# Question 75068

Feb 26, 2017

So the original numbers are: 15 girls and 12 boys

The addition of 3 more boys gives: 15 girls and 15 boys

#### Explanation:

$\textcolor{b l u e}{\text{Expansion of Taimur's solution}}$

Let original count of girls be $g$
Let original count of boys be $b$

Using ratio in fractional format

$\left(\text{girls")/("boys}\right) \to \frac{g}{b} \to \frac{5}{4}$

color(green)(5/4color(red)(xx1) " "-=" "5/4color(red)(xx3/3)" "=" "15/12
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If we were to add 3 more boys then the numbers would become:

("girls")/("boys")->5/4 ->15/(12+3) = 15/15=1/1 larr" new ratio as required"#

$\textcolor{p u r p \le}{\text{As the final ration of "1/1" was achieved the previous ratio of "15/12" is correct}}$

So we have: 15 girls and 12 boys as the original count

$\textcolor{w h i t e}{.}$

$\textcolor{w h i t e}{.}$

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$\textcolor{b l u e}{\text{Solved using substitution}}$

$\frac{g}{b} = \frac{5}{4}$.....................Equation(1) $\leftarrow \text{ 1st ratio}$

$\frac{g}{b + 3} = \frac{1}{1}$.................Equation(2) $\leftarrow \text{ 2nd ratio}$

Consider Equation(2)

$\implies g = b + 3$..................Equation(3)

Substitute for $g$ in Equation(2) using Equation(3)

$\frac{g}{b} = \frac{5}{4} \text{ "->" } \frac{b + 3}{b} = \frac{5}{4}$

$\implies 4 b + 12 = 5 b$

$b = 12$

By substitution of $b$ in Equation(1) givies

$\frac{g}{b} = \frac{5}{4} \text{ "->" } \frac{g}{12} = \frac{5}{4}$

$g = \frac{5 \times {\cancel{12}}^{3}}{{\cancel{4}}^{1}} = 15$