Question

In a school hall, \[460\] students were sitting in rows and columns in such a way that the number of students sitting in each column was three more than the number of students sitting in each row. The number of students in each column was.
A) \[20\]
B) \[23\]
C) \[24\]
D) None of these

Answer 1

Hint: Here we have to assume the number of students sitting in each row as one variable and also assume the number of students sitting in each column as another variable. Then using the given statements find the value of the variables.

Complete step by step solution:
Let \[x\] and \[y\] be the number of students sitting in each row and each column.
Since, \[460\] students in a school were sitting in rows and columns in such a way that the number of students sitting in each column was three more than the number of students sitting in each row.
Hence, \[xy = 460\]---(1) and
\[y = x + 3\]---(2)
Put \[y = x + 3\] in the equation (1), we get
\[x\left( {x + 3} \right) = 460\]
\[ \Rightarrow \]\[{x^2} + 3x - 460 = 0\]---(3)
Since the equation (3) is in the quadratic form \[a{x^2} + bx + c = 0\]. We know that the solutions of \[a{x^2} + bx + c = 0\] is obtained by the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
In the equation (2), \[a = 1\],\[b = 3\] and \[c = - 460\]
Then \[x = \dfrac{{ - 3 \pm \sqrt {9 - 4\left( 1 \right)\left( { - 460} \right)} }}{2} = \dfrac{{ - 3 \pm \sqrt {9 + 1840} }}{2} = \dfrac{{ - 3 \pm 43}}{2} = 20, - 23\].
Since, \[x\] must be positive. Hence, \[x = 20\]
Then \[y = 23\]
Hence, the number of students in each column was \[23\]. So, Option (B) is correct.

Note:
Note that the arithmetic operations of addition, subtraction, multiplication, and division help us solve mathematical problems. Algebra deals with these concepts and can be considered as generalized arithmetic. An equation is a mathematical sentence with an equal sign.