# CRE 3 - Working Alternately | Arithmetic - Time & Work

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**CRE 3 - Working Alternately | Arithmetic - Time & Work**

A and B can complete a work in 10 and 15 days respectively each. How long will it take to finish the work when they work on alternate days?

- (a)
6 days

- (b)
12 days

- (c)
4 days

- (d)
None of these

Answer: Option B

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**Explanation** :

1 day’s work of A = 1/10 and that of B = 1/15.

Since they work on alternate days, they would work in cycles of 2 days.

Work done by (A + B) in 1 cycle of two days = 1/10 + 1/15 = 1/6.

Hence, they take 1/(1/6) = 6 cycles to complete the work.

[Since the number of cycles comes out to be an integer, it would not matter who starts the work.]

Therefore, they would finish the work in 6 × 2 = 12 days irrespective of who starts the work.

Hence, option (b).

Workspace:

**CRE 3 - Working Alternately | Arithmetic - Time & Work**

A and B can complete a work in 10 and 100 days respectively each. How long will it take to finish the work when they work on alternate days starting with B?

Answer: 19

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**Explanation** :

Let the total work to be done = LCM(10, 100) = 100 units.

Efficiency of A = 100/10 = 10 units/day

Efficiency of B = 100/100 = 1 unit/day.

Since they work on alternate days, they would work in cycles of 2 days.

Work done by (A + B) in 1 cycle of two days = 10 + 1 = 11 units/cycle.

Hence, they take 100/11 =9(1/11) cycles to complete the work.

Since, the no. of cycles is a fraction, we will have to analyze the last incomplete cycle.

Work done in 9 complete cycles = 9 × 11 = 99.

Time that has passed till 9 cycles = 9 × 2 = 18 days.

Hence, work left for the 10th incomplete cycle = 100 – 99 = 1.

Since B starts the work, he would work on 19th day and would complete 1 unit of the work. Hence the work gets completed at the end of 19th day.

The work gets completed in 19 days.

Hence, 19.

Workspace:

**CRE 3 - Working Alternately | Arithmetic - Time & Work**

In the previous question, if A starts the work and they work alternately, how long will it take to complete the work?

- (a)
19

- (b)
18.5

- (c)
18.1

- (d)
None of these

Answer: Option C

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**Explanation** :

Let the total work to be done = LCM(10, 100) = 100 units.

Efficiency of A = 100/10 = 10 units/day

Efficiency of B = 100/100 = 1 unit/day.

Since they work on alternate days, they would work in cycles of 2 days.

Work done by (A + B) in 1 cycle of two days = 10 + 1 = 11 units/cycle.

Hence, they take 100/11 =9 (1/11) cycles to complete the work.

Since, the no. of cycles is a fraction, we will have to analyze the last incomplete cycle.

Work done in 9 complete cycles = 9 × 11 = 99.

Time that has passed till 9 cycles = 9 × 2 = 18 days.

Hence, work left for the 10th incomplete cycle = 100 – 99 = 1.

Since A starts the work, he would work on 19th day and would complete 1 unit of the work in 1/10 day. Hence the work gets completed before the end of 19th day.

The work gets completed in 18 (1/10) days.

Hence, option (c).

Workspace:

**CRE 3 - Working Alternately | Arithmetic - Time & Work**

A and B can complete a work in 10 and 20 days respectively each. At most how long will it take to finish the work when they work on alternate days?

Answer: 13.5

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**Explanation** :

1 day’s work of A = 1/10 and that of B = 1/20.

Since they work on alternate days, they would work in cycles of 2 days.

Work done by (A + B) in 1 cycle of two days = 1/10 + 1/20 = 3/20.

Hence, they take $\frac{1}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}}$ = $\frac{20}{3}$ = $6\frac{2}{3}$ cycles to complete the work.

Since, the no. of cycles is a fraction, we will have to analyze the last incomplete cycle.

Work done in 6 complete cycles = 6 × (3/20) = 18/20 = 9/10

Time that has passed till 6 cycles = 6 × 2 = 12 days

Hence, work left for the 7th incomplete cycle = 1 – 9/10 = 1/10

Case I: If A starts the work, he would work on 13th day and take $\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}$ = 1 day to complete the work. Hence, the work gets completed in 13 days.

Case II: If B starts the work, he would work on 13th day and would complete 1/20 of the work. Work still left = 1/10 – 1/20 = 1/20. This work would finally be completed by A on 14th day. Time taken by A = $\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}$ = 1/2 day. Hence the work gets completed in 13.5 days.

The work, at most can be completed in 13.5 days.

Hence, 13.5.

Workspace:

**CRE 3 - Working Alternately | Arithmetic - Time & Work**

A, B and C can complete a piece of work in 10, 20 and 40 days respectively. They work on a rotation basis with A working on the first day, B on the second. C the third. Then again A on the fourth day and so on. In how many days the work gets completed?

Answer: 16.5

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**Explanation** :

Let the total units of work be LCM (10, 20, 40) = 40.

A completes 40/10 = 4 units per day; Similarly, B and C will complete 2 and 1 units respectively.

So, at the end of one cycle (first 3 days) they will have completed 4 + 2 + 1 = 7 units of work.

They can carry out [40/7] = 5 such cycles thereby completing 35 units of work.

The next day A completes 4 units of work leaving 1 unit of work.

The next day, B will complete the work in ½ part of the day.

Total time = 5 × 3 + 1 + ½ = 16.5 days.

Hence, 16.5.

Workspace:

**CRE 3 - Working Alternately | Arithmetic - Time & Work**

A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?

- (a)
2

- (b)
3

- (c)
5

- (d)
11

Answer: Option B

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**Explanation** :

Let efficiency of each worker by 1 unit/day.

**Case 1**: When a worked leaves every day.

Since n workers start on day 1 and on last day only 1 worker will remain, hence, the work will last n days.

Day | 1 | 2 | 3 | ... | n |

Number of workers | n | n - 1 | n - 2 | ... | n - 3 |

Workdone/day | n × 1 | (n - 1) × 1 | (n - 2) × 1 | ... | (n - 3) × 1 |

⇒ Total work done = n × 1 + (n - 1) × 1 + (n - 2) × 1 + … + n × 1 = n(n+1)/2 …(1)

**Case 2**: When no workers leave

Time taken is 2/3rd of n days.

⇒ Total work done = n × 2/3 n = 2/3 × n^{2} …(2)

Since same work is done in both the cases, we can equate (1) and (2).

n(n + 1)/2 = 2/3 × n^{2}

⇒ n = 3.

Hence, option (b).

Workspace:

**CRE 3 - Working Alternately | Arithmetic - Time & Work**

A and B can completely build a wall in 20 and 30 days respectively while C can demolish the entire wall in 60 days. If A, B and C work one after the other starting with A, then B and then C. On which day from the start will the work get completed.

- (a)
42

- (b)
43

- (c)
44

- (d)
45

- (e)
46

Answer: Option C

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**Explanation** :

Let the work to be done = LCM (20, 30, 60) = 60 units.

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

Efficiency of B = 60/30 = 2 units/day

Efficiency of C = 60/60 = -1 units/day

Since they work on alternate days, they would work in cycles of 3 days.

Work done by A on first day of the cycle = 3 units

Work done by B on second day of the cycle = 2 units

Work done by C on third day of the cycle = -1 units

Work done in 1 cycle = 3 + 2 - 1 = 4 units

Number of cycles required = 60/4 = 15 cycles.

Now since work done by C is negative we need to analyse last couple of cycles.

Work done by the end of 13 cycles (39 days) = 13 × 4 = 52.

**Work done in 14 ^{th} cycle**:

on 40

^{th}day = 3. Total work done = 3 + 52 = 55

on 41

^{st}day = 2. Total work done = 2 + 55 = 57

on 42

^{nd}day = -1. Total work done = -1 + 57 = 56

**Work done in 15 ^{th} cycle**:

on 43

^{rd}day = 3. Total work done = 3 + 56 = 59

on 44

^{th}day = 2. Total work done = 2 + 59 = 61

Hence, the wall gets built on 44^{th} day itself.

Hence, 44.

Workspace:

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