Question: Two water taps together can fill a tank in 9(\\frac{3}{8})) hours. The tap of larger dimensions takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Video Explanation
Explanatory Answer
Step 1: Frame equations based on the information about the time taken by the two taps to fill the tank.
Let the time taken by the tap of larger dimension be t hours.
∴ time taken by the tap of smaller dimension = (t + 10) hours.
The larger tap will fill \\frac{1}{t}) of the tank in 1 hour
The smaller tap will fill \\frac{1}{t + 10}) of the tank in 1 hour
Together the 2 taps will fill \\frac{1}{t}) + \\frac{1}{t + 10}) of the tank in 1 hour ----------(1)
Statement: Two water taps fill a tank in 9\\frac{3}{8}) hrs
i.e., in \\frac{75}{8}) hours
So, the taps will fill \\frac{1}{75/8})= \\frac{8}{75}) of the tank in 1 hour ----------(2)
Equate (1) and (2)
\\frac{1}{t}) + \\frac{1}{t + 10}) = \\frac{8}{75})
Because t ≠ 0 and t cannot be -10, multiply the entire equation by t(t + 10)
(t + 10) + t = \\frac{8}{75})(t (t + 10))
2t + 10 = \\frac{8}{75})(t2 + 10t)
75(2t + 10) = 8t2 + 80 t
150t + 750 = 8t2 + 80 t
8t2 - 70t - 750 = 0
Divide the equation by 2 :
4t2 - 35t - 375 = 0
Step 2: Factorize to find the values of ‘t’
4t2 – 60t + 25t – 375 = 0
4t(t – 15) + 25 (t – 15) = 0
(4t + 25) (t – 15) = 0
t = \\frac{-25}{4}) and t = 15
Time taken is positive
∴ t = 15 hrs and t + 10 = 15 + 10 = 25 hrs