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

Given:

Rahul's time to complete the project = 20 days

Ratio of Rahul : Priya = 2 : 1 → So Priya's time = 10 days

Rahul finishes remaining work in 8 days

Let k = number of days Priya worked with Rahul at the beginning

Work done is 1 (whole project)

Work rates:

Rahul’s 1 day work = 1/20

Priya’s 1 day work = 1/10

Together, in 1 day, they do:

(1/10 + 1/20) = (2 + 1)/20 = 3/20 of the work per day

Work done in first k days:

Work done together in k days =

k × 3/20 = 3k/20

Remaining work =

1 − 3k/20 = (20 − 3k)/20

Rahul finishes remaining work in 8 days:

Rahul’s 1-day work = 1/20

So in 8 days = 8/20 = 2/5

Now equate:

(20 − 3k)/20 = 2/5

Multiply both sides by 20:

20 − 3k = 8

⇒ 3k = 12

⇒ k = 4

Thus, the correct answer is 4 days.

Given:

2 men or 9 women can complete the work in 14 days

2 men work for 9 days and then leave

Formula used:

Work = Number of persons × Days × Efficiency

Calculation:

2 men = 9 women ⇒ 1 man = 4.5 women

Total work in woman-days = 9 women × 14 days = 126 woman-days

Work done by 2 men in 9 days:

⇒ 2 men = 9 women

⇒ Work = 9 women × 9 days = 81 woman-days

Remaining work = 126 - 81 = 45 woman-days

Let required number of women = x

⇒ x × 9 = 45

⇒ x = 5

∴ The number of women required is 5

Given:

A's time to complete work = 57 hours

B and C's time to complete work = 28 hours

A and C's time to complete work = 19 hours

Formula Used:

Work Rate = \(\dfrac{1}{\text{Time taken}}\)

Combined Work Rate = Sum of individual work rates

Time taken = \(\dfrac{1}{\text{Work Rate}}\)

Calculations:

A's Work Rate \((R_A)\) = \(\dfrac{1}{57}\) work/hour

B and C's Combined Work Rate \((R_B + R_C)\) = \(\dfrac{1}{28}\) work/hour

A and C's Combined Work Rate \((R_A + R_C)\) = \(\dfrac{1}{19}\) work/hour

⇒ C's Work Rate \((R_C)\) = \((R_A + R_C) - R_A\)

⇒ \(R_C = \dfrac{1}{19} - \dfrac{1}{57}\)

⇒ \(R_C = \dfrac{2}{57}\) work/hour

⇒ B's Work Rate \((R_B)\) = \((R_B + R_C) - R_C\)

⇒ \(R_B = \dfrac{1}{28} - \dfrac{2}{57}\)

⇒ \(R_B = \dfrac{57}{1596} - \dfrac{56}{1596}\)

⇒ \(R_B = \dfrac{1}{1596}\) work/hour

⇒ Time for B alone = \(\dfrac{1}{R_B}\)  = \(\dfrac{1}{\frac{1}{1596}}\)

⇒ Time for B alone = 1596 hours.

Given:

9 men or 8 women can complete the job in 19 days.

9 men worked for 9 days and left.

Remaining work needs to be completed by women in 8 days.

Formula Used:

Work done = Efficiency × Time

Total Work = LCM of efficiencies × Total Time

Calculation:

Efficiency of 9 men = Efficiency of 8 women

Efficiency of 1 man = Efficiency of 8 women / 9

Efficiency of 1 woman = Efficiency of 9 men / 8

Total Work = 9 men × 19 days = 8 women × 19 days

Total Work = LCM of efficiencies × 19

Work completed by 9 men in 9 days:

Work completed = 9 men × 9 days

Remaining Work = Total Work - Work completed by 9 men

Remaining Work = (9 × 19) - (9 × 9)

Remaining Work = 171 - 81

Remaining Work = 90

Work to be done by women in 8 days:

Efficiency of 1 woman = Total Work / (8 × 19)

Efficiency of 1 woman = 171 / 152

Number of women required:

Work done by women = Number of women × Efficiency of 1 woman × 8 days

⇒ 90 = Number of women × (171 / 152) × 8

⇒ Number of women = 90 / ((171 / 152) × 8)

⇒ Number of women = (90 × 152) / (171 × 8)

⇒ Number of women = (13680) / (1368)

⇒ Number of women = 10

Correct Option: Option 4

Solution:

10 women are required to complete the remaining work in 8 days.

Given:

A can do the work in = 8 days

B can do the work in = 24 days

A + B + C can do the work in = 3 days

Formula used:

Work done per day = 1 / Number of days

Total work = LCM of all given days

Calculation:

let the total work = LCM of 8, 24, 3 = 24 units

A’s 1-day work = 24 ÷ 8 = 3 units

B’s 1-day work = 24 ÷ 24 = 1 unit

(A + B + C)’s 1-day work = 24 ÷ 3 = 8 units

⇒ C’s 1-day work = 8 − (3 + 1) = 8 − 4 = 4 units

⇒ Time taken by C alone = 24 ÷ 4 = 6 days

∴ C alone can do the job in 6 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.