Two water taps can fill a tank in 75/8 hours. The tap of larger diameter takes 10 hours less than small tap. Find the time which each tap fill tank separately
Answers
Answer:
Time taken by smaller tap to fill the tank alone = 25 hours
Time taken by larger tap to fill the tank alone = 25 - 10 = 15 hours
Step-by-step explanation:
Time taken by both the taps together = 75/8 hours
Time taken by smaller tap to fill the tank alone = x hours
Time taken by larger tap to fill the tank alone = x - 10 hours
Now,
1/x + 1/x-10 = 8/75
x-10 + x/ x^2-10x = 8/75
2x - 10/ x^2 - 10x = 8/75
Cross multiplying,
8(x^2 - 10x) = 75(2x -10)
8x^2 - 80x = 150x - 750
8x^2 - 80x - 150x + 750 = 0
8x^2 - 230x + 750 = 0
D = b^2 - 4ac
= (-230)^2 - 4 * 8 * 750
= 52900 - 24000
= 28900
x = -b +or- root D/ 2a
x = -(-230) +or- root 28900/2*8
= 230 +or- root 28900/16
= 230 +or- 170/16
Therefore,
x = 230 + 170/16
= 400/16 = 25
x = 230 - 170/16
= 60/16 = 15/4
Therefore,
Time taken by smaller tap to fill the tank alone = 25 hours
Time taken by larger tap to fill the tank alone = 25 - 10 = 15 hours