Difference between revisions of "2017 AMC 12B Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | Kymbrea has <math>30</math> comic books initially and every month, she adds two. This can be represented as <math>30 + 2x</math> where x is the number of months elapsed. LaShawn's collection, similarly, is <math>10 + 6x</math>. To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation <math>2(2x + 30) = 6x + 10</math> and get <math>x = | + | Kymbrea has <math>30</math> comic books initially and every month, she adds two. This can be represented as <math>30 + 2x</math> where x is the number of months elapsed. LaShawn's collection, similarly, is <math>10 + 6x</math>. To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation <math>2(2x + 30) = 6x + 10</math> and get <math>x = \boxed{\textbf{(E) } 25}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=2}} | {{AMC12 box|year=2017|ab=B|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:50, 17 February 2017
Problem 1
Kymbrea's comic book collection currently has 
comic books in it, and she is adding to her collection at the rate of 
comic books per month. LaShawn's collection currently has 
comic books in it, and he is adding to his collection at the rate of 
comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?
Solution
Kymbrea has comic books initially and every month, she adds two. This can be represented as 
where x is the number of months elapsed. LaShawn's collection, similarly, is 
. To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation 
and get 
.
See Also
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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. 
