Work Efficiency MCQ Quiz - Objective Question with Answer for Work Efficiency - Download Free PDF

Calculation

One day work of X, Y & Z together = 13/P

One day work of X & Y together = 10/ P

One day work of Z = 13 / P – 10/ P = 3 /P

Time taken by Z alone to complete the work = P/3 days

One day work of X & Y together = 10/P

One day work Y = 4 / P

One day work X = 10 /P - 4 /P = 6 /P

Time taken by X alone to complete the work = P / 6 days

ATQ, P/ 3 − P /6 = 20

Or, P/ 6 = 20

So, P = 120

Calculation

Amit's 1-day work = 1/45

Shivani's 1-day work = 1/60

Combined 1-day work = 1/45 + 1/60 = [4 + 3] / 180 =7/180

Work done in 18 days = [7/180] × 18 = 7/10

​Remaining work = 1− [7/10] = 3/10

Given:

A and B together can complete a work in 12 days.

A and C together can complete the same work in 8 days.

B and C together can complete the same work in 6 days.

Formula Used:

Total Work = Time × Efficiency

Calculation:

Let the total work be LCM(12, 8, 6) = 24 units

Efficiency and A and B = 24/12 = 2

Efficiency and B and C = 24/8 = 3

Efficiency and C and A = 24/6 = 4

Efficiency of A, B and C = 9/2

Efficiency of B alone = 9/2 - 4 = 1/2

Time taken by B to complete the total work = 24/(1/2) = 48 days

∴ Option 2 is the correct answer.

Calculation

Calculation

Total work =  LCM of 20 and 30 = 60

Efficiency of B is 60 /30 = 2

Total Efficiency of A and B together is 60/20 = 3

So, efficiency of A is 3 -2 = 1

So, A alone complete the work in 60 / 1 = 60 days

Given:

A and B together can do a piece of work in 50 days.

A is 40% less efficient than B

Concept used:

Total work = Efficiency of the workers × time taken by them

Calculation:

Let the efficiency of B be 5a

So, efficiency of A = 5a × 60%

⇒ 3a

So, total efficiency of them = 8a

Total work = 8a × 50

⇒ 400a

Now,

60% of the work = 400a × 60%

⇒ 240a

Now,

Required time = 240a/3a

⇒ 80 days

∴ A can complete 60% of the work working alone in 80 days.

Shortcut Trick

​We know 40% = 2/5, Efficiency of B = 5 and A = 3

So, Total work = (5 + 3) × 50 = 400 units

So, 60% of the total work = 60% of 400 = 240 units

So A alone can do the work in 240/3 = 80 days

Given:

A can finish in 15 days, B can finish it in 25 days.

They work together for 5 days.

Concept used:

Efficiency = (Total work / Total time taken)

Efficiency = work done in a single day 

Calculation:

Let total work be 75 units ( LCM of 15 and 25 is 75)

The efficiency of A

⇒ 75 /15 = 5 units

The efficiency of B  

⇒ 75 / 25 = 3 units

The efficiency of A+B,

⇒ (5 + 3) units = 8 units

In 5 days total work done is 8 × 5 = 40 units

Remaining work 75 - 40 = 35 units

In the last 4 days, A does 4 × 5 = 20 units

Remaining work 35 - 20 = 15 units done by C in 4 days

So C does 75 units in (75 / 15) × 4 = 20 days

∴ The correct option is 3

Given:

23 people could do a piece of work in 18 days.

After 6 days 8 of the workers left. 

Concept used:

Total work = Men needed × Days needed to finish it entirely

Calculation:

Total work = 23 × 18 = 414 units

In 6 days, total work done = 23 × 6 = 138 units

Remaining work = (414 - 138) = 276 units

Time taken to complete the remaining work = 276 ÷ (23 - 8) = 18.4 days

∴ 18.4 days it will take to finish the work.

Given:

Efficiency of A, B and C = 2 : 3 : 5

A alone can complete the work in = 50 days

Formula:

Total work = Efficiency × Time

Calculation:

Let efficiency of A be 2 units/day

Efficiency of A, B and C = 2 : 3 : 5

Total work = 2 × 50 = 100 units

Work done by A, B and C in 5 days = (2 + 3 + 5) × 5 = 10 × 5 = 50 units

Remaining work = 100 – 50 = 50 units

∴ Time taken by A and B to complete the remaining work = 50/(2 + 3) = 50/5 = 10 days

Given:

A can do a piece of work = 30 days

B can do a piece of work = 40 days

C can do a piece of work = 50 days

Formula used:

Total work = efficiency × time

Calculation:

According to the question:

⇒ (20 + 15 + 12) = 47 units = 3 days

⇒ 47 × 12 = 564 units = 3 × 12 = 36 days

⇒ (564 + 20 + 15) = 599 units = 38 days

Total work = 600 units = 38 + (1/12) = 38\(1\over12\) days.

∴ The correct answer is 38\(1\over12\) days.

Given:

A is 6 times more efficient than B, & B takes 32 days to complete the task.

Formula used:

Total work = Efficiency × Time taken

Calculation:

A is 6 times more efficient than B

Efficiency of A ∶ Efficiency of B = 7 ∶ 1

Total work = Efficiency of B × Time taken 

⇒ 1 × 32 = 32 units

Number of days required to finish the whole work by (A + B)  = Total work/Efficiency of (A+ B)

⇒ 32/8

⇒ 4 

∴ The total number of days required to finish the whole work by (A + B) is 4 days.

There is a difference in "Efficient" and " More efficient"

A is 6 times efficient than B means if B is 1 then, A will be 6

A is 6 times more efficient than B means if B is 1 then, A will be (1 + 6) = 7

In the question, it is given that A is 6 times more efficient which means if B is 1, then A will (1 + 6) times = 7 times efficient

So, Total efficiency of A and B = (1 + 7) = 8 units/day

Time taken to complete the work together = 32/8 days

⇒ 4 days and this is the answer. 

Given A alone can complete the work in 21 days

A and B together can complete the same work in 12 days

⇒ Total work = L.C.M of (12, 21) = 84

⇒ One day work of A = 4

⇒ One day work of (A + B) = 7

⇒ One day work of B = 3

Let A worked for x days and B worked for 16 days

⇒ 4x + 3 × 16 = 84

⇒ x = 9 days

Given:

A can complete the work in 24 days

A and B work on alternate days with B beginning the work, can complete the work in \(11 \frac{1}{3}\) days

Formula Used:

Total Work = Efficiency × Time

Calculation:

Let the total work be 24 units

⇒ Efficiency of A = 24/24 = 1 unit

According to the question,

A works on 2nd, 4th, 6th, 8th, 10th and (1/3) of 12th day

⇒ A works for 5(1/3) = 16/3 days

⇒ B works for = \(11 \frac{1}{3}\) - \(\frac{16}{3}\) = \(\frac{34-16}{3}\) = 6 days

⇒ Work done by A in 16/3 days = 16/3 units

Remaining work = 24 - (16/3) = 56/3 units

⇒ 56/3 units are completed by B in 6 days

(7/9)th part of 24 = (24 × 7)/9 = 56/3 units

∴ B alone will complete (7/9)th of the original work in 6 days.

Alternate Method

Given:

Time taken by A to finish a task alone = 24 days 

Calculation:

Let the total work be = 1 

A alone can finish the task in 24 days 

⇒ A's one-day work = 1/24 

A and B complete the whole task in = \(11 \frac{1}{3}\) days 

A and B work on alternate days, with B beginning so, we can say B will work only 6 days 

⇒ A will work only \(11 \frac{1}{3}\) - 6 = \(5 \frac{1}{3}\) days 

If A's one day work = 1/24 of work A completes in 1 day

⇒ A's \(5 \frac{1}{3}\) days work = 1/24 × \(5 \frac{1}{3}\) = 1/24 × 16/3 

⇒ 2/9 

Remaining work = 1 - 2/9 = 7/9 

∴ B does the 7/9th part of the work in 6 days. 

Note-

B, A, B, A, B, A, B, A, B, A, B, A/3

B completely 6-day work

That's why we have taken the 6 days work by B alone.

Given:

A = 2C

A + B in 20 days

B + C in 24 days

Concept used:

Total work = LCM of time taken by the workers

Calculation:

LCM of 20 and 24 is 120

So, efficiency of A and B = 120/20 = 6 and efficiency of B and C = 120/24 = 5

Now 2C + B = 6 and B + C = 5

So, C = 1

B = 4

40% of the work = 120 × 2/5 = 48 units

So, B will take 48/4 = 12 days

∴ B alone do 40% of the same work in 12 days

Shortcut Trick

Now 2C + B = 6 and B + C = 5

So, C = 1

B = 4

So,

A, B and C = 2, 4, and 1

40% of the work = 120 × 2/5 = 48 units

So, B will take 48/4 = 12 days

Formula Used:

Total work = Efficiency × Time taken

Calculation

Ajay can finish the work alone in 32 days

A’s one day work = 1/32

A and B complete the whole work in = 8 days

Ajay and Bharat work on alternate days, with Bharat starting the work on the first day so, we can say B will work only 4 days A will work only:

= 8 - 4 = 4 days

If A’s 4-day work = 4/32 = 1/8

Remaining work = 1 – [1/8] = 7/8

B complete 7/8 work in = 4 days

B complete whole work in = 4 × [8/7] = 32/7 days

B alone can finish 7 times the same work in = [32/7] × 7 = 32 days

B alone can finish 7 times the same work in 32 days.