Two water taps together can fll a tank in 9hours. The tap of larger diameter
takes 10 hours less than the smaller one to fillthe tank separately. Find the time
in which each tap can separately fill the tank.
Answers
Answer:
Let us consider the time taken by the smaller diameter tap = x
The time is taken by the larger diameter tap = x – 10
Total time is taken to fill a tank = 9\frac{3}{8}
Total time is taken to fill a tank = 75/8
In one hour portion filled by smaller diameter tap = 1/x
In one hour portion filled by larger diameter tap = 1/(x – 10)
In one hour portion filled by taps = 8/75
Step-by-step explanation:
⇒ 1x+1x−10=875 x−10+xx(x−10)=875 2x−10x2−10x=87575 (2x-10) = 8(x2-10x)150x – 750 = 8x2 – 80x8x2 − 230x + 750 = 04x2−115x + 375 = 04x2 − 100x −15x + 375 = 04x(x−25)−15(x−25) = 0(4x−15)(x−25) = 04x−15 = 0 or x – 25 = 0x = 15/4 or x = 25Case 1: When x = 15/4Then x – 10 = 15/4 – 10⇒ 15-40/4⇒ -25/4Time can never be negative so x = 15/4 is not possible.Case 2: When x = 25 thenx – 10 = 25 – 10 = 15∴ The tap of smaller diameter can separately fill the tank in 25 hours
Answer:
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