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

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==Solution 2==
 
==Solution 2==
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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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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 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 <math>9 + 8 = \boxed{\textbf{(B)}\ 17}</math>. ~DY, edited by SuperVince1
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==

Latest revision as of 10:46, 26 May 2024

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 denominator, 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 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 $9 + 8 = \boxed{\textbf{(B)}\ 17}$. ~DY, edited by SuperVince1

Video Solution by WhyMath

https://youtu.be/HseEFQDuh_c

~savannahsolver

See Also

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

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