Question

I'm in 10class .. please give me solution of 9 question in exercises 4.3

Answer 1

there is no 9th problem

Answer 2

Let the time taken by the smaller tap to fill the tank be x hr.
Time taken by the larger tap = (x − 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x
Part of tank filled by larger pipe in 1 hour = 1/x-10
It is given that the tank can be filled in 9 3/8 = 75/8 hours by both the taps together. Therefore,
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 – 230x + 750 = 0
= 8x2 – 200x – 30x +750 = 0
= 8x(x – 25) – 30(x – 25) = 0
=(x – 25 )(8x – 30) = 0
= x = 25, 30/8

Time taken by the smaller pipe cannot be 3/8 = 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 and the larger pipe will be 25 and 25 − 10 =15 hours respectively.