# Exercise : 6

(b) 15 min

(c) 20 min

(d) Cannot be emptied

(e) None of these

**Ans.c**

Proportion of the volume of the tank filled by both the

pipes in 4 min = 4(^{1}⁄_{15} + ^{1}⁄_{10}) = ^{2}⁄_{3}rd of the tank

Volume of the tank filled by all the pipes working together = ^{1}⁄_{15} + ^{1}⁄_{10} - ^{1}⁄_{5} = -^{1}⁄_{30}

i.e. ^{1}⁄_{30} tank is emptied in 1 min.

∴ ^{2}⁄_{3} rd of the tank can be emptied in \(\frac{2 \times 30}{3} = 20 \; min \)

^{1}⁄

_{7}min.

(b) 3

^{1}⁄

_{3}min.

(c) 5 min.

(d) 3 min.

(e) None of these

**Ans.a**

Let cistern will be full in x min. Then,

part filled by B in x min + part filled by A in (x – 4)min = 1

⇒ \(\frac{x}{16} + \frac{x - 4}{12} = 1\)

⇒ x = ^{64}⁄_{7} = 9^{1}⁄_{7} hours.

(b) 10 min

(c) 12 min

(d) 7 min

(e) None of these

**Ans.a**

Let A was turned off after x min. Then,

cistern filled by A in x min + cistern filled by B in (x + 23) min = 1

⇒ \(\frac{x}{45} + \frac{x + 23}{40} = 1\)

⇒ 17x + 207 = 360 ⇒ x = 9 min.

^{1}⁄

_{9}min.

(b) 4

^{1}⁄

_{2}min.

(c) 3

^{3}⁄

_{4}min.

(d) 3 min.

(e) None of these

**Ans.a**

Let cistern will be full in x min. Then,

part filled by A in x min + part filled by B in (x – 1) min + part filled by C in (x – 2)min = 1

⇒ \(\frac{x}{3} + \frac{x - 1}{4} + \frac{x - 2}{6} = 1\)

⇒ 9x = 19 ⇒ x = ^{19}⁄_{9} = 2^{1}⁄_{9} min.

(b) Filled; 8 min

(c) Emptied; 8 min

(d) Filled; 12 min

(e) None of these

**Ans.c**

If both the pumps are opened together, then the tank will be emptied because the working efficiency of pump empting is more than that of the pump filling it. Thus in 1 min net proportion of the volume of tank filled

= (^{1}⁄_{8} - ^{1}⁄_{16}) = ^{1}⁄_{16}

or the tank will be emptied in 16 min

⇒ ^{1}⁄_{2} tank will be emptied in 8 min.