Pupil's of 10th standard of a school had arranged for a function at a total cost of Rs.1,000 which was to be shared equally among them. Since 10 of them failed to join the function each of them had to pay Rs.5 more. Find the number of pupils in the class?
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x = number of pupils
x - 10 = number of pupils without 10
If all pay, then for one is 1000 / x Rs
x - 10 pupils pay 5 Rs. more: 1000 / x + 5
(x - 10) · (1000 / x + 5) = 1000
1000 + 5x - 10000 / x - 50 = 1000
5x - 10000 / x = 50
Dividing the sides by 5:
x - 2000 / x = 10
Multiplying sides of the equation by x, but x must be ≠0
x² - 2000 = 10x
x² - 10x - 2000 = 0
We have the quadratic equation.
The discriminant of equation is:
Δ = (-10)² - 4 · 1 · (-2000) = 8100
√Δ = 90
x₁ = (10 - 90) / 2 = -40, this is false, because x > 0
x₂ = (10 + 90) / 2 = 50
Chek:
50 · 20 = 1000 Rs. - if all pay
(50 - 10) · (20 + 5) = 1000 - if 40 pupils pay 25 Rp. (= more 5 Rs.)
The number of 10th class pupils in the school is 50.
x - 10 = number of pupils without 10
If all pay, then for one is 1000 / x Rs
x - 10 pupils pay 5 Rs. more: 1000 / x + 5
(x - 10) · (1000 / x + 5) = 1000
1000 + 5x - 10000 / x - 50 = 1000
5x - 10000 / x = 50
Dividing the sides by 5:
x - 2000 / x = 10
Multiplying sides of the equation by x, but x must be ≠0
x² - 2000 = 10x
x² - 10x - 2000 = 0
We have the quadratic equation.
The discriminant of equation is:
Δ = (-10)² - 4 · 1 · (-2000) = 8100
√Δ = 90
x₁ = (10 - 90) / 2 = -40, this is false, because x > 0
x₂ = (10 + 90) / 2 = 50
Chek:
50 · 20 = 1000 Rs. - if all pay
(50 - 10) · (20 + 5) = 1000 - if 40 pupils pay 25 Rp. (= more 5 Rs.)
The number of 10th class pupils in the school is 50.
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