Question

Rs.69 were divided into 115 students so that each girl gets 50 paisa less than a boy. Thus, each boy received twice the point as each girl received. The number of girl in the class is

Answer 1

Hint: We will start by assuming that, number of boys are x and number of girls are y as we have total students as 115. So, we get $\text{x}+\text{y}=\text{115}$, then we let amount each boy have as z so girl will have (z-0.5) rupees, then we have boys money is twice the girls money so we compare z with $\text{2}\left( \text{z}-0.\text{5} \right)$ to find value of z, once we have value of z, we get total money as $\text{x}+0.\text{5y}=\text{69}$. Then, we use elimination methods to find the value of x and y.

Complete step-by-step answer:
We are asked to find the number of boys and girls.
Let us start with the assumption that there are x boys and y girls in total.
Now we are given that, total students are 115. So, we get:
\[\begin{align}
  & \text{Boys}+\text{Girls}=\text{115} \\
 & \text{x}+\text{y}=\text{115 }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }.\text{ }\left( \text{1} \right) \\
\end{align}\]
Now, let each boy get 2 rupees. We have that girls get 50 paisa less than boys. So, girls get (z-0.5) rupees as 50 paisa = 0.5 rupees.
We also have that boys' money is twice the girls' money. So, we get:
\[z=2\left( z-0.5 \right)\]
Solving further we get:
\[\begin{align}
  & z-2z=-2\times 0.5 \\
 & -z=-1 \\
 & z=1 \\
\end{align}\]
So each boy get 1 rupees and each girl gets \[\text{z}-0.\text{5}=\text{1}-0.\text{5}=0.\text{5 rupees}\]
Now, the total money x boys and y girls will have will be $1\times x+0.5y$ but we have total money as 69.
\[\Rightarrow x+0.5y=69\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( 2 \right)\]
Now, subtracting (2) from (1) we get:
\[\begin{align}
  & x+y=115 \\
 & x+0.5y=69 \\
 & \begin{matrix}
   - & - & - \\
\end{matrix} \\
 & \overline{+0.5y=46} \\
 & y=\dfrac{46}{0.5}=92 \\
\end{align}\]
So, y = 92.
So, the number of girls is 92. Now putting this in (1) we get:
\[\begin{align}
  & x+92=115 \\
 & x=115-92 \\
 & x=23 \\
\end{align}\]
So the number of boys is 23.
So we get, girls are 92 while boys are 23.

Note: While comparing $z=\text{2}\left( \text{z}-0.\text{5} \right)$ mistake happens like $z=2z-0.5$ one could forget to multiply 2 with each term of (z-0.5) and this leads to wrong solution. While dividing a number with decimal, we solve as follows: \[\Rightarrow \dfrac{46}{0.5}=\dfrac{46\times 10}{0.5\times 10}=\dfrac{460}{5}=92\]
While subtracting x+y=115 and x+0.5y=69 mistakes occur,
\[\begin{align}
  & x+y=115 \\
 & x+0.5y=69 \\
 & \begin{matrix}
   - & - & - \\
\end{matrix} \\
 & \overline{0.5=46} \\
\end{align}\]
y gets cancelled by y which is wrong.