> <br /> ===Solution 2(Contest Solution)=== <br /> <br /> We draw a square as shown:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. Because the inner square has an area of &lt;math&gt;(5-2)3&lt;/math&gt; so it has length &lt;math&gt;\sqrt{9}=3&lt;/math&gt; and because the height of the red triangle is 1(because the gray squares have length one), the area of one triangle is namely &lt;math&gt;\frac{3 \cdot 1}{2}&lt;/math&gt;. Thus, the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. Furthermore, the area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> So the area of the square we need is &lt;math&gt;25- (4+6) = 15\implies \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/WNiJWmKCfj0 - Happytwin<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=136537 2015 AMC 8 Problems/Problem 25 2020-11-04T12:22:55Z <p>Clapatron7568: /* Solution 3 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown:<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the &lt;math&gt;4&lt;/math&gt; big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by &lt;math&gt;\mathrm{AA}&lt;/math&gt; similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be &lt;math&gt;x&lt;/math&gt;; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;. Note that the other solution we got, namely, &lt;math&gt;x=\dfrac{5+\sqrt 5} 2&lt;/math&gt;, is the length of the segment &lt;math&gt;5-x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;5-x&lt;/math&gt; together sum to &lt;math&gt;5&lt;/math&gt;, the side of the length of the large square, and similarly, the sum of the solutions is &lt;math&gt;5&lt;/math&gt;. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be &lt;math&gt;x&lt;/math&gt;, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> Thus the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> Don't use this in a contest scenario, only use when practicing math skills :)<br /> <br /> ===Solution 2(Contest Solution)=== <br /> <br /> We draw a square as shown:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. Because the inner square has an area of &lt;math&gt;(5-2)3&lt;/math&gt; so it has length &lt;math&gt;\sqrt{9}=3&lt;/math&gt; and because the height of the red triangle is 1(because the gray squares have length one), the area of one triangle is namely &lt;math&gt;\frac{3 \cdot 1}{2}&lt;/math&gt;. Thus, the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. Furthermore, the area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> So the area of the square we need is $25- (4+6) = 15\implies \boxed{\textbf{ (C})<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/WNiJWmKCfj0 - Happytwin<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=136536 2015 AMC 8 Problems/Problem 25 2020-11-04T12:22:41Z <p>Clapatron7568: /* Solution 3 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown:<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the &lt;math&gt;4&lt;/math&gt; big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by &lt;math&gt;\mathrm{AA}&lt;/math&gt; similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be &lt;math&gt;x&lt;/math&gt;; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;, because by symmetry, &lt;math&gt;x &lt; \dfrac52&lt;/math&gt;. Note that the other solution we got, namely, &lt;math&gt;x=\dfrac{5+\sqrt 5} 2&lt;/math&gt;, is the length of the segment &lt;math&gt;5-x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;5-x&lt;/math&gt; together sum to &lt;math&gt;5&lt;/math&gt;, the side of the length of the large square, and similarly, the sum of the solutions is &lt;math&gt;5&lt;/math&gt;. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be &lt;math&gt;x&lt;/math&gt;, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.<br /> <br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> Thus the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;<br /> <br /> Don't use this in a contest scenario, only use when practicing math skills :)<br /> <br /> ===Solution 2(Contest Solution)=== <br /> <br /> We draw a square as shown:<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br /> path arc = arc((2.5,4),1.5,0,90); <br /> pair P = intersectionpoint(arc,(0,5)--(5,5)); <br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br /> draw(P--Pp--Ppp--Pppp--cycle); <br /> filldraw((1,4)--P--(4,4)--cycle,red);<br /> filldraw((4,4)--Pppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Ppp--(4,1)--cycle,red);<br /> filldraw((1,1)--Pp--(1,4)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. Because the inner square has an area of &lt;math&gt;(5-2)3&lt;/math&gt; so it has length &lt;math&gt;\sqrt{9}=3&lt;/math&gt; and because the height of the red triangle is 1(because the gray squares have length one), the area of one triangle is namely &lt;math&gt;\frac{3 \cdot 1}{2}&lt;/math&gt;. Thus, the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. Furthermore, the area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=15\implies \boxed{\textbf{(C)}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> <br /> Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> So the area of the square we need is$25- (4+6) = 15\implies \boxed{\textbf{(C})<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/WNiJWmKCfj0 - Happytwin<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems&diff=136478 1985 AJHSME Problems 2020-11-03T11:55:31Z <p>Clapatron7568: /* Problem 27 */</p> <hr /> <div>{{AJHSME Problems<br /> |year = 1985<br /> }}<br /> == Problem 1 ==<br /> <br /> &lt;math&gt;\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50&lt;/math&gt;<br /> <br /> == Problem 2 ==<br /> <br /> &lt;math&gt;90+91+92+93+94+95+96+97+98+99=&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045&lt;/math&gt;<br /> <br /> == Problem 3 ==<br /> <br /> &lt;math&gt;\frac{10^7}{5\times 10^4}=&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000&lt;/math&gt;<br /> <br /> == Problem 4 ==<br /> <br /> The area of polygon &lt;math&gt;ABCDEF&lt;/math&gt;, in square units, is<br /> <br /> &lt;math&gt;\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle);<br /> label(&quot;A&quot;,(0,9),NW);<br /> label(&quot;B&quot;,(6,9),NE);<br /> label(&quot;C&quot;,(6,0),SE);<br /> label(&quot;D&quot;,(2,0),SW);<br /> label(&quot;E&quot;,(2,4),NE);<br /> label(&quot;F&quot;,(0,4),SW);<br /> label(&quot;6&quot;,(3,9),N);<br /> label(&quot;9&quot;,(6,4.5),E);<br /> label(&quot;4&quot;,(4,0),S);<br /> label(&quot;5&quot;,(0,6.5),W);<br /> &lt;/asy&gt;<br /> <br /> == Problem 5 ==<br /> <br /> &lt;asy&gt;<br /> unitsize(13);<br /> draw((0,0)--(20,0));<br /> draw((0,0)--(0,15));<br /> draw((0,3)--(-1,3));<br /> draw((0,6)--(-1,6));<br /> draw((0,9)--(-1,9));<br /> draw((0,12)--(-1,12));<br /> draw((0,15)--(-1,15));<br /> fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black);<br /> fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black);<br /> fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black);<br /> fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black);<br /> fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black);<br /> label(&quot;A&quot;,(2.5,-.5),S);<br /> label(&quot;B&quot;,(4.5,-.5),S);<br /> label(&quot;C&quot;,(6.5,-.5),S);<br /> label(&quot;D&quot;,(8.5,-.5),S);<br /> label(&quot;F&quot;,(10.5,-.5),S);<br /> label(&quot;Grade&quot;,(15,-.5),S);<br /> label(&quot;$1$&quot;,(-1,3),W);<br /> label(&quot;$2$&quot;,(-1,6),W);<br /> label(&quot;$3$&quot;,(-1,9),W);<br /> label(&quot;$4$&quot;,(-1,12),W);<br /> label(&quot;$5$&quot;,(-1,15),W);<br /> &lt;/asy&gt;<br /> <br /> The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory?<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{1}{2} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{4}{5} \qquad \text{(E)}\ \frac{9}{10}&lt;/math&gt;<br /> <br /> == Problem 6 ==<br /> <br /> A stack of paper containing &lt;math&gt;500&lt;/math&gt; sheets is &lt;math&gt;5&lt;/math&gt; cm thick. Approximately how many sheets of this type of paper would there be in a stack &lt;math&gt;7.5&lt;/math&gt; cm high?<br /> <br /> &lt;math&gt;\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250&lt;/math&gt;<br /> <br /> == Problem 7 ==<br /> <br /> A &quot;stair-step&quot; figure is made of alternating black and white squares in each row. Rows &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;4&lt;/math&gt; are shown. All rows begin and end with a white square. The number of black squares in the &lt;math&gt;37\text{th}&lt;/math&gt; row is<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(7,0)--(7,1)--(0,1)--cycle);<br /> draw((1,0)--(6,0)--(6,2)--(1,2)--cycle);<br /> draw((2,0)--(5,0)--(5,3)--(2,3)--cycle);<br /> draw((3,0)--(4,0)--(4,4)--(3,4)--cycle);<br /> fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black);<br /> fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black);<br /> fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black);<br /> fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black);<br /> fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black);<br /> fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38&lt;/math&gt;<br /> <br /> == Problem 8 ==<br /> <br /> If &lt;math&gt;a = - 2&lt;/math&gt;, the largest number in the set &lt;math&gt; - 3a, 4a, \frac {24}{a}, a^2, 1&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1&lt;/math&gt;<br /> <br /> == Problem 9 ==<br /> <br /> The product of the 9 factors &lt;math&gt;\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}&lt;/math&gt;<br /> <br /> == Problem 10 ==<br /> <br /> The fraction halfway between &lt;math&gt;\frac{1}{5}&lt;/math&gt; and &lt;math&gt;\frac{1}{3}&lt;/math&gt; (on the number line) is<br /> <br /> &lt;asy&gt;<br /> unitsize(12);<br /> draw((-1,0)--(20,0),EndArrow);<br /> draw((0,-.75)--(0,.75));<br /> draw((10,-.75)--(10,.75));<br /> draw((17,-.75)--(17,.75));<br /> label(&quot;$0$&quot;,(0,-.5),S);<br /> label(&quot;$\frac{1}{5}$&quot;,(10,-.5),S);<br /> label(&quot;$\frac{1}{3}$&quot;,(17,-.5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{1}{4} \qquad \text{(B)}\ \frac{2}{15} \qquad \text{(C)}\ \frac{4}{15} \qquad \text{(D)}\ \frac{53}{200} \qquad \text{(E)}\ \frac{8}{15}&lt;/math&gt;<br /> <br /> == Problem 11 ==<br /> <br /> A piece of paper containing six joined squares labeled as shown in the diagram is folded along the edges of the squares to form a cube. The label of the face opposite the face labeled &lt;math&gt;\text{X}&lt;/math&gt; is <br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,1)--(2,1)--(2,2)--(3,2)--(3,0)--(2,0)--(2,-2)--(1,-2)--(1,0)--cycle);<br /> draw((1,0)--(1,1));<br /> draw((2,0)--(2,1));<br /> draw((1,0)--(2,0));<br /> draw((1,-1)--(2,-1));<br /> draw((2,1)--(3,1));<br /> label(&quot;U&quot;,(.5,.3),N);<br /> label(&quot;V&quot;,(1.5,.3),N);<br /> label(&quot;W&quot;,(2.5,.3),N);<br /> label(&quot;X&quot;,(1.5,-.7),N);<br /> label(&quot;Y&quot;,(2.5,1.3),N);<br /> label(&quot;Z&quot;,(1.5,-1.7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ \text{Z} \qquad \text{(B)}\ \text{U} \qquad \text{(C)}\ \text{V} \qquad \text{(D)}\ \ \text{W} \qquad \text{(E)}\ \text{Y}&lt;/math&gt;<br /> <br /> == Problem 12 ==<br /> <br /> A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are &lt;math&gt;6.2 \text{ cm}&lt;/math&gt;, &lt;math&gt;8.3 \text{ cm}&lt;/math&gt; and &lt;math&gt;9.5 \text{ cm}&lt;/math&gt;. The area of the square is<br /> <br /> &lt;math&gt;\text{(A)}\ 24\text{ cm}^2 \qquad \text{(B)}\ 36\text{ cm}^2 \qquad \text{(C)}\ 48\text{ cm}^2 \qquad \text{(D)}\ 64\text{ cm}^2 \qquad \text{(E)}\ 144\text{ cm}^2&lt;/math&gt;<br /> <br /> == Problem 13 ==<br /> <br /> If you walk for &lt;math&gt;45&lt;/math&gt; minutes at a rate of &lt;math&gt;4 \text{ mph}&lt;/math&gt; and then run for &lt;math&gt;30&lt;/math&gt; minutes at a rate of &lt;math&gt;10\text{ mph,}&lt;/math&gt; how many miles will you have gone at the end of one hour and &lt;math&gt;15&lt;/math&gt; minutes?<br /> <br /> &lt;math&gt;\text{(A)}\ 3.5\text{ miles} \qquad \text{(B)}\ 8\text{ miles} \qquad \text{(C)}\ 9\text{ miles} \qquad \text{(D)}\ 25\frac{1}{3}\text{ miles} \qquad \text{(E)}\ 480\text{ miles}&lt;/math&gt;<br /> <br /> == Problem 14 ==<br /> <br /> The difference between a &lt;math&gt;6.5\% &lt;/math&gt; sales tax and a &lt;math&gt;6\% &lt;/math&gt; sales tax on an item priced at &lt;math&gt;\$20&lt;/math&gt; before tax is<br /> <br /> &lt;math&gt;\text{(A)}&lt;/math&gt; &lt;math&gt;\$.01&lt;/math&gt; <br /> <br /> &lt;math&gt;\text{(B)}&lt;/math&gt; &lt;math&gt;\$.10&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(C)}&lt;/math&gt; &lt;math&gt;\$ .50&lt;/math&gt; <br /> <br /> &lt;math&gt;\text{(D)}&lt;/math&gt; &lt;math&gt;\$1&lt;/math&gt; <br /> <br /> &lt;math&gt;\text{(E)}&lt;/math&gt; &lt;math&gt;\$10&lt;/math&gt;<br /> <br /> == Problem 15 ==<br /> <br /> How many whole numbers between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;400&lt;/math&gt; contain the digit &lt;math&gt;2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148&lt;/math&gt;<br /> <br /> == Problem 16 ==<br /> <br /> The ratio of boys to girls in Mr. Brown's math class is &lt;math&gt;2:3&lt;/math&gt;. If there are &lt;math&gt;30&lt;/math&gt; students in the class, how many more girls than boys are in the class?<br /> <br /> &lt;math&gt;\text{(A)}\ 10 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 2&lt;/math&gt;<br /> <br /> == Problem 17 ==<br /> <br /> If your average score on your first six mathematics tests was &lt;math&gt;84&lt;/math&gt; and your average score on your first seven mathematics tests was &lt;math&gt;85&lt;/math&gt;, then your score on the seventh test was<br /> <br /> &lt;math&gt;\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92&lt;/math&gt;<br /> <br /> == Problem 18 ==<br /> <br /> Nine copies of a certain pamphlet cost less than &lt;math&gt;\$10.00&lt;/math&gt; while ten copies of the same pamphlet (at the same price) cost more than &lt;math&gt;\$11.00&lt;/math&gt;. How much does one copy of this pamphlet cost?<br /> <br /> &lt;math&gt;\text{(A)}&lt;/math&gt; &lt;math&gt;\$1.07&lt;/math&gt; <br /> <br /> &lt;math&gt;\text{(B)}&lt;/math&gt; &lt;math&gt;\$1.08&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(C)}&lt;/math&gt; &lt;math&gt;\$1.09&lt;/math&gt; <br /> <br /> &lt;math&gt;\text{(D)}&lt;/math&gt; &lt;math&gt;\$1.10&lt;/math&gt; <br /> <br /> &lt;math&gt;\text{(E)}&lt;/math&gt; &lt;math&gt;\$1.11&lt;/math&gt;<br /> <br /> == Problem 19 ==<br /> <br /> If the length and width of a rectangle are each increased by &lt;math&gt;10\% &lt;/math&gt;, then the perimeter of the rectangle is increased by<br /> <br /> &lt;math&gt;\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\% &lt;/math&gt;<br /> <br /> == Problem 20 ==<br /> <br /> In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January &lt;math&gt;1&lt;/math&gt; fall that year?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Tuesday} \qquad \text{(C)}\ \text{Wednesday} \qquad \text{(D)}\ \text{Friday} \qquad \text{(E)}\ \text{Saturday}&lt;/math&gt;<br /> <br /> == Problem 21 ==<br /> <br /> Mr.Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> == Problem 22 ==<br /> <br /> Assume every 7-digit whole number is a possible telephone number except those that begin with &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;. What fraction of telephone numbers begin with &lt;math&gt;9&lt;/math&gt; and end with &lt;math&gt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{1}{63} \qquad \text{(B)}\ \frac{1}{80} \qquad \text{(C)}\ \frac{1}{81} \qquad \text{(D)}\ \frac{1}{90} \qquad \text{(E)}\ \frac{1}{100}&lt;/math&gt;<br /> <br /> ''Note: All telephone numbers are 7-digit whole numbers.''<br /> <br /> == Problem 23 ==<br /> <br /> King Middle School has &lt;math&gt;1200&lt;/math&gt; students. Each pupil takes &lt;math&gt;5&lt;/math&gt; classes a day. Each teacher teaches &lt;math&gt;4&lt;/math&gt; classes. Each class has &lt;math&gt;30&lt;/math&gt; students and &lt;math&gt;1&lt;/math&gt; teacher. How many teachers are there at King Middle School? <br /> <br /> &lt;math&gt;\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50&lt;/math&gt;<br /> <br /> == Problem 24 ==<br /> <br /> In a magic triangle, each of the six whole numbers &lt;math&gt;10-15&lt;/math&gt; is placed in one of the circles so that the sum, &lt;math&gt;S&lt;/math&gt;, of the three numbers on each side of the triangle is the same. The largest possible value for &lt;math&gt;S&lt;/math&gt; is<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),1));<br /> draw(dir(60)--6*dir(60));<br /> draw(circle(7*dir(60),1));<br /> draw(8*dir(60)--13*dir(60));<br /> draw(circle(14*dir(60),1));<br /> draw((1,0)--(6,0));<br /> draw(circle((7,0),1));<br /> draw((8,0)--(13,0));<br /> draw(circle((14,0),1));<br /> draw(circle((10.5,6.0621778264910705273460621952706),1));<br /> draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176));<br /> draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40&lt;/math&gt;<br /> <br /> <br /> == Problem 25 ==<br /> <br /> <br /> Five cards are lying on a table as shown.<br /> <br /> &lt;cmath&gt;\begin{matrix} &amp; \qquad &amp; \boxed{\tt{P}} &amp; \qquad &amp; \boxed{\tt{Q}} \\<br /> \\<br /> \boxed{\tt{3}} &amp; \qquad &amp; \boxed{\tt{4}} &amp; \qquad &amp; \boxed{\tt{6}} \end{matrix}&lt;/cmath&gt;<br /> <br /> Each card has a letter on one side and a whole number on the other side. Jane said, &quot;If a vowel is on one side of any card, then an even number is on the other side.&quot; Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?<br /> <br /> &lt;math&gt;\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ \text{P} \qquad \text{(E)}\ \text{Q}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AJHSME box|year=1985|before=First&lt;br&gt;AJHSME|after=[[1986 AJHSME Problems|1986 AJHSME]]}}<br /> * [[AJHSME]]<br /> * [[AJHSME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_4&diff=136419 2019 AMC 8 Problems/Problem 4 2020-11-02T17:22:26Z <p>Clapatron7568: /* Solution 2 */</p> <hr /> <div>== Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> <br /> == Solution 1 ==<br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,5)--(12,0));<br /> draw((12,0)--(0,-5));<br /> draw((0,-5)--(-12,0));<br /> draw((0,0)--(12,0));<br /> draw((0,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,-5));<br /> dot((-12,0));<br /> dot((0,5));<br /> dot((12,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(12,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> label(&quot;E&quot;,(0,0),SW);<br /> &lt;/asy&gt;<br /> <br /> A rhombus has sides of equal length. Because the perimeter of the rhombus is &lt;math&gt;52&lt;/math&gt;, each side is &lt;math&gt;\frac{52}{4}=13&lt;/math&gt;. In a rhombus, diagonals are perpendicular and bisect each other, which means &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt; = &lt;math&gt;\overline{EC}&lt;/math&gt;.<br /> <br /> Consider one of the right triangles:<br /> <br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,5));<br /> dot((-12,0));<br /> dot((0,5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;E&quot;,(0,0),SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\overline{AB}&lt;/math&gt; = &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt;. Using Pythagorean theorem, we find that &lt;math&gt;\overline{BE}&lt;/math&gt; = &lt;math&gt;5&lt;/math&gt;.<br /> &quot;You may recall the famous pythagorean triple, (5, 12, 13), that's how I did it&quot;-Zack2008<br /> <br /> Thus the values of the two diagonals are &lt;math&gt;\overline{AC}&lt;/math&gt; = &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; = &lt;math&gt;10&lt;/math&gt;.<br /> The area of a rhombus is = &lt;math&gt;\frac{d_1\cdot{d_2}}{2}&lt;/math&gt; = &lt;math&gt;\frac{24\cdot{10}}{2}&lt;/math&gt; = &lt;math&gt;120&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 120}&lt;/math&gt; ~phoenixfire<br /> <br /> Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&amp;t=2s<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=3|num-a=5}}<br /> <br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_4&diff=136418 2019 AMC 8 Problems/Problem 4 2020-11-02T17:19:30Z <p>Clapatron7568: /* Solution 1 */</p> <hr /> <div>== Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> <br /> == Solution 1 ==<br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,5)--(12,0));<br /> draw((12,0)--(0,-5));<br /> draw((0,-5)--(-12,0));<br /> draw((0,0)--(12,0));<br /> draw((0,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,-5));<br /> dot((-12,0));<br /> dot((0,5));<br /> dot((12,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(12,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> label(&quot;E&quot;,(0,0),SW);<br /> &lt;/asy&gt;<br /> <br /> A rhombus has sides of equal length. Because the perimeter of the rhombus is &lt;math&gt;52&lt;/math&gt;, each side is &lt;math&gt;\frac{52}{4}=13&lt;/math&gt;. In a rhombus, diagonals are perpendicular and bisect each other, which means &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt; = &lt;math&gt;\overline{EC}&lt;/math&gt;.<br /> <br /> Consider one of the right triangles:<br /> <br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,5));<br /> dot((-12,0));<br /> dot((0,5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;E&quot;,(0,0),SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\overline{AB}&lt;/math&gt; = &lt;math&gt;13&lt;/math&gt;, and &lt;math&gt;\overline{AE}&lt;/math&gt; = &lt;math&gt;12&lt;/math&gt;. Using Pythagorean theorem, we find that &lt;math&gt;\overline{BE}&lt;/math&gt; = &lt;math&gt;5&lt;/math&gt;.<br /> &quot;You may recall the famous pythagorean triple, (5, 12, 13), that's how I did it&quot;-Zack2008<br /> <br /> Thus the values of the two diagonals are &lt;math&gt;\overline{AC}&lt;/math&gt; = &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; = &lt;math&gt;10&lt;/math&gt;.<br /> The area of a rhombus is = &lt;math&gt;\frac{d_1\cdot{d_2}}{2}&lt;/math&gt; = &lt;math&gt;\frac{24\cdot{10}}{2}&lt;/math&gt; = &lt;math&gt;120&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 120}&lt;/math&gt; ~phoenixfire<br /> <br /> Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&amp;t=2s<br /> <br /> == Solution 2 ==<br /> &lt;asy&gt;<br /> draw((-12,0)--(0,5));<br /> draw((0,5)--(12,0));<br /> draw((12,0)--(0,-5));<br /> draw((0,-5)--(-12,0));<br /> draw((0,0)--(12,0));<br /> draw((0,0)--(0,5));<br /> draw((0,0)--(-12,0));<br /> draw((0,0)--(0,-5));<br /> dot((-12,0));<br /> dot((0,5));<br /> dot((12,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-12,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(12,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> label(&quot;E&quot;,(0,0),SW);<br /> &lt;/asy&gt;<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=3|num-a=5}}<br /> <br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=136417 2019 AMC 8 Problems/Problem 3 2020-11-02T17:18:03Z <p>Clapatron7568: /* Problem 3 */</p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Each one is in the form &lt;math&gt;\frac{x+4}{x}&lt;/math&gt; so we are really comparing &lt;math&gt;\frac{4}{11}, \frac{4}{15},&lt;/math&gt; and &lt;math&gt;\frac{4}{13}&lt;/math&gt; where you can see &lt;math&gt;\frac{4}{11}&gt;\frac{4}{13}&gt;\frac{4}{15}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We take a common denominator:<br /> &lt;cmath&gt;\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> ~ dolphin7 - I took your idea and made it an explanation.<br /> <br /> ==Solution 3==<br /> When &lt;math&gt;\frac{x}{y}&gt;1&lt;/math&gt; and &lt;math&gt;z&gt;0&lt;/math&gt;, &lt;math&gt;\frac{x+z}{y+z}&lt;\frac{x}{y}&lt;/math&gt;. Hence, the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> ~ ryjs<br /> <br /> This is also similar to Problem 20 on the AMC 2012.<br /> <br /> ==Solution 4 (probably won't use this solution)==<br /> We use our insane mental calculator to find out that &lt;math&gt;\frac{15}{11} \approx 1.36&lt;/math&gt;, &lt;math&gt;\frac{19}{15} \approx 1.27&lt;/math&gt;, and &lt;math&gt;\frac{17}{13} \approx 1.31&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ~~ by an insane math guy.<br /> <br /> ==Solution 5==<br /> Suppose each fraction is expressed with denominator &lt;math&gt;2145&lt;/math&gt;: &lt;math&gt;\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}&lt;/math&gt;. Clearly &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}} The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=136416 2019 AMC 8 Problems/Problem 3 2020-11-02T17:16:59Z <p>Clapatron7568: /* Solution 4(probably won't use this solution) */</p> <hr /> <div>==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> ==Solution 1==<br /> Each one is in the form &lt;math&gt;\frac{x+4}{x}&lt;/math&gt; so we are really comparing &lt;math&gt;\frac{4}{11}, \frac{4}{15},&lt;/math&gt; and &lt;math&gt;\frac{4}{13}&lt;/math&gt; where you can see &lt;math&gt;\frac{4}{11}&gt;\frac{4}{13}&gt;\frac{4}{15}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We take a common denominator:<br /> &lt;cmath&gt;\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; it follows that the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> -xMidnightFirex<br /> <br /> ~ dolphin7 - I took your idea and made it an explanation.<br /> <br /> ==Solution 3==<br /> When &lt;math&gt;\frac{x}{y}&gt;1&lt;/math&gt; and &lt;math&gt;z&gt;0&lt;/math&gt;, &lt;math&gt;\frac{x+z}{y+z}&lt;\frac{x}{y}&lt;/math&gt;. Hence, the answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> ~ ryjs<br /> <br /> This is also similar to Problem 20 on the AMC 2012.<br /> <br /> ==Solution 4 (probably won't use this solution)==<br /> We use our insane mental calculator to find out that &lt;math&gt;\frac{15}{11} \approx 1.36&lt;/math&gt;, &lt;math&gt;\frac{19}{15} \approx 1.27&lt;/math&gt;, and &lt;math&gt;\frac{17}{13} \approx 1.31&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)}\frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}}&lt;/math&gt;.<br /> <br /> ~~ by an insane math guy.<br /> <br /> ==Solution 5==<br /> Suppose each fraction is expressed with denominator &lt;math&gt;2145&lt;/math&gt;: &lt;math&gt;\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}&lt;/math&gt;. Clearly &lt;math&gt;2717&lt;2805&lt;2925&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}} The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=136359 2019 AMC 8 Problems 2020-11-01T19:38:46Z <p>Clapatron7568: /* Problem 20 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> &lt;asy&gt;<br /> unitsize(0.4 cm);<br /> <br /> pair transx, transy;<br /> int i, j;<br /> int x, y;<br /> <br /> transx = (13,0);<br /> transy = (0,-9);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> for (j = 0; j &lt;= 1; ++j) {<br /> if (i &lt;= 1 || j &lt;= 0) {<br /> for (x = 1; x &lt;= 10; ++x) {<br /> draw(shift(i*transx + j*transy)*((x,0)--(x,5)),gray(0.7) + dashed);<br /> }<br /> for (y = 1; y &lt;= 5; ++y) {<br /> draw(shift(i*transx + j*transy)*((0,y)--(10,y)),gray(0.7) + dashed);<br /> }<br /> draw(shift(i*transx + j*transy)*((0,0)--(11,0)),Arrow(6));<br /> draw(shift(i*transx + j*transy)*((0,0)--(0,6)),Arrow(6));<br /> label(&quot;time&quot;, (5,-0.5) + i*transx + j*transy);<br /> label(rotate(90)*&quot;distance&quot;, (-0.5,2.5) + i*transx + j*transy);<br /> }<br /> }}<br /> <br /> draw((0,0)--(1.5,2.5)--(7.5,2.5)--(9,5),linewidth(1.5*bp));<br /> draw((0,0)--(10,5),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(2.5,2.5)--(7.5,2.5)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(2.5,3)--(7,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(2.5,2.5)--(6.5,2.5)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(7.5,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(2.5,2)--(7.5,3)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> <br /> label(&quot;(A)&quot;, (-1,6));<br /> label(&quot;(B)&quot;, (-1,6) + transx);<br /> label(&quot;(C)&quot;, (-1,6) + 2*transx);<br /> label(&quot;(D)&quot;, (-1,6) + transy);<br /> label(&quot;(E)&quot;, (-1,6) + transx + transy);<br /> &lt;/asy&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt; <br /> <br /> [[2019 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume of one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> &lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> [[2019 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eight graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> &lt;asy&gt;<br /> unitsize(2 cm);<br /> <br /> pair x, y, z, trans;<br /> int i;<br /> <br /> x = dir(-5);<br /> y = (0.6,0.5);<br /> z = (0,1);<br /> trans = (2,0);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle));<br /> draw(shift(i*trans)*((x + z)--x));<br /> draw(shift(i*trans)*((x + z)--(x + y + z)));<br /> draw(shift(i*trans)*((x + z)--z));<br /> }<br /> <br /> label(rotate(-3)*&quot;$R$&quot;, (x + z)/2);<br /> label(rotate(-5)*slant(0.5)*&quot;$B$&quot;, ((x + z) + (y + z))/2);<br /> label(rotate(35)*slant(0.5)*&quot;$G$&quot;, ((x + z) + (x + y))/2);<br /> <br /> label(rotate(-3)*&quot;$W$&quot;, (x + z)/2 + trans);<br /> label(rotate(50)*slant(-1)*&quot;$B$&quot;, ((x + z) + (y + z))/2 + trans);<br /> label(rotate(35)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (x + y))/2 + trans);<br /> <br /> label(rotate(-3)*&quot;$P$&quot;, (x + z)/2 + 2*trans);<br /> label(rotate(-5)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (y + z))/2 + 2*trans);<br /> label(rotate(-85)*slant(-1)*&quot;$G$&quot;, ((x + z) + (x + y))/2 + 2*trans);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> Alice has 24 apples. In how many ways can she share them with Becky and<br /> Chris so that each of the three people has at least two apples?<br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=136358 2019 AMC 8 Problems 2020-11-01T19:38:09Z <p>Clapatron7568: /* Problem 20 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> &lt;asy&gt;<br /> unitsize(0.4 cm);<br /> <br /> pair transx, transy;<br /> int i, j;<br /> int x, y;<br /> <br /> transx = (13,0);<br /> transy = (0,-9);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> for (j = 0; j &lt;= 1; ++j) {<br /> if (i &lt;= 1 || j &lt;= 0) {<br /> for (x = 1; x &lt;= 10; ++x) {<br /> draw(shift(i*transx + j*transy)*((x,0)--(x,5)),gray(0.7) + dashed);<br /> }<br /> for (y = 1; y &lt;= 5; ++y) {<br /> draw(shift(i*transx + j*transy)*((0,y)--(10,y)),gray(0.7) + dashed);<br /> }<br /> draw(shift(i*transx + j*transy)*((0,0)--(11,0)),Arrow(6));<br /> draw(shift(i*transx + j*transy)*((0,0)--(0,6)),Arrow(6));<br /> label(&quot;time&quot;, (5,-0.5) + i*transx + j*transy);<br /> label(rotate(90)*&quot;distance&quot;, (-0.5,2.5) + i*transx + j*transy);<br /> }<br /> }}<br /> <br /> draw((0,0)--(1.5,2.5)--(7.5,2.5)--(9,5),linewidth(1.5*bp));<br /> draw((0,0)--(10,5),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(2.5,2.5)--(7.5,2.5)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(2.5,3)--(7,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(2.5,2.5)--(6.5,2.5)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(7.5,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(2.5,2)--(7.5,3)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> <br /> label(&quot;(A)&quot;, (-1,6));<br /> label(&quot;(B)&quot;, (-1,6) + transx);<br /> label(&quot;(C)&quot;, (-1,6) + 2*transx);<br /> label(&quot;(D)&quot;, (-1,6) + transy);<br /> label(&quot;(E)&quot;, (-1,6) + transx + transy);<br /> &lt;/asy&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt; <br /> <br /> [[2019 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume of one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> &lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> [[2019 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eight graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> &lt;asy&gt;<br /> unitsize(2 cm);<br /> <br /> pair x, y, z, trans;<br /> int i;<br /> <br /> x = dir(-5);<br /> y = (0.6,0.5);<br /> z = (0,1);<br /> trans = (2,0);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle));<br /> draw(shift(i*trans)*((x + z)--x));<br /> draw(shift(i*trans)*((x + z)--(x + y + z)));<br /> draw(shift(i*trans)*((x + z)--z));<br /> }<br /> <br /> label(rotate(-3)*&quot;$R$&quot;, (x + z)/2);<br /> label(rotate(-5)*slant(0.5)*&quot;$B$&quot;, ((x + z) + (y + z))/2);<br /> label(rotate(35)*slant(0.5)*&quot;$G$&quot;, ((x + z) + (x + y))/2);<br /> <br /> label(rotate(-3)*&quot;$W$&quot;, (x + z)/2 + trans);<br /> label(rotate(50)*slant(-1)*&quot;$B$&quot;, ((x + z) + (y + z))/2 + trans);<br /> label(rotate(35)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (x + y))/2 + trans);<br /> <br /> label(rotate(-3)*&quot;$P$&quot;, (x + z)/2 + 2*trans);<br /> label(rotate(-5)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (y + z))/2 + 2*trans);<br /> label(rotate(-85)*slant(-1)*&quot;$G$&quot;, ((x + z) + (x + y))/2 + 2*trans);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\qquad\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> Alice has 24 apples. In how many ways can she share them with Becky and<br /> Chris so that each of the three people has at least two apples?<br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=136357 2019 AMC 8 Problems 2020-11-01T19:30:55Z <p>Clapatron7568: /* Problem 25 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> &lt;asy&gt;<br /> unitsize(0.4 cm);<br /> <br /> pair transx, transy;<br /> int i, j;<br /> int x, y;<br /> <br /> transx = (13,0);<br /> transy = (0,-9);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> for (j = 0; j &lt;= 1; ++j) {<br /> if (i &lt;= 1 || j &lt;= 0) {<br /> for (x = 1; x &lt;= 10; ++x) {<br /> draw(shift(i*transx + j*transy)*((x,0)--(x,5)),gray(0.7) + dashed);<br /> }<br /> for (y = 1; y &lt;= 5; ++y) {<br /> draw(shift(i*transx + j*transy)*((0,y)--(10,y)),gray(0.7) + dashed);<br /> }<br /> draw(shift(i*transx + j*transy)*((0,0)--(11,0)),Arrow(6));<br /> draw(shift(i*transx + j*transy)*((0,0)--(0,6)),Arrow(6));<br /> label(&quot;time&quot;, (5,-0.5) + i*transx + j*transy);<br /> label(rotate(90)*&quot;distance&quot;, (-0.5,2.5) + i*transx + j*transy);<br /> }<br /> }}<br /> <br /> draw((0,0)--(1.5,2.5)--(7.5,2.5)--(9,5),linewidth(1.5*bp));<br /> draw((0,0)--(10,5),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(2.5,2.5)--(7.5,2.5)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(2.5,3)--(7,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(2.5,2.5)--(6.5,2.5)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(7.5,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(2.5,2)--(7.5,3)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> <br /> label(&quot;(A)&quot;, (-1,6));<br /> label(&quot;(B)&quot;, (-1,6) + transx);<br /> label(&quot;(C)&quot;, (-1,6) + 2*transx);<br /> label(&quot;(D)&quot;, (-1,6) + transy);<br /> label(&quot;(E)&quot;, (-1,6) + transx + transy);<br /> &lt;/asy&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt; <br /> <br /> [[2019 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume of one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> &lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> [[2019 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eight graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> &lt;asy&gt;<br /> unitsize(2 cm);<br /> <br /> pair x, y, z, trans;<br /> int i;<br /> <br /> x = dir(-5);<br /> y = (0.6,0.5);<br /> z = (0,1);<br /> trans = (2,0);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle));<br /> draw(shift(i*trans)*((x + z)--x));<br /> draw(shift(i*trans)*((x + z)--(x + y + z)));<br /> draw(shift(i*trans)*((x + z)--z));<br /> }<br /> <br /> label(rotate(-3)*&quot;$R$&quot;, (x + z)/2);<br /> label(rotate(-5)*slant(0.5)*&quot;$B$&quot;, ((x + z) + (y + z))/2);<br /> label(rotate(35)*slant(0.5)*&quot;$G$&quot;, ((x + z) + (x + y))/2);<br /> <br /> label(rotate(-3)*&quot;$W$&quot;, (x + z)/2 + trans);<br /> label(rotate(50)*slant(-1)*&quot;$B$&quot;, ((x + z) + (y + z))/2 + trans);<br /> label(rotate(35)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (x + y))/2 + trans);<br /> <br /> label(rotate(-3)*&quot;$P$&quot;, (x + z)/2 + 2*trans);<br /> label(rotate(-5)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (y + z))/2 + 2*trans);<br /> label(rotate(-85)*slant(-1)*&quot;$G$&quot;, ((x + z) + (x + y))/2 + 2*trans);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\times{(A) }4\qquad\textbf{(B) }1\qquad\textbf{(C) }64\qquad\textbf{(e) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> Alice has 24 apples. In how many ways can she share them with Becky and<br /> Chris so that each of the three people has at least two apples?<br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=136356 2019 AMC 8 Problems 2020-11-01T19:28:38Z <p>Clapatron7568: /* Problem 24 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> &lt;asy&gt;<br /> unitsize(0.4 cm);<br /> <br /> pair transx, transy;<br /> int i, j;<br /> int x, y;<br /> <br /> transx = (13,0);<br /> transy = (0,-9);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> for (j = 0; j &lt;= 1; ++j) {<br /> if (i &lt;= 1 || j &lt;= 0) {<br /> for (x = 1; x &lt;= 10; ++x) {<br /> draw(shift(i*transx + j*transy)*((x,0)--(x,5)),gray(0.7) + dashed);<br /> }<br /> for (y = 1; y &lt;= 5; ++y) {<br /> draw(shift(i*transx + j*transy)*((0,y)--(10,y)),gray(0.7) + dashed);<br /> }<br /> draw(shift(i*transx + j*transy)*((0,0)--(11,0)),Arrow(6));<br /> draw(shift(i*transx + j*transy)*((0,0)--(0,6)),Arrow(6));<br /> label(&quot;time&quot;, (5,-0.5) + i*transx + j*transy);<br /> label(rotate(90)*&quot;distance&quot;, (-0.5,2.5) + i*transx + j*transy);<br /> }<br /> }}<br /> <br /> draw((0,0)--(1.5,2.5)--(7.5,2.5)--(9,5),linewidth(1.5*bp));<br /> draw((0,0)--(10,5),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(2.5,2.5)--(7.5,2.5)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(2.5,3)--(7,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(2.5,2.5)--(6.5,2.5)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(7.5,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(2.5,2)--(7.5,3)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> <br /> label(&quot;(A)&quot;, (-1,6));<br /> label(&quot;(B)&quot;, (-1,6) + transx);<br /> label(&quot;(C)&quot;, (-1,6) + 2*transx);<br /> label(&quot;(D)&quot;, (-1,6) + transy);<br /> label(&quot;(E)&quot;, (-1,6) + transx + transy);<br /> &lt;/asy&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt; <br /> <br /> [[2019 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume of one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> &lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> [[2019 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eight graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> &lt;asy&gt;<br /> unitsize(2 cm);<br /> <br /> pair x, y, z, trans;<br /> int i;<br /> <br /> x = dir(-5);<br /> y = (0.6,0.5);<br /> z = (0,1);<br /> trans = (2,0);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle));<br /> draw(shift(i*trans)*((x + z)--x));<br /> draw(shift(i*trans)*((x + z)--(x + y + z)));<br /> draw(shift(i*trans)*((x + z)--z));<br /> }<br /> <br /> label(rotate(-3)*&quot;$R$&quot;, (x + z)/2);<br /> label(rotate(-5)*slant(0.5)*&quot;$B$&quot;, ((x + z) + (y + z))/2);<br /> label(rotate(35)*slant(0.5)*&quot;$G$&quot;, ((x + z) + (x + y))/2);<br /> <br /> label(rotate(-3)*&quot;$W$&quot;, (x + z)/2 + trans);<br /> label(rotate(50)*slant(-1)*&quot;$B$&quot;, ((x + z) + (y + z))/2 + trans);<br /> label(rotate(35)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (x + y))/2 + trans);<br /> <br /> label(rotate(-3)*&quot;$P$&quot;, (x + z)/2 + 2*trans);<br /> label(rotate(-5)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (y + z))/2 + 2*trans);<br /> label(rotate(-85)*slant(-1)*&quot;$G$&quot;, ((x + z) + (x + y))/2 + 2*trans);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\times{(A) }4\qquad\textbf{(B) }1\qquad\textbf{(C) }64\qquad\textbf{(e) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> == Problem 25==<br /> Alice has 24 apples. In how many ways can she share them with Becky and<br /> Chris so that each of the three people has at least two apples?<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems&diff=136355 2019 AMC 8 Problems 2020-11-01T19:26:24Z <p>Clapatron7568: /* Problem 23 */</p> <hr /> <div>==Problem 1==<br /> <br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Three identical rectangles are put together to form rectangle &lt;math&gt;ABCD&lt;/math&gt;, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(3,0));<br /> draw((0,0)--(0,2));<br /> draw((0,2)--(3,2));<br /> draw((3,2)--(3,0));<br /> dot((0,0));<br /> dot((0,2));<br /> dot((3,0));<br /> dot((3,2));<br /> draw((2,0)--(2,2));<br /> draw((0,1)--(2,1));<br /> label(&quot;A&quot;,(0,0),S);<br /> label(&quot;B&quot;,(3,0),S);<br /> label(&quot;C&quot;,(3,2),N);<br /> label(&quot;D&quot;,(0,2),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Which of the following is the correct order of the fractions &lt;math&gt;\frac{15}{11},\frac{19}{15},&lt;/math&gt; and &lt;math&gt;\frac{17}{13},&lt;/math&gt; from least to greatest? <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{15}{11}&lt; \frac{17}{13}&lt; \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}&lt; \frac{19}{15}&lt;\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}&lt;\frac{19}{15}&lt;\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}&lt;\frac{15}{11}&lt;\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}&lt;\frac{17}{13}&lt;\frac{15}{11}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is a rhombus with perimeter &lt;math&gt;52&lt;/math&gt; meters. The length of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; is &lt;math&gt;24&lt;/math&gt; meters. What is the area in square meters of rhombus &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((-13,0)--(0,5));<br /> draw((0,5)--(13,0));<br /> draw((13,0)--(0,-5));<br /> draw((0,-5)--(-13,0));<br /> dot((-13,0));<br /> dot((0,5));<br /> dot((13,0));<br /> dot((0,-5));<br /> label(&quot;A&quot;,(-13,0),W);<br /> label(&quot;B&quot;,(0,5),N);<br /> label(&quot;C&quot;,(13,0),E);<br /> label(&quot;D&quot;,(0,-5),S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance &lt;math&gt;d&lt;/math&gt; traveled by the two animals over time &lt;math&gt;t&lt;/math&gt; from start to finish?<br /> &lt;asy&gt;<br /> unitsize(0.4 cm);<br /> <br /> pair transx, transy;<br /> int i, j;<br /> int x, y;<br /> <br /> transx = (13,0);<br /> transy = (0,-9);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> for (j = 0; j &lt;= 1; ++j) {<br /> if (i &lt;= 1 || j &lt;= 0) {<br /> for (x = 1; x &lt;= 10; ++x) {<br /> draw(shift(i*transx + j*transy)*((x,0)--(x,5)),gray(0.7) + dashed);<br /> }<br /> for (y = 1; y &lt;= 5; ++y) {<br /> draw(shift(i*transx + j*transy)*((0,y)--(10,y)),gray(0.7) + dashed);<br /> }<br /> draw(shift(i*transx + j*transy)*((0,0)--(11,0)),Arrow(6));<br /> draw(shift(i*transx + j*transy)*((0,0)--(0,6)),Arrow(6));<br /> label(&quot;time&quot;, (5,-0.5) + i*transx + j*transy);<br /> label(rotate(90)*&quot;distance&quot;, (-0.5,2.5) + i*transx + j*transy);<br /> }<br /> }}<br /> <br /> draw((0,0)--(1.5,2.5)--(7.5,2.5)--(9,5),linewidth(1.5*bp));<br /> draw((0,0)--(10,5),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(2.5,2.5)--(7.5,2.5)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(2.5,3)--(7,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(2*transx)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(2.5,2.5)--(6.5,2.5)--(9,5)),linewidth(1.5*bp));<br /> draw(shift(transy)*((0,0)--(7.5,2)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(2.5,2)--(7.5,3)--(10,5)),linewidth(1.5*bp));<br /> draw(shift(transx + transy)*((0,0)--(9,5)),linewidth(1.5*bp));<br /> <br /> label(&quot;(A)&quot;, (-1,6));<br /> label(&quot;(B)&quot;, (-1,6) + transx);<br /> label(&quot;(C)&quot;, (-1,6) + 2*transx);<br /> label(&quot;(D)&quot;, (-1,6) + transy);<br /> label(&quot;(E)&quot;, (-1,6) + transx + transy);<br /> &lt;/asy&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> There are &lt;math&gt;81&lt;/math&gt; grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point &lt;math&gt;P&lt;/math&gt; is in the center of the square. Given that point &lt;math&gt;Q&lt;/math&gt; is randomly chosen among the other &lt;math&gt;80&lt;/math&gt; points, what is the probability that the line &lt;math&gt;PQ&lt;/math&gt; is a line of symmetry for the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,8));<br /> draw((0,8)--(8,8));<br /> draw((8,8)--(8,0));<br /> draw((8,0)--(0,0));<br /> dot((0,0));<br /> dot((0,1));<br /> dot((0,2));<br /> dot((0,3));<br /> dot((0,4));<br /> dot((0,5));<br /> dot((0,6));<br /> dot((0,7));<br /> dot((0,8));<br /> <br /> dot((1,0));<br /> dot((1,1));<br /> dot((1,2));<br /> dot((1,3));<br /> dot((1,4));<br /> dot((1,5));<br /> dot((1,6));<br /> dot((1,7));<br /> dot((1,8));<br /> <br /> dot((2,0));<br /> dot((2,1));<br /> dot((2,2));<br /> dot((2,3));<br /> dot((2,4));<br /> dot((2,5));<br /> dot((2,6));<br /> dot((2,7));<br /> dot((2,8));<br /> <br /> dot((3,0));<br /> dot((3,1));<br /> dot((3,2));<br /> dot((3,3));<br /> dot((3,4));<br /> dot((3,5));<br /> dot((3,6));<br /> dot((3,7));<br /> dot((3,8));<br /> <br /> dot((4,0));<br /> dot((4,1));<br /> dot((4,2));<br /> dot((4,3));<br /> dot((4,4));<br /> dot((4,5));<br /> dot((4,6));<br /> dot((4,7));<br /> dot((4,8));<br /> <br /> dot((5,0));<br /> dot((5,1));<br /> dot((5,2));<br /> dot((5,3));<br /> dot((5,4));<br /> dot((5,5));<br /> dot((5,6));<br /> dot((5,7));<br /> dot((5,8));<br /> <br /> dot((6,0));<br /> dot((6,1));<br /> dot((6,2));<br /> dot((6,3));<br /> dot((6,4));<br /> dot((6,5));<br /> dot((6,6));<br /> dot((6,7));<br /> dot((6,8));<br /> <br /> dot((7,0));<br /> dot((7,1));<br /> dot((7,2));<br /> dot((7,3));<br /> dot((7,4));<br /> dot((7,5));<br /> dot((7,6));<br /> dot((7,7));<br /> dot((7,8));<br /> <br /> dot((8,0));<br /> dot((8,1));<br /> dot((8,2));<br /> dot((8,3));<br /> dot((8,4));<br /> dot((8,5));<br /> dot((8,6));<br /> dot((8,7));<br /> dot((8,8));<br /> label(&quot;P&quot;,(4,4),NE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{5}\qquad\textbf{(B) }\frac{1}{4} \qquad\textbf{(C) }\frac{2}{5} \qquad\textbf{(D) }\frac{9}{20} \qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt; <br /> <br /> [[2019 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> Gilda has a bag of marbles. She gives &lt;math&gt;20\%&lt;/math&gt; of them to her friend Pedro. Then Gilda gives &lt;math&gt;10\%&lt;/math&gt; of what is left to another friend, Ebony. Finally, Gilda gives &lt;math&gt;25\%&lt;/math&gt; of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are &lt;math&gt;6&lt;/math&gt; cm in diameter and &lt;math&gt;12&lt;/math&gt; cm high. Felicia buys cat food in cylindrical cans that are &lt;math&gt;12&lt;/math&gt; cm in diameter and &lt;math&gt;6&lt;/math&gt; cm high. What is the ratio of the volume of one of Alex's cans to the volume one of Felicia's cans?<br /> <br /> &lt;math&gt;\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually &lt;math&gt;21&lt;/math&gt; participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?<br /> &lt;asy&gt;<br /> unitsize(2mm);<br /> defaultpen(fontsize(8bp));<br /> real d = 5;<br /> real t = 0.7;<br /> real r;<br /> int[] num = {20,26,16,22,16};<br /> string[] days = {&quot;Monday&quot;,&quot;Tuesday&quot;,&quot;Wednesday&quot;,&quot;Thursday&quot;,&quot;Friday&quot;};<br /> for (int i=0; i&lt;30;<br /> i=i+2) { draw((i,0)--(i,-5*d),gray);<br /> }for (int i=0;<br /> i&lt;5;<br /> ++i) { r = -1*(i+0.5)*d;<br /> fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray);<br /> label(days[i],(-1,r),W);<br /> }for(int i=0;<br /> i&lt;32;<br /> i=i+4) { label(string(i),(i,1));<br /> }label(&quot;Number of students at soccer practice&quot;,(14,3.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median does not change.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; The mean increases by &lt;math&gt;1&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; The mean increases by &lt;math&gt;5&lt;/math&gt; and the median increases by &lt;math&gt;5&lt;/math&gt;.<br /> <br /> [[2019 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eight graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> &lt;asy&gt;<br /> unitsize(2 cm);<br /> <br /> pair x, y, z, trans;<br /> int i;<br /> <br /> x = dir(-5);<br /> y = (0.6,0.5);<br /> z = (0,1);<br /> trans = (2,0);<br /> <br /> for (i = 0; i &lt;= 2; ++i) {<br /> draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle));<br /> draw(shift(i*trans)*((x + z)--x));<br /> draw(shift(i*trans)*((x + z)--(x + y + z)));<br /> draw(shift(i*trans)*((x + z)--z));<br /> }<br /> <br /> label(rotate(-3)*&quot;$R$&quot;, (x + z)/2);<br /> label(rotate(-5)*slant(0.5)*&quot;$B$&quot;, ((x + z) + (y + z))/2);<br /> label(rotate(35)*slant(0.5)*&quot;$G$&quot;, ((x + z) + (x + y))/2);<br /> <br /> label(rotate(-3)*&quot;$W$&quot;, (x + z)/2 + trans);<br /> label(rotate(50)*slant(-1)*&quot;$B$&quot;, ((x + z) + (y + z))/2 + trans);<br /> label(rotate(35)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (x + y))/2 + trans);<br /> <br /> label(rotate(-3)*&quot;$P$&quot;, (x + z)/2 + 2*trans);<br /> label(rotate(-5)*slant(0.5)*&quot;$R$&quot;, ((x + z) + (y + z))/2 + 2*trans);<br /> label(rotate(-85)*slant(-1)*&quot;$G$&quot;, ((x + z) + (x + y))/2 + 2*trans);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\times{(A) }4\qquad\textbf{(B) }1\qquad\textbf{(C) }64\qquad\textbf{(e) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> == Problem 1<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> [[2019 AMC 8 Problems/Problem 24|Solution]]</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=136093 2017 AMC 8 Problems/Problem 19 2020-10-30T00:32:19Z <p>Clapatron7568: /* Solution 1 */</p> <hr /> <div>==Problem 19==<br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> Factoring out &lt;math&gt;98!+99!+100!&lt;/math&gt;, we have &lt;math&gt;98!(1+99+99*100)&lt;/math&gt; which is &lt;math&gt;98!(10000)&lt;/math&gt; Next, &lt;math&gt;98!&lt;/math&gt; has &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;. The &lt;math&gt;19&lt;/math&gt; is because of all the multiples of &lt;math&gt;5&lt;/math&gt;. Now &lt;math&gt;10,000&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;, so there are a total of &lt;math&gt;22 + 4 = \boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The number of &lt;math&gt;5&lt;/math&gt;'s in the factorization of &lt;math&gt;98! + 99! + 100!&lt;/math&gt; is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by &lt;math&gt;5&lt;/math&gt;, until you can't divide by &lt;math&gt;5&lt;/math&gt; anymore. Factorizing &lt;math&gt;98! + 99! + 100!&lt;/math&gt;, you get &lt;math&gt;98!(1+99+9900)=98!(10000)&lt;/math&gt;. To find the number of trailing zeroes in 98!, we do &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22&lt;/math&gt;. Now since &lt;math&gt;10000&lt;/math&gt; has 4 zeroes, we add &lt;math&gt;22 + 4&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=136092 2017 AMC 8 Problems/Problem 19 2020-10-30T00:30:55Z <p>Clapatron7568: /* Solution 1 */</p> <hr /> <div>==Problem 19==<br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> Factoring out &lt;math&gt;98!+99!+100!&lt;/math&gt;, we have &lt;math&gt;98!(1+99+99*100)&lt;/math&gt; which is &lt;math&gt;98!(1000)&lt;/math&gt; Next, &lt;math&gt;98!&lt;/math&gt; has &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;. The &lt;math&gt;19&lt;/math&gt; is because of all the multiples of &lt;math&gt;5&lt;/math&gt;. Now &lt;math&gt;10,000&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;, so there are a total of &lt;math&gt;22 + 4 = \boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The number of &lt;math&gt;5&lt;/math&gt;'s in the factorization of &lt;math&gt;98! + 99! + 100!&lt;/math&gt; is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by &lt;math&gt;5&lt;/math&gt;, until you can't divide by &lt;math&gt;5&lt;/math&gt; anymore. Factorizing &lt;math&gt;98! + 99! + 100!&lt;/math&gt;, you get &lt;math&gt;98!(1+99+9900)=98!(10000)&lt;/math&gt;. To find the number of trailing zeroes in 98!, we do &lt;math&gt;\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22&lt;/math&gt;. Now since &lt;math&gt;10000&lt;/math&gt; has 4 zeroes, we add &lt;math&gt;22 + 4&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(D)}\ 26}&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_4&diff=135882 2020 AMC 10A Problems/Problem 4 2020-10-26T20:20:05Z <p>Clapatron7568: /* Solution */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #3]] and [[2020 AMC 10A Problems|2020 AMC 10A #4]]}}<br /> <br /> ==Problem==<br /> <br /> A driver travels for &lt;math&gt;2&lt;/math&gt; hours at &lt;math&gt;60&lt;/math&gt; miles per hour, during which her car gets &lt;math&gt;30&lt;/math&gt; miles per gallon of gasoline. She is paid &lt;math&gt;\$0.50&lt;/math&gt; per mile, and her only expense is gasoline at &lt;math&gt;\$2.00&lt;/math&gt; per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the driver travels 60 miles per hour and each hour she uses 2 gallons of gasoline, she spends \$4 per hour on gas. If she gets \$0.50 per mile, then she gets \\$30 per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is &lt;math&gt;\boxed{\textbf{(E)}\ 26}&lt;/math&gt;.<br /> ~mathsmiley<br /> <br /> ==Video Solution==<br /> https://youtu.be/WUcbVNy2uv0<br /> <br /> ~IceMatrix<br /> <br /> https://www.youtube.com/watch?v=7-3sl1pSojc<br /> <br /> ~bobthefam<br /> <br /> https://youtu.be/Dj_DFoZO-xw<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=3|num-a=5}}<br /> {{AMC12 box|year=2020|ab=A|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Clapatron7568 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_9&diff=135700 2013 AMC 8 Problems/Problem 9 2020-10-23T21:55:34Z <p>Clapatron7568: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9^\text{th} \qquad \textbf{(B)}\ 10^\text{th} \qquad \textbf{(C)}\ 11^\text{th} \qquad \textbf{(D)}\ 12^\text{th} \qquad \textbf{(E)}\ 13^\text{th}&lt;/math&gt;<br /> <br /> ==Solution==<br /> This is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that &lt;math&gt;2^{10}=1024&lt;/math&gt;.<br /> <br /> However, because the first term is &lt;math&gt;2^0=1&lt;/math&gt; and not &lt;math&gt;2^1=2&lt;/math&gt;, the solution to the problem is &lt;math&gt;10-0+1=\boxed{\textbf{(C)}\ 11^{\text{th}}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can also solve this problem by listing out how far the Hulk jumps on each jump: on the first jump, he goes 1 meter, the second jump 2 meters, and so on. Listing out these numbers, we get:<br /> <br /> &lt;math&gt;1, 2, 4, 8, 16, 32, 64, 128, 256, 512, \boxed{\textbf{1024}}&lt;/math&gt;<br /> <br /> On the 11th jump, the Hulk jumps 1024 meters &gt; 1000 meters (1 kilometer), so our answer is the 11th jump, or &lt;math&gt;\boxed{\textbf{(C)}}.&lt;/math&gt;<br /> <br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Clapatron7568