Difference between revisions of "1985 AHSME Problems/Problem 4"
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==Problem== | ==Problem== | ||
− | A large bag of coins contains pennies, dimes | + | A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is |
− | <math> \mathrm{(A)\ } | + | <math> \mathrm{(A)\ } \$306 \qquad \mathrm{(B) \ } \$333 \qquad \mathrm{(C)\ } \$342 \qquad \mathrm{(D) \ } \$348 \qquad \mathrm{(E) \ } \$360 </math> |
==Solution== | ==Solution== | ||
− | Let there be <math> | + | Let there be <math> x </math> pennies. Therefore, there are <math> 2x </math> dimes and <math> 3(2x)=6x </math> quarters. Since pennies are \$<math> 0.01 </math>, dimes are \$<math> 0.10 </math>, and quarters are \$<math> 0.25 </math>, the total amount of money is \$<math> 0.01x+(2)(0.10x)+(6)(0.25x)= </math>\$<math> 1.71x </math>. Therefore, any amount of money must be a multiple of \$<math> 1.71 </math>. |
− | Since the answer choices are all whole dollar amounts, we multiply by <math> 100 </math> to get a whole dollar number: | + | Since the answer choices are all whole dollar amounts, we multiply by <math> 100 </math> to get a whole dollar number: \$<math>171 </math>. Notice that twice this is \$<math> 342 </math>, which is answer choice <math> \boxed{\text{C}} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=3|num-a=5}} | {{AHSME box|year=1985|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:32, 3 April 2018
Problem
A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is
Solution
Let there be pennies. Therefore, there are dimes and quarters. Since pennies are $, dimes are $, and quarters are $, the total amount of money is $$. Therefore, any amount of money must be a multiple of $.
Since the answer choices are all whole dollar amounts, we multiply by to get a whole dollar number: $. Notice that twice this is $, which is answer choice .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.