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> | + | <math> \mathrm{(A)\ } </math>\$<math>306 \qquad \mathrm{(B) \ } </math>\$<math>333 \qquad \mathrm{(C)\ } </math>\$<math>342 \qquad \mathrm{(D) \ } </math>\$<math>348 \qquad \mathrm{(E) \ } </math>\$<math>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 23:30, 2 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. 
