Question

‘Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank(This is a challenge to math lovers) no calculator shall be used answers approved till 30 minutes after the question has been asked(remember no spam to be allowed)

Answer 1

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

Answer 2

→ 25 hours and 15 hours .

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 .

→ Part filled by the larger tap in 1 hr = 1/( x - 10 ) .

→ Part filled by both the taps in 1 hr = 8/75 .

▶ Now,

$frac{1}{x} + \frac{1}{(x - 10)} = \frac{8}{75}. \\ \\ \sf \implies \frac{(x - 10) + x}{x(x - 10)} = \frac{8}{75} . \\ \\ \sf \implies \frac{(2x - 10)}{x(x - 10)} = \frac{8}{75} . \\ \\ \sf \implies75(2x - 10) = 8x(x - 10). \: \: \: \{by \: cross \: multiplication \} \\ \\ \sf \implies150x - 750 = 8 {x}^{2} - 80x. \\ \\ \sf \implies8 {x}^{2} - 230x + 750 = 0. \\ \\ \sf \implies4 {x}^{2} - 115x + 375 = 0. \\ \\ \sf \implies4 {x}^{2} - 100x - 15x + 375 = 0. \\ \\ \sf \implies4x(x - 25) - 15(x - 25) = 0. \\ \\ \sf \implies(x - 25)(4x - 15) = 0. \\ \\ \sf \implies x - 25 = 0. \: \: \green{or} \: \: 4x - 15 = 0. \\ \\ \sf \implies x = 25 \: \: \green{or} \: \: x = \frac{15}{4} . \\ \\ \huge \pink{ \boxed{ \tt \implies x = 25. }} \\ \\ \tt \bigg( \because x = \frac{15}{4} \implies(x - 10) < 0.$

Hence, the time taken by the smaller tap to fill the tank = 25 hours .

And , the time taken by the larger tap to fill the tank = ( 25 - 10 ) = 15 hours .