# CRE 2 - Working in Shifts | Arithmetic - Time & Work

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**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

A alone can do a piece of work in 15 days and B alone can do it in 20 days. Both A and B worked together for 4 days and then A left. In how many more days will B finish the remaining work?

- (a)
$9\frac{4}{11}$

- (b)
$10\frac{5}{8}$

- (c)
$10\frac{2}{3}$

- (d)
$9\frac{1}{8}$

Answer: Option C

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**Explanation** :

**Method 1**:

(A + B)’s 4 days work = $4\left(\frac{1}{15}+\frac{1}{20}\right)$ = $\frac{7}{15}$

Remaining work = (1 - 7/15) = 8/15;

Now, $\frac{1}{20}$th work is finished by B in 1 day.

∴ $\frac{8}{15}$th work will be finished by B in $\frac{8/15}{1/20}$ = $\frac{8\times 20}{15}$ = $\frac{32}{3}$ = $10\frac{2}{3}$

**Method 2**:

Let the total work to be done = LCM(15, 20) = 60 units.

Efficiency of A = 60/15 = 4 units/day

Efficiency of B = 60/20 = 3 units/day

A and B work for 4 days and then let B finish the remaining work in x days.

∴ Total work done = Work done by A and B in 4 days + Work done by B in x days

⇒ 60 = (4 + 3) × 4 + 3 × x

⇒ 60 = 28 + 3x

⇒ 3x = 32

⇒ x = $\frac{32}{3}$ = $10\frac{2}{3}$

**Method 3**:

Let the total work to be done = LCM(15, 20) = 60 units.

Efficiency of A = 60/15 = 4 units/day

Efficiency of B = 60/20 = 3 units/day

A works for 4 days and B work for 4 + x days

∴ Total work done = Work done by A in 4 days + Work done by B in 4 + x days

⇒ 60 = 4 × 4 + (4 + x) × 3

⇒ 60 = 28 + 3x

⇒ 3x = 32

⇒ x = $\frac{32}{3}$ = $10\frac{2}{3}$

Hence, option (c).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

A alone can do a certain job in 25 days while B alone can do the same in 20 days. A started the work and was joined by B after 5 days. The number of days taken to complete the entire work will be:

- (a)
50/9

- (b)
140/9

- (c)
125/9

- (d)
130/9

Answer: Option C

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**Explanation** :

A taken 25 days alone and B takes 20 days alone to complete a certain task.

Let the work to be done = LCM (25, 20) = 100 units

∴ Efficiency of A = 100/25 = 4 units/day, and

Efficiency of B = 100/20 = 5 units/day

First 5 days A was working alone, hence work done by A = 5 × 4 = 20 units.

Remaining work = 100 - 20 = 80 units

If A & B work together, they can complete (5 + 4 =) 9 units/day; so to complete 80 units they will take 80/9 days

Total time required = 80/9 + 5 = 125/9 days

Hence, option (c).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

It takes Mr. Aakash ‘a’ hours to complete typing a manuscript. After 3 hours, he was called away. What fractional part of the assignment was left incomplete?

- (a)
1/a

- (b)
(a - 2)/a

- (c)
a/2

- (d)
a - 2

- (e)
(a - 3)/a

Answer: Option E

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**Explanation** :

Aakash completes the assignment in a hours

So, in 1 hour he completes 1/a^{th} of the assignment

So, in 3 hours he will complete 3(1/a) = 3/a of the assignment

Left assignment = 1 – (3/a) = (a - 3)/a

Hence, option (e).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

A and B can do a piece of work in 21 and 24 days respectively. They start the work together and after some days A leaves the work and B completes the remaining work in 9 days. After how many days did A leave?

- (a)
5

- (b)
7

- (c)
8

- (d)
6

Answer: Option B

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**Explanation** :

In 9 days, B will do 9/24 = 3/8^{th} of the work.

Hence, A and B together did 1 – 3/8 = 5/8^{th} of the work together.

Now, A and B can complete;

1/21 + 1/24 = 5/56^{th} of the work in one day.

Hence, they will complete, 5/8^{th} of the work in = 5/8 × 56/5 = 7 days.

**Alternately**,

Let the work to be done = LCM (21, 24) = 168 units

∴ Efficiency of A = 168/21 = 8 units/day, and

Efficiency of B = 168/24 = 7 units/day

Let A and B work together for x days after which B completes the remaining work in 9 days.

∴ Total work done = Work done by A and B in 4 days + Work done by B in x days

⇒ 168 = (8 + 7) × x + 7 × 9

⇒ 168 = 15x + 63

⇒ 105 = 15x

⇒ x = 7 days

∴ A leaves after 7 days.

Hence, option (b).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

A can build a wall in 24 days, B can do the same in 32 days and C can do in 48 days. Totally 12 walls are to be built. 'A' does the work for first 64 days and 'B' continues the work for the next 24 days. How many days should C work?

- (a)
412

- (b)
332

- (c)
50

- (d)
64

- (e)
None of these

Answer: Option A

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**Explanation** :

Let the amount of work done to build one wall = LCM (24, 32, 48) = 96 units.

Efficiency of A = 96/24 = 4 units/day

Efficiency of B = 96/32 = 3 units/day

Efficiency of C = 96/48 = 2 units/day

Total 12 walls are to be built. i.e., total work to be done = 12 × 96 = 1152 units

Now,

Work done by A in 1^{st} 64 days = 4 × 64 = 256

Work done by B (next 24 days) = 3 × 24 = 72

Total work done by A and B = 256 + 72 = 328 units

Work remaining = 1152 – 328 = 824 units

So, Work done by C = 824 = 2 × no. of days C has to work

∴ No. of days C has to work = 824/2 = 412 days

Hence, option (a).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

If 1/8^{th} out of x members are absent, then work increased for each person is:

- (a)
x/7

- (b)
1/7x

- (c)
7/x

- (d)
2x/7

- (e)
2/7x

Answer: Option B

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**Explanation** :

Let the total work to be done = 1 unit

Number of persons originally scheduled to work = x

Work done by each person (w) = $\frac{1}{x}$

Given 1/8 members are absent, so number of persons present = x – $\frac{x}{8}$ = x $\left(1-\frac{1}{8}\right)$ = $\frac{7x}{8}$

∴ Work done by each person in second case (w’) = $\frac{1}{7x/8}$ = $\frac{8}{7x}$

Extra work by each person in second case = w’ – w = $\frac{8}{7x}$ – $\frac{1}{x}$ = $\frac{1}{7x}$

Hence, option (b).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

A alone can complete a task in 27 days while B alone can complete the same task in 45 days. Both of them started the work together, but A left 5 days before actual completion of work. Find how many days did a work?

- (a)
20 days

- (b)
15 days

- (c)
25 days

- (d)
5 days

- (e)
10 days

Answer: Option B

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**Explanation** :

Let the work to be done = LCM (27, 45) = 135 units.

∴ Efficiency of A = 135/27 = 5 units/day, and

Efficiency of B = 135/45 = 3 units/day

Let A and B together work for 'x' days after which B completes the remaining work in 5 days.

∴ Total work done = Work done by A and B in x days + Work done by B in last 5 days.

⇒ 135 = (5 + 3) × x + 3 × 5

⇒ 135 = 8x + 15

⇒ 120 = 8x

⇒ x = 120/8 = 15 days.

∴ A and B together worked for 15 days.

Hence, option (b).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

A alone can complete a task in 56 days while B alone can complete the same task in 42 days. A and B. A and B took upon the work to complete it together. However, A left 12 days before the scheduled completion of work. Find how many days did B work alone?

- (a)
12

- (b)
9

- (c)
21

- (d)
Cannot be determined

- (e)
None of these

Answer: Option C

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**Explanation** :

Let the work to be done = LCM (56, 42) = 168 units.

∴ Efficiency of A = 168/56 = 3 units/day, and

Efficiency of B = 168/42 = 4 units/day

If A and B worked together, the work should have finished in 168/(3 + 4) = 24 days.

However, A and B together work for 12 days after which A leaves and B completes the remaining work.

Let B complete the remaining work in x days.

Work done by A and B in first 12 days = 12 × (3 + 4) = 84 units.

∴ Work remaining for B = 168 - 84 = 84

⇒ Time taken by B to complete 84 units of work = 84/4 = 21 days.

∴ B worked alone for 21 days.

Hence, option (c).

Workspace:

**Answer the next 2 questions based on the information given below.**

A is twice as efficient as B and B is twice as efficient as C. If A, B and C work together, a piece of work will be completed in 12 days. A and B start it together and continue for 8 days. After that, A is replaced by C. For his work, A gets paid Rs. 2000.

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

What amount will B get paid?

- (a)
Rs. 2000

- (b)
Rs. 2500

- (c)
Rs. 2650

- (d)
Rs. 2733.33

- (e)
Rs. 2800

Answer: Option B

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**Explanation** :

Assume C can do x work in 1 day,

B can do 2x work in 1 day, and

A can do 4x work in 1 day.

Total work = 12(x + 2x + 4x) = 84x

Now, A and B work for 8 days, therefore 8(2x + 4x) = 48x of work will be completed. B and C can do 3x work in 1 day.

So, the remaining 36x of work can be completed in 12 days.

A gets Rs. 2000 for his 32x of work.

B has done 40x of work.

So, B will get (2000 × 40)/32 = Rs. 2500.

Hence, option (b).

Workspace:

**CRE 2 - Working in Shifts | Arithmetic - Time & Work**

How many more days are required to complete the work?

- (a)
8 days

- (b)
10 days

- (c)
12 days

- (d)
15 days

- (e)
16 days

Answer: Option C

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**Explanation** :

Assume, C can do x work in 1 day; B can do 2x work in 1 day; and A can do 4x work in 1 day.

Total work = 12(x + 2x + 4x) = 84x

A and B work for 8 days, 8(2x + 4x) = 48x will be completed.

B and C can do 3x work in 1 day.

So, the remaining 36x can be completed in 12 days.

Hence, option (c).

Workspace:

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