# PE 2 - Time & Work | Arithmetic - Time & Work

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**PE 2 - Time & Work | Arithmetic - Time & Work**

10 horses and 15 cows eat grass of 5 acres in a certain time. How many acres will feed 15 horses and 10 cows for the same time , supposing a horse eats as much as 3 cows.

- (a)
40/7

- (b)
39/8

- (c)
40/11

- (d)
55/9

Answer: Option D

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**Explanation** :

10 horses and 15 cows can graze 5 acres of land in say d days.

15 horses and 10 cows can graze x acres of land in say d days.

∴ $\frac{(10h+15c)\times d}{5}$ = $\frac{(15h+10c)\times d}{x}$

Given, 1 horse eats as much as 3 cows: h = 3c

⇒ $\frac{(30c+15c)}{5}$ = $\frac{(45c+10c)}{x}$

⇒ x = 55/9 acres

Hence, option (d).

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**PE 2 - Time & Work | Arithmetic - Time & Work**

A man, a woman or a boy can do a job in 20 days, 30 days or 60 days respectively. How many boys must assist 4 men and 5 women to do the work in 2 days?

- (a)
16

- (b)
13

- (c)
15

- (d)
None of these

Answer: Option D

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**Explanation** :

Let the total work to be done = 60 units

∴ Efficiency of a man = 60/20 = 3 units/day

∴ Efficiency of a woman = 60/30 = 2 units/day

∴ Efficiency of a boy = 60/60 = 1 units/day

Let x boys assist 4 men and 5 women.

∴ (4m + 5w + xb) × 2 = 60

⇒ (12 + 10 + x) = 30

⇒ x = 8

Hence, option (d).

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**PE 2 - Time & Work | Arithmetic - Time & Work**

A group of workers engaged in plastering a wall completed half of the work on one day and quarter of the remaining work on the next day. If still 60 square meters of the wall remained to be plastered, find the area of the wall.

- (a)
160 sq.m.

- (b)
120 sq.m.

- (c)
125 sq.m.

- (d)
None of these

Answer: Option A

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**Explanation** :

On 1^{st} day $\frac{1}{2}$ work was completed and on 2^{nd} day $\frac{1}{4}$ × $\left(\frac{1}{2}\right)$ work i.e. $\frac{1}{8}$^{th} work

So, work remaining =1 - $\left(\frac{1}{2}+\frac{1}{8}\right)$ = $\frac{3}{8}$^{th}

⇒ $\frac{3}{8}$^{th} work is 60m^{2}

⇒ comlete work = 60 × $\frac{8}{3}$ = 160 m^{2}

Hence, option (a).

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

A number of men undertook to dig a ditch and could have finished the job in 14 hours, if they had all begun working simultaneously. But only one of them starts working. A 2nd person starts working after x hours. 3rd person starts working x hours after 2nd person starts working. Similarly each successive person starts working x hours after the previous worker starts. After the last worker had begun working the job was finished in x hours, each one of the participants working till the completion of the job. How long did they work, if the first worker to begin worked 6 times as long as the last one to begin?

- (a)
18 hours

- (b)
25 hours

- (c)
24 hours

- (d)
30 hours

Answer: Option C

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**Explanation** :

Let the number of men be ‘n’ and equal time durations be ‘x’ and the efficiency of each person be 1 unit/hour.

Had all of them worked together the work would’ve been completed in 14 hours.

∴ Total work done = n × 1 × 14 = 14n units.

Now, each of them joins x hours after the previous man and the last man works for x hours.

∴ The first person works for a total of nx hours.

Since the 1^{st} person worked 6 times as long as the last one i.e., he works for 6x hours.

⇒ nx = 6x

⇒ n = 6

∴ The works lasts for a total of 6x hours. The 1st person works for 6x hours, 2nd for 5x hours and so on till the last person who works for x hours.

∴ Total work done = 6x × 1 + 5x × 1 + 4x × 1 + 3x × 1 + 2x × 1 + x × 1 = 21x

Now, 21x = 14n = 14 × 6 = 84

⇒ x = 4 hours

So, total time taken = 6x = 6 × 4 = 24 hrs

Hence, option (c).

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

A tank can be filled in 30 minutes by 20 pumps. If five pumps go out of order, what time will be taken by the remaining pumps?

- (a)
40 mins

- (b)
38 mins

- (c)
32 mins

- (d)
30 mins

Answer: Option A

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**Explanation** :

We know,

Time ∝ 1/no. of pumps

∴ Time × (Number of pumps) = constant.

⇒ 20 × 30 = 15 × x

⇒ x = 20 × 30/15 = 40 mins

Hence, option (a).

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is 3/4^{th} full, a leak develops in, through which one-fourth of water supplied by both the pipes goes out. What is the total time taken to fill the tank?

- (a)
18 hours

- (b)
14 hours

- (c)
15 hours

- (d)
13 hours

Answer: Option D

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**Explanation** :

Time to completely fill the tank by the two pipe:

1/20 + 1/30= 1/n

⇒ n = 12 hours

So, 3/4^{th} of the tank will be filled in 3/4 × 12 = 9 hours.

Remaining time = 12 – 9 = 3 hours.

But, for the remaining 1/4^{th} of the tank, the combined efficiency drops to 3/4^{th} (1/4^{th} is getting leaked),

∴ Time required will be come 4/3 times, i.e. 4/3 × 3 = 4 hours.

Hence, total time taken to fill the tank = 9 + 4 = 13 hours.

Hence, option (d).

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

An empty cistern has three taps A, B and C. A and B can fill it in 3 and 4 hours respectively and C can empty it in six hours. If the taps C, B and A are opened at 1 p.m., 2 p.m. and 3 p.m. respectively, find when will the cistern be filled (approximately)?

- (a)
4 p.m.

- (b)
5 p.m.

- (c)
6 p.m.

- (d)
7 p.m.

Answer: Option B

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**Explanation** :

Let the total work to be done = 12 units.

∴ Efficiency of Pipe A = 12/3 = 4 units/hour.

∴ Efficiency of Pipe B = 12/4 = 3 units/hour.

∴ Efficiency of Pipe c = 12/6 = -2 units/hour. (-ve indicates that it is an emptying pipe)

Between 1 p.m. till 3 p.m.,

For the first 1 hour pipe C would not do any work as the tank is empty initially.

Work done by pipe C from 2 – 3 p.m. = 1 × -2 = -2 units.

Work done by pipe B from 2 – 3 p.m. = 1 × 3 = 3 units.

Total work done till 3 p.m. = 3 – 2 = 1 unit.

Work left = 12 – 1 = 11 units.

Now all three of them work and their combined efficiency = 4 + 3 – 2 = 5 units/hour.

∴ Time taken after 3 p.m. = 11/5 = 2.2 hours i.e., 2 hours 12 mins.

Hence, the tank gets filled 11/5 hours after 3 pm i.e. at 5:12 p.m.

Hence, option (b).

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

3 men and 4 boys do a piece of work in 8 days, while 4 men and 4 boys finish it in 6 days. 3 men and 6 boys will finish it in :

- (a)
8 days

- (b)
10 days

- (c)
12 days

- (d)
14 days

Answer: Option A

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**Explanation** :

Let m and b be 1 day’s work of a man and a boy respectively.

∴ 3m + 4b = 1/8 …(1)

also, 4m + 4b = 1/6 …(2)

Solving (1) and (2)

⇒ m = 1/24 and b = 0

Hence, 1 day’s work of 3 men and 6 boys = 3 × 1/24 + 6 × 0 = 1/8

∴ They will complete the work in 8 days.

Hence, option (a).

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

In a military camp there is sufficient food supply for 200 soldiers for 40 days. After 10 days, certain number of men join and everyone eats 50% more than earlier. Now the food lasts for another 10 days. How many soldiers joined the camp after 10 days?

Answer: 200

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**Explanation** :

Let eating capacity of each soldier initially = 1 unit/day.

Initial food supply = 200 × 1 × 40 = 8000 units.

Supply left after 10 days = 200 × 1 × 30 units.

Let ‘s’ numbers of soldiers joined the camp.

New capacity of each soldier = 1.5 units/day

∴ The remaining supply is consumed by (200 + s) solders in another 10 days.

∴ 200 × 1 × 30 = (200 + s) × 1.5 × 10

⇒ 400 = 200 + s

⇒ s = 200

Hence, 200.

Workspace:

**PE 2 - Time & Work | Arithmetic - Time & Work**

4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish it in 10 days. 40 women working alone will finish it in :

- (a)
6 days

- (b)
8 days

- (c)
9 days

- (d)
10 days

Answer: Option D

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**Explanation** :

Let the total work to be done = LCM (8, 10) = 40 units.

Also, let efficiency of each man and woman be m and w respectively.

∴ (4m + 6w) × 8 = 40

⇒ 4m + 6w = 5 …(1)

Similarly, (3m + 7w) × 10 = 40

⇒ 3m + 7w = 4 …(2)

Solving (1) and (2) we get,

m = $\frac{11}{10}$ and w = $\frac{1}{10}$

Now, time taken by 1 woman to complete the work = $\frac{40}{1/10}$ = 400 days.

∴ Time taken by 40 women to complete the work = 10 days.

Hence, option (d).

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