2008 AMC 8 Problems/Problem 20
Problem
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
Solution
Let be the number of boys and be the number of girls.
For and to be integers, must cancel out with the denominator, and the smallest possible value is . This yields boys. The minimum number of students is .
Solution 2
We know that or . So, the ratio of the number of boys to girls is 8:9 (Not 9:8 because the numbers 8 and 9 are in the denominators of the fractions). The smallest total number of students is . ~DY, edited by SuperVince1
Video Solution by WhyMath
~savannahsolver
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AJHSME/AMC 8 Problems and Solutions |
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