Difference between revisions of "2017 AMC 10A Problems/Problem 14"
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If we replace <math>A</math> with <math>100</math> we get a system of equations, and the sum of the values of <math>M</math> and <math>S</math> is the percentage of <math>A</math>. | If we replace <math>A</math> with <math>100</math> we get a system of equations, and the sum of the values of <math>M</math> and <math>S</math> is the percentage of <math>A</math>. | ||
Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>. | Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>. | ||
− | Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math>. | + | Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>(D)</math>. |
-Harsha12345 | -Harsha12345 |
Revision as of 01:34, 27 January 2021
Problem
Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was dollars. The cost of his movie ticket was
of the difference between
and the cost of his soda, while the cost of his soda was
of the difference between
and the cost of his movie ticket. To the nearest whole percent, what fraction of
did Roger pay for his movie ticket and soda?
Solution
Let = cost of movie ticket
Let = cost of soda
We can create two equations:
Substituting we get:
which yields:
Now we can find s and we get:
Since we want to find what fraction of did Roger pay for his movie ticket and soda, we add
and
to get:
Solution 2
We have two equations from the problem:
and
If we replace
with
we get a system of equations, and the sum of the values of
and
is the percentage of
.
Solving, we get
and
.
Adding, we get
, which is closest to
which is
.
-Harsha12345
Video Solution
~savannahsolver
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.