Question

In an Auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of row was doubled and the number of seat in each row was reduced by 10, total number of seats increased by 300. Find :
a) ​​​​​​ the number of rows in the original arrangement.
b) ​​​​​​ the number of seats in the Auditorium after rearrangement​

Answer 1

Answer:

a) 30

b) 1200

Step-by-step explanation:

$let \: the \: no. \: of \: rows \: and \: column \: be \: x \: \\ as \: the \: no. \: of \: \: rows \: and \: column \: are \: equal \: so \: \\ both \: be \: x \\ so \: the \: no. \: of \: seats \: before \: arrangement \: was \: \\ x \times x = {x}^{2} \\ no. \: of \: seats \: after \: arrangment \: = \\ no.of \: rows \: \times \: no. \: of \: column \: \\ no. \: of \: rows \: = \: 2 \times no. \: of \: rows \: earlier \: \\ so \: 2 \times x = 2x \: \\ and \: no. \: of \: columns \: = x - 10 \\ no. \: of \: seats \: = 2x \times (x - 10) \\ {2x}^{2} - 20x \\ and \: we \: know \: the \: no. \: of \: seats \: after \: arrangement \\ \: = {x}^{2} + 300 \\ so \: {x}^{2} + 300 = {2x}^{2} - 20x \\ {x}^{2} - 20x - 300 = 00 \\ (x - 30)(x - 10) = 0 \\ x - 30 = 0 \\ x = 30 \\ no. \: of \: seats \: after \: arrangement \: = {2x}^{2} - 20x \\ 2 \times {30}^{2} - 20 \times 30 \\ 2 \times 900 - 20 \times 30 \\ 1800 - 600 \\ = 1200$

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