Difference between revisions of "2006 AMC 12A Problems/Problem 9"
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== Problem == | == Problem == | ||
− | + | Three pencils and one pen costs <math>\$1.20</math>. Seven pencils and two pens cost <math>\$2.50</math>. No prices include tax. In cents, what is the cost of a pencil? | |
<math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math> | <math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math> |
Revision as of 15:41, 17 April 2020
Problem
Three pencils and one pen costs . Seven pencils and two pens cost . No prices include tax. In cents, what is the cost of a pencil?
Solution
Let the price of a pencil be and an eraser . Then with . Since and are positive integers, we must have and .
Considering the equation modulo 3 (that is, comparing the remainders when both sides are divided by 3) we have so leaves a remainder of 1 on division by 3.
Since , possible values for are 4, 7, 10 ....
Since 13 pencils cost less than 100 cents, . is too high, so must be 4 or 7.
If then and so giving . This contradicts the pencil being more expensive. The only remaining value for is 7; then the 13 pencils cost cents and so the 3 erasers together cost 9 cents and each eraser costs cents.
Thus one pencil plus one eraser cost cents, which is answer choice .
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 |
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