Time and Work Exercise 2

Let Sunil finishes the job in x hours.

Then, Ramesh will finish the job in x/2 hours.

We have,

Therefore, Sunil finishes the job in 6 hours and Ramesh in 3 hours.

Work done by both of them in 1 hour

=

They together finish the piece of work in 2 hours.

The part of job that Suresh completes in 9 hours

=

Remaining job

=

Remaining job can be done by Ashutosh in

⅖ × 10 = 4 hours

15 women’s work of a day

part

∴ for 1 whole part a woman will take

= 15 × 15 = 225 days.

Sunil takes 5 days and Pradeep takes 15 days to do the work.

In a day they would complete

i.e., work.

The remaining 11/15th work would be completed by Pradeep in

i.e. 11 days.

Suresh, working alone 42 days = 1 unit of work.

Mahesh is 1/5 times more efficient that Suresh.

So Mahesh is 6/5 times as efficient as Suresh.

Hence Mahesh should require 5/6th of the time, the time taken by Suresh.

Therefore time taken by Mahesh = 5/6 × 42 = 35 days.

Ganpat’s day work = 1/15 of the total.

Yogesh’s 1 day work = 1/20 of the total.

Bhagwat’s 1 day work = 1/30 of the total.

They can do of the total work in 1 day.

⇒ Total work can be finished in

=

=

= days.

Given 12 men ≡ 15 women ≡ 18 boys

∴ 1 Man ≡ 1.5 boys, 1 woman = 6/5 boys.

Now, 5W + 6B = 12B.

Required answer is calculated as follows :

Total no. of boys required

= 18 × [(15/16) × (8/9)] = 15 boys

The number of boys already present = 12.

Hence, 3 boys more required.

But 3 boys = 2 men.

So, 2 men are required.

Given 6 BSF ≡ 10 CRPF

⇒ 4 BSF + 9 CRPF

= 4 + (9 × 6/10) BSF

= BSF

Now work = 6 × 2 BSF days

= BSF days

We have 6 × 2 ≡

⇒ X = 1.27 days

1 horse = 2 cows, 10 horses = 20 cows.

⇒ 10 horses + 15 cows = 20 + 15 = 35 cows.

15 horses + 10 cows = 40 cows.

Now 35 cows eat 5 acres.

⇒ 40 cows eat 5 × ⁴⁰⁄₃₅ = 5⁵⁄₇ acres.

Here we have converted everything in terms of cows, you can work in terms of horses also.

Let 1 man’s 1 days’ work = x and 1 boy’s 1 day’s work = y

Then, 2x + 3y = 1/10 and 3x + 2y = 1/8

Solving, we get :

and

∴ (2 men + 1 boy)’s 1 day’s work

=

So, 2 men and 1 boy together can finish the work in 12½ days.

A, B and C’s 1 day’s work = 1/10

i.e. ...(1)

Also, only C’s 1 day’s work = 3/100

i.e. ... (2)

From the given condition,

.... (3)

By comparing the ratio given in equ (1) and (2),

We can say C is the lowest worker.

Also, from equation (1) and (3), B is the fastest worker.

∴ We have,

{from (1), (2), (3)}

Hence, B completes the entire work in 20 days.

Using M₁T₁W₂ = M₂T₂W₁, we get

⇒ x + 10 =

⇒ x = 18 – 10 = 8

Hence, 8 men must be associated.

10 men finishes a work in 10 days and 12 women finishes in 10 days.

∴ 10 men and 12 women finishes a work in 10 days

∴ 15 men and 6 women will complete the work in

days

i.e., in 5 days.

1st man can do in 3 days = ³⁄₆ part of the work

2nd man can do in 3 days = ³⁄₈ part of the work

Boy can do in 3 days

part of the work

∴ Ratio of their wages

= 4 : 3 : 1

Boy’s share

1st man can do in 3 days =³⁄₇ part of the work

2nd man can do in 3 days ³⁄₈ part of the work

Boy can do in 3 days

=

part of the work

∴ Ratio of their wages

= 24 : 21 : 11

∴ 1st man’s share

=

2nd man’s share

Boy’s share

Remaining work

4 men + 10 women do 1 work in 12 days.

6 men + 12 women do 1 work in 9 days.

48 men + 120 women = 54 men + 108 women

⇒ 6 men = 12 women

⇒ 1 men = 2 women

∴ In 12 days 1 work requires 9 men

∴ In 1 day 1 work requires 9 × 12 men

∴ In 3 days 1 work requires

men

∴ In 3 days ⁴⁄₉ work requires

men

There are 6 men and 12 women or (12 men equivalent)

So, 4 men equivalent is required additionally

∴ 8 women are needed to finish the work.

In one hr. B finishes ¹⁄₂₀ of the work.

In one hr. A finishes

of the work.

A+B finish

of the work in 1 hr.

Both of them will take 8 hrs. to finish the work.

Let the man alone do the work in x days.

then, the woman alone do the work in 2x days.

Their one day’s work = ⅛th part of whole work

i.e.

⇒ x = 12 days

∴ man takes 12 days and woman 2x = 24 days.

X’s one day’s work = ¹⁄₂₅th part of whole work.

Y’s one day’s work = ¹⁄₃₀th part of whole work.

Their one day’s work

= part of whole work.

Now, work is done in 5 days

of whole work

∴ Remaining work

= of whole work

Now, ¹⁄₃₀th work is done by y in one day.

∴ ¹⁹⁄₃₀th work is done by y in

A’s one day’s work = ¹⁄₁₆ th work

B’s one day’s work = ¹⁄₁₂th work

Let B has worked alone = x days. Then,

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

⇒ x = 5 days

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

A’s 1 day’s work = ¹⁄₂₃th;

B’s 1 days work = ¹⁰⁄₂₉₉

(A + B)’s 1 day’s work

=

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

Man’s two day’s work

= 2×¹⁄₂₀th work

Woman’s two days’s work

Boy’s two day’s work

= 2×¹⁄₆₀th work = ¹⁄₃₀th 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

⇒ x = 8 boys

Suppose that X men must be discharged at the end of the 18th day.

100 × 10 + 150 × 1 + 200 × 7 + (200 – X) × 5

= 100 × 30

5X = 550

⇒ X = 110 men

15 W = 10 M

Now, 5W + 4M

=

= 5W + 6W = 11 W

If 15 women can complete the project in 55 days,

11 women can complete the same project in

Let if both A and B work together, they take x days.

∴ (A + B)’s 1 days’s work = 1/xth work.

A’s 1 day’s work

B’s 1 day’s work

Now,

⇒ x(2x + 9 – 2x + 16) = (x + 8)(2x + 9)

⇒ 4x² + 25x = 2x² + 25x + 72

⇒ x² = 36 ⇒ x = 6 days