# The ratio of the number of boys in the class to the number of girls in that class is 3 to 2. how...

## Question:

The ratio of the number of boys in the class to the number of girls in that class is 3 to 2. how many boys and how many girls could there be? give at least 4 possible answers.

## Ratios: Proportionality Constant:

Given a set two different quantities {eq}a{/eq} and {eq}b{/eq} with a specific ratio {eq}a : b = c{/eq}, a proportionality constant {eq}k{/eq} should be introduced in order to find the total sum of the two quantities. Rewriting the ratio as {eq}ka : kb = c{/eq}, the total sum of the set would be equal to {eq}S = ka + kb = k(a + b) {/eq}

Our problem states that:

$$a : b = 3 : 2 \\$$

Let's introduce a proportionality constant {eq}k{/eq}:

$$ka : kb = 3k : 2k$$

The total size of the class as a result is then:

$$S = 3k + 2k = 5k$$

The number of girls and boys in the class respectively are:

\begin{align} g &= \dfrac{3k}{5k}S = \dfrac{3}{5}\cdot 5k\\[0.2cm] b &= \dfrac{2k}{5k}S = \dfrac{2}{5}\cdot 5k \end{align}

Let's take natural values of {eq}k{/eq}. Let's say that {eq}k = 1{/eq}:

\begin{align} k &= 1\\[0.2cm] g &= \dfrac{3}{5}\cdot 5\cdot 1 = 3\\[0.2cm] b &= \dfrac{3}{5}\cdot 5\cdot 1 = 2 \end{align}

Let's say that {eq}k = 2{/eq}:

\begin{align} k &= 2\\[0.2cm] g &= \dfrac{3}{5}\cdot 5\cdot 2 = 6\\[0.2cm] b &= \dfrac{3}{5}\cdot 5\cdot 2 = 4 \end{align}

Let's say that {eq}k = 3{/eq}:

\begin{align} k &= 3\\[0.2cm] g &= \dfrac{3}{5}\cdot 5\cdot 3 = 9\\[0.2cm] b &= \dfrac{3}{5}\cdot 5\cdot 3 = 6 \end{align}

Let's say that {eq}k = 4{/eq}:

\begin{align} k &= 4\\[0.2cm] g &= \dfrac{3}{5}\cdot 5\cdot 4 = 12\\[0.2cm] b &= \dfrac{3}{5}\cdot 5\cdot 4 = 8 \end{align} \\

Thus, four possible sets of girls and boys {eq}(g, b){/eq} are:

$$\boxed{(3, 2),~(6, 4),~(9, 6),~(12, 8)}$$