# Exercise : 2

(b) 8 hours

(c) 18 hours

(d) 12 hours

(e) None of these

**Ans.d**

Required time to fill the tank

= \(\frac{1}{\left(\frac{1}{4} + \frac{1}{6} \right) - \frac{1}{3}} = \frac{1}{\frac{5}{12} - \frac{1}{3}} = \frac{1}{\frac{1}{12}} = 12 h\)

(b) 24

(c) 36

(d) 16

(e) None of the above

**Ans.b**

24 men complete the work in 16 days

∴ 16 men complete (^{16}⁄_{24} × ^{12}⁄_{16}) = ^{1}⁄_{2} part of work in 12 days

32 women complete the work in 24 days

∴ 16 women complete ^{16}⁄_{32} × ^{14}⁄_{24}) = ^{7}⁄_{24} part of work in (12 + 2 ) = 14 days

So, the remaining part of the work which is done by sixteen men + sixteen women and the reqd additional no. of men in 2 days

= 1 - (^{1}⁄_{2} + ^{7}⁄_{24}) = ^{1}⁄_{2} - ^{7}⁄_{24} = ^{5}⁄_{24} (part)

Now, in 2 days ^{5}⁄_{24} part of the work is done by

24 × ^{16}⁄_{2} × ^{5}⁄_{24} = 40 men

Hence, the reqd. additional no. of men

= 40 – 16 = 24 men.

(b) 8,000

(c) 9,000

(d) 6,500

(e) None of these

**Ans.b**

4M + 2W = 46000;

Again, W = M + 500

or, M = W – 500

∴ (W – 500) + 2W = 46000

or, 6W = 46000 + 2000 = 48000

∴ W = 8000

(b) 7

^{2}⁄

_{3}

(c) 6

^{2}⁄

_{3}

(d) 6

^{1}⁄

_{3}

(e) None of these

**Ans.c**

10 men + 15 women in 1 day do ^{1}⁄_{15} + ^{1}⁄_{12} = ^{9}⁄_{60} work

\ Time taken = ^{60}⁄_{9} days = 6^{2}⁄_{3} days

(b) 8

(c) 6

(d) 4

(e) None of these

**Ans.a**

Work done by ‘A’ in 3 days

= ^{1}⁄_{12} × 3 = ^{1}⁄_{4}

∴ Remaining work = 1 - ^{1}⁄_{4} = ^{3}⁄_{4}

Work done by A and B together = \(\frac{12 \times 15}{27} = \frac{20}{3}\)

∴ Remaining work done by A and B together in

= ^{3}⁄_{4} × ^{20}⁄_{3} = 5 days

(b) 20 days

(c) 46 days

(d) 21 days

(e) None of these

**Ans.e**

Remaining pages to read = 445 – 157 = 288

∴ Reqd. number of days = ^{288}⁄_{24} = 12

(b) 36

(c) 34

(d) 32

(e) None of these

**Ans.d**

m_{1} × d_{1} × t_{1} × w_{2} = m_{2} × d_{2} × t_{2} × w_{1}

24 × 10 × 8 × 1 = m2 × 6 × 10 × 1

⇒ m_{2} = \(\frac{24 \times 10 \times 8}{6 \times 10}\) = 32 men

(b) 19 days

(c) 20 days

(d) 21 days

(e) None of these

**Ans.b**

X’s one day’s work = 1.25 th part of whole work.

Y’s one day’s work = ^{1}⁄_{30} th part of whole work.

Their one day’s work = ^{1}⁄_{25} + ^{1}⁄_{30} = ^{1}⁄_{150} th part of whole work.

Now, work is done in 5 days = ^{11}⁄_{150} × 5 = ^{11}⁄_{30} th of whole work

∴ Remaining work = 1 - ^{11}⁄_{30} = ^{19}⁄_{30} th of whole work

Now, ^{1}⁄_{30} th work is done by Y in one day.

∴ ^{19}⁄_{30} th work is done by Y in \(\frac{1}{\frac{1}{30}} \times \frac{19}{30}\) = 19 days.

(b) 4 days

(c) 5 days

(d) 7 days

(e) None of these

**Ans.c**

A’s one day’s work = ^{1}⁄_{16} th work

B’s one day’s work ^{1}⁄_{12} th work

Let B has worked alone = x days. Then,

A’s amount of work + B’s amount of work = 1

⇒ 4(^{1}⁄_{16}) + (x + 4)(^{1}⁄_{12}) = 1

⇒ \(\frac{1}{4} + \frac{x + 4}{12}\) = 1 ⇒ x = ^{3}⁄_{4} × 12 - 4

⇒ x = 5 days.

(b) 30

(c) 35

(d) 20

(e) None of these

**Ans.a**

50 men complete 0.4 work in 25 days.

Applying the work rule, m_{1} × d_{1} × w_{2} = m_{2} × d_{2} × w_{1}

we have,

50 × 25 × 0.6 = m_{2} × 25 × 0.4

or m_{2} = \(\frac{50 \times 25 \times 0.6}{25 \times 0.4}\) = 75 men