Question

Students of a school were made to stand in rows for drill. If 3 students less were standing in each row, 10 more rows would be required and if 5 students more were standing in each row, then the number of rows would be reduced by 10. Find the number of students participating in the drill. \[\]
A.400\[\]
B.500\[\]
C.600\[\]
D.700\[\]

Answer 1

Hint: We assume us assume the number of students in each row as $x$ and total number of rows as$y$. We use the given information and two linear equations in two variables $10x-3y=30,-10x+5y=50$ which we solve by method of elimination to get $x,y$. We find the total number of students as $x\times y$.\[\]

Complete step-by-step answer:
We are given the question that students of a school were made to stand in rows for drill. Let us assume the number of students in each row as $x$ and total number of rows as $y$. So the total number of students standing is
\[\text{Number of students in 1 row}\times \text{ Number of rows }=x\times y=xy\]
We are also given if 3 less students were standing in each row, 10 more rows would be required. So the number of students in each row will decrease by 3 and the number of rows will increase by 10. So we have,
\[\begin{align}
  & \left( x-3 \right)\left( y+10 \right)=xy \\
 & \Rightarrow xy+10x-3y-30=xy \\
 & \Rightarrow 10x-3y=30.......\left( 1 \right) \\
\end{align}\]
We are further given the question if 5 students more were standing in each row, then the number of rows would be reduced by 10. So the number of students in one row will increase by 5 and number of rows will decrease by 10, We have,
\[\begin{align}
  & \left( x+5 \right)\left( y-10 \right)=xy \\
 & \Rightarrow xy-10x+5y-50=xy \\
 & \Rightarrow -10x+5y=50.......\left( 2 \right) \\
\end{align}\]
We add corresponding sides of equation (1) and equation (2) to have
\[\begin{align}
  & 10x+5y-\left( 10x-3y \right)=50-30 \\
 & \Rightarrow 2y=80 \\
 & \Rightarrow y=\dfrac{80}{2} \\
 & \therefore y=40 \\
\end{align}\]
We put the above obtained value of $y$ in equation to get,
\[\begin{align}
  & 10x-3\times 40=30 \\
 & \Rightarrow 10x=150 \\
 & \Rightarrow x=\dfrac{150}{10} \\
 & \therefore x=15 \\
\end{align}\]
So the total number of students standing in the drill is
\[xy=15\times 40=600\]
So the correct option is C.\[\]

So, the correct answer is “Option C”.

Note: We note that the key in this problem is to understand that the total number of students $xy$ will not change after increase or decrease in number of students per row or increase or decrease in number of rows. So we do not need to add or subtract in $xy$. We can also use method of substation to solve the linear equations