two water taps can fill a tank in 15/8 hours .The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately . Find the time in which each tap can fill the tank separately
Answers
Answer:
Time taken by large tap individually = 3 hrs.
Time taken by small tap individually = 5 hrs
Step-by-step explanation:
Let the time taken by the tap with large dia = Tl = t hrs
Time taken by other tap = Ts = t+2 hrs
Rate of water by each tap per hour will be,
Rate of 1st tap Rl = 1/t
Rate of 2nd tap Rs = 1/(t+2)
When both taps are on, the time taken = 15/8 hrs.
∴ The rate of tank filling per hour = 1/time taken = 1 / (15/8)
∴ Rate of tank filling Rt = 8 / 15 per hour
This rate is same as the addition of rate of individual taps
∴ Rt = Rl + Rs
∴ 8/15 = 1/t + 1/(t+2) .... (1)
Solving above equation for t
∴ (8/15) = [(t+2) + t] / [t×(t+2)]
∴ 8×t×(t+2) = 15(2t+2)
∴ 8t² + 16t = 30t + 30
divide by 2
∴ 4t² + 8t = 15t + 15
∴ 4t² - 7t - 15 = 0
∴ 4t² - 12t + 5t -15 = 0
∴ 4t(t-3) + 5(t-3) = 0
∴ (t-3) × (4t+5) = 0
∴ t-3 = 0 .... or .... 4t+5 = 0
∴ t = 3 ..... or t = (-5/4)
Buttime taken cannot be negative,
∴ t = 3 hrs
Time taken by large tap individually = 3 hrs.
Time taken by small tap individually = 3+2 = 5 hrs