CRE 3 - Working Alternately | Time & Work

Answer: Option B

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.

Therefore, they would finish the work in 6 × 2 = 12 days

Hence, option (b).

Answer: 19

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.

Answer: Option 18.1

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, 18.1.

Answer: 13.5

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.

Answer: 16.5

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.

Answer: Option B

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.

⇒ 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 × n²   …(2)

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

n(n + 1)/2 = 2/3 × n²

⇒ n = 3.

Hence, option (b).

Answer: Option C

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.