Question

Two water taps together can fill a tank in
$\tt9 \ \frac{3}{8} \: hours$
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.​

Answer 1

$\huge\underline{\overline{\mid{\bold{ \orange{Given}}\mid}}}$

• Two water taps together can fill a tank in$\rm \:9 \frac{3}{8} \: hours$
• The tap of larger diameter takes = 10 hours

• the smaller one to fill the tank separately

$\huge\underline{\overline{\mid{\bold{ \red{To \: Find}}\mid}}}$

• the time in which each tap can separately fill the tank

$\huge\underline{\overline{\mid{\bold{ \purple{Solution}}\mid}}}$

Let the time taken by the smaller pipe to fill the tank be x hr

• Time taken by the larger pipe = (x - 10) hr

• Part of tank filled by smaller pipe in 1 hour = $\rm \frac{1}{x}$

• Part of tank filled by larger pipe in 1 hour = $\rm \frac{1}{x - 10}$

• It is given that the tank can be filled in$\rm9 \ \frac{3}{8} = \frac{75}{8}$

• hours by both the pipes together

$\huge\underline{\overline{\mid{\bold{ \green{Therefore,}}\mid}}}$

$\rm \implies \frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$

$\rm \implies \frac{x - 10 + x}{x( x - 10)} = \frac{8}{75}$

$\rm \implies \frac{2x - 10}{x(x - 10)} = \frac{8}{75}$

$\rm \implies75(2x - 10) = 8 {x}^{2} - 80x$

$\rm \implies150x - 750 = 8 {x}^{2} - 80x$

$\rm \implies8 {x}^{2} - 230x + 750 = 0$

$\rm \implies8 {x}^{2} - 200x - 30x + 750 = 0$

$\rm \implies8 {x}^{2} (x - 25) - 30(x - 25) = 0$

$\rm \implies(x - 25)(8x - 30) = 0$

$\rm \implies x = 25,\frac{30}{8}$

Time taken by the smaller pipe cannot be$\rm \frac{30}{8} = 3.75$hours

As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

$\huge\underline{\overline{\mid{\bold{ \blue{Hence,}}\mid}}}$

time taken individually by the smaller pipe and the larger pipe will be 25 and 25 - 10 = 15 hours respectively.

Answer 2

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

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

Totaltimeistakentofillatank=938

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

⇒ 1x+1x−10=875 x−10+xx(x−10)=875 2x−10x2−10x=875

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