Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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==Problem 13== | ==Problem 13== | ||
− | + | Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of his coins. How many 10-cent coins does Claudia have? | |
<math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math> | <math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math> |
Revision as of 17:11, 25 February 2021
Problem 13
Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of his coins. How many 10-cent coins does Claudia have?
Solution 1
Let Ashraful have 5-cent coins and 10-cent coins. It is easily observed that any multiple of between and inclusive can be obtained by a combination of coins. Thus, combinations can be made, so . But the answer is not because we are asked for the number of 10-cent coins, which is
Solution 2
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of To have exactly different multiples of we will need to make up to cents. If all twelve coins were 5-cent coins, we will have cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain cents, and as we need to gain cents, the answer is
For small scale substitutions of 5 cent for 10 cent, like in this problem, it is quite fast, however, if the problem is a little bit more complex, knowing that you need 85 cents, explained above, it is also possible to use a system of equations. It would be Solving this (5-cent coins) and (10-cent coins) , so again the answer is
Video Solution
~savannahsolver
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
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