✔!!.Two Water taps together can fill a tank in 75/8 hours. The taps of larger diameter takes 10hrs less than the smaller one to fill the tank separately. Find the time which each tap can separately fill the tank.......✔✔ ❤❤❤❤❤❤❤❤❤❤
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Answers
Answer:
25 hours, 15 hours
Step-by-step explanation:
Let the time taken by the smaller tap be 'x' hours.
Given that taps of larger diameter takes 10 hrs less then smaller one.
Then the time taken by the larger tap be (x - 10) hours.
Now,
∴ Portion of the tank filled by the smaller tap in 1 hour = (1/x).
∴ Portion of the tank filled by the larger tap in 1 hour = (1/x - 10).
Given that Portion of the tank filled by both tanks in 1 hour = 75/8 hours.
∴ (1/x) + (1/x - 10) = 75/8
⇒ (x - 10 + x)/x(x - 10) = 75/8
⇒ (2x - 10)/x(x - 10) = 75/8
⇒ 75(2x - 10) = 8[x(x - 10)]
⇒ 150x - 750 = 8[x² - 10x]
⇒ 150x - 750 = 8x² - 80x
⇒ 8x² - 230x + 750 = 0
⇒ 8x² - 200x - 30x + 750 = 0
⇒ 8x(x - 25) - 30(x - 25) = 0
⇒ (x - 25)(8x - 30) = 0
⇒ x = 25, 80/8{Rejected ∴ x ≠ ≤ 10}
⇒ x = 25.
When x = 25:
= x - 10
= 25 - 10
= 15.
Therefore:
Time taken by the larger tap to fill the tank = 15 hours.
Time taken by the smaller tap to fill the tank = 25 hours.
Hope it helps!
Answer:
it's answer was 16 CM
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