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> | ||
− | ==Solution== | + | ==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. | ||
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<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 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Dp1i5sCWN_c | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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+ | {{MAA Notice}} |
Latest revision as of 08:22, 10 January 2023
Problem
A ream of paper containing sheets is cm thick. Approximately how many sheets of this type of paper would there be in a stack cm high?
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 . So by multiplying by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
is
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 where is the number we are looking for. Next, we cross-multiply to get so which is
~GrantStar
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.