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Difference between revisions of "1985 AHSME Problems/Problem 4"

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==Problem==
 
==Problem==
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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
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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
  
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<math> \mathrm{(A)\ } </math>&#036;<math>306 \qquad \mathrm{(B) \ }  </math>&#036;<math>333 \qquad \mathrm{(C)\ } </math>&#036;<math>342 \qquad \mathrm{(D) \  }  </math>&#036;<math>348 \qquad \mathrm{(E) \  }  </math>&#036;<math>360  </math>
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<math> \mathrm{(A)\ } \$306 \qquad \mathrm{(B) \ }  \$333 \qquad \mathrm{(C)\ } \$342 \qquad \mathrm{(D) \  }  \$348 \qquad \mathrm{(E) \  }  \$360  </math>
  
 
==Solution==
 
==Solution==
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Let there be <math> p </math> pennies. Therefore, there are <math> 2x </math> dimes and <math> 3(2x)=6x </math> quarters. Since pennies are &#036;<math> 0.01 </math>, dimes are &#036;<math> 0.10 </math>, and quarters are &#036;<math> 0.25 </math>, the total amount of money is &#036;<math> 0.01x+(2)(0.10x)+(6)(0.25x)= </math>&#036;<math> 1.71x </math>. Therefore, any amount of money must be a multiple of &#036;<math> 1.71 </math>.  
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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>.  
  
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Since the answer choices are all whole dollar amounts, we multiply by <math> 100 </math> to get a whole dollar number: &#036;<math>171 </math>. Notice that twice this is &#036;<math> 342 </math>, which is answer choice <math> \boxed{\text{C}} </math>.
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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}}
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{{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

$\mathrm{(A)\ } $306 \qquad \mathrm{(B) \ } $333 \qquad \mathrm{(C)\ } $342 \qquad \mathrm{(D) \ } $348 \qquad \mathrm{(E) \ } $360$

Solution

If there are $x$ pennies in the bag, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$, dimes are $$0.10$, and quarters are $$0.25$, the total amount of money in the bag is \[$ \left(0.01x+(0.10)(2x)+(0.25)(6x)\right) = $1.71x.\] Therefore, the possible amounts of money are precisely the integer multiples of $$1.71$.

Since the answer choices are all integer numbers of dollars, we multiply by $100$ to deduce that the answer must be an integer multiple of $$171$. The only such multiple among the answer choices is $$(2 \cdot 171) = \boxed{\text{(C)} \ $342}$.

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

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