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:

20

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23

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24

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None of these

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Solution

The correct option is C $23$
Let the number of students in each row be $x$ .
Hence, $x (x + 3) = 460$
$x^{2} + 3 x = 460$
$x^{2} + 3 x - 460 = 0$
$(x + 23) (x - 20) = 0$
$x = - 23$ and $x = 20$
Hence, the number students in each column is $20 + 3 = 23$ .

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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:

The correct option is C 23Let the number of students in each row be x.Hence, x(x+3)=460x2+3x=460x2+3x−460=0(x+23)(x−20)=0x=−23 and x=20Hence, the number students in each column is 20+3=23.

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:

The correct option is C $23$
Let the number of students in each row be $x$ .
Hence, $x (x + 3) = 460$
$x^{2} + 3 x = 460$
$x^{2} + 3 x - 460 = 0$
$(x + 23) (x - 20) = 0$
$x = - 23$ and $x = 20$
Hence, the number students in each column is $20 + 3 = 23$ .