Question

School ‘A’ has $ 30\% $ more students than school ‘B’. If 120 more students join school ‘B’, the two schools will have the same number of students. What is the sum of the number of students in school ‘A’ and school ‘B’ initially?

Answer 1

Hint:
 It is given that students in school ‘A’ are 30% more than students in ‘B’, so we form the first equation from here by considering initial students in school ‘A’ as ‘x’ and in school ‘B’ as ‘y’.
Then we use that adding 120 students to school ‘A’ will make the number equal, so using this we get another equation.
Then using the equation we solve for ‘x’ and ‘y’ and the sum of them to find the required solution.

Complete step by step answer:
We are given that school ‘A’ has 30% more students than school ‘B’, also if 120 more are added to school ‘B’ then they have the same number of students.
We are asked to find the sum of students in school ‘A’ and school ‘b’ initially.
To solve our problem, let us start with the assumption that our school ‘A’ has ‘x’ students initially while our school ‘B’ will have ‘y’ students initially.
Now, as we are given that school ‘A’ has 30% more student than school ‘B’/
So, it simply means that –
Student of school ‘A’ in sum of a student at school ‘B’ along with 30% more of the student at school ‘B’
As a student in school ‘A’ is ‘x’ and student in school ‘B’ is ‘y’ so we get –
 $ x=y+30\%\text{of y} $
Simplify, we get –
 $ x=y+\dfrac{30}{100}\times y $
By solving this, we get –
 $ x=\dfrac{130}{100}y $
And we get –
 $ x=\dfrac{13}{10}y $ …………………. (i)
Now, we are also given that if we add 120 students more to the number of students of school ‘B’ then the number of students is the same as the number of students of the school ‘A’.
So, we get –
 $ x=120+y $ ……………….. (ii)
Now solving (i) and (ii0 we will solve further
In equation (2) we use $ x=\dfrac{13}{10}y $ ,
So we get –
 $ \dfrac{13}{10}y=120+y $
By cross multiplying we get –
 $ 13y=10\left( 120+y \right) $
Simplifying, we get –
 $ 13y=1200+10y $
Solving for ‘y’ we have –
 $ 13y-10y=1200 $
 $ \Rightarrow 3y=1200 $
So, we have –
 $ y=400 $
Means number of students in school ‘B’ is 400 as from equation (2)
 $ x=y+120 $
So using $ y=400 $ , we get –
 $ x=400+120 $
We get –
 $ x=520 $
So, we have that number of students in school ‘A’ in 520.
So, the total number of students in both schools is $ 520+400=920 $ .
Hence, the solution is 920 students were there initially.

Note:
 Short way to deal with this is to look for some trick or shortcut,
We can see there adding 120 or adding 30% of students of school ‘B’ doing the same work.
So, we have 30% of students in school $ B=120 $.
So, we first solve this, we get –
 $ \dfrac{30}{100}\times y=1200 $
Simplifying, we get –
 $ y=\dfrac{1200\times 10}{30} $
So, we get –
 $ y=400 $
School ‘B’ has 400 students.
As we are given in school ‘A’ student is 120 more than school ‘B’,
So, $ x=120+400=520 $
So, total student in ‘A’ and ‘B’ $ =x+y $
 $ \begin{align}
  & =400+520 \\
 & =920 \\
\end{align} $