__Answers :__

**Answer (d).**

Rate of filling of Pipe A – 1/8 of tank in 1 hour

Rate of filling of Pipe B – 1/6 of tank in 1 hour

Rate of filling when both are opened together

1/8 + 1/6 = 7/24 of tank in 1 hour

Since both are open for 2 hours, 2 x(7/24) = 7/12 of tank gets filled

Left over portion i.e. 5/12 of tank is to be filled by B, which fills 1/6 ( or 2/12) tank in 1 hour.

So it takes 2 ½ hours to fill 5/12 of tank.

**Answer (c).**

Rate of filling of the pump without the open tap – 1/8 of tank in 1 hour

Rate of filling of the pump with the open tap – 1/10 of tank in 1 hour

Rate of emptying of the tap (1/8 – 1/10) of tank in 1 hour

Or, the tap empties 1/40 of the tank in 1 hour

Thus the tap would empty a full tank in 40 hours.

**Answer (a).**

Rate of filling when both the pipes are open

1/15 + 1/20 or 7/60 of the tank in 1 minute

Rate of emptying the tank by the leak – 1/12 of the tank in 1 minute

So with both the taps open and the active leak, the rate of fill would be

(7/60 – 1/12) = 2/60 or 1/30 of the tank in 1 minute.

Thus, the tank would be full in 30 minutes.

**Answer (c).**

Let Pipe A fill the cistern in H hours

&there4 Pipe B can fill it in H + 6 hours

⇒ both the pipes can fill the ^{1}/_{H} + ^{1}/_{(H + 6)} of the cistern in 1 hour

^{1}/_{H} + ^{1}/_{H + 6} = ^{1}/_{4}

or ^{H + 6 + H}/_{H(H + 6)} = ^{1}/_{4}

or H^{2} – 2H – 24 = 0

(H – 6)(H + 4) = 0 or H = 6

Pipe A can fill the cistern in 6 hours.

**Answer (a).**

If the faster tap can fill the drum in M minutes, the slower tap fills it in 3M minutes

Both the pipes can fill (1/M + 1/3M) of the drum in 1 minute

^{1}/_{M} + ^{1}/_{3M} = ^{1}/_{36} or ^{4}/_{3M} = ^{1}/_{36} or 3M = 144 or M = 48

Faster tap can fill the drum in 48 minutes.

Slower tap can fill the drum in 3 x 48 = 144 minutes.

**Answer (c).**

The leak can empty ^{1}/_{10} of the cistern in 1 hour

With the tap open, the ^{1}/_{15} cistern is emptied in 1 hour

Thus, the rate of filling of the tap is ^{1}/_{10} – ^{1}/_{15} = ^{1}/_{30} of the cistern in 1 hour

In other words the tap can fill the cistern in 30 hours or 1800 minutes

Since the tap supplies 8 litres of water per minute, the capacity of the cistern is 1800 x 8 = 14,400 litres

**Answer (d).**

Taps A and B can fill ^{1}/_{40} of the tank in 1 minute

Tap A can fill ^{1}/_{60} of the tank in 1 minute

Therefore Tap B can fill ^{1}/_{40} – ^{1}/_{60} of the tank in 1 minute

Or Tap B can fill ^{1}/_{120} of the tank in 1 minute

Therefore Tap B can fill the tank in 120 minutes

**Answer (c).**

8 taps can fill a tank in 30 minutes

6 taps can fill the tank in M minutes

M = (8 x 30)/6 = 40 minutes

**Answer (a).**

When both the pipes are open (^{1}/_{30} + ^{1}/_{45}) of the tank is filled in 1 minute.

In the first 10 minutes (^{1}/_{30} + ^{1}/_{45}) x 10 = ^{10}/_{18} of the tank gets filled ....(1).

In the next 10 minutes, ^{10}/_{45} of the tank is filled by pipe B ....(2)

Thus, adding (1) and (2) we get ^{7}/_{9}

Thus, ^{2}/_{9} of the tank is to be filled by A

A takes ^{2}/_{9} x 30 = 6 ^{2}/_{3} minutes to fill the tank

**Answer (c).**

Pipe A fills ^{1}/_{12} of the tank in 1 hour

Pipe B can empty ^{1}/_{36} in 1 hour or ^{1}/_{144} in 15 minutes

From 8 a.m. to 9 a.m. ^{1}/_{12} of the tank gets filled

Thus, ^{11}/_{12} of the tank is empty

From 9 a.m. to 10 a.m. (^{1}/_{12} – ^{1}/_{144}) = ^{11}/_{144} of the tank gets filled

i.e. in 1 hour ^{11}/_{144} of the tank gets filled

So to fill, ^{11}/_{12} of the tank it would take ^{11}/_{12} x ^{144}/_{11} = 12 hours

Total time required is 1 + 12 = 13 hours

The tank would be full at 8 a.m. + 13 = 9 p.m.