Two water taps can fill a tank together in 75/8 hours. The tap with the smaller diameter takes 10 hours more than the tap with the larger diameter to fill the tank seperately. Find the time taken by each tap to fill the tank seperately.
Answers
Answer:
Step-by-step explanation:
Answer:
15 hours, 25 hours
Step-by-step explanation:
Let the time taken by the larger pipe to fill the tank be 'x' hours.
Then,Time taken by the smaller pipe = (x + 10) hr.
∴ Part of tank filled by larger pipe = (1/x)
∴ Part of tank filled by smaller pipe = (1/x + 10).
Given that taps can fill a tank together in 75/8 hours.
⇒ (1/x) + (1/x + 10) = 8/75
⇒ (x + 10 + x)/x(x + 10) = 8/75
⇒ (2x + 10)/x(x + 10) = 8/75
⇒ 75(2x + 10) = 8x² + 80x
⇒ 150x + 750 = 8x² + 80x
⇒ 8x² - 70x - 750 = 0
⇒ 4x² - 35x - 375 = 0
⇒ 4x² - 60x + 25x - 375 = 0
⇒ 4x(x - 15) + 25(x - 15) = 0
⇒ (4x + 25)(x - 15) = 0
⇒ x = -25/4, 15.
x ≠ -25/4 as time cannot be negative.
∴ x = 15.
∴ x + 10 = 25 hours.
Therefore:
Time taken by larger tap = 15 hours.
Time taken by smaller tap = 25 hours.
Hope it helps!