Difference between revisions of "2015 AMC 10A Problems/Problem 6"
BeastX-Men (talk | contribs) m (→See also: Added See Also section.) |
|||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #4]] and [[2015 AMC 10A Problems|2015 AMC 10A #6]]}} | ||
==Problem== | ==Problem== | ||
| Line 15: | Line 16: | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=5|num-a=7}} | {{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}} | {{MAA Notice}} | ||
Revision as of 20:43, 4 February 2015
- 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?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Let 
be the bigger number and 
be the smaller.
.
Solving gives 
, so the answer is 
.
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. 
