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>. | ||
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| + | ==Solution 2== | ||
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| + | Without loss of generality, let the two numbers be <math>3</math> and <math>2</math>, as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is <math>\boxed{\textbf{(B) }\frac32}</math>. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/GLfDY92_td0 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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| + | |||
| + | |||
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| + | ==Video Solution== | ||
| + | https://youtu.be/JLKoh-Nb0Os | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 22:05, 26 June 2023
- 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 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 (CREATIVE THINKING)
~Education, the Study of Everything
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. 
