Question 2792
Let the total number of students supposed to be contributing be 'x'.

Let cost of each student be M = 20/x.

Now,actual number of students contributing = (x-4).

Let actual cost of each student be N = 20/(x-4).

When there were 4 students less,they paid 25 cents more.

This means that N > M by 25 cents.

Therefore,

 N-M=25 Cents or $(1/4)

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Then we get,

{{{ (20/(x-4)) - (20/x) = (1/4) }}}

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Taking the LCM we get,

{{{ (20(x) - 20(x-4))/(x(x-4)) = (1/4) }}}

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{{{ (20x - 20x + 80)/(x^2 - 4x) = (1/4) }}}

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{{{ 80/(x^2 - 4x) = (1/4) }}}

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On cross multiplying we get,

{{{ 4(80) = (x^2 - 4x) }}}
{{{ (320) = (x^2 - 4x) }}}

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Taking 320 to the other side we get,

x^2 - 4x - 320 = 0

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Factorizing 320 to get a difference of 4 we get,

x^2 - 16x + 20x - 320 = 0
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x(x-16) + 20(x-16) = 0
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(x-16)(x+20) = 0
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Then we get,
x=16 OR x=-20
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As number of students cannot be negative,
x=16 is the required value of x.
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Then number of students contributing is (x-4) = 12 students.

ANS: 12 students contributed.

Hope this helped,
Best of luck.