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9. Two water taps together can fill a tank in 9 – hours. The tap of larger diameter takes 10
hours less than the smaller one to fill the tank separately. Find the time in which each tap
can separately fill the tank.
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Answers
Step-by-step explanation:
Let the smaller tap fill the tank in x hours . → Then, the larger tap fills it in ( x - 10 ) hours . → Time taken by both together to fill the tank = 75/8 hours. → Part filled by the smaller tap in 1 hr = 1/x hours
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Answer:
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Step-by-step explanation:
suppose that the smaller tap fills the tank in x hours.
then, the larger tap will fill the tank in (x-10) hours.
since the smaller tap takes x hours to fill the tank.
•
. . portion of the tank filled by the smaller tap in one hour = 1
x
—> portion of the tank filled by the smaller tap in 9⅜ i.e., 75 hours.
8
similarly, We have,
portion of the tank filled by larger tap in 75 hours = ( 75 × 1.
8 8 x-10 )
= 75
8(x-10)
It is given that the two taps fill the tank in 75 hours. 8
75 + 75 = 1
8x 8(x-10)
—> 1 + 1 = 8
x x-10 75
—> x-10+x = 8
x(x-10) 75
—> 2x-10 = 8
x²-10x 75
—> (2x-10)×75 = 8(x²-10x)
—> 150x - 750 = 8x² - 80x
—> 8x² - 8x - 150x + 750 = 0
—> 8x² - 230x + 750 = 0
—> 2(4x² - 115x + 375) = 0
—> 4x² - 115x + 375 = 0
2
—> 4x² - 115x + 375 = 0
—> 4x² - 100x - 15x + 375 = 0
—> 4x(x-25) -15(x-25) = 0
—> (4x-15) (x-25) = 0
—> x = 15 ,. x = 25
4
Time taken by the smaller pipe cannot be 15 =
4
3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe will be 25 hours. Then, the time taken by the larger pipe will be 25-10 = 15 hours.