Difference between revisions of "2016 AMC 10A Problems/Problem 3"

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If Ben paid <math>\$ 12.50</math> more than David, then he paid <math>\frac{12.5}{.25}= \$ 50.00</math>. Thus, David paid <math>\$ 37.50</math>, and they spent <math>50.00+37.50 =\$ 87.50 \implies \boxed{\textbf{(C) }\$ 87.50}</math>.
 
If Ben paid <math>\$ 12.50</math> more than David, then he paid <math>\frac{12.5}{.25}= \$ 50.00</math>. Thus, David paid <math>\$ 37.50</math>, and they spent <math>50.00+37.50 =\$ 87.50 \implies \boxed{\textbf{(C) }\$ 87.50}</math>.
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==Video Solution==
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https://youtu.be/VIt6LnkV4_w?t=81
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~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=2|num-a=4}}
 
{{AMC10 box|year=2016|ab=A|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:31, 5 May 2020

Problem

For every dollar Ben spent on bagels, David spent $25$ cents less. Ben paid $$12.50$ more than David. How much did they spend in the bagel store together?

$\textbf{(A)}\ $37.50 \qquad\textbf{(B)}\ $50.00\qquad\textbf{(C)}\ $87.50\qquad\textbf{(D)}\ $90.00\qquad\textbf{(E)}\ $92.50$

Solution

If Ben paid $$ 12.50$ more than David, then he paid $\frac{12.5}{.25}= $ 50.00$. Thus, David paid $$ 37.50$, and they spent $50.00+37.50 =$ 87.50 \implies \boxed{\textbf{(C) }$ 87.50}$.

Video Solution

https://youtu.be/VIt6LnkV4_w?t=81

~IceMatrix

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 AMC 10 Problems and Solutions

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