Question

3
8
Example 32 Two water pipes together can fill a tank in 9 hours. The pipe of larger
diameter takes 10 hours less than the smaller one to fill the tank separately. Find the
time in which each pipe can fill the tank separately.
75​

Answer 1

Answer:Let us consider the time taken by the smaller diameter tap = x

The time is taken by the larger diameter tap = x – 10

Total time is taken to fill a tank = 9\frac{3}{8}

Total time is taken to fill a tank = 75/8

In one hour portion filled by smaller diameter tap = 1/x

In one hour portion filled by larger diameter tap = 1/(x – 10)

In one hour portion filled by taps = 8/75

⇒ \frac{1}{x} + \frac{1}{x-10} = \frac{8}{75} \frac{x-10+x}{x(x-10)}= \frac{8}{75} \frac{2x-10}{x^{2}-10x}= \frac{8}{75}

75 (2x-10) = 8(x2-10x)

150x – 750 = 8x2 – 80x

8x2 − 230x + 750 = 0

4x2−115x + 375 = 0

4x2 − 100x −15x + 375 = 0

4x(x−25)−15(x−25) = 0

(4x−15)(x−25) = 0

4x−15 = 0 or x – 25 = 0

x = 15/4 or x = 25

Case 1: When x = 15/4

Then x – 10 = 15/4 – 10

⇒ 15-40/4

⇒ -25/4

Time can never be negative so x = 15/4 is not possible.

Case 2: When x = 25 then

x – 10 = 25 – 10 = 15

∴ The tap of smaller diameter can separately fill the tank in 25 hours

$\huge\red{@alurringbabe}$