Difference between revisions of "2008 AMC 8 Problems/Problem 20"

Revision as of 13:15, 4 January 2023

Problem

The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ 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?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 17\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution

Let $b$ be the number of boys and $g$ be the number of girls.

$\frac23 b = \frac34 g \Rightarrow b = \frac98 g$

For $g$ and $b$ to be integers, $g$ must cancel out with the numerator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\boxed{\textbf{(B)}\ 17}$ .

Solution 2

We know that $\frac23 B = \frac34 G$ or $\frac69 B = \frac68 G$ . So, the ratio of the number of boys to girls is 9:8. The smallest total number of students is $9 + 8 = \boxed{\textbf{(B)}\ 17}$ . ~DY

Video Solution by WhyMath

https://youtu.be/HseEFQDuh_c

~savannahsolver

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 15: / Line 15: @@
 For <math>g</math> and <math>b</math> to be integers, <math>g</math> must cancel out with the numerator, and the smallest possible value is <math>8</math>. This yields <math>9</math> boys. The minimum number of students is <math>8+9=\boxed{\textbf{(B)}\ 17}</math>.
-==Solution==
+==Solution 2==
 We know that <math>\frac23 B = \frac34 G</math>  or  <math>\frac69 B = \frac68 G</math>. So, the ratio of the number of boys to girls is 9:8. The smallest total number of students is <math>9 + 8 = \boxed{\textbf{(B)}\ 17}</math>. ~DY

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