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- Question 1 of 5
##### 1. Question

Two pipes can fill a tank in 8 hrs & 6 hrs respectively. If they are opened on alternate hours and if pipe A gets opened first, then in how many hours, the tank will be full?

Correctanswer: (b)

**Explanation**

Pipe A’s work in 1 hr = 1/8

Pipe B’s work in 1 hr = 1/6

Pipes (A+B)’s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24

Now,

In 4 hrs they fill : 2 X (7/24) = 7/12

In 6 hrs they fill : 3 X (7/24) = 7/8

After 6 hrs, part left empty = 1/8

Now it is A’s turn to open up.

In one hr it fills 1/8 of the tankSo, the tank will be full in = 6 hrs + 1 hr = 7 hrs

Incorrectanswer: (b)

**Explanation**

Pipe A’s work in 1 hr = 1/8

Pipe B’s work in 1 hr = 1/6

Pipes (A+B)’s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24

Now,

In 4 hrs they fill : 2 X (7/24) = 7/12

In 6 hrs they fill : 3 X (7/24) = 7/8

After 6 hrs, part left empty = 1/8

Now it is A’s turn to open up.

In one hr it fills 1/8 of the tankSo, the tank will be full in = 6 hrs + 1 hr = 7 hrs

- Question 2 of 5
##### 2. Question

Two pipes A & B can fill a tank in 5 min & 10 min respectively. Both the pipes are opened together but after 2 min, pipe A is turned off. What is the total time required to fill the tank?

CorrectCorrect answer (b)

**Hint:**

Assume the total time required = x + 2 min

Part filled by pipe A in 1min = 1/5 & part filled by B in 1 min = 1/10

After 2 min, part filed by A & B together = 2 [ 1/5 + 1/10] = 3/5

Remaining part = 1 – 3/5 = 2/5

Part filled by pipe B in 1 min = 1/10As pipe A is turned off after 2 minutes, we get the relation as,

**1/10 : 2/5 :: 1: x**x = (2/5) x 1 x 10 = 4 min

**Tank will be full in ( 2 min + 4 min) = 6 min**IncorrectCorrect answer (b)

**Hint:**

Assume the total time required = x + 2 min

Part filled by pipe A in 1min = 1/5 & part filled by B in 1 min = 1/10

After 2 min, part filed by A & B together = 2 [ 1/5 + 1/10] = 3/5

Remaining part = 1 – 3/5 = 2/5

Part filled by pipe B in 1 min = 1/10As pipe A is turned off after 2 minutes, we get the relation as,

**1/10 : 2/5 :: 1: x**x = (2/5) x 1 x 10 = 4 min

**Tank will be full in ( 2 min + 4 min) = 6 min** - Question 3 of 5
##### 3. Question

It is observed that the pipe A can fill the tank in 15 hrs and the same tank is filled by pipe B in 20 hrs. The third pipe C can vacant the tank in 25 hrs. If all the pipes get opened initially and after 10 hrs, the pipe C is closed, then how long will it take to fill the tank?

CorrectCorrect answer: (c)

**Hint:**

Assume that the tank will be full in ’10 + x’ hrs.Part filled by pipe A in 1 hr = 1/15

Part filled by pipe B in 1 hr = 1/20

Part emptied by pipe C in 1 hr = 1/ 25(A+ B)’s part filled in 1 hr = 1/15 + 1/20 = 7 /60

As Pipe C is closed after 10 hours, let us find the part of tank filled in 10 hrs.

Tank filled in 10 hrs = 10 (part filled by A in 1 hr + part filled by B in 1 hr– part emptied by C in 1 hr)

= 10 [1/15 + 1/20 – 1/25] = 23 /30Remaining part = 1 – part filled in 10 hours

= 1 – 23/30 = 7/30We know that,

So, we get the ratio as ,

**7/60 : 7/30 :: 1 : x**So, the value of x = 2/30 x 1 x 60/7 = 2 hrs.

**Hence, the tank will be full in x + 2 = 10 + 2 hrs = 12 hrs**IncorrectCorrect answer: (c)

**Hint:**

Assume that the tank will be full in ’10 + x’ hrs.Part filled by pipe A in 1 hr = 1/15

Part filled by pipe B in 1 hr = 1/20

Part emptied by pipe C in 1 hr = 1/ 25(A+ B)’s part filled in 1 hr = 1/15 + 1/20 = 7 /60

As Pipe C is closed after 10 hours, let us find the part of tank filled in 10 hrs.

Tank filled in 10 hrs = 10 (part filled by A in 1 hr + part filled by B in 1 hr– part emptied by C in 1 hr)

= 10 [1/15 + 1/20 – 1/25] = 23 /30Remaining part = 1 – part filled in 10 hours

= 1 – 23/30 = 7/30We know that,

So, we get the ratio as ,

**7/60 : 7/30 :: 1 : x**So, the value of x = 2/30 x 1 x 60/7 = 2 hrs.

**Hence, the tank will be full in x + 2 = 10 + 2 hrs = 12 hrs** - Question 4 of 5
##### 4. Question

Two pipes fills a cistern in 15 hrs and 20 hrs respectively. The pipes are opened simultaneously and it is observed that it took 26 min more to fill the cistern because of leakage at the bottom. If the cistern is full, then in what time will the leak empty it?

CorrectCorrect answer: (c)

**Hint:**Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.The time taken by the leak to empty the full tank = xy hrs y – x But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60

Hence, the time taken by these pipes to fill the tank = 60/7 hrs = 8 .57 hrs = 8 hrs 34 min ——–( by multiplying ‘0.57’ hrs x 60 = 34 minutes)

Due to leakage, time taken = 8 hrs 34 min + 26 min = 8 hrs 60 min = 9 hrs ———-( because 60 min = 1 hr)

Thus, work done by (two pipes + leak) in 1 hr = 1/9

Hence, work done by leak in 1 hr = work done by two pipes – 1/9

= 7 /60 – 1/9 = 3/ 540 = 1/180Therefore, leak will empty the full cistern in 180 hours

IncorrectCorrect answer: (c)

**Hint:**Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.The time taken by the leak to empty the full tank = xy hrs y – x But, direct values of x & y are not given. So, we need to find the work done by the two pipes in 1 hr = (1/15) + (1/20) = 7/60

Hence, the time taken by these pipes to fill the tank = 60/7 hrs = 8 .57 hrs = 8 hrs 34 min ——–( by multiplying ‘0.57’ hrs x 60 = 34 minutes)

Due to leakage, time taken = 8 hrs 34 min + 26 min = 8 hrs 60 min = 9 hrs ———-( because 60 min = 1 hr)

Thus, work done by (two pipes + leak) in 1 hr = 1/9

Hence, work done by leak in 1 hr = work done by two pipes – 1/9

= 7 /60 – 1/9 = 3/ 540 = 1/180Therefore, leak will empty the full cistern in 180 hours

- Question 5 of 5
##### 5. Question

An electric pump can fill a tank in 4 hours. Due to leakage in the tank, it took 4(1/2) hrs to fill the tank. If the tank is full, how much time will the leak take to empty the full tank?

CorrectCorrect answer: (d)

**Hint:**Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.The time taken by the leak to empty the full tank = xy hrs y – x Time taken to empty the tank by the leak = 4 x (9/2) / (9/2 ) – 4 = (36/2 ) / ½ = 18/ ½ = 18 x 2 = 36 hrs

IncorrectCorrect answer: (d)

**Hint:**Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.The time taken by the leak to empty the full tank = xy hrs y – x Time taken to empty the tank by the leak = 4 x (9/2) / (9/2 ) – 4 = (36/2 ) / ½ = 18/ ½ = 18 x 2 = 36 hrs

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