__Question No. 01__

**Two pipes A
and B can fill a cistern in 37½ minutes
and 45 minutes respectively. Both pipes are opened. The cistern will be filled
in just half an hour, if the B is turned off after:**

(A) 5 min

(B) 9 min

(C) 10 min

(D) 15 min

Answer:
Option B

__Explanation:__

Let B be
turned off after *x* minutes. Then,

Part filled
by (A + B) in *x* min. + Part filled by A in (30 - *x*)
min. = 1.

∴ *x
*(^{2}/_{75}
+ ^{1}/_{45}) + [(30 -
*x*). ^{2}/_{75} = 1

=> (11*x*/225) + [(60 - 2*x*)/75] = 1

=> 11*x* +
180 - 6*x* = 225

=> *x* =
9

__Question No. 02__

**Two pipes
can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3
gallons per minute. All the three pipes working together can fill the tank in
15 minutes. The capacity of the tank is:**

(A) 60
gallons

(B) 100
gallons

(C) 120
gallons

(D) 180
gallons

Answer:
Option C

__Explanation:__

Work done by
the waste pipe in 1 minute = ^{1}/_{15 }- (^{1}/_{20
}+ ^{1}/_{24})

= (^{1}/_{15}
- ^{11}/_{120})

= - (^{1}/_{40}) [-ve sign means emptying]

∴ Volume
of ^{1}/_{40} part = 3 gallons.

Volume of
whole = (3 × 40) gallons = 120 gallons.

__Question No. 03__

**Two pipes ‘***A*’ and ‘*B*’ can fill a tank in 15 minutes and 20 minutes respectively. Both
the pipes are opened together but after 4 minutes, pipe ‘*A*’ is turned off. What is the total time required to fill the tank?

(A) 10 min.
20 sec.

(B) 11 min.
45 sec.

(C) 12 min.
30 sec.

(D) 14 min.
40 sec.

Answer:
Option D

__Explanation:__

Part filled
in 4 minutes = 4 {(1/15) + (1/20)} = (7/15)

Remaining
part = {1 - (7/15)} = (8/15)

Part filled
by B in 1 minute = (1/20)

∴ (1/20) : (8/15) **::** 1 : *x*

*x* =
{(8/15) × 1 × 20} = 10⅔ min = 10 min. 40 sec.

∴ The
tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

__Question No. 04__

**A large
tanker can be filled by two pipes ‘***A*’
and ‘*B*’ in 60 minutes and 40 minutes
respectively. How many minutes will it take to fill the tanker from empty state
if ‘*B*’ is used for half the time and ‘*A*’ and ‘*B*’ fill it together for the other half?

(A) 15 min

(B) 20 min

(C) 27.5 min

(D) 30 min

Answer:
Option D

__Explanation:__

Part filled
by (A + B) in 1 minute = {(1/60) + (1/40)} = (1/24)

Suppose the
tank is filled in *x* minutes.

Then, (*x*/2) × {(1/24) + (1/40)} = 1

=> (*x*/2) × (1/15) = 1

=> *x* =
30 min.

__Question No. 05__

**A tap can
fill a tank in 6 hours. After half the tank is filled, three more similar taps are
opened. What is the total time taken to fill the tank completely?**

(A) 3 hrs 15
min

(B) 3 hrs 45
min

(C) 4 hrs

(D) 4 hrs 15
min

Answer:
Option B

__Explanation:__

Time taken
by one tap to fill half of the
tank = 3 hrs.

Part filled
by the four taps in 1 hour = {4 × (1/6)} = 2/3

Remaining
part = {1 - (1/2)} = 1/2

∴ (2/3) : (1/2) **::** 1 : *x*

=> *x* = [(1/2) × 1 × (3/2)] = (3/4) hours *i.e.,* 45
mins.

So, total
time taken = 3 hrs. 45 mins.

__Question No. 06__

**Three pipes ***A*, *B*
and *C* can fill a tank in 6 hours.
After working at it together for 2 hours, ‘*C*’
is closed and ‘*A*’ and ‘*B*’ can fill the remaining part in 7
hours. The number of hours taken by ‘*C*’
alone to fill the tank is:

(A) 10

(B) 12

(C) 14

(D) 16

Answer:
Option C

__Explanation:__

Part filled
in 2 hours = 2/6 = 1/3

Remaining
part = [1 - (1/3)] = 2/3

∴ (A
+ B)'s 7 hour's work = 2/3

∴ C's
1 hour's work = {(A + B + C)'s 1 hour's work} - {(A + B)'s 1 hour's work}

= {(1/6) -
(2/21)} = 1/14

∴ C
alone can fill the tank in 14 hours.

__Question No. 07__

**Three pipes ***A*, *B*
and *C* can fill a tank from empty to
full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is
empty, all the three pipes are opened. *A*,
*B* and *C* discharge chemical solutions *P*,
*Q* and *R* respectively. What is the proportion of the solution ‘*R*’ in the liquid in the tank after 3
minutes?

(A) 5/11

(B) 6/11

(C) 7/11

(D) 8/11

Answer:
Option B

__Explanation:__

Part filled
by (A + B + C) in 3 minutes = 3 [(1/30) + (1/20) + (1/10)] = 3 × (11/60) =
11/20

Part filled
by C in 3 minutes = 3/10

∴ Required
ratio = (3/10) × (20/11) = 6/11

__Question No. 08__

**A tank is
filled by three pipes with uniform flow. The first two pipes operating
simultaneously fill the tank in the same time during which the tank is filled
by the third pipe alone. The second pipe fills the tank 5 hours faster than the
first pipe and 4 hours slower than the third pipe. The time required by the
first pipe is:**

(A) 6 hours

(B) 10 hours

(C) 15 hours

(D) 30 hours

Answer:
Option C

__Explanation:__

Suppose, first
pipe alone takes *x* hours to fill the tank

Then, second
and third pipes will take (*x* -5) and (*x* - 9) hours
respectively to fill the tank.

∴ ^{1}/_{x}
+ ^{1}/_{(x - 5)} = ^{1}/_{(x - 9)}

=> ^{(x
- 5 + x)}/_{x (x - 5)} = ^{1}/_{(x - 9)}

=> (2*x* -
5)(*x* - 9) = *x*(*x* - 5)

=>* x*² - 18*x* +
45 = 0

=> (*x* -
15)(*x* - 3) = 0

=>* x* = 15.
[Neglecting *x* = 3]

__Question No. 09__

**Two pipes ***A* and *B* together can fill a cistern in 4 hours. Had they been opened
separately, then ‘*B*’ would have taken
6 hours more than ‘*A*’ to fill the
cistern. How much time will be taken by ‘*A*’
to fill the cistern separately?

(A) 1 hour

(B) 2 hours

(C) 6 hours

(D) 8 hours

Answer:
Option C

__Explanation:__

Let the
cistern be filled by pipe A alone in *x* hours.

Then, pipe B
will fill it in (*x* + 6) hours.

∴ (1/*x*)
+ 1/(*x*+6) = 1/4

=> [(*x* + 6 + *x*)/* x*(*x* + 6)]* *= 1/4

=> *x*^{2} - 2*x* -
24 = 0

=> (*x* -6)(*x* +
4) = 0

=> *x* =
6. [Neglecting the negative value of *x*]

__Question No. 10__

**A tank is
filled in 5 hours by three pipes ***A*, *B* and *C*. The pipe ‘*C’* is twice
as fast as ‘*B*’ and ‘*B*’ is twice as fast as ‘*A*’. How much time will pipe ‘*A*’ alone take to fill the tank?

(A) 20 hours

(B) 25 hours

(C) 35 hours

(D) None of
these

Answer:
Option C

__Explanation:__

Suppose pipe
A alone takes *x* hours to fill the tank.

Then, pipes *B* and *C* will take *x*/2 and *x*/4 hours respectively to fill the tank

∴ (1/*x*)
+ (2/*x*) + (4/*x*) = (1/5)