Question

From a group of boys and girls, 15 girls leave. There are then left two boys for each girl. After this 45 boys leave. There are then 5 girls for each boy. The number of girls in the beginning was
A. 40
B. 43
C. 29
D. 50
E. None of these

Answer 1

Hint: Suppose the initial number of girls and boys as two different variables. Now, use the given condition to get the changed number of girls and boys. And the meaning of “two boys for each girl” and “5 girls for each boy” is that the ratio of boys to girls is changed to $2:1$ and $1:5$ respectively. So, form two equations in two variables by equating the calculated and given ratios. And hence solve them to get a number of girls.

Complete Step-by-Step solution:
Let the number of girls and boys in the group initially be x and y.
So, it is given that there will be two boys for each girl, if 15 girls will leave. In other words the ratio of boys and girl will be $2:1$ if 15 girls will leave from the group.
So, the number of boys will not change by changing in the number of girls, but there will be $\left( x-15 \right)$ after 15 girls left the group.
So, number of girls now $=x-15$
number of girls now $=y$
As the ratio of boys and girls is $2:1$ when the number of girls and boys are $x-15$ and y respectively. So, we can get equation as
$\dfrac{y}{x-15}=\dfrac{2}{1}$
On cross multiplying the above equation, we get
$y=2\left( x-15 \right)$ …………………………………………(i)
Now, it is given that the ratio will become $1:5$ (Boys : Girls) if 45 boys leave the group after leaving 15 girls. It means number of boys and girls will become
Number of boys now $=y-45$
Number of girls $=x-15$
Hence, we can get an equation as
$\dfrac{y-45}{x-15}=\dfrac{1}{5}$
On cross-multiplying the above equation, we can get one more relation in ‘x’ and ‘y’. So, we get
$5\left( y-45 \right)=1\left( x-15 \right)$
$\Rightarrow 5y-225=x-15$
$5y=x-15+225$
$5y=x+210$ …………………………………….(ii)
Now, we can solve the equations (i) and (ii) by substituting the value of ‘y’ from the equation (i) to the (ii). So, we get
$5\left( 2\left( x-15 \right) \right)=x+210$
$10\left( x-15 \right)=x+210$
$10x-150=x+210$
$9x=150+210=360$
$x=\dfrac{360}{9}=40$
Now, we can get the value of ‘y’ from the equation (i) by substituting the value of $x=40$ in it. So, we get
$y=2\left( 40-15 \right)=2\times 25$
$y=50$
Hence, the number of boys and girls are 50 and 40 respectively.
So, option (A) is the correct answer.

Note: Another approach for the question would be that we can suppose the girls and boys as x and $2x$ (after 15 girls leave the group) and similarly suppose the number of girls and boys as 5y and y (after further leaving of 45 boys). Now, we can equate the actual values of boys and girls. As total girls in terms of ‘x’ initially would be $\left( x+15 \right)$ and boys would be $2x$ . Similarly, the number of boys in terms of ‘y’ would be $y+45$ and the number of girls would be $5y+15$. So, we can get equations as
$x+15=5y+15$ and $2x=y+45$.