Question

Two water taps together can fill a tank in 9 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

Given :-

• Two water taps together can fill a tank in= 9(3/8) hours. = (75/8) Hours.
• The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately.

To Find :-

• Time by each tap separately ?

Solution :-

Let us Assume That, Tap with Smaller diameter can fill the tank alone in x Hours.

So,

→ The larger diameter tap fills the tank alone in (x - 10) hours.

Therefore,

→ Efficiency of Tap with smaller diameter = (1/x) / Hr.

→ Efficiency of Tap with Larger diameter = 1/(x - 10) / Hr.

A/q,

→ Both Tap can fill in = (75/8) Hours.

Hence,

→ Per Hour They will fill = (8/75) units of water.

So we can say That,

→ 1/x + 1/(x - 10) = 8/75

→ (x - 10 + x) /x(x - 10) = 8/75

→ (2x - 10) / (x² - 10x) = 8/75

→ 75(2x - 10) = 8(x² - 10x)

→ 150x - 750 = 8x² - 80x

→ 8x² - 80x - 150x + 750 = 0

→ 8x² - 230x + 750 = 0

→ 2(4x² - 115x + 375) = 0

→ 4x² - 115x + 375 = 0

→ 4x² - 100x - 15x + 375 = 0

→ 4x ( x - 25) - 15( x - 25) = 0

→ (4x - 15)( x - 25) = 0

Putting Both Equal to Zero Now,

→ 4x - 15 = 0

→ 4x = 15

→ x = (15/4) ≠ As x is Greater Than 10.

And,

→ x - 25 = 0

→ x = 25. (Ans.)

Hence :-

→ Smaller Diameter of the tank can fill the tank Alone in = x Hr. = 25 Hours.

→ Larger Diameter of the tank can fill the tank Alone in = x - 10 Hr. = 15 Hours.

Answer 2

Given :

• Two water taps can fill tank in 9 (3/8) hours together.
• Tap of larger diameter takes 10 hours less than the smaller tap.

To find :-

• Time in which each tap can fill tank separately.

Solution :-

They both can fill tank in 9 (3/8) hours

⇒ 75/8 hours.

∴ Efficiency of both taps to fill tank = (8/75) hours.

Let the time that the larger diameter tap can fill tank be y hours & time the smaller tap can fill tank be (y + 10) hours.

⇒ Efficiency of larger tap = 1/y hours.

⇒ Efficiency of smaller tap = 1/(y + 10) hours.

According to question :

⇒ 1/y + 1/(y + 10) = 8/75

⇒ (y + 10 + y)/[ y(y + 10)] = 8/75

⇒ (2y + 10)/(y² + 10y) = 8/75

⇒ 2 [(y + 5)/(y² + 10y)] = 8/75

⇒ (y + 5)/(y² + 10y) = 8/75 × 1/2

⇒ (y + 5)/(y² + 10y) = 8/150

⇒ (y + 5)/(y² + 10y) = 4/75

⇒ 4(y² + 10y) = 75(y + 5)

⇒ 4y² + 40y = 75y + 375

⇒ 4y² + 40y - 75y - 375 = 0

⇒ 4y² - 35y - 375 = 0

⇒ 4y² - 60y + 25y - 375 = 0

⇒ 4y(y - 15) + 25(y - 15) = 0

⇒ (4y + 25)(y - 15) = 0

⇒ 4y = -25 or, y = 15

⇒ y = -25/4 or, y = 15

∵ Time can't be negative in this respect.

∴ y = 15 hours.

So, time taken by larger tap to fill tank = y = 15 hours.

And, time taken by smaller tap to fill tank = y + 10 = 15 + 10 = 25 hours.

Therefore,

Larger tap can fill tank in 15 hours & smaller one can fill tank in 25 hours separately.