https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Junkmail&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T07:21:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_7&diff=489252000 AMC 8 Problems/Problem 72012-10-27T16:29:36Z<p>Junkmail: /* Solution */ wrong number in solution box</p>
<hr />
<div>==Problem==<br />
<br />
What is the minimum possible product of three different numbers of the set <math>\{-8,-6,-4,0,3,5,7\}</math>?<br />
<br />
<math>\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0</math><br />
<br />
==Solution==<br />
<br />
The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices<br />
exist: <math>(-8)\times(-6)\times(-4) = (-8)\times(24) = -192</math> and <math>(-8)\times5\times7 = (-8)\times35 = -280</math>.<br />
The latter is smaller, so <math>\boxed{\text{(B) -280}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=6|num-a=8}}</div>Junkmailhttps://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_9&diff=489111999 AMC 8 Problems/Problem 92012-10-24T02:12:31Z<p>Junkmail: Removed superfluos "okay"</p>
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<div>Given that the area common to A and C has 100 plants, and C has a total of 350 plants, there are 250 plants in the area exclusive to C. Area B has a total of 450 plants, but 50 of those are shared with Area A, so the number of plants exclusive to Area B is <math>450 - 50 = 400</math>. Lastly, in all of Area A, there are 500 plants, but 50 of those are shared with Area C, and 100 of those plants are shared with Area B. So the number of plants in the area exclusive to Area A is <math>500 - 100 - 50 = 400 - 50 = 350</math>. None of the plants are double-counted, so to find the total, the number of plants in all of the zones are added together: <math>250 + 100 + 350 + 50 + 400 = 350 + 400 + 400 = 750 + 400 = 1150</math> plants total.</div>Junkmailhttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_20&diff=488191986 AJHSME Problems/Problem 202012-10-10T00:59:49Z<p>Junkmail: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The value of the expression <math>\frac{(304)^5}{(29.7)(399)^4}</math> is closest to<br />
<br />
<math>\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30</math><br />
<br />
==Solution==<br />
<br />
<cmath> \frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}</cmath><br />
Which is closest to <math>3\rightarrow\boxed{\text{D}}</math>.<br />
<br />
(The original expression is approximately equal to <math>3.44921198</math>.)<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=19|num-a=21}}<br />
[[Category:Introductory Algebra Problems]]</div>Junkmailhttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_14&diff=488171986 AJHSME Problems/Problem 142012-10-09T01:16:30Z<p>Junkmail: /* Solution */ removed the word "obviously" where unnecessary.</p>
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<div>==Problem==<br />
<br />
If <math>200\leq a \leq 400</math> and <math>600\leq b\leq 1200</math>, then the largest value of the quotient <math>\frac{b}{a}</math> is<br />
<br />
<math>\text{(A)}\ \frac{3}{2} \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 300 \qquad \text{(E)}\ 600</math><br />
<br />
==Solution==<br />
<br />
<math>\frac{b}{a}</math> will be largest if <math>b</math> is the largest it can be, and <math>a</math> is the smallest it can be.<br />
<br />
Since <math>b</math> can be no larger than <math>1200</math>, <math>b = 1200</math>. Since <math>a</math> can be no less than <math>200</math>, <math>a = 200</math>. <math>\frac{1200}{200} = 6</math><br />
<br />
<math>6</math> is <math>\boxed{\text{C}}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=13|num-a=15}}<br />
[[Category:Introductory Algebra Problems]]</div>Junkmailhttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_13&diff=488161986 AJHSME Problems/Problem 132012-10-09T01:15:16Z<p>Junkmail: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
The perimeter of the polygon shown is <br />
<br />
<center><br />
<asy><br />
draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle);<br />
label("$6$",(0,3),W);<br />
label("$8$",(4,6),N);<br />
draw((0.5,0)--(0.5,0.5)--(0,0.5));<br />
draw((0.5,6)--(0.5,5.5)--(0,5.5));<br />
draw((7.5,6)--(7.5,5.5)--(8,5.5));<br />
draw((7.5,3)--(7.5,3.5)--(8,3.5));<br />
draw((2.2,0)--(2.2,0.5)--(2.7,0.5));<br />
draw((2.7,2.5)--(3.2,2.5)--(3.2,3));<br />
</asy><br />
</center><br />
<br />
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48</math><br />
<br />
<math>\text{(E)}\ \text{cannot be determined from the information given}</math><br />
<br />
==Solution==<br />
<br />
===Solution 1===<br />
<br />
For the segments parallel to the side with side length 8, let's call those two segments <math>a</math> and <math>b</math>, the longer segment being <math>b</math>, the shorter one being <math>a</math>.<br />
<br />
For the segments parallel to the side with side length 6, let's call those two segments <math>c</math> and <math>d</math>, the longer segment being <math>d</math>, the shorter one being <math>c</math>.<br />
<br />
So the perimeter of the polygon would be...<br />
<br />
<math>8 + 6 + a + b + c + d</math><br />
<br />
Note that <math>a + b = 8</math>, and <math>c + d = 6</math>.<br />
<br />
Now we plug those in:<br />
<cmath>\begin{align*}<br />
8 + 6 + a + b + c + d &= 8 + 6 + 8 + 6 \\<br />
&= 14 \times 2 \\<br />
&= 28 \\<br />
\end{align*}</cmath><br />
<br />
28 is <math>\boxed{\text{C}}</math>.<br />
<br />
===Solution 2===<br />
<br />
<asy><br />
unitsize(12);<br />
draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle);<br />
label("$6$",(0,3),W);<br />
label("$8$",(4,6),N);<br />
draw((8,3)--(8,0)--(2.7,0),dashed);<br />
</asy><br />
<br />
The perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion. This makes the answer <math>2(6+8)=28\rightarrow \boxed{\text{C}}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=12|num-a=14}}<br />
[[Category:Introductory Geometry Problems]]</div>Junkmailhttps://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_12&diff=488151986 AJHSME Problems/Problem 122012-10-09T01:14:27Z<p>Junkmail: /* Solution */ Removed unnecessary text.</p>
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<div>==Problem==<br />
<br />
The table below displays the grade distribution of the <math>30</math> students in a mathematics class on the last two tests. For example, exactly one student received a 'D' on Test 1 and a 'C' on Test 2 (see circled entry). What percent of the students received the same grade on both tests?<br />
<br />
<asy><br />
draw((2,0)--(7,0)--(7,5)--(2,5)--cycle);<br />
draw((3,0)--(3,5));<br />
draw((4,0)--(4,5));<br />
draw((5,0)--(5,5));<br />
draw((6,0)--(6,5));<br />
draw((2,1)--(7,1));<br />
draw((2,2)--(7,2));<br />
draw((2,3)--(7,3));<br />
draw((2,4)--(7,4));<br />
draw((.2,6.8)--(1.8,5.2));<br />
draw(circle((4.5,1.5),.5),linewidth(.6 mm));<br />
label("0",(2.5,.2),N);<br />
label("0",(3.5,.2),N);<br />
label("2",(4.5,.2),N);<br />
label("1",(5.5,.2),N);<br />
label("0",(6.5,.2),N);<br />
label("0",(2.5,1.2),N);<br />
label("0",(3.5,1.2),N);<br />
label("1",(4.5,1.2),N);<br />
label("1",(5.5,1.2),N);<br />
label("1",(6.5,1.2),N);<br />
label("1",(2.5,2.2),N);<br />
label("3",(3.5,2.2),N);<br />
label("5",(4.5,2.2),N);<br />
label("2",(5.5,2.2),N);<br />
label("0",(6.5,2.2),N);<br />
label("1",(2.5,3.2),N);<br />
label("4",(3.5,3.2),N);<br />
label("3",(4.5,3.2),N);<br />
label("0",(5.5,3.2),N);<br />
label("0",(6.5,3.2),N);<br />
label("2",(2.5,4.2),N);<br />
label("2",(3.5,4.2),N);<br />
label("1",(4.5,4.2),N);<br />
label("0",(5.5,4.2),N);<br />
label("0",(6.5,4.2),N);<br />
label("F",(1.5,.2),N);<br />
label("D",(1.5,1.2),N);<br />
label("C",(1.5,2.2),N);<br />
label("B",(1.5,3.2),N);<br />
label("A",(1.5,4.2),N);<br />
label("A",(2.5,5.2),N);<br />
label("B",(3.5,5.2),N);<br />
label("C",(4.5,5.2),N);<br />
label("D",(5.5,5.2),N);<br />
label("F",(6.5,5.2),N);<br />
label("Test 1",(-.5,5.2),N);<br />
label("Test 2",(2.6,6),N);<br />
</asy><br />
<br />
<math>\text{(A)}\ 12\% \qquad \text{(B)}\ 25\% \qquad \text{(C)}\ 33\frac{1}{3}\% \qquad \text{(D)}\ 40\% \qquad \text{(E)}\ 50\% </math><br />
<br />
==Solution==<br />
<br />
We need to find the number of those who did get the same on both tests over 30 (the number of students in the class).<br />
<br />
So, we have <cmath>\frac{2 + 4 + 5 + 1}{30}</cmath><br />
<br />
Which simplifies to <cmath>\frac{12}{30} = \frac{4}{10} = \frac{40}{100} = 40 \%</cmath><br />
<br />
<math>\boxed{\text{D}}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=11|num-a=13}}<br />
[[Category:Introductory Algebra Problems]]</div>Junkmail