# Exercise : 3

(b) 13 days

(c) 20

^{3}⁄

_{17}days

(d) 12 days

(e) None of these

**Ans.b**

Ratio of times taken by A and B = 100 : 130 = 10 : 13.

Suppose B takes x days to do the work.

Then, 10 : 13 : : 23 : x ⇒ x = \(\frac{\left(23 \times 13 \right)}{10}\) ⇒ x = ^{299}⁄_{10}

A’s 1 day’s work = ^{1}⁄_{23}; B’s 1 days work = ^{10}⁄_{299}

(A + B)’s 1 day’s work = (^{1}⁄_{23} + ^{10}⁄_{299}) = ^{23}⁄_{299} = ^{1}⁄_{13}

∴ A and B together can complete the job in 13 days.

(b) 15 days

(c) 45 days

(d) 24 days

(e) None of these

**Ans.c**

Let C completes the work in x days.

Work done by (A + B) in 1 day = ^{1}⁄_{10}

Work done by (B + C) in 1 day = ^{1}⁄_{18}

A’s 5 days’ work + B’s 10 days’ work + C’s 15 days’work = 1

or (A + B)’s 5 days’ work + (B + C)’s 5 days’ work + C’s 10 days’ work = 1

or ^{5}⁄_{10} + ^{5}⁄_{18} + ^{10}⁄_{x} = 1 or x = 45 days

(b) 12 days

(c) 15 days

(d) 9 days

(e) None of these

**Ans.d**

In 1 day, work done by 12 men = ^{1}⁄_{18}

In 6 days, work done by 12 men = ^{6}⁄_{18} = ^{1}⁄_{3}

Remaining work = ^{2}⁄_{3}

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

or 12 × 18 × ^{2}⁄_{3} = 16 × d_{2} × 1

or d_{2} = \(\frac{4 \times 18 \times 2}{16} = 9 \; days\)

^{1}⁄

_{2}minutes

(b) 3

^{1}⁄

_{2}minutes

(c) 3

^{3}⁄

_{5}minutes

(d) 4

^{1}⁄

_{4}minutes

(e) None of these

**Ans.c**

1 minute’s work of both the punctures = (^{1}⁄_{9} + ^{1}⁄_{6}) = ^{5}⁄_{18}

So, both the punctures will make the tyre flat in ^{18}⁄_{5} = 3^{3}⁄_{5} min.

(b) 8 boys

(c) 10 boys

(d) 11 boys

(e) None of these

**Ans.b**

Man’s two day’s work = 2 × ^{1}⁄_{20} th work = ^{1}⁄_{10} th work

Woman’s two days’s work

= 2 × ^{1}⁄_{30} th work = ^{1}⁄_{15} th work

Boy’s two day’s work = 2 × ^{1}⁄_{60} th work = ^{1}⁄_{30} th work

Now, let 2 men, 8 women and x boys can complete work in 2 days. Then ,

2 men’s work + 8 women’s work + x boy’s work = 1

2(^{1}⁄_{10}) + 8(^{1}⁄_{15}) + x(^{1}⁄_{30}) = 1

⇒ x = (1 - ^{1}⁄_{5} - ^{8}⁄_{15}) × 30 ⇒ x = 8 boys

(b) 8

(c) 13

^{1}⁄

_{3}

(d) 5

^{1}⁄

_{2}

(e) None of these

**Ans.b**

B alone can do a work in 20 hours.

∴ A alone can do ^{3}⁄_{2} of the work in 20 hours.

i.e., A alone can do the same work in ^{40}⁄_{3} hours

∴ (A + B)’s one hour’s work = ^{3}⁄_{40} + ^{1}⁄_{20} = ^{5}⁄_{40} = ^{1}⁄_{8}

⇒ A and B together can finish the whole work in 8 hours.

(b) 8

(c) 9

(d) 85

(e) None of these

**Ans.a**

15 W = 10 M

Now, 5W + 4M = 5W + \(\frac{4 \times 15}{10}W\) = 5W + 6W = = 11 W

If 15 women can complete the project in 55 days,

11 women can complete the same project in

(b) 12 noon

(c) 12 : 30 pm

(d) 1 pm

(e) None of these

**Ans.d**

(P + Q + R)’s 1 hour’s work = (^{1}⁄_{8} + ^{1}⁄_{10} + ^{1}⁄_{12}) = ^{37}⁄_{120}

Work done by P, Q and R in 2 hours = (^{37}⁄_{120} × 2) = ^{37}⁄_{60}

Remaining work = (1 - ^{37}⁄_{60}) = ^{23}⁄_{60}

(Q + R)’s 1 hour’s work = (^{1}⁄_{10} + ^{1}⁄_{12}) = ^{11}⁄_{60}

Now, ^{11}⁄_{60} work is done by Q and R in 1 hour.

So, ^{23}⁄_{60} work will be done by Q and R in

(^{60}⁄_{11} × ^{23}⁄_{60}) = ^{23}⁄_{11} hours ≈ 2 hours.

So, the work will be finished approximately 2 hours after 11 a.m., i.e., around 1 p.m.

(b) 6

^{1}⁄

_{3}

(c) 6

^{2}⁄

_{3}

(d) 7

^{2}⁄

_{3}

(e) None of these

**Ans.c**

10 men’s 1 day’s work = ^{1}⁄_{15}

15 women’s 1 day’s work = ^{1}⁄_{12}

(10 men + 15 women)’s 1 day’s work

= (^{1}⁄_{15} + ^{1}⁄_{12}) = ^{9}⁄_{60} = ^{3}⁄_{20}

∴ 10 men and 15 women will complete the work in ^{20}⁄_{3} = 6^{2}⁄_{3} days.

(b) 250, 100, 100

(c) 200, 150, 100

(d) 175, 175, 100

(e) None of these

**Ans.a**

Work done by A and B in 5 days = (^{1}⁄_{10} + ^{1}⁄_{15}) × 5 = ^{5}⁄_{6}

Work remaining = 1 - ^{5}⁄_{6} = ^{1}⁄_{6}

∴ C alone can do the work in 6 × 2 = 12 days

Ratio of their share work = ^{5}⁄_{10} : ^{5}⁄_{15} : ^{2}⁄_{12} = 3 : 2 : 1

Share of wages = 225, 150, 75.