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Difference between revisions of "1985 AJHSME Problems/Problem 6"

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<math>\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250</math>
 
<math>\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250</math>
  
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==Solution==
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==Solution 1==
−
 
 
 
We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
 
We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
  
−
Let's say that <math>\frac{500 \text{ sheets of paper}}{5 \text{ cm}} = 1</math>. So by multiplying <math>7.5 \text{ cm}</math> by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
+
Let's say that <math>500\text{ sheets}=5\text{ cm}\Rightarrow \frac{500 \text{ sheets}}{5 \text{ cm}} = 1</math>. So by multiplying <math>7.5 \text{ cm}</math> by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{7.5 \times 500}{5} &= 7.5 \times 100 \\
 
\frac{7.5 \times 500}{5} &= 7.5 \times 100 \\
−
&= 750 sheets \\
+
&= 750 \text{ sheets} \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
 
<math>750</math> is <math>\boxed{\text{D}}</math>
 
<math>750</math> is <math>\boxed{\text{D}}</math>
 +
 +
==Solution 2==
 +
We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness.
 +
Our proportion is <cmath>\frac{5}{500}=\frac{7.5}{x}</cmath> where <math>x</math> is the number we are looking for.
 +
Next, we cross-multiply to get <math>5x=500 \times 7.5</math> so <math>x=750</math> which is <math>\boxed{\text{D}}</math>
 +
 +
~GrantStar
 +
 +
==Solution 3 (Very quick, use this to do the problem quickly)==
 +
We immediately see <math>500</math> sheets of paper per <math>5</math>cm is <math>\frac{500}{5}</math>, which can be simplified to <math>\frac{100}{1}</math>. The denominator is  supposed to be <math>7.5</math>. We can now just multiply the numerator, <math>100</math>, by <math>7.5</math>, to get <math>\boxed{\textbf{(D)}\ 750}</math>.
 +
~[[shunyipanda]]
 +
 +
==Video Solution==
 +
https://youtu.be/Dp1i5sCWN_c
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
  
−
[[1985 AJHSME Problems]]
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{{AJHSME box|year=1985|num-b=5|num-a=7}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
 
 +
 
 +
{{MAA Notice}}

Latest revision as of 18:13, 23 October 2025

Problem

A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?

$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$

Solution 1

We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.

Let's say that $500\text{ sheets}=5\text{ cm}\Rightarrow \frac{500 \text{ sheets}}{5 \text{ cm}} = 1$. So by multiplying $7.5 \text{ cm}$ by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets

\begin{align*} \frac{7.5 \times 500}{5} &= 7.5 \times 100 \\ &= 750 \text{ sheets} \\ \end{align*}

$750$ is $\boxed{\text{D}}$

Solution 2

We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness. Our proportion is \[\frac{5}{500}=\frac{7.5}{x}\] where $x$ is the number we are looking for. Next, we cross-multiply to get $5x=500 \times 7.5$ so $x=750$ which is $\boxed{\text{D}}$

~GrantStar

Solution 3 (Very quick, use this to do the problem quickly)

We immediately see $500$ sheets of paper per $5$cm is $\frac{500}{5}$, which can be simplified to $\frac{100}{1}$. The denominator is supposed to be $7.5$. We can now just multiply the numerator, $100$, by $7.5$, to get $\boxed{\textbf{(D)}\ 750}$. ~shunyipanda

Video Solution

https://youtu.be/Dp1i5sCWN_c

~savannahsolver

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions


These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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