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

Rahul's work rate: 1/15 task per day

Sohan's work rate: 1/25 task per day

Priya's work rate: 1/12 task per day

Let:

x = Number of days Rahul and Priya worked.

Total time taken to complete the task = T days.

Rahul: Worked for x days.

Work done = x/15.

Sohan: Worked for the entire T days.

Work done = T/25

Priya: Worked for x days but 3 days before the task was completed.

Work done = (x - 3)/12

The total work done is equal to 1 the entire task. So:

x/15 + T/25 + (x - 3)/12 = 1

Since Rahul and Priya worked for the same number of days x, and Priya 3 days before the task was completed:

T = x + (x - 3) = 2x - 3

Substitute T = (2x - 3) into the Work Equation

x/15 + (2x - 3)/25 + (x - 3)/12 = 1

Multiply through by 300 LCM of 15, 25, and 12 to eliminate denominators:

20x + 12(2x - 3) + 25(x - 3) = 300

20x + 24x - 36 + 25x - 75 = 300

69x - 111 = 300

69x = 411

x = 411 / 69 = 6

T = 2x - 3 = 2(6) - 3 = 9

Sohan worked for the entire T = 9 days.

Work done by Sohan = 9/25

The amount of work completed by Sohan is 9/25.

Original Distance: Let the distance be D.

Original Time: 15 days.

Resting Time: 6 hours per day.

Walking Time per Day: 24 - 6 = 18 hours.

Total walking time in 15 days = 15 × 18 = 270 hours.

Speed = D/270 distance per hour.

New Distance: 3D.

New Speed: 1.5 × D/270 = 1.5D/270 = D/180

New Resting Time: 2 × 6 = 12 hours per day.

New Walking Time per Day: 24 - 12 = 12 hours.

Time required to walk 3D at speed D/180:

Time = 3D / (D/180) = 540 hours.

Number of days required:

Days = 540 / 12 = 45 days.

The person will take 45 days to walk three times the distance, 1.5 times as fast, and resting twice as long each day.

[A + B]'s one days work is 1/12

[A +B+C]' one days work is 1/6

so, C's one days work is 1/6 - 1/12 = 1/12

so, C alone complete the work alone in 12 days.

12 = R + 8

so, R = 4 

Let A and B alone complete the work in x and 3x days respectively.

so, 1/x + 1/3x = 1/12

or, 4/3x = 1/12

or, x = 16

so, A alone complete the work in 16 days.

16 = R + 12

Time taken by A to complete the job: \( 16 \) hours

Time taken by B to complete the job: \( 8 \text{ to } 12 \) hours (Since it is mentioned that B completes the work in the morning session, it can be at \( 8:00 \) a.m., \( 12:00 \) noon, or any time in between.)

Maximum time taken by A and B together:

\( \frac{1}{16} + \frac{1}{12} = \frac{7}{48} \Rightarrow 6 \text{ hours } 51 \text{ minutes} \)

Minimum time taken by A and B together:

\( \frac{1}{16} + \frac{1}{8} = \frac{3}{16} \Rightarrow 5 \text{ hours } 20 \text{ minutes} \)

Thus, together, A and B can complete the work at any time between \( 2:20 \) p.m. and \( 3:51 \) p.m. on the same day.

Only option (4), satisfies the given time frame.

Given:

10 pipes can fill a tank in 24 minutes.

2 pipes go out of order.

Formula used:

Time taken is inversely proportional to the number of pipes.

Calculation:

Initially, 10 pipes take 24 minutes to fill the tank.

⇒ Total time with 10 pipes = 24 minutes

Let the time taken by the remaining 8 pipes be x minutes.

⇒ 10 pipes × 24 minutes = 8 pipes × x minutes

⇒ 10 × 24 = 8 × x

⇒ x = (10 × 24) / 8

⇒ x = 30 minutes

∴ The correct answer is option 1.

Shortcut Trick

If both pipes are open, total efficiency = (A + B) = 5 + (-8) = -3 units

According to question,

Amount of water in the tank = (1/5) × 80 = 16 units

Time taken to empty the tank = work/efficiency = 16/((-3)) = 5.33 hours

Alternate Method

GIVEN :

Time by which pipe A can fill the tank = 16 hours

Time by which pipe B can empty the tank = 10 hours 

The cistern is (1/5)th full.

CONCEPT :

Total work = time × efficiency

CALCULATION :

Negative efficiency indicates pipe B is emptying the tank.

If both pipes are open, total efficiency = (A + B) = 5 + (-8) = -3 units

From the total efficiency it is clear that when both are opened, the tank is being emptied. 

Amount of water in the tank = (1/5) × 80 = 16 units

The water level will not rise as the total action is emptying when both are opened together.

Time taken to empty the tank = work/efficiency = 16/((-3)) = 5.33 hours

∴ Time taken to empty the tank is 5.33 hours.

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:

First pipe can fill the cistern = 3 hours

Second pipe can fill the cistern = 4 hours

Third pipe can drain the cistern = 8 hours

Calculation:

Let the total amount of work in filling a cistern be 24 units. (LCM of 3, 4 and 8)

Work done by pipe 1 in 1 hour = 24/3 = 8 units.

Work done by pipe 2 in 1 hour = 24/4 = 6 units.

Work done by pipe 3 in 1 hour = 24/ (-8) = -3 units

Total work done in 1 hour = 8 + 6 – 3 = 11 units

The time required to complete 11/12^th of the work = 11/12 × 24/ 11 = 2 hours

∴ The correct answer is 2 hours.

Given:

No of days taken by Harish and Bimal = 20 

Formula used:

No of days taken = Work/Efficiency 

Calculation:

Let the total work be = 1 

One day work done by Harish and Bimal = 1/20 

Work done by Harish and Bimal in 15 days = 1/20 × 15 = 3/4

⇒ Remaining work = 1 - 3/4 = 1/4 

Harish did remaining work in 10 days alone.

⇒ One day work done by Harish  = 1/4 ÷ 10 = 1/40 

∴ Time taken by Harish to do the entire task alone = 1 ÷ 1/40 = 40 days

 Shortcut TrickFraction of work done by Harish & Bimal in 15 days = 15/20 = 3/4 

The remaining 1/4th (25%) of work was done by Harish in 10 days. 

∴ The 100% work would be done by Harish in (10 × 4) 40 days.

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:

Difference between wages of A and C = Rs. 7600

Formula Used:

Share in wages = Work done/Total work × Total wages

Calculation:

Let total work be 60 unit,

Work done by A and B = 13/15 × 60 = 52 unit

⇒ Work done by C = 60 – 52 = 8 unit

Work done by B and C = 11/20 × 60 = 33 unit

⇒ Work done by A = 60 – 33 = 27 unit

Work done by B = 60 – 27 – 8 = 25 unit

According to the question,

27 – 8 = 19 unit = 7600

⇒ 1 unit = 400

Total wages of A and C = (27 + 8) = 35 units = 35 × 400 = Rs. 14000

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.