Explore Exams
Login Signup
Sign in Open App
All Exams  >   Interview Preparation  >   Quantitative Aptitude (Quant)  >   All Questions

All questions of Time & Work for Interview Preparation Exam

Clear all your doubts with EduRev

P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in
  • a)
    30 days
  • b)
    25 days
  • c)
    20 days
  • d)
    15 days
Correct answer is option 'B'. Can you explain this answer?

Future Foundation Institute answered
Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,
Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 
∴ d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t2 + 15t
⇒ 3t2 − 5t − 50 = 0
⇒ 3t2 + 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
⇒ t = 5 (ignoring -ve value) 
Thus, Total time taken by slower horse = 5 + 5 = 10 hr
So Option B is correct

Anup can dig a well in 10 days. but particularly in difficult time the work is such that due to fatigue every subsequent day efficiency of a worker falls by 10%.If Anup is given a task of digging one such well in the difficult time, then in how many days will he finish the work?
  • a)
    12th day
  • b)
    15 th day
  • c)
    11th day
  • d)
    Never
Correct answer is option 'D'. Can you explain this answer?

Wizius Careers answered
Correct Answer :- d
Explanation : The total no. of days in which Anoop can dig the well is 10 days.
Anoop's one day efficiency is 10%.
On day one Anoop performs 10% of his work efficiency, then the next day he won't be able to perform because as per the question the efficiency of a worker falls by 10%.
thus, 10%-10% = 0.

Sheldon had to cover a distance of 60 km. However, he started 6 minutes later than his scheduled time and raced at a speed 1 km/h higher than his originally planned speed and reached the finish at the time he would reach it if he began to race strictly at the appointed time and raced with the assumed speed. Find the speed at which he travelled during the journey described.
  • a)
    25 km/h 
  • b)
    10 km/h 
  • c)
    6 km/h
  • d)
    15 km/h
Correct answer is option 'A'. Can you explain this answer?

Asf Institute answered
Solve this question through options.
⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
⇒ The time difference can be seen to be 6 minutes in this case = 60 / 24 – 60 / 25 = 0.1 hrs = 6 mins
Thus, 25 km/h is the correct answer. 
So Option A is correct

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A  take to finish the task if she works alone at her usual efficiency?
(CAT 2019)
  • a)
    18
  • b)
    36
  • c)
    24
  • d)
    12
Correct answer is option 'A'. Can you explain this answer?

Meera Rana answered
let usual efficiency of A = 2a and that of B = b
So total work = 12*(2a+b) = 24a+12b——1)
In 2nd case when efficiency of A is halved and efficiency of B get tripled , new efficiency of A = 2a/2 = a
New efficiency of B = b*3 =3b
Total work = 9(a+3b) = 9a+27b ——2)
From eq 1) and 2)
24a+12b = 9a + 27b
15a = 15b
Or a=b
So total work = 24a+12b = 24a+12a = 36a
Time taken by A alone to complete the work = 36a/2a = 18 days

Hence Option A is correct

For Complete syllabus of Quant for CAT click on the link given below:

P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
  • a)
    8/15
  • b)
    7/15
  • c)
    11/15
  • d)
    2/11
Correct answer is option 'A'. Can you explain this answer?

Academic Studio answered
Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).
Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4
So Option A is correct

A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
  • a)
    30 days
  • b)
    40 days
  • c)
    60 days
  • d)
    70 days
Correct answer is option 'C'. Can you explain this answer?

Sameer Rane answered
Let A's 1 day's work = x and B's 1 day's work = y.

Then, x + y =1/30 and 16x + 44y = 1.

Solving these two equations, we get: x = 1/60 and y = 1/60

 B's 1 day's work = 1/60.

Hence, B alone shall finish the whole work in 60 days.

10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
  • a)
    3
  • b)
    5
  • c)
    7
  • d)
    Cannot be determined
Correct answer is option 'C'. Can you explain this answer?

Dhruv Mehra answered
1 woman's 1 day's work = 1/70
1 child's 1 day's work = 1/140
(5 women + 10 children)'s day's work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7
5 women and 10 children will complete the work in 7 days.

A can do a piece of work in 90 days, B in 40 days and C in 12 days. They work for a day each in turn, i.e., first day A does it alone, second day B does it alone and 3rd day C does it alone. After that the cycle is repeated till the work is finished. They get Rs 240 for this job. If the wages are divided in proportion to the work each had done. Find the amount A will get?
  • a)
    14
  • b)
    24
  • c)
    34
  • d)
    36
Correct answer is 'B'. Can you explain this answer?

Ishani Rane answered
Work done by A in 1 day = 1/90
Work done by B in 1 day = 1/40
Work done by C in 1 day = 1/12
Work done in 3 days = 1/90 + 1/40 + 1/12 = 43/360
in 8 * 3 = 24 days , work completed = 8 * 43/360 = 344/360
Remaining work = 1 - 344/360 = 16/360
in 25th day, A works and completes 1/90 work .
Remaining work = 16/360 - 1/90 = 12/360
in 26th day, B works and completes 1/40 work .
Remaining work = 12/360 - 1/40 = 1/120
in 27th day, C works and completes this entire 1/120 work
A worked 9 days by doing 1/90 work each day. Total work done by A = 9 * 1/90 = 1/10
B worked 9 days by doing 1/40 work each day. Total work done by B = 9 * 1/40 = 9/40
C worked 9 days by doing 1/12 work in the initial 8 days and 1/120 work in the 9th day.
Total work done by C = 8 * 1/12 + 1/120 = 81/120
Work done by A : Work done by B : Work done by C
= 1/10 : 9/40 : 81/120 
= 12 : 27 : 81
Total amount that they get = 240
Amount that A get = 240 * 12/(12+27+81) = Rs.24

P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. how many days does P alone need to finish the remaining work?
  • a)
    8
  • b)
    5
  • c)
    4
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?

Spectrum Coaching Institute answered
Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.
So Option D is correct

Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:
  • a)
    15
  • b)
    16
  • c)
    18
  • d)
    25
Correct answer is option 'B'. Can you explain this answer?

Raghavendra Sharma answered
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x     x = (4 * 20/5)
 x = 16 days.
Hence, Tanya takes 16 days to complete the work.

A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in :
  • a)
    8 days
  • b)
    10 days
  • c)
    12 days
  • d)
    15 days
Correct answer is option 'C'. Can you explain this answer?

Gowri Chakraborty answered
Let total work be 30 units (LCM of 10 and 15).
In one day, A can do 2 units of work and B can do 3 units of work.
In one day, both A and B can do 5 units of work.
In two days, A and B will complete 10 units of work. Remaining 20 units can be completed by A in 10 days (at rate of 2 units per day).
Hence, whole work will be completed in 12 days.

In what time would a cistern be filled by three pipes of diameter of 1 cm, 2 cm and 3 cm if the largest pipe alone can fill the cistern in 49 minutes, the amount of water flowing through each pipe being proportional to the square of its diameter?
  • a)
    31.5 minutes
  • b)
    63 minutes
  • c)
    126 minutes
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Naroj Boda answered
Since the amount of water flowing through each pipe is proportional to square of its diameter so if efficiency of longest pipe (3 cm) = 1/49
Then efficiency of pipe (2 cm) = 4/(49 x 9)
and efficiency of pipe (1 cm) = 1/ (49 x 9) 
Now let cistern is filled by all three pipes in x minutes.

Directions for Question: Study the following and answer the questions that follow.
A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas.
The amount of heat generated is equal to 1 kcal per cc of gas.
However, there is wastage of the heat as per follows:

@ (Include higher extremes)
Q. For Question 3, if burner A had been opened only 25% and burner B had been opened 50%, the amount of heat available for cooking would be
  • a)
    780 kcal
  • b)
    800 kcal
  • c)
    750 kcal
  • d)
    Cannot be determined
Correct answer is option 'A'. Can you explain this answer?

Divey Sethi answered
First, let's find the rate at which each burner discharges gas when opened at the given percentages.

For burner A, opened at 25%:
Rate of discharge = 1 cc/min * 25% = 0.25 cc/min

For burner B, opened at 50%:
Rate of discharge = 2 cc/min * 50% = 1 cc/min

Now, let's find the wastage of heat for each burner based on the given discharge rates.

For burner A, with a discharge rate of 0.25 cc/min:
Wastage = 10% (since it falls within the 0 - 0.5 cc/min range)

For burner B, with a discharge rate of 1 cc/min:
Wastage = 25% (since it falls within the 1 - 1.5 cc/min range)

Now, we can find the effective heat generated by each burner after accounting for wastage.

For burner A:
Effective heat = 1 kcal/cc * (1 - 10%) = 1 kcal/cc * 90% = 0.9 kcal/cc

For burner B:
Effective heat = 1 kcal/cc * (1 - 25%) = 1 kcal/cc * 75% = 0.75 kcal/cc

Next, we need to find the total amount of gas discharged by both burners until the gas cylinder is empty. Since the capacity of the cylinder is 1000 cc, we can find the time it takes for both burners to empty the cylinder.

Total rate of discharge = 0.25 cc/min (burner A) + 1 cc/min (burner B) = 1.25 cc/min

Time to empty the cylinder = 1000 cc / 1.25 cc/min = 800 minutes

Now, we can find the total heat generated by each burner during these 800 minutes.

Heat generated by burner A = 0.9 kcal/cc * 0.25 cc/min * 800 min = 180 kcal

Heat generated by burner B = 0.75 kcal/cc * 1 cc/min * 800 min = 600 kcal

Finally, we can find the total heat available for cooking by adding the heat generated by both burners.

Total heat available = 180 kcal (burner A) + 600 kcal (burner B) = 780 kcal

So, the amount of heat available for cooking would be 780 kcal.

A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in
  • a)
    4 days
  • b)
    6 days
  • c)
    8 days
  • d)
    18 days
Correct answer is option 'A'. Can you explain this answer?

BT Educators answered
Ratio of rates of working of A and B = 2 : 1
So, ratio of time taken = 1 : 2
B's 1 day's work = 1/12
∴ A's one day work = 1/3 (2 times of B's work)
(A + B)'s 1 day's work = (1/6) + (1/12) = 3/12 = 1/4
So, A and B together can finish work in 4 days.

Chetan is thrice as efficient as Mamta and together they can finish a piece of work in 60 days. Mamta will take how many days to finish this work alone?
  • a)
    80
  • b)
    160
  • c)
    240
  • d)
    320
Correct answer is option 'C'. Can you explain this answer?

Ritika Choudhury answered
  • Chetan is thrice as efficient as Mamta.
  • Let, Mamta takes 3x days and Chetan takes x days to complete the work.
  • ∴ 1/x + 1/3x = 1/60 ⇒ x = 80.
  • ∴ Mamta will take 80 × 3 = 240 days to complete the work.

A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in:
  • a)
    1/24
  • b)
    7/24
  • c)
    24/7
  • d)
    4
Correct answer is option 'C'. Can you explain this answer?

Divey Sethi answered
A’s 1 day work = 1/24
B’s 1 day work = 1/6
C’s 1 day work = 1/12
(A+B+C)’s 1 day work = 1/24 + 1/6 + 1/12 = 7/24
The work will be completed by them is 24/7 days.

6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days.
  • a)
    4 days
  • b)
    6 days
  • c)
    2 days
  • d)
    8 days
Correct answer is option 'A'. Can you explain this answer?

Vikas Choudhury answered
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b 
Work done by 6 men and 8 women in 1 day = 1/10 
=> 6m + 8b = 1/10
=> 60m + 80b = 1    (1)
Work done by 26 men and 48 women in 1 day = 1/2 
=> 26m + 48b =1/2
=> 52m + 96b = 1    (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day 
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days

Mayank can do 50% more work than Shishu in the same time. Shishu alone can do a piece of work in 30 hours. Shishu starts working and he had already worked for 12 hours when Mayank joins him. How many hours should Shishu and Mayank work together to complete the remaining work?
  • a)
    6
  • b)
    12
  • c)
    4.8
  • d)
    7.2
Correct answer is option 'D'. Can you explain this answer?

Ishani Rane answered
Shishu alone does the work in 30 hours 

So in 1 hour he does 1/30 of the work 

Mayank in 1 hour does 1/30 + 1/2*1/30= 1/30 +1/60 = 3/60 = 1/20 of the work 

Together in 1 hour they do 1/30 +1/20 = 5/60 = 1/12 of the work 

Together they can finish the work in 12 hours 

Shishu in 12 hours does 12/ 30 = 2/5 

Remaining work = 3/5 

3/5 X 12 = 36/5 = 7.2 hours

Read the passage below and solve the questions based on it.
The tank at a water supply station is filled with water by several pumps. At first three pumps of Ihe same capacity are turned on: 2.5 hours later, two more pumps (both the same) of a different capacity are set into operation. After 1 hour, the additional pumps were set into operation; the tank was almost filled to its capacity (15 m3 were still lacking): in another hour the tank was full. One of the two additional pumps could have filled the tank in 40 hours
Q. What is the volume of the tank?
  • a)
    60 m3
  • b)
    80 m3
  • c)
    75 m3
  • d)
    90 m3
Correct answer is option 'A'. Can you explain this answer?

Bhavya Saha answered
Let us assume that, first three pumps fills the tank in x hours .
so,
→ Efficiency of each pump = (1/x) m³ / hour .
then,
→ Efficiency of three pump = (3/x) m³ / hour .
 
now,
→ First three pumps works for = 2.5h + 1h + 1h = 4.5 hours.
 
so,
→ Water filled by 3 pumps in 4.5 hours = 4.5 * (3/x) = (13.5/x) m³ .
 
now, given that,
→ Time taken by additional pump to fill the tank = 40 hours.
so,
→ Efficiency of 2 additional tanks = 2 * (1/40) = (1/20) m³ / h .
 
and,
→ Additional pumps work for = 1 + 1 = 2 hours.
 
so,
→ Water filled by additional pumps in 2 hours = 2 * (1/20) = (1/10) m³ .
 
therefore,
→ (13.5/x) + (1/10) = 1
→ (13.5/x) = 1 - (1/10)
→ (13.5/x) = (9/10)
→ x = 135/9 = 15 hours.
 
since given that, in last 1 hour they filled 15 m³ .
 
hence,
→ 3 * (1/15) + (1/20) = 15 m³
→ (1/5) + (1/20) = 15
→ (4 + 1)/20 = 15
→ (5/20) = 15
→ (1/4) = 15
→ 1 = 60 m³ (Ans.) (Option A)

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on ?
(CAT 2018)
Correct answer is '10'. Can you explain this answer?

Alok Verma answered
Let the rate of each filling pipe be 'f ' litres/hr and the rate of each draining pipes be 'd' litres/hr
As per the first condition given in the question, Capacity of tank will be = (6f - 5d ) x 6 …(i)
And as per the second condition given in the question, Capacity of tank will be = (5f - 6d ) x 60 …(ii)
Since the capacity of tank is same in both the cases, on equating (i) and (ii), we have
    ► (6f - 5d) x 6 = (5f - 6d) x 60
    ► or, 6f - 5d = 50f - 60d
    ► or, 44f = 55d
    ► or, 4f = 5d
Therefore, f = 1.25d , putting this value in eq. (i),
Capacity of the tank = (6f - 5d ) x 6 = (7.5d - 5d) x 6 = 15d
When 2 filling and 1 draining pipes work simultaneously, the effective rate of filling the tank = (2f - d ),
Now putting f =1.25d in this eq. we have
Effective rate = (2.5d - d ) = 1.5d
Therefore time required to fill the empty tank = 15d / 1.5d = 10 hours.

Directions for Question: Study the following and answer the questions that follow.
A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas.
The amount of heat generated is equal to 1 kcal per cc of gas.
However, there is wastage of the heat as per follows :

@ (Include higher extremes)
Q. If both burners are opened simultaneously such that the first is opened to 90% of its capacity and the second is opened to 80% of its capacity, the amount of time in which the gas cylinder will be empty (if it was half full at the start) will be:
  • a)
    250 minutes
  • b)
    400 minutes
  • c)
    200 minutes
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Talent Skill Learning answered

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?
  • a)
    16
  • b)
    22
  • c)
    20
  • d)
    18
Correct answer is option 'C'. Can you explain this answer?

Sarthak Iyer answered
Understanding the problem:
Given that B needs 25% more time to finish a job than A does when working alone.

Given information:
- A and B finish the job in 13 days
- A works alone till half the job is done
- A and B work together for four days
- B works alone to complete the remaining 5% of the job

Solution:
Let's assume that A alone can finish the job in x days, then B alone can finish the job in (1.25x) days.

Calculating the work rate:
Let the total work be 100 units.
- A can complete 100/x units of work in a day
- B can complete 100/(1.25x) = 80/x units of work in a day

Calculating the work done:
- A works alone for 6.5 days to complete half the job (50 units)
- A and B work together for 4 days, completing 4*(100/x + 80/x) = 720/x units of work
- B works alone for the remaining 2.5 days, completing 5 units of work

Equation:
A's work + A&B's work + B's work = Total work
50 + 720/x + 10 = 100
720/x = 40
x = 18

Calculating B's time to finish the job alone:
B can finish the job alone in 1.25x = 1.25*18 = 22.5 days
Therefore, B alone can finish the entire job in 22.5 days, which is closest to option (c) 20 days.

A finishes 6 / 7th of the work in 2z hours, B works twice as fast and finishes the remaining work. For how long did B work?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Amit Kumar answered
Since A finishes 6 / 7th of the work in 2z hours.
B would finish 12 / 7 of the work in 2z hours.
Thus, to do 1/7th of the work (which represents the remaining work), B would require
     ► 2z /12 = z / 6 hours.

Directions for Question: Study the following and answer the questions that follow.
A gas cylinder can discharge gas at the rate of 1 cc/minute from burner A and at the rate of 2 cc/minute from burner B (maximum rates of discharge). The capacity of the gas cylinder is 1000 cc of gas.
The amount of heat generated is equal to 1 kcal per cc of gas.
However, there is wastage of the heat as per follows:

@ (Include higher extremes)
Q. The maximum amount of heat with fastest speed of cooking that can be utilised for cooking will be when:
  • a)
    The first burner is opened upto 50% of it's aperture
  • b)
    The second burner is opened upto 25% of it's aperture
  • c)
    Either (a) or (b)
  • d)
    None of these 
Correct answer is option 'C'. Can you explain this answer?

Mahevish Ali answered
Correct ans. is option 'C'

If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?
  • a)
    60
  • b)
    45
  • c)
    40
  • d)
    30
Correct answer is option 'A'. Can you explain this answer?

Sagar Sharma answered
Explanation:

Given data:
- A+B can complete work in 15 days
- B alone can complete work in 20 days

Let's calculate:

Efficiency:
Let's assume the total work to be done is 60 units (LCM of 15 and 20)
- A+B's efficiency: 60/15 = 4 units/day
- B's efficiency: 60/20 = 3 units/day

Work done by A alone:
As A+B's efficiency is 4 units/day and B's efficiency is 3 units/day, A's efficiency will be the difference between the two, which is 1 unit/day.

Time taken by A alone to complete the work:
Now, as A's efficiency is 1 unit/day, A will take 60 units of work in 60 days.
Therefore, A alone can complete the work in 60 days.
So, the correct answer is option A) 60.

Refer to the data below and answer the questions that follow.
Anoop was writing the reading comprehension sections in Lhe DOG entrance examinations, There were four passages of exactly equal length in terms of number of words and die four passages had 5, 8, 8 and 6 questions following each of them respectively. It is known that Anoop can answer exactly 12 questions in the time he takes to read any one of the four passages. Assume that his rate of reading and answering questions remains the same throughout the section.
Q.
By what per cent should Anoop increase his reading speed if he has to cut down on his total time spent on the section by 20%? Assume that the time spent on answering the questions is constant and as given in the directions.
  • a)
    36.36%
  • b)
    54.54%
  • c)
    50.50%
  • d)
    45.45%
Correct answer is option 'D'. Can you explain this answer?

Maulik Rane answered
To solve this problem, let's first find out the total time Anoop takes to read all four passages and answer all the questions.

Let the time he takes to read one passage be T. Since there are four passages, he takes 4T time to read all the passages. It is given that he can answer 12 questions in the time he takes to read one passage. So, the time he takes to answer one question is T/12.

There are a total of 5+8+8+6 = 27 questions. The time he takes to answer all the questions is 27 * (T/12) = 27T/12 = 9T/4.

Now, the total time spent on the section is the sum of the time spent on reading all the passages and answering all the questions: 4T + 9T/4 = 25T/4.

To cut down on his total time spent on the section by 20%, the new total time should be 80% of the original time, which is 0.8 * (25T/4) = 5T.

Since the time spent on answering the questions remains constant, the time spent on reading should reduce to 5T - 9T/4 = 11T/4. The new time he takes to read one passage is (11T/4) / 4 = 11T/16.

Now, let's find out the percentage increase in reading speed. The original time to read one passage is T, and the new time is 11T/16. Since speed is inversely proportional to time, the new speed will be 16/11 times the original speed.

The percentage increase in speed is [(16/11 - 1) * 100] = [(5/11) * 100] = 45.45%.

So, Anoop should increase his reading speed by 45.45% to cut down on his total time spent on the section by 20%.

Direction for Question: Refer to the data below and answer the questions that follow: 
Anoop was writing the reading comprehension sections in the SIP entrance examinations. There were four passages of exactly equal length in terms of number of words and the four passages had 5, 8, 8 and 6 questions following each of them, respectively. It is known that Anoop can answer exactly 12 questions in the time he takes to read any one of the the four passages. Assume that his rate of reading and answering questions remains the same throughout the section.
Q. Anoop took 13 min more to finish the first three passages than the time he took to finish the last passage. Assuming that Anoop answered all the questions in each passage, what percentage of the total time did he spent on the first passage?
  • a)
    24.5%
  • b)
    25.4%
  • c)
    22.6%
  • d)
    26.2%
Correct answer is option 'C'. Can you explain this answer?

Amrutha Desai answered
►Let us assume that time to read one passage = x
►So, according to the question, 3x + 21( x / 12) - 13 ⇒ x = 4
►Therefore, total time taken for answering all questions = 4 x 4 + (4 x 27 / 12) = 25 min
►Time spent on first passage = 4 + (4 x 5 / 12) = 17 / 3 min
►Required percentage = 22 .6%

A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
  • a)
    Rs. 375
  • b)
    Rs. 400
  • c)
    Rs. 600
  • d)
    Rs. 800
Correct answer is 'B'. Can you explain this answer?

Sameer Rane answered
Payment is always directly proportional to efficiency, that is more the efficiency higher the payLet the total work be eating 24 chocolates (LCM of 6 and 8)Therefore in one day A can eat = 24/6 = 4B can eat = 24/8 = 3All three together in one day can = 24/3 = 8This means C can eat 1 chocolate in 1 daySo efficiency ratio of A:B:C = 4:3:1Therefore payment will also be in same ratioC gets 1/8 th of the amount = 1/8 * 3200 = 400 Rs

A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
  • a)
    15 days
  • b)
    20 days
  • c)
    25 days
  • d)
    30 days
Correct answer is option 'C'. Can you explain this answer?

Anirban Khanna answered
LCM(10,50)=50
 
Suppose total work is 50 units.

Work done by A and B together in 1 day 

= 50/10 = 5 unit.

Work done by C in 1 day

= 50/50 = 1 unit.

Work done by A,B,C together in 1 day 

= 5 + 1 = 6 unit.

Since work done by A is equal to work done by B and C together,
work done by B and C together in 1 day 

= 6/2 = 3 unit.

Work done by B in 1 day 

= 3 − 1 = 2 unit.

Therefore, B alone can do the work in 50/2 = 25 days.

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?
  • a)
    35
  • b)
    40
  • c)
    45
  • d)
    50
Correct answer is option 'B'. Can you explain this answer?

Ishani Rane answered
Let 1 man's 1 day work = x and 1 woman's 1 day work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10
Solving these two equations, we get:
x = 11/400 and y = 1/400
10 woman's 1 day work = (1/400 x 10) = 1/40.
Hence, 10 women will complete the work in 40 days.

Two pipes can separately fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is 3/4th full, a leak develops in, through which one-fourth of water supplied by both the pipes goes out. What is the total time taken to fill the tank?
  • a)
    18 hours
  • b)
    14 hours
  • c)
    15 hours
  • d)
    13 hours
Correct answer is option 'D'. Can you explain this answer?

Tanvi nair answered
Understanding the problem:
- Pipe 1 can fill the tank in 20 hours.
- Pipe 2 can fill the tank in 30 hours.
- When the tank is 3/4th full, a leak develops and one-fourth of the water supplied by both pipes goes out.

Solution:
1. Let's calculate the filling rate of each pipe:
- Pipe 1 can fill 1/20 of the tank in 1 hour.
- Pipe 2 can fill 1/30 of the tank in 1 hour.
2. When the tank is 3/4th full, the total amount of water in the tank is 3/4 of the tank's capacity.
3. At this point, the leak starts and one-fourth of the water supplied by both pipes goes out. This means only 3/4 of the water supplied by both pipes remains in the tank.
4. Let the total time taken to fill the tank be x hours. In x hours, the combined filling rate of both pipes is 1/x of the tank's capacity.
5. The effective filling rate after the leak starts is (3/4) * (1/20 + 1/30) = (3/4) * (1/20 + 1/30) = (3/4) * (1/12) = 1/16 of the tank's capacity.
6. Equating the effective filling rate to the combined filling rate:
1/x = 1/16
x = 16 hours
Therefore, the total time taken to fill the tank is 16 hours, which corresponds to option 'D'.

Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty cone in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?
(XAT 2015)
  • a)
    Pipe A was open for 19 hours
  • b)
    Pipe A was open for 19 hours 30 minutes
  • c)
    Pipe B was open for 19 hours 30 minutes
  • d)
    Pipe C was open for 19 hours 50 minutes
  • e)
    The situation is not possible
Correct answer is option 'C'. Can you explain this answer?

Akanksha Sen answered


Chapter doubts & questions for Time & Work - Quantitative Aptitude (Quant) 2025 is part of Interview Preparation exam preparation. The chapters have been prepared according to the Interview Preparation exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Interview Preparation 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Time & Work - Quantitative Aptitude (Quant) in English & Hindi are available as part of Interview Preparation exam. Download more important topics, notes, lectures and mock test series for Interview Preparation Exam by signing up for free.

Quantitative Aptitude (Quant)

164 videos|133 docs|118 tests

Top Courses Interview Preparation

© EduRev
Education Revolution
Signup to see your scores go up
within 7 days!
Takes less than 10 seconds to signup