Difference between revisions of "1993 AJHSME Problems/Problem 6"
Math Kirby (talk | contribs) (Created page with "== Problem 6 == 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...") |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
| − | == Problem | + | == 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? | ||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | A can of soup will feed <math>5</math> children so <math>15</math> children are feed by <math>3</math> cans of soup. Therefore, there are <math> 5-3=2 </math> cans for adults, so <math> 3 \times 2 =\boxed{\ | + | A can of soup will feed <math>5</math> children so <math>15</math> children are feed by <math>3</math> cans of soup. Therefore, there are <math> 5-3=2 </math> cans for adults, so <math> 3 \times 2 =\boxed{\textbf{(B)}\ 6} </math> adults are fed. |
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AJHSME box|year=1993|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:11, 4 July 2013
Problem
A can of soup can feed 
adults or 
children. If there are cans of soup and 
children are fed, then how many adults would the remaining soup feed?
Solution
A can of soup will feed children so children are feed by cans of soup. Therefore, there are 
cans for adults, so 
adults are fed.
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
