Difference between revisions of "2015 AMC 10A Problems/Problem 6"
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<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> | <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 1== |
Let <math>a</math> be the bigger number and <math>b</math> be the smaller. | Let <math>a</math> be the bigger number and <math>b</math> be the smaller. | ||
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<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 | + | Multiplying out gives <math>a + b = 5a - 5b</math> and rearranging gives <math>4a = 6b</math> and factorised into <math>2a = 3b</math> and then solving gives |
<math>\frac{a}{b} = \frac32</math>, so the answer is <math>\boxed{\textbf{(B) }\frac32}</math>. | <math>\frac{a}{b} = \frac32</math>, so the answer is <math>\boxed{\textbf{(B) }\frac32}</math>. | ||
Revision as of 23:56, 16 June 2022
- The following problem is from both the 2015 AMC 12A #4 and 2015 AMC 10A #6, so both problems redirect to this page.
Problem
The sum of two positive numbers is 
times their difference. What is the ratio of the larger number to the smaller number?
Solution 1
Let 
be the bigger number and 
be the smaller.
.
Multiplying out gives 
and rearranging gives 
and factorised into 
and then solving gives
, so the answer is 
.
Solution 2
Without loss of generality, let the two numbers be 
and 
, as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller 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. 
