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Difference between revisions of "1978 AHSME Problems/Problem 5"

(Created page with "== Problem 5 == Four boys bought a boat for <math>\textdollar 60</math>. The first boy paid one half of the sum of the amounts paid by the other boys; the second boy paid on...")
 
 
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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
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==See Also==
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{{AHSME box|year=1978|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 11:00, 13 February 2021

Problem 5

Four boys bought a boat for $\textdollar 60$. 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?

$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$

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 $\frac{1}{3}$ 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 $\frac{1}{4}$ 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 $\frac{1}{5}$ of the total.

Summing it up, we get $\textdollar 20 + \textdollar 15 + \textdollar 12 = \textdollar 47$. Therefore, our answer is $\textdollar 60 - \textdollar 47 = \boxed{\textbf{(C) }\textdollar 13}$ ~awin

See Also

1978 AHSME (ProblemsAnswer KeyResources)
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

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