Difference between revisions of "1993 AJHSME Problems/Problem 6"

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== Problem 6 ==
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== Problem ==
  
 
A can of soup can feed <math>3</math> adults or <math>5</math> children.  If there are <math>5</math> cans of soup and <math>15</math> children are fed, then how many adults would the remaining soup feed?
 
A can of soup can feed <math>3</math> adults or <math>5</math> children.  If there are <math>5</math> cans of soup and <math>15</math> children are fed, then how many adults would the remaining soup feed?

Revision as of 21:31, 22 December 2012

Problem

A can of soup can feed $3$ adults or $5$ children. If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution

A can of soup will feed $5$ children so $15$ children are feed by $3$ cans of soup. Therefore, there are $5-3=2$ cans for adults, so $3 \times 2 =\boxed{\textbf{(B)}\ 6}$ adults are fed.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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
All AJHSME/AMC 8 Problems and Solutions