Question

There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
A. 16
B. 20
C. 25
D. 32

Answer 1

Hint: In this question use the given information to make the equation and also remember that to distribute the children equally in row and column ${x^2} = 500$ Here x is the number of rows and columns, using these instructions can help you to approach the solution of the question.

Complete step by step answer:
According to the given information we have 500 students the students are divided in such manner that in row and column the numbers of students are equal in row and manner we know that when the value of elements are equal in each row and column then the multiple of row and column is equal to the sum of the all the elements in row and column.
Let suppose x be the total number of rows since the number of rows is equal to the number of columns.
Therefore numbers of columns are equal to x.
Thus for distributing the equal numbers of students in rows and column ${x^2} = 500$
Finding the perfect square root of 500 let’s use the division method
Let’s taking the 00 of 500 as a first pair and 5 as a second pair
Taking 5 as a dividend
Now taking 2 as a divisor and quotient
Now considering 100 as dividend
And 42 as divisor and 2 as a quotient
Thus the remainder is 16

	22
2	\[5\overline {00} \]
	4
42	1 00
	84
	16

Since the 16 is the remainder therefore the 16 children will be left out in the arrangement.

So, the correct answer is “Option A”.

Note: In the above questions we used the division method the steps to use these methods are;
The first step to find the square root of the given number is to break the digits into pairs from right to left where each pair consists of 2 digits of a number.
In the second step we start dividing the leftmost pair of digits where we take the divisor and quotient the largest number which has a square equal to or less than the leftmost pair of digits.
For the step third we subtract the square of the divisor from the leftmost pair of digits and to have a new dividend we bring the next pair of digits at the right side of remainder.
In fourth step we consider the first part of new divisor as 2 times of the previous quotient and taking the second part such that the product of complete divisor and second part is equal or less than the new dividend and in this step the quotient is equal to the second part of new divisor.