Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1985 AHSME Problems/Problem 4"
m (Improved wording)
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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
+
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>&#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>
+
<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> 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>.  
+
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 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>.
+
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

$\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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_4&oldid=217087"