https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Herobrine-india&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-24T06:53:37Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_12&diff=152981 2020 AMC 10A Problems/Problem 12 2021-04-30T06:17:53Z <p>Herobrine-india: </p> <hr /> <div>== Problem 12 == <br /> Triangle &lt;math&gt;AMC&lt;/math&gt; is isosceles with &lt;math&gt;AM = AC&lt;/math&gt;. Medians &lt;math&gt;\overline{MV}&lt;/math&gt; and &lt;math&gt;\overline{CU}&lt;/math&gt; are perpendicular to each other, and &lt;math&gt;MV=CU=12&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0, 3.6), S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> == Solution 0 (High IQ) ==<br /> &lt;math&gt;[\square MUVC] = 72&lt;/math&gt;. Let intersection of line &lt;math&gt;AP&lt;/math&gt; and base &lt;math&gt;MC&lt;/math&gt; be &lt;math&gt;B&lt;/math&gt; &lt;cmath&gt; [AUV]=[MUB]=[UVB]=[BVC] \implies \left[\frac{\triangle AMC}{4}\right] = \left[\frac{\square MUVC}{3}\right] \implies [AMC] = 96&lt;/cmath&gt;<br /> <br /> ~herobrine-india <br /> <br /> <br /> == Solution 1 ==<br /> Since quadrilateral &lt;math&gt;UVCM&lt;/math&gt; has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that &lt;math&gt;\triangle AUV&lt;/math&gt; has &lt;math&gt;\frac 14&lt;/math&gt; the area of triangle &lt;math&gt;AMC&lt;/math&gt; by similarity, so &lt;math&gt;[UVCM]=\frac 34\cdot [AMC].&lt;/math&gt; Thus,<br /> &lt;cmath&gt;\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]&lt;/cmath&gt;<br /> &lt;cmath&gt;72=\frac 34\cdot [AMC]&lt;/cmath&gt;<br /> &lt;cmath&gt;[AMC]=96\rightarrow \boxed{\textbf{(C)}}.&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Trapezoid)==<br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> draw((-2,6)--(2,6));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0, 3.6), S);<br /> &lt;/asy&gt;<br /> <br /> We know that &lt;math&gt;\triangle AUV \sim \triangle AMC&lt;/math&gt;, and since the ratios of its sides are &lt;math&gt;\frac{1}{2}&lt;/math&gt;, the ratio of of their areas is &lt;math&gt;\left(\frac{1}{2}\right)^2=\frac{1}{4}&lt;/math&gt;. <br /> <br /> If &lt;math&gt;\triangle AUV&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;\triangle AMC&lt;/math&gt;, then trapezoid &lt;math&gt;MUVC&lt;/math&gt; is &lt;math&gt;\frac{3}{4}&lt;/math&gt; the area of &lt;math&gt;\triangle AMC&lt;/math&gt;. <br /> <br /> Let's call the intersection of &lt;math&gt;\overline{UC}&lt;/math&gt; and &lt;math&gt;\overline{MV}&lt;/math&gt; &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;\overline{UP}=x&lt;/math&gt;. Then &lt;math&gt;\overline{PC}=12-x&lt;/math&gt;. Since &lt;math&gt;\overline{UC} <br /> \perp \overline{MV}&lt;/math&gt;, &lt;math&gt;\overline{UP}&lt;/math&gt; and &lt;math&gt;\overline{CP}&lt;/math&gt; are heights of triangles &lt;math&gt;\triangle MUV&lt;/math&gt; and &lt;math&gt;\triangle MCV&lt;/math&gt;, respectively. Both of these triangles have base &lt;math&gt;12&lt;/math&gt;. <br /> <br /> Area of &lt;math&gt;\triangle MUV = \frac{x\cdot12}{2}=6x&lt;/math&gt;<br /> <br /> Area of &lt;math&gt;\triangle MCV = \frac{(12-x)\cdot12}{2}=72-6x&lt;/math&gt;<br /> <br /> Adding these two gives us the area of trapezoid &lt;math&gt;MUVC&lt;/math&gt;, which is &lt;math&gt;6x+(72-6x)=72&lt;/math&gt;.<br /> <br /> This is &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the triangle, so the area of the triangle is &lt;math&gt;\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}&lt;/math&gt; ~quacker88, diagram by programjames1<br /> <br /> ==Solution 3 (Medians)==<br /> Draw median &lt;math&gt;\overline{AB}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> draw((0,12)--(0,0));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0.5, 4), E);<br /> label(&quot;B&quot;, (0, 0), S);<br /> &lt;/asy&gt;<br /> <br /> Since we know that all medians of a triangle intersect at the centroid, we know that &lt;math&gt;\overline{AB}&lt;/math&gt; passes through point &lt;math&gt;P&lt;/math&gt;. We also know that medians of a triangle divide each other into segments of ratio &lt;math&gt;2:1&lt;/math&gt;. Knowing this, we can see that &lt;math&gt;\overline{PC}:\overline{UP}=2:1&lt;/math&gt;, and since the two segments sum to &lt;math&gt;12&lt;/math&gt;, &lt;math&gt;\overline{PC}&lt;/math&gt; and &lt;math&gt;\overline{UP}&lt;/math&gt; are &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, respectively.<br /> <br /> Finally knowing that the medians divide the triangle into &lt;math&gt;6&lt;/math&gt; sections of equal area, finding the area of &lt;math&gt;\triangle PUM&lt;/math&gt; is enough. &lt;math&gt;\overline{PC} = \overline{MP} = 8&lt;/math&gt;.<br /> <br /> The area of &lt;math&gt;\triangle PUM = \frac{4\cdot8}{2}=16&lt;/math&gt;. Multiplying this by &lt;math&gt;6&lt;/math&gt; gives us &lt;math&gt;6\cdot16=\boxed{\textbf{(C) }96}&lt;/math&gt; <br /> <br /> ~quacker88<br /> <br /> ==Solution 4 (Triangles)==<br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> draw((-2,6)--(2,6));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0, 3.6), S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AU = UM&lt;/math&gt;, &lt;math&gt;AV = VC&lt;/math&gt;, so &lt;math&gt;UV = \frac{1}{2} MC&lt;/math&gt;.<br /> <br /> As &lt;math&gt;\angle UPM = \angle VPC = 90&lt;/math&gt;, we can see that &lt;math&gt;\triangle UPM \cong \triangle VPC&lt;/math&gt; and &lt;math&gt;\triangle UVP \sim \triangle MPC&lt;/math&gt; with a side ratio of &lt;math&gt;1 : 2&lt;/math&gt;.<br /> <br /> So &lt;math&gt;UP = VP = 4&lt;/math&gt;, &lt;math&gt;MP = PC = 8&lt;/math&gt;.<br /> <br /> With that, we can see that &lt;math&gt;[\triangle UPM] = 16&lt;/math&gt;, and the area of trapezoid &lt;math&gt;MUVC&lt;/math&gt; is 72.<br /> <br /> As said in solution 1, &lt;math&gt;[\triangle AMC] = 72 / \frac{3}{4} = \boxed{\textbf{(C) } 96}&lt;/math&gt;.<br /> <br /> -QuadraticFunctions, solution 1 by ???<br /> <br /> ==Solution 5 (Only Pythagorean Theorem)==<br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> draw((0,12)--(0,0));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0.5, 4), E);<br /> label(&quot;B&quot;, (0, 0), S);<br /> <br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;AB&lt;/math&gt; be the height. Since medians divide each other into a &lt;math&gt;2:1&lt;/math&gt; ratio, and the medians have length 12, we have &lt;math&gt;PC=MP=8&lt;/math&gt; and &lt;math&gt;UP=VP=4&lt;/math&gt;. From right triangle &lt;math&gt;\triangle{MUP}&lt;/math&gt;, &lt;cmath&gt;MU^2=MP^2+UP^2=8^2+4^2=80,&lt;/cmath&gt; so &lt;math&gt;MU=\sqrt{80}=4\sqrt{5}&lt;/math&gt;. Since &lt;math&gt;CU&lt;/math&gt; is a median, &lt;math&gt;AM=8\sqrt{5}&lt;/math&gt;. From right triangle &lt;math&gt;\triangle{MPC}&lt;/math&gt;, &lt;cmath&gt;MC^2=MP^2+PC^2=8^2+8^2=128,&lt;/cmath&gt; which implies &lt;math&gt;MC=\sqrt{128}=8\sqrt{2}&lt;/math&gt;. By symmetry &lt;math&gt;MB=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}&lt;/math&gt;. <br /> <br /> Applying the Pythagorean Theorem to right triangle &lt;math&gt;\triangle{MAB}&lt;/math&gt; gives &lt;math&gt;AB^2=AM^2-MB^2=8\sqrt{5}^2-4\sqrt{2}^2=288&lt;/math&gt;, so &lt;math&gt;AB=\sqrt{288}=12\sqrt{2}&lt;/math&gt;. Then the area of &lt;math&gt;\triangle{AMC}&lt;/math&gt; is &lt;cmath&gt;\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}&lt;/cmath&gt;<br /> <br /> ==Solution 6 (Drawing)==<br /> By similarity, the area of &lt;math&gt;AUV&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{4}&lt;/math&gt;.<br /> <br /> The area of &lt;math&gt;UVCM&lt;/math&gt; is equal to 72.<br /> <br /> Assuming the total area of the triangle is S, the equation will be : &lt;math&gt;\frac{3}{4}&lt;/math&gt;S = 72.<br /> <br /> S = &lt;math&gt;\boxed{96}&lt;/math&gt;<br /> <br /> ==Solution 7==<br /> Given a triangle with perpendicular medians with lengths &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, the area will be &lt;math&gt;\frac{2xy}{3}=\boxed{\textbf{(C) }96}&lt;/math&gt;.<br /> <br /> ==Solution 8 (Fastest)==<br /> Connect the line segment &lt;math&gt;UV&lt;/math&gt; and it's easy to see quadrilateral &lt;math&gt;UVMC&lt;/math&gt; has an area of the product of its diagonals divided by &lt;math&gt;2&lt;/math&gt; which is &lt;math&gt;72&lt;/math&gt;. Now, solving for triangle &lt;math&gt;AUV&lt;/math&gt; could be an option, but the drawing shows the area of &lt;math&gt;AUV&lt;/math&gt; will be less than the quadrilateral meaning the the area of &lt;math&gt;AMC&lt;/math&gt; is less than &lt;math&gt;72*2&lt;/math&gt; but greater than &lt;math&gt;72&lt;/math&gt;, leaving only one possible answer choice, &lt;math&gt;\boxed{\textbf{(C) } 96}&lt;/math&gt;.<br /> <br /> -Rohan S.<br /> <br /> ==Solution 9==<br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> draw((0,12)--(0,0));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0.5, 4), E);<br /> label(&quot;B&quot;, (0, 0), S);<br /> &lt;/asy&gt;<br /> <br /> Connect &lt;math&gt;AP&lt;/math&gt;, and let &lt;math&gt;B&lt;/math&gt; be the point where &lt;math&gt;AP&lt;/math&gt; intersects &lt;math&gt;MC&lt;/math&gt;. &lt;math&gt;MB=CB&lt;/math&gt; because all medians of a triangle intersect at one point, which in this case is &lt;math&gt;P&lt;/math&gt;. &lt;math&gt;MP:PV=2:1&lt;/math&gt; because the point at which all medians intersect divides the medians into segments of ratio &lt;math&gt;2:1&lt;/math&gt;, so &lt;math&gt;MP=8&lt;/math&gt; and similarly &lt;math&gt;CP=8&lt;/math&gt;. We apply the Pythagorean Theorem to triangle &lt;math&gt;MPC&lt;/math&gt; and get &lt;math&gt;MC=\sqrt{128}=8\sqrt{2}&lt;/math&gt;. The area of triangle &lt;math&gt;MPC&lt;/math&gt; is &lt;math&gt;\dfrac{MP\cdot CP}{2}=32&lt;/math&gt;, and that must equal to &lt;math&gt;\dfrac{MC\cdot BP}{2}&lt;/math&gt;, so &lt;math&gt;BP=\dfrac{8}{\sqrt{2}}=4\cdot\sqrt{2}&lt;/math&gt;. &lt;math&gt;BP=\dfrac{1}{3}BA&lt;/math&gt;, so &lt;math&gt;BA=12\sqrt{2}&lt;/math&gt;. The area of triangle &lt;math&gt;AMC&lt;/math&gt; is equal to &lt;math&gt;\dfrac{MC\cdot BA}{2}=\dfrac{8 \cdot \sqrt{2} \cdot 12 \cdot \sqrt{2}}{2}=\boxed{\textbf{(C)} 96}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Video Solution 1==<br /> <br /> Education, The Study of Everything<br /> <br /> https://youtu.be/0TslJ3aDXac<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution 3==<br /> https://youtu.be/7ZvKOYuwSnE<br /> <br /> ~savannahsolver<br /> <br /> == Video Solution 4 ==<br /> https://youtu.be/4_x1sgcQCp4?t=2067<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Herobrine-india https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10D&diff=152980 2021 AMC 10D 2021-04-30T05:55:49Z <p>Herobrine-india: </p> <hr /> <div>The 2021 AMC 10D will be administered in November 2021.</div> Herobrine-india https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_5&diff=152477 2020 AMC 12B Problems/Problem 5 2021-04-23T06:39:54Z <p>Herobrine-india: </p> <hr /> <div>==Problem==<br /> <br /> Teams &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team &lt;math&gt;A&lt;/math&gt; has won &lt;math&gt;\tfrac{2}{3}&lt;/math&gt; of its games and team &lt;math&gt;B&lt;/math&gt; has won &lt;math&gt;\tfrac{5}{8}&lt;/math&gt; of its games. Also, team &lt;math&gt;B&lt;/math&gt; has won &lt;math&gt;7&lt;/math&gt; more games and lost &lt;math&gt;7&lt;/math&gt; more games than team &lt;math&gt;A.&lt;/math&gt; How many games has team &lt;math&gt;A&lt;/math&gt; played?<br /> <br /> &lt;math&gt;\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> First, let us assign some variables. Let<br /> <br /> &lt;cmath&gt;A_w=2x, A_l=x, A_g=3x,&lt;/cmath&gt;<br /> &lt;cmath&gt;B_w=5y, B_l=3y, B_g=8y,&lt;/cmath&gt;<br /> <br /> where &lt;math&gt;X_w&lt;/math&gt; denotes number of games won, &lt;math&gt;X_l&lt;/math&gt; denotes number of games lost, and &lt;math&gt;X_g&lt;/math&gt; denotes total games played for &lt;math&gt;X\in \{A, B\}&lt;/math&gt;. Using the given information, we can set up the following two equations:<br /> <br /> &lt;cmath&gt;B_w=A_w+7\implies 5y=2x+7,&lt;/cmath&gt; <br /> &lt;cmath&gt;B_l=A_l+7\implies 3y=x+7.&lt;/cmath&gt;<br /> <br /> We can solve through substitution, as the second equation can be written as &lt;math&gt;x=3y-7&lt;/math&gt;, and plugging this into the first equation gives &lt;math&gt;5y=6y-7\implies y=7&lt;/math&gt;, which means &lt;math&gt;x=3(7)-7=14&lt;/math&gt;. Finally, we want the total number of games team &lt;math&gt;A&lt;/math&gt; has played, which is &lt;math&gt;A_g=3(14)=\boxed{\textbf{(C) } 42}&lt;/math&gt;.<br /> <br /> ~Argonauts16<br /> <br /> <br /> ==Solution 2==<br /> <br /> Using the information from the problem, we can note that team A has lost &lt;math&gt;\frac{1}{3}&lt;/math&gt; of their matches. Using the answer choices, we can find the following list of possible win-lose scenarios for &lt;math&gt;A&lt;/math&gt;, represented in the form &lt;math&gt;(w, l)&lt;/math&gt; for convenience:<br /> <br /> &lt;cmath&gt;A \implies (14, 7)&lt;/cmath&gt;<br /> &lt;cmath&gt;B \implies (18, 9)&lt;/cmath&gt;<br /> &lt;cmath&gt;C \implies (28, 14)&lt;/cmath&gt;<br /> &lt;cmath&gt;D \implies (32, 16)&lt;/cmath&gt;<br /> &lt;cmath&gt;E \implies (42, 21)&lt;/cmath&gt;<br /> <br /> Thus, we have 5 matching &lt;math&gt;B&lt;/math&gt; scenarios, simply adding 7 to &lt;math&gt;w&lt;/math&gt; and &lt;math&gt;l&lt;/math&gt;. We can then test each of the five &lt;math&gt;B&lt;/math&gt; scenarios for &lt;math&gt;\frac{w}{w+l} = \frac{5}{8}&lt;/math&gt; and find that &lt;math&gt;(35, 21)&lt;/math&gt; fits this description. Then working backwards and subtracting 7 from &lt;math&gt;w&lt;/math&gt; and &lt;math&gt;l&lt;/math&gt; gives us the point &lt;math&gt;(28, 14)&lt;/math&gt;, making the answer &lt;math&gt;\boxed{\textbf{C}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 3 (Answer Choices)==<br /> <br /> Let's say that team &lt;math&gt;A&lt;/math&gt; plays &lt;math&gt;n&lt;/math&gt; games in total. Therefore, team &lt;math&gt;B&lt;/math&gt; must play &lt;math&gt;n + 14&lt;/math&gt; games in total (7 wins, 7 losses) Since the ratio of &lt;math&gt;A&lt;/math&gt; is &lt;cmath&gt;\frac{2}{3} \implies n \equiv 0 \pmod{3}&lt;/cmath&gt; Similarly, since the ratio of &lt;math&gt;B&lt;/math&gt; is &lt;cmath&gt;\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}&lt;/cmath&gt; Now, we can go through the answer choices and see which ones work:<br /> <br /> &lt;cmath&gt;\textbf{(A) } 21 \implies 21 + 14 = 35 \not \equiv \pmod{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;\textbf{(B) } 27 \implies 27 + 14 = 41 \not \equiv \pmod{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;\textbf{(C) } 42 \implies 42 + 14 = 56 \equiv \pmod{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;\textbf{(D) } 48 \implies 48 + 14 = 62 \not \equiv \pmod{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv \pmod{8}&lt;/cmath&gt;<br /> <br /> So we can see &lt;math&gt;\boxed{\textbf{(C) } 42.0} \text{ \tiny nice}&lt;/math&gt; is the only valid answer.<br /> <br /> ~herobrine-india<br /> <br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/WfTty8Fe5Fo<br /> <br /> ~IceMatrix<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Herobrine-india https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_4&diff=142949 2012 AMC 10B Problems/Problem 4 2021-01-22T06:03:19Z <p>Herobrine-india: </p> <hr /> <div>== Problem 4 ==<br /> <br /> When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be leftover?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> <br /> == Solution 1 ==<br /> <br /> <br /> In total, there were &lt;math&gt;3+4=7&lt;/math&gt; marbles left from both Ringo and Paul.We know that &lt;math&gt;7 \equiv 1 \pmod{6}&lt;/math&gt;. This means that there would be &lt;math&gt;1&lt;/math&gt; marble leftover, or &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> == Solution 2 (modulo) ==<br /> Let &lt;math&gt;r&lt;/math&gt; be the number of marbles Ringo has and let &lt;math&gt;p&lt;/math&gt; be the number of marbles Paul has. we have the following equations:<br /> &lt;cmath&gt; r \equiv 4 \mod{6} &lt;/cmath&gt; <br /> &lt;cmath&gt; p \equiv 3 \mod{6} &lt;/cmath&gt;<br /> Adding these equations we get:<br /> &lt;cmath&gt; p + r \equiv 7 \mod{6} &lt;/cmath&gt;<br /> We know that &lt;math&gt;7 \equiv 1 \mod{6}&lt;/math&gt; so therefore:<br /> &lt;cmath&gt; p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6} &lt;/cmath&gt;<br /> Thus when Ringo and Paul pool their marbles, they will have &lt;math&gt;\boxed{\textbf{(A)}\ 1}&lt;/math&gt; marbles left over.<br /> <br /> ~ herobrine-india<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Herobrine-india https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_4&diff=115177 2011 AMC 10A Problems/Problem 4 2020-01-23T06:05:57Z <p>Herobrine-india: </p> <hr /> <div>==Problem==<br /> Let X and Y be the following sums of arithmetic sequences: <br /> <br /> &lt;cmath&gt; \begin{eqnarray*}X &amp;=&amp; 10+12+14+\cdots+100,\\ Y &amp;=&amp; 12+14+16+\cdots+102.\end{eqnarray*} &lt;/cmath&gt;<br /> <br /> What is the value of &lt;math&gt;Y - X? &lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We see that both sequences have equal numbers of terms, so reformat the sequence to look like: <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> Y = \ &amp;12 + 14 + \cdots + 100 + 102\\<br /> X = 10 \ + \ &amp;12 + 14 + \cdots + 100\\<br /> \end{align*}&lt;/cmath&gt;<br /> From here it is obvious that &lt;math&gt;Y - X = 102 - 10 = \boxed{92 \ \mathbf{(A)}}&lt;/math&gt;.<br /> <br /> ===Note===<br /> Another way to see this is to let the sum &lt;math&gt;12+14+16+...+100=x.&lt;/math&gt; So, the sequences become<br /> &lt;cmath&gt;\begin{align*}<br /> X = 10+x \\<br /> Y= x+102 \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Like before, the difference between the two sequences is &lt;math&gt;Y-X=102-12=92.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be:<br /> &lt;math&gt;46\cdot 2=\boxed{92}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> X&amp;=10+12+14+\cdots +100 \\<br /> Y&amp;=X-10+102 = X+92 \\<br /> Y-X &amp;= (X+92)-X \\<br /> Y-X &amp;= X-X+92 \\<br /> Y-X &amp;= 0+92 \\<br /> Y-X &amp;= \boxed{92} \quad \quad \textbf{(A)}\\<br /> \end{align*} &lt;/cmath&gt;<br /> &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> - &lt;math&gt;\text{herobrine-india}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> {{AMC10 box|year=2011|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Herobrine-india https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=110494 2002 AMC 8 Problems/Problem 25 2019-10-21T06:01:32Z <p>Herobrine-india: /* Solution */</p> <hr /> <div>==Problem==<br /> Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br /> <br /> &lt;math&gt; \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since Ott gets equal amounts of money from each friend, we can say that he gets &lt;math&gt;x&lt;/math&gt; dollars from each friend. This means that Moe has &lt;math&gt;5x&lt;/math&gt; dollars, Loki has &lt;math&gt;4x&lt;/math&gt; dollars, and Nick has &lt;math&gt;3x&lt;/math&gt; dollars. The total amount is &lt;math&gt;12x&lt;/math&gt; dollars, and since Ott gets &lt;math&gt;3x&lt;/math&gt; dollars total, &lt;math&gt;\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}&lt;/math&gt;.<br /> &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2002|num-b=24|after=Last &lt;br /&gt; Problem}}<br /> {{MAA Notice}}</div> Herobrine-india https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_25&diff=110493 2002 AMC 8 Problems/Problem 25 2019-10-21T06:00:18Z <p>Herobrine-india: /* Solution */</p> <hr /> <div>==Problem==<br /> Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br /> <br /> &lt;math&gt; \text{(A)}\ \frac{1}{10}\qquad\text{(B)}\ \frac{1}{4}\qquad\text{(C)}\ \frac{1}{3}\qquad\text{(D)}\ \frac{2}{5}\qquad\text{(E)}\ \frac{1}{2} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since Ott gets equal amounts of money from each friend, we can say that he gets &lt;math&gt;x&lt;/math&gt; dollars from each friend. This means that Moe has &lt;math&gt;5x&lt;/math&gt; dollars, Loki has &lt;math&gt;4x&lt;/math&gt; dollars, and Nick has &lt;math&gt;3x&lt;/math&gt; dollars. The total amount is &lt;math&gt;12x&lt;/math&gt; dollars, and since Ott gets &lt;math&gt;3x&lt;/math&gt; dollars total, &lt;math&gt;\frac{3x}{12x}= \frac{3}{12} = \boxed{\text{(B)}\ \frac14}&lt;/math&gt;.<br /> \blacksquare<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2002|num-b=24|after=Last &lt;br /&gt; Problem}}<br /> {{MAA Notice}}</div> Herobrine-india