Difference between revisions of "2015 AMC 10A Problems/Problem 6"
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+ | {{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #4]] and [[2015 AMC 10A Problems|2015 AMC 10A #6]]}} | ||
==Problem== | ==Problem== | ||
The sum of two positive numbers is <math> 5 </math> times their difference. What is the ratio of the larger number to the smaller number? | The sum of two positive numbers is <math> 5 </math> times their difference. What is the ratio of the larger number to the smaller number? | ||
− | <math> \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D) | + | <math> \textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2} </math> |
==Solution== | ==Solution== | ||
Line 11: | Line 12: | ||
<math>a + b = 5(a - b)</math>. | <math>a + b = 5(a - b)</math>. | ||
− | + | Multiplying out gives <math>a + b = 5a - 5b</math> and rearranging gives <math>4a = 6b</math> and then solving gives | |
+ | |||
+ | <math>\frac{a}{b} = \frac32</math>, so the answer is <math>\boxed{\textbf{(B) }\frac32}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/JLKoh-Nb0Os | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=5|num-a=7}} | ||
+ | {{AMC12 box|year=2015|ab=A|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Revision as of 16:08, 29 June 2020
- The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #6, so both problems redirect to this page.
Contents
Problem
The sum of two positive numbers is times their difference. What is the ratio of the larger number to the smaller number?
Solution
Let be the bigger number and be the smaller.
.
Multiplying out gives and rearranging gives and then solving gives
, so the answer is .
Video Solution
~savannahsolver
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.