# PE 1 - Time and Work | Time and Work

**PE 1 - Time and Work | Time and Work**

Some staff promised to do a job in 18 days, but 6 of them went on leave. So, the remaining men took 20 days to complete the job. How many men were there originally?

- A.
55

- B.
62

- C.
56

- D.
60

Answer: Option D

**Explanation** :

Let the number of staff that were supposed to come = x

∴ x men were supposed to complete the work in 18 days

Whereas, (x – 6) men completed the same work in 20 days.

⇒ x × 18 = (x – 6) × 20

⇒ 18x = 20x – 120

⇒ 120 = 2x

⇒ x = 60

Hence, option (d).

Workspace:

**PE 1 - Time and Work | Time and Work**

5 men and 5 women earn Rs. 660 in 3 days. 10 men and 20 women earn Rs. 3500 in 5 days. In how many days can 6 men and 4 women earn Rs. 1060?

- A.
5 days

- B.
10 days

- C.
6 days

- D.
12 days

Answer: Option A

**Explanation** :

Let the earnings of a man and a woman be m and q per day respectively.

5 men and 5 women in 3 days earn Rs. 660

∴ 5 men and 5 women in 1 day earn Rs. 220

⇒ 5m + 5w = 220 …(1)

10 men and 20 women in 5 days earn Rs. 3500

∴ 10 men and 20 women in 1 day earn Rs. 700

⇒ 10m + 20w = 700 …(2)

Solving (1) and (2), we get

m = Rs. 18 and w = Rs. 26.

∴ 6 men and 4 women in 1 day earn = 6m + 4w = 6 × 18 + 4 × 26 = Rs. 212

⇒ To earn Rs. 1060 they will take = 1060/212 = 5 days.

Hence, option (a).

Workspace:

**PE 1 - Time and Work | Time and Work**

A track of 100 m can be built by 7 men or 10 women in 10 days. How many days will 14 men and 20 women take to build a track of 600 m?

- A.
15

- B.
20

- C.
25

- D.
30

Answer: Option A

**Explanation** :

Let efficiency of each man and woman be m and w units/day.

Given, 7m × 10 = 10w × 10

⇒ 7m = 10w

Now,

⇒ $\frac{7m\times 10}{100}=\frac{(14m+20w)\times d}{600}$

⇒ $\frac{7m\times 10}{1}=\frac{(14m+14w)\times d}{6}$

⇒ 7m × 10 × 6 = 28m × d

⇒ d = 15 days

Hence, option (a).

Workspace:

**PE 1 - Time and Work | Time and Work**

A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone would do the work in

- A.
5 days

- B.
6 days

- C.
9 days

- D.
10 days

Answer: Option B

**Explanation** :

Let the time taken by A and B alone be ‘A’ and ‘B’ days respectively.

Together they can complete the work in 3 days.

⇒ $\frac{1}{A}+\frac{1}{B}=\frac{1}{3}$ …(1)

Now, they work for 2 days together and rest of the work is completed by A alone in 2 more day.

⇒ 1 = $\left(\frac{1}{A}+\frac{1}{B}\right)$ × 2 + $\frac{1}{A}$× 2

⇒ 1 = $\frac{1}{3}$× 2+ $\frac{1}{A}$× 2

⇒ $\frac{1}{3}$ = $\frac{2}{A}$

⇒ A = 3 days

Substituting this in (1), we get

B = 6 days.

Hence, option (b).

Workspace:

**PE 1 - Time and Work | Time and Work**

P can complete 1/4^{th} of a work in 10 days, Q can complete 40% of the same work in 15 days, R can complete 1/3^{rd} of the work in 13 days and S can complete 1/6^{th} of the work in 7 days, Who will be able to complete the work first?

- A.
P

- B.
Q

- C.
R

- D.
S

Answer: Option B

**Explanation** :

P can complete 1/4^{th} of a work in 10 days.

∴ P can complete the entire work in 4 × 10 = 40 days.

Q can complete 40% i.e., 2/5^{th} of the same work in 15 days.

∴ Q can complete the entire work in 15 × 5/2 = 37.5 days.

R can complete 1/3^{rd} of the work in 13 days.

∴ R can complete the entire work in 13 × 3 = 39 days.

S can complete 1/6^{th} of the work in 7 days.

∴ S can complete the entire work in 7 × 6 = 42 days.

⇒ Q completes the work first.

Hence, option (b).

Workspace:

**PE 1 - Time and Work | Time and Work**

A can do a work in 20 days and B in 40 days. If they work on it together for 5 days. Then fraction of the work that is left, is:

- A.
5/8

- B.
1/3

- C.
7/15

- D.
1/10

Answer: Option A

**Explanation** :

Fraction of work done in 5 days = $\left(\frac{1}{20}+\frac{1}{40}\right)$× 5 = $\frac{3}{40}$× 5 = $\frac{3}{8}$^{th}.

∴ Fraction of work left = 1 – $\frac{3}{8}$ = $\frac{5}{8}$

Hence, option (a).

Workspace:

**PE 1 - Time and Work | Time and Work**

A is 50% as efficient as B. C does half of the work done by A and B together in same time. If C alone does the work in 20 days, then A, B and C together can do work in:

- A.
$2\frac{5}{3}$ days

- B.
$6\frac{2}{3}$ days

- C.
6 days

- D.
7 days

Answer: Option B

**Explanation** :

C’s efficiency = 1/20

C’s efficiency is half that of A and B combined.

⇒ A’s efficiency + B’s efficiency = 1/10

A is 50% as efficient as B or B’s efficiency is twice that of A.

⇒ A’s efficiency + 2 × A’s efficiency = 1/10

⇒ A’s efficiency = 1/30

⇒ Time taken by A alone to complete the work = 30 days.

∴ Time taken by B alone to complete the work = 15 days.

Now, let Time taken by A, B and C together = N days

⇒ $\frac{1}{N}$ = $\frac{1}{30}+\frac{1}{15}+\frac{1}{20}$ = $\frac{2+4+3}{60}$ = $\frac{9}{60}$ = $\frac{3}{20}$

⇒ N = 20/3 = $6\frac{2}{3}$ days.

Hence, option (b).

Workspace:

**PE 1 - Time and Work | Time and Work**

If 12 men or 18 women can make a wall in 14 days, then working at the same rate, 8 men and 16 women can make the same wall in:

- A.
9 days

- B.
5 days

- C.
7 days

- D.
8 days

Answer: Option A

**Explanation** :

Let the efficiency of a man and a woman be m and w units/day respectively.

∴ Work done = 12m × 14 = 18w × 14 = (8m + 16w) × d

⇒ 12m × 14 = 18w × 14

⇒ 2m = 3w

Now, 18w × 14 = (8m + 16w) × d

⇒ 18w × 14 = (12w + 16w) × d

⇒ 18w × 14 = 28w × d

⇒ d = 9 days

Hence, option (a).

Workspace:

**PE 1 - Time and Work | Time and Work**

A certain number of men complete a work in 160 days. If there were 18 men more, the work could be finished in 20 days less. How many men were originally there?

- A.
116

- B.
122

- C.
124

- D.
126

Answer: Option D

**Explanation** :

Let x men can complete the work in 160 days, whereas

(x + 18) men can complete the work in 140 days.

⇒ x × 160 = (x + 18) × 140

⇒ 8x = 7(x + 18)

⇒ x = 7 × 18 = 126

Hence, option (d).

Workspace:

**PE 1 - Time and Work | Time and Work**

A team of 30 men is supposed to do a work in 38 days. After 25 days, 5 more men were employed and the work was finished one day earlier. How many days would it have been delayed if 5 more men were not employed?

- A.
1 day

- B.
2 days

- C.
3 days

- D.
4 days

Answer: Option A

**Explanation** :

Suppose the original work would have been completed in d days.

Total work to be done = 30 × d

Work done in first 25 days = 30 × 25

Work done in next 12 days = 35 × 12

⇒ 30d = 30 × 25 + 35 × 12

⇒ 30d = 750 + 420

⇒ 30d = 1170

⇒ d = 39

∴ Had 5 more men not been employed the work would have got delayed by (39 – 38 =) 1 day.

Hence, option (a).

Workspace:

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