Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter take 10 hours less than the the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank
Answers
SOLUTION :
Given : Two water taps together can fill a tank in 9 ⅜ = 75/8 h
Let the smaller tap fill the tank in ‘x’ hours
The tap of larger diameter fills the tank in ‘x – 10’ hours.
In 1 hour the smaller tap fills the portion of the tank : 1/x
In 75/8 hour the smaller tap fills the portion of the tank : 75/8 × 1/x = 75/8x
In 1 hour the tap of larger diameter fills the portion of the tank : 1/(x - 10)
In 75/8 hour the tap of larger diameter fills the portion of the tank : 75/8 × 1/(x - 10) = 75/8(x - 10)
A.T.Q
75/8x + 75/8(x - 10) = 1
75/8[1/x + 1/(x - 10)] = 1
1/x + 1/(x - 10) = 8/75
(x - 10 + x)/(x) (x - 10) = 8/75
[By taking LCM]
{x - 10 + x}/(x² - 10x) = 8/75
75(x + x – 10) = 8(x² – 10x)
75(x + x – 10) = 8x² – 80x
75(2x– 10) = 8x² – 80x
150x – 750 = 8x² – 80x
8x² – 80x - 150x + 750 = 0
8x² – 230x + 750 = 0
2(4x² – 115x + 375) = 0
4x² – 115x + 375 = 0
4x² – 100x – 15x + 375 = 0
[By middle term splitting]
4x(x – 25) – 15(x – 25) = 0
(4x – 15)(x – 25) = 0
(4x – 15) = 0 (x – 25) = 0
x = 15/4 or x = 25
Value of x can’t be 15/4 as (x – 10) will be negative. Since, time cannot be negative,
Therefore, x = 25
The smaller tap fill the tank = x = 25 hours
The tap of larger diameter fills the tank = (x – 10) = 25 - 10 = 15 h
Hence, the time taken by the tap of larger diameter be 15 h & time taken by smaller tap be 25 hours.
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