Model 3 Opening both taps and leaks Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Let the pipe B fill the tank in x minutes.

Part of the tank filled by pipes A and B in 1 minute = $1/36$

Part of the tank filled by pipe A in 1 minute

= $1/36 - 1/x$

According to the question,

30 × $1/x + 40(1/36 - 1/x) = 1$

$30/x + 10/9 - 40/x$ = 1

$40/x - 30/x = 10/9 - 1$

$10/x = 1/9 ⇒ x = 90$ minutes

Answer: (a)

Part of the tank filled in 4 hours by pipe A = $4/6 = 2/3$

Remaining part = ${1 - 2}/3 = 1/3$

Time taken by pipe B in filling $1/3$ part = $8/3$ hours

Using Rule 8,

Here, x = 6, y = 8, t = 4

Required time = $[y(1 –{t/x})]$ hours

= $[8(1 - 4/6)]$ hours = $8/3$ hours

Answer: (d)

Since, P < q,

On opening pipe and sink together,

Part of the tub filled in 1 hour

= $1/P - 1/q$

Clearly, $1/P - 1/q = 1/r$

Answer: (a)

Let the capacity of the tank = x litres

According to the question,

Quantity of water emptied by the leak in 1 hour = $x/10$ litres

Qunatity of water filled by the tap in 1 hour = 240 litres

According to the question,

$x/10 - x/15$ = 240

${3x - 2x}/30 = 240$

$x/30$ = 240

$x$ = 240 × 30 = 7200 litres

Answer: (b)

A, B and C together fill the tank in 6 hours.

Part of the tank filled in 1 hour by (A + B + C) = $1/6$

Part of the tank filled in 2 hours by all three pipes

= $2/6 = 1/3$

Remaining empty part

= $1 - 1/3 = 2/3$

This $2/3$ part is filled by (A + B).

Time taken by (A + B) to fill the fully empty tank

= ${7 × 3}/2 = 21/2$ hours

Part of tank filled by C in 1 hour

= $1/6 - 2/21 = {7 - 4}/42 = 3/42 = 1/14$

∴ Required time = 14 hours.

Answer: (a)

A tap can fill the tank in 6 hours. In filling the tank to its half, time required = 3 hours.

Remaining part = $1/2$

Since, 1 tap takes 6 hours to fill the tank

Time taken by 4 taps take to fill $1/2$ of the tank

= $6/4 × 1/2 = 3/4$ hour

Total time = $3 + 3/4$

= $3{3}/4$ hours = 3 hours 45 minutes