Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter tajes 10 hours less than the smaller one to fill the tank seperately. Find the time in which each tap can seperately fill the tank
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Answer:
Let us consider the time taken by the smaller diameter tap = x
The time is taken by the larger diameter tap = x – 10
Totaltimeistakentofillatank=938
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
⇒ 1x+1x−10=875 x−10+xx(x−10)=875 2x−10x2−10x=875
75 (2x-10) = 8(x2-10x)
150x – 750 = 8x2 – 80x
8x2 − 230x + 750 = 0
4x2−115x + 375 = 0
4x2 − 100x −15x + 375 = 0
4x(x−25)−15(x−25) = 0
(4x−15)(x−25) = 0
4x−15 = 0 or x – 25 = 0
x = 15/4 or x = 25
Case 1: When x = 15/4
Then x – 10 = 15/4 – 10
⇒ 15-40/4
⇒ -25/4
Time can never be negative so x = 15/4 is not possible.
Case 2: When x = 25 then
x – 10 = 25 – 10 = 15
∴ The tap of smaller diameter can separately fill the tank in 25 hours