Difference between revisions of "1978 AHSME Problems/Problem 5"
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Therefore, our answer is <math>\textdollar 60 - \textdollar 47 = \boxed{\textbf{(C) }\textdollar 13}</math> | Therefore, our answer is <math>\textdollar 60 - \textdollar 47 = \boxed{\textbf{(C) }\textdollar 13}</math> | ||
~awin | ~awin | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1978|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:00, 13 February 2021
Problem 5
Four boys bought a boat for 
. The first boy paid one half of the sum of the amounts paid by the other boys;
the second boy paid one third of the sum of the amounts paid by the other boys;
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?
Solution 1
If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid 
of the total.
If the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid 
of the total.
If the third boy paid one fourth of the sum of the amounts paid by the other boys, that means that he paid 
of the total.
Summing it up, we get 
.
Therefore, our answer is 
~awin
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
