Answer
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Hint: Our question is derived into two parts. In the first part we have to find the number of rows in the original arrangement. So, assume the no. of rows to be a variable. Then formulate linear equations in two variables using the information provided in the question then solve to find your answer.
Complete step-by-step answer:
Let, no. of rows in the auditorium $ = x$
$\therefore $No. of seats in the auditorium $ = x$
$\therefore $Total seats in the auditorium $ = x.x$$ = {x^2}$.
After re-arrangement, it is given that the no. of rows was doubled and seats reduced by \[10\]in each row.\[\therefore \]Now, no. of rows \[ = 2x\].
And, no. of seats \[ = x - 10\].
Given, the total no. of seats increased by \[300\].
\[\therefore \]Total no. of seats \[ = {x^2} + 300\][$\because $Total no. of seat$ = $no. of rows$ \times $no. of seats in each row]
\[{
\Rightarrow \left( {2x} \right) + \left( {x - 10} \right) = {x^2} + 300 \\
\Rightarrow 2{x^2} - 20x = {x^2} + 300 \\
\Rightarrow 2{x^2} - {x^2} - 20x - 300 = 0 \\
\Rightarrow {x^2} - 20x - 300 = 0 \\
\Rightarrow {x^2} - 30x + 10x - 300 = 0 \\
\Rightarrow x(x - 30) + 10(x - 30) = 0 \\
\Rightarrow \left( {x + 10} \right)\left( {x - 30} \right) = 0 \\
} \]
[Simplifying the above equation]
[ taking all the terms in one side]
[by middle term factorization to get factors those are multiples of $300$]
[ taking $(x - 30)$as common multiple]
Either,
\[{
x + 10 = 0 \\
\therefore x = - 10 \\
} \]
Or, \[{
x - 30 = 0 \\
\therefore x = 30 \\
} \]
Since, no. of rows cannot have negative value.
\[\therefore x = 30\].
\[\because \]No. of seats in each row \[ = 30 - 10\]\[ = 20\]
\[\therefore \]Total no. of seats in the auditorium
\[ = \]No. of rows\[ \times \]no. of seats in each row
\[{
= 30 \times 20 \\
= 600 \\
} \]
Hence, no. of seats in the auditorium is \[600\].\[\]
Note: In the above question we used the factorization method for solving quad ratio equations. You also can use Sridhar acharya’s formula to solve the equation.
Using Sridhar acharya’s formula with the help of a formula we can directly find the roots of the equation.
Complete step-by-step answer:
Let, no. of rows in the auditorium $ = x$
$\therefore $No. of seats in the auditorium $ = x$
$\therefore $Total seats in the auditorium $ = x.x$$ = {x^2}$.
After re-arrangement, it is given that the no. of rows was doubled and seats reduced by \[10\]in each row.\[\therefore \]Now, no. of rows \[ = 2x\].
And, no. of seats \[ = x - 10\].
Given, the total no. of seats increased by \[300\].
\[\therefore \]Total no. of seats \[ = {x^2} + 300\][$\because $Total no. of seat$ = $no. of rows$ \times $no. of seats in each row]
\[{
\Rightarrow \left( {2x} \right) + \left( {x - 10} \right) = {x^2} + 300 \\
\Rightarrow 2{x^2} - 20x = {x^2} + 300 \\
\Rightarrow 2{x^2} - {x^2} - 20x - 300 = 0 \\
\Rightarrow {x^2} - 20x - 300 = 0 \\
\Rightarrow {x^2} - 30x + 10x - 300 = 0 \\
\Rightarrow x(x - 30) + 10(x - 30) = 0 \\
\Rightarrow \left( {x + 10} \right)\left( {x - 30} \right) = 0 \\
} \]
[Simplifying the above equation]
[ taking all the terms in one side]
[by middle term factorization to get factors those are multiples of $300$]
[ taking $(x - 30)$as common multiple]
Either,
\[{
x + 10 = 0 \\
\therefore x = - 10 \\
} \]
Or, \[{
x - 30 = 0 \\
\therefore x = 30 \\
} \]
Since, no. of rows cannot have negative value.
\[\therefore x = 30\].
\[\because \]No. of seats in each row \[ = 30 - 10\]\[ = 20\]
\[\therefore \]Total no. of seats in the auditorium
\[ = \]No. of rows\[ \times \]no. of seats in each row
\[{
= 30 \times 20 \\
= 600 \\
} \]
Hence, no. of seats in the auditorium is \[600\].\[\]
Note: In the above question we used the factorization method for solving quad ratio equations. You also can use Sridhar acharya’s formula to solve the equation.
Using Sridhar acharya’s formula with the help of a formula we can directly find the roots of the equation.
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