Difference between revisions of "1960 AHSME Problems/Problem 3"

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==See Also==
 
==See Also==
{{AHSME 40p box|year=1960 |before=[[Problem 2]]|after=[[Problem 4]]}}
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Latest revision as of 20:11, 10 May 2018

Problem

Applied to a bill for $\textdollar{10,000}$ the difference between a discount of $40$% and two successive discounts of $36$% and $4$%, expressed in dollars, is:

$\textbf{(A)}0\qquad \textbf{(B)}144\qquad \textbf{(C)}256\qquad \textbf{(D)}400\qquad \textbf{(E)}416$

Solution

Taking the discount of $40$% means you're only paying $60$% of the bill. That results in $10,000\cdot0.6=\textdollar{6,000}$.

Likewise, taking two discounts of $36$% and $4$% means taking $64$% of the original amount and then $96$% of the result. That results in $10,000\cdot0.64\cdot0.96=\textdollar{6,144}$.

Taking the difference results in $6,144-6,000=\textdollar{144}$, or answer choice $\boxed{\textbf{(B)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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