Difference between revisions of "1985 AHSME Problems/Problem 4"
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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 | 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== | ||
| − | + | If there are <math>x</math> pennies in the bag, then 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 in the bag is <cmath>\$ \left(0.01x+(0.10)(2x)+(0.25)(6x)\right) = \$1.71x.</cmath> Therefore, the possible amounts of money are precisely the integer multiples of <math>\$1.71</math>. | |
| − | Since the answer choices are all | + | Since the answer choices are all integer numbers of dollars, we multiply by <math>100</math> to deduce that the answer must be an integer multiple of <math>\$171</math>. The only such multiple among the answer choices is <math>\$(2 \cdot 171) = \boxed{\text{(C)} \ \$342}</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}} | ||
Latest revision as of 17:47, 19 March 2024
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
If there are 
pennies in the bag, then there are 
dimes and 
quarters. Since pennies are 
, dimes are 
, and quarters are 
, the total amount of money in the bag is ![\[$ \left(0.01x+(0.10)(2x)+(0.25)(6x)\right) = $1.71x.\]](http://latex.artofproblemsolving.com/2/6/5/265fd948ce338422abfceaa1a1b75e28cf47bda6.png)
Therefore, the possible amounts of money are precisely the integer multiples of 
.
Since the answer choices are all integer numbers of dollars, we multiply by 
to deduce that the answer must be an integer multiple of 
. The only such multiple among the answer choices is 
.
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. 
