Difference between revisions of "2016 AMC 10A Problems/Problem 3"
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<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math> | <math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | If for every dollar Ben spent on bagels David spent <math>25</math> cents less, then the ratio of the money that Ben spent to David spent is <math>4:3</math>. We know that Ben paid <math>\$12.50</math> more than David, so we can write <math>\$12.50\cdot (4+3)=\$ 87.50.</math> Then the answer is <math>\boxed{\text{(C)}}.</math> | ||
| + | |||
| + | ~@azure123456 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 08:59, 26 July 2021
Problem
For every dollar Ben spent on bagels, David spent 
cents less. Ben paid 
more than David. How much did they spend in the bagel store together?
Solution 1
If Ben paid 
more than David, then he paid 
. Thus, David paid 
, and they spent 
.
Solution 2
If for every dollar Ben spent on bagels David spent cents less, then the ratio of the money that Ben spent to David spent is 
. We know that Ben paid more than David, so we can write 
Then the answer is 
~@azure123456
Video Solution
https://youtu.be/VIt6LnkV4_w?t=81
~IceMatrix
~savannahsolver
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
