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All questions of Work & Time for Mechanical Engineering Exam
Charlie and Alan run a race between points A and B, 5 km apart. Charlie starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Alan starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed. At what time do Charlie and Alan first meet each other?
- a)10 a.m.
- b)10:20 a.m.
- c)10:10 a.m
- d)10:30 a.m.
Correct answer is option 'C'. Can you explain this answer?
Charlie and Alan run a race between points A and B, 5 km apart. Charlie starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Alan starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed. At what time do Charlie and Alan first meet each other?
a)
10 a.m.
b)
10:20 a.m.
c)
10:10 a.m
d)
10:30 a.m.
|
|
Madhujya Chutia answered |
15 men could finish a piece of work in 210 days. But at the end of 100 days, 15 additional menare employed. In how many more days will the work be complete?- a)80 days
- b)60 days
- c)55 days
- d)50 days
Correct answer is option 'C'. Can you explain this answer?
15 men could finish a piece of work in 210 days. But at the end of 100 days, 15 additional menare employed. In how many more days will the work be complete?
a)
80 days
b)
60 days
c)
55 days
d)
50 days
|
Ishani Rane answered |
15 men took 210 days.
Therefore,
1 man requires 15 * 210 = 3150 days to complete the same work
So, in 1 day 1 man complete 1/3150 of the total work
Thus,
In 100 days 15 men completes (15 * 1/3150 * 100) = 10/21 of the total work
After 100 days there are 30 men,
They will complete the remaining task i.e. 1 - 10/21 of the total work in 1/30 * 11/21 * 3150 = 55 days.
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?
a)
30 days
b)
25 days
c)
20 days
d)
15 days
|
Varun Kapoor answered |
Work done by P and Q in 1 day = 1/10
Work done by R in 1 day = 1/50
Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as
Work done by P in 1 day * 2 = 6/50
=> Work done by P in 1 day = 3/50
=> Work done by Q and R in 1 day = 3/50
Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25
So Q alone can do the work in 25 days
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?
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
|
Shalini Patel answered |
Let us assume that digging one well = 40 unit
Efficiency of Anup = 40/10 = 4 unit/day
Now efficiency is falling by 10%
So, this is case of GP
⇒ 4,18/5,81/25,…… where common ratio = 9/10
Now, sum of infinite GP = a / (1 – r)
⇒ 4/ (1 – 9/10) = 40
Hence, it is clear that Anup will takes infinite time, so he will never finish the work.
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?
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
|
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
⇒ 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
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?
a)
8/15
b)
7/15
c)
11/15
d)
2/11
|
|
Varun Kapoor answered |
Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 x (7/60) = 7/15
Fraction of work left = 1 – 7/15= 8/15
Read the passage below and solve the questions based on it There are three taps A,B. C ami an outlet pipe D. A, B and C can fill the tank in the Panikam locality in 10, 20 and 25 h respectively. The outlet pipe can empty the same tank in 100 h. There are 2,000 houses in the locality. The tank has a capacity of 50.000 litresQ.If all the taps and the outlet pipe are opened simultaneously, how much water is thrown into the tank every hour?- a)8000 litres
- b)9000 litres
- c)1000 litres
- d)Cannot be determined
Correct answer is 'B'. Can you explain this answer?
Read the passage below and solve the questions based on it There are three taps A,B. C ami an outlet pipe D. A, B and C can fill the tank in the Panikam locality in 10, 20 and 25 h respectively. The outlet pipe can empty the same tank in 100 h. There are 2,000 houses in the locality. The tank has a capacity of 50.000 litres
Q.
If all the taps and the outlet pipe are opened simultaneously, how much water is thrown into the tank every hour?
a)
8000 litres
b)
9000 litres
c)
1000 litres
d)
Cannot be determined
|
Ishita Reddy answered |
Total time required to fill the tank: 1/10 + 1/20 + 1/25 - 1/100 = 18/100
Time required to fill the tank in 1hr = capacity/total time
= 50000*18/100
= 9000litre
Time required to fill the tank in 1hr = capacity/total time
= 50000*18/100
= 9000litre
Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
- a)3 pm
- b)2 pm
- c)1:00 pm
- d)11 am
Correct answer is option 'C'. Can you explain this answer?
a)
3 pm
b)
2 pm
c)
1:00 pm
d)
11 am
|
Kiran Mehta answered |
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?
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
|
|
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
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?
a)
4 days
b)
6 days
c)
2 days
d)
8 days
|
|
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
Refer to the data below and answer the questions that follow Anoop and Sandeep can dig a well each in 10 and 5 days respectively. But on a particularly difficult terrain the work is such that due to fatigue every' subsequent day the efficiency of a worker falls by 10%.Q.If both Anoop and Sandeep work together to finish two such wells then, in how many days will the work finish?- a)11th day
- b)12th day
- c)8th day
- d)Never
Correct answer is option 'C'. Can you explain this answer?
Refer to the data below and answer the questions that follow Anoop and Sandeep can dig a well each in 10 and 5 days respectively. But on a particularly difficult terrain the work is such that due to fatigue every' subsequent day the efficiency of a worker falls by 10%.
Q.
If both Anoop and Sandeep work together to finish two such wells then, in how many days will the work finish?
a)
11th day
b)
12th day
c)
8th day
d)
Never
|
Anupam Guduri answered |
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?
a)
8
b)
5
c)
4
d)
6
|
Aruna Singh answered |
Work done by P in 1 day = 1/18
Work done by Q in 1 day = 1/15
Work done by Q in 10 days = 10/15 = 2/3
Remaining work = 1 – 2/3 = 1/3
Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6
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?
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
|
|
Amita Das answered |
Since the amount of water flowing through each pipe is proportional to the square of its diameter , 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 the cistern is filled by all the three pipes in x minutes.
So , 1 / 49 + 4 / 49 x 9 + 1 / 49 x 9 = 1 / x
= x = 31.5 minutes.
A and B can complete a task in 30 days when working together. After A and B have been working together for 11 days, B is called away and A, all by himself completes the task in the next 28 days. Had A been working alone from begining, the number of days taken by him to complete the task would have been:- a)
- b)
- c)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
A and B can complete a task in 30 days when working together. After A and B have been working together for 11 days, B is called away and A, all by himself completes the task in the next 28 days. Had A been working alone from begining, the number of days taken by him to complete the task would have been:
a)
b)
c)
d)
None of these
|
Nipun Tuteja answered |
Work done in 11 days = 11 / 30
Rest work = 19 / 30
1 day's work of A =
Total number of days required to complete the whole work alone.
Rest work = 19 / 30
1 day's work of A =
Total number of days required to complete the whole work alone.
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?
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
|
|
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
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?
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
|
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.
To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?- a)30 days
- b)35 days
- c)40 days
- d)45 days
Correct answer is option 'A'. Can you explain this answer?
To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?
a)
30 days
b)
35 days
c)
40 days
d)
45 days
|
|
Rajesh Khatri answered |
We have,
B = 3/2*A
→
A = 2/3*B
One day's work, (A+B) = 1/18
(2/3*B+B) = 1/18
5/3*B = 1/18
One day's work of B = 3/90
B alone can complete the work in,
= 90/3
= 30 days.
Pipe A can fill the tank in 4 hours, while pipe B can fill it in 6 hours working separately. Pipe C can empty the whole tank in 4 hours. Ramesh opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was half-filled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. What is the time difference between both the cases, to fill the tank fully.- a)48 min.
- b)54 min.
- c)30 min.
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Pipe A can fill the tank in 4 hours, while pipe B can fill it in 6 hours working separately. Pipe C can empty the whole tank in 4 hours. Ramesh opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was half-filled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. What is the time difference between both the cases, to fill the tank fully.
a)
48 min.
b)
54 min.
c)
30 min.
d)
None of these
|
|
Stuti Choudhury answered |
In ideal case :
Time taken to fill the half tank by A and B = 50 / 41.66 = 6 / 5 hours
Time taken by A,B and C to fill rest half of the tank = 50 / 16.66 = 3 hours
Total time =
In second case :
Time taken to fill 3 / 4 tank by A and B = 75 / 41.66 = 9 / 5 hours
Time taken by A,B and C to fill rest 1 / 4 tank = 25 / 16.66 = 3 / 2 hours
Total time = 9 / 5 + 3 / 2 = 3 hours 18 minutes
Therefore, difference in time = 54 minutes
Time taken to fill the half tank by A and B = 50 / 41.66 = 6 / 5 hours
Time taken by A,B and C to fill rest half of the tank = 50 / 16.66 = 3 hours
Total time =
In second case :
Time taken to fill 3 / 4 tank by A and B = 75 / 41.66 = 9 / 5 hours
Time taken by A,B and C to fill rest 1 / 4 tank = 25 / 16.66 = 3 / 2 hours
Total time = 9 / 5 + 3 / 2 = 3 hours 18 minutes
Therefore, difference in time = 54 minutes
Twenty workers can finish a piece of work in 30 days. After how many days should 5 workersleave the job so that the work is completed in 35 days?- a)5 days
- b)10 days
- c)15 days
- d)20 days
Correct answer is option 'C'. Can you explain this answer?
Twenty workers can finish a piece of work in 30 days. After how many days should 5 workersleave the job so that the work is completed in 35 days?
a)
5 days
b)
10 days
c)
15 days
d)
20 days
|
|
Ishani Rane answered |
rate of 1 man per day=1/(20*30)=1/600
let d=number of days with 20 workers
20d/600+[15(35-d)]/600=1
d=15 days
A, B, C, D and E are five taps. The capacity of B is 2 times that of A, the capacity of C is 3 times that of A. Capacities of D and E are 4 and 5 times that of A respectively. In the first case A, C and E act as input pipes and B and D act as output pipes.In the second case, C, D, E act as input pipes and A and B act as output pipes.If A and B working together as input pipes can fill the tank in 4 hours, then what is the difference in the time required to fill the tank in the first and second case stated above?
- a)4/3 hours
- b)3/3 hours
- c)2/2 hours
- d)1 hours
Correct answer is option 'A'. Can you explain this answer?
A, B, C, D and E are five taps. The capacity of B is 2 times that of A, the capacity of C is 3 times that of A. Capacities of D and E are 4 and 5 times that of A respectively. In the first case A, C and E act as input pipes and B and D act as output pipes.In the second case, C, D, E act as input pipes and A and B act as output pipes.If A and B working together as input pipes can fill the tank in 4 hours, then what is the difference in the time required to fill the tank in the first and second case stated above?
a)
4/3 hours
b)
3/3 hours
c)
2/2 hours
d)
1 hours
|
|
Aisha Gupta answered |
There are two pipes in a tank. Pipe A is for filling the tank and Pipe B is for emptying the tank. If A can fill the tank in 10 hours and B can empty the tank in 15 hours then find how many hours will it take to completely fill a half empty tank?- a)30 hours
- b)15 hours
- c)20 hours
- d)33.33 hours
Correct answer is option 'B'. Can you explain this answer?
There are two pipes in a tank. Pipe A is for filling the tank and Pipe B is for emptying the tank. If A can fill the tank in 10 hours and B can empty the tank in 15 hours then find how many hours will it take to completely fill a half empty tank?
a)
30 hours
b)
15 hours
c)
20 hours
d)
33.33 hours
|
Rithika Chakraborty answered |
A’s work = 10%
B’s negative work = 6.66%
(A + B)’s work = 3.33%
To fill a half empty tank, they would take 50/3.33 = 15 hours.
B’s negative work = 6.66%
(A + B)’s work = 3.33%
To fill a half empty tank, they would take 50/3.33 = 15 hours.
Sashi can do a piece of work in 25 days and Rishi can do it in 20 days. They work for 5 days andthen Sashi goes away. In how many more days will Rishi finish the work?- a)10 days
- b)12 days
- c)14 days
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Sashi can do a piece of work in 25 days and Rishi can do it in 20 days. They work for 5 days andthen Sashi goes away. In how many more days will Rishi finish the work?
a)
10 days
b)
12 days
c)
14 days
d)
None of these
|
|
Aarav Sharma answered |
Given:
Sashi can do the work in 25 days
Rishi can do the work in 20 days
They worked together for 5 days
To find:
In how many more days will Rishi finish the work?
Solution:
Let's assume that the total work is 100 units.
So, Sashi can do 4 units of work in a day, and Rishi can do 5 units of work in a day.
In the first 5 days, the total work done by Sashi and Rishi together is:
Sashi's work = 4 units/day * 5 days = 20 units
Rishi's work = 5 units/day * 5 days = 25 units
Total work done = 20+25 = 45 units
Remaining work = 100 - 45 = 55 units
Now, only Rishi is working on the remaining work. So, the time taken by Rishi to complete the remaining work will be:
Time = Remaining work / Rishi's efficiency
Time = 55 units / 5 units/day = 11 days
Therefore, the total time taken by Rishi to complete the work is:
Time taken = 5 days (when Sashi and Rishi worked together) + 11 days (when only Rishi worked)
= 16 days
Hence, the correct answer is option D, none of these.
Sashi can do the work in 25 days
Rishi can do the work in 20 days
They worked together for 5 days
To find:
In how many more days will Rishi finish the work?
Solution:
Let's assume that the total work is 100 units.
So, Sashi can do 4 units of work in a day, and Rishi can do 5 units of work in a day.
In the first 5 days, the total work done by Sashi and Rishi together is:
Sashi's work = 4 units/day * 5 days = 20 units
Rishi's work = 5 units/day * 5 days = 25 units
Total work done = 20+25 = 45 units
Remaining work = 100 - 45 = 55 units
Now, only Rishi is working on the remaining work. So, the time taken by Rishi to complete the remaining work will be:
Time = Remaining work / Rishi's efficiency
Time = 55 units / 5 units/day = 11 days
Therefore, the total time taken by Rishi to complete the work is:
Time taken = 5 days (when Sashi and Rishi worked together) + 11 days (when only Rishi worked)
= 16 days
Hence, the correct answer is option D, none of these.
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?
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
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
|
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 can do a piece of work in 20 days. He works at it for 5 days and then B finishes it in 10 moredays. In how many days will A and B together finish the work?- a)8 days
- b)10 days
- c)12 days
- d)6 days
Correct answer is option 'A'. Can you explain this answer?
A can do a piece of work in 20 days. He works at it for 5 days and then B finishes it in 10 moredays. In how many days will A and B together finish the work?
a)
8 days
b)
10 days
c)
12 days
d)
6 days
|
Gowri Chakraborty answered |
Work done by A in 5 days = 1/20*5
=1/4
remaining work=(1-1/4)
=3/4
now,3/4 work is done by B in 10 days
whole work will be done by B in 10*4/3= 40/3 days
as 1 day's work of A = 1/20
1 day's work of B =3/40
(A+B)'s 1 day's work= 1/20+3/40
= 1/8
so,both finished the work in 8 day's.
In a fort there was sufficient food for 200 soldiers for 31 days. After 27 days 120 soldiers left thefort. For how many extra days will the rest of the food last for the remaining soldiers?- a)12 days
- b)10 days
- c)8 days
- d)6 days
Correct answer is option 'D'. Can you explain this answer?
In a fort there was sufficient food for 200 soldiers for 31 days. After 27 days 120 soldiers left thefort. For how many extra days will the rest of the food last for the remaining soldiers?
a)
12 days
b)
10 days
c)
8 days
d)
6 days
|
|
Prerna Choudhary answered |
After 27 days, food left = 4 × 200 = 800 soldier days worth of food. Since, now there are only 80
soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts
= 10 – 4 = 6 days.
soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts
= 10 – 4 = 6 days.
If 10 persons can do a job in 20 days, then 20 person with twice the efficiency can do the same job in:- a)5 days
- b)40 days
- c)10 days
- d)20 days
Correct answer is option 'A'. Can you explain this answer?
If 10 persons can do a job in 20 days, then 20 person with twice the efficiency can do the same job in:
a)
5 days
b)
40 days
c)
10 days
d)
20 days
|
Tejas Das answered |
M x D = 10 x 20 = 200
Man-days New Man-days = (20 x 2) x x
► 200 = 20 x 2 x x
► x = 5 days
or M1D1 = M2D2
► 10 x 20 = (20 x 2) x x
► ⇒ x = 5
Man-days New Man-days = (20 x 2) x x
► 200 = 20 x 2 x x
► x = 5 days
or M1D1 = M2D2
► 10 x 20 = (20 x 2) x x
► ⇒ x = 5
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?
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%
|
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%.
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%.
A man, a woman and a girl worked for a contractor for the same period. A man is twice efficient as a woman and a woman is thrice efficient as a girl. ₹ 10000 were given to all of them. What is the sum of money received by a woman and a girl together?- a)₹ 5500
- b)₹ 4500
- c)₹ 4000
- d)₹ 6000
Correct answer is option 'C'. Can you explain this answer?
A man, a woman and a girl worked for a contractor for the same period. A man is twice efficient as a woman and a woman is thrice efficient as a girl. ₹ 10000 were given to all of them. What is the sum of money received by a woman and a girl together?
a)
₹ 5500
b)
₹ 4500
c)
₹ 4000
d)
₹ 6000
|
Ayush Sharma answered |
Efficiency of a man : woman : girl = 6 : 3 : 1
∴ Share of a woman and girl =
=
∴ Share of a woman and girl =
=
Manoj takes twice as much time as Anjay and thrice as much as Vijay to finish a piece of work.Together they finish the work in 1 day. What is the time taken by Manoj to finish the work?- a)6 days
- b)3 days
- c)2 days
- d)4 days
Correct answer is option 'A'. Can you explain this answer?
Manoj takes twice as much time as Anjay and thrice as much as Vijay to finish a piece of work.Together they finish the work in 1 day. What is the time taken by Manoj to finish the work?
a)
6 days
b)
3 days
c)
2 days
d)
4 days
|
|
Prerna Choudhary answered |
Let Anjay take 3t days, Vijay take 2t days and Manoj take 6t days in order to complete the work.
Then we get:
1/3t + 1/2t + 1/6t = 1 Æ t = 1. Thus, Manoj would take 6t = 6 days to complete the work.
Then we get:
1/3t + 1/2t + 1/6t = 1 Æ t = 1. Thus, Manoj would take 6t = 6 days to complete the work.
A can work twice as fast as B. A and C together can work thrice as fast as B. If A, B and C complete a job in 30 days working together, in how many days can each of them complete the work?- a)60, 120, 120
- b)50, 100, 120
- c)60, 100, 80
- d)40, 80, 100
Correct answer is option 'A'. Can you explain this answer?
A can work twice as fast as B. A and C together can work thrice as fast as B. If A, B and C complete a job in 30 days working together, in how many days can each of them complete the work?
a)
60, 120, 120
b)
50, 100, 120
c)
60, 100, 80
d)
40, 80, 100
|
|
Shilpa Choudhury answered |
A, B and C complete a job in 30 days working together,
⇒ 1/A + 1/B + 1/C = 1/30
A can work twice as fast as B,
⇒ 1/B = 1/2A
A and C together can work thrice as fast as B,
⇒ 1/B = 1/3(1/A + 1/C)
Solving,
⇒ 3/B = 1/A + 1/C
⇒ 3/B = 1/30 – 1/B
⇒ 1/B = 120
Then,
⇒ 1/A = 1/60
⇒ 1/C = 1/120
∴ A, B and C can complete work in 60, 120 and 120 days respectively.
12 men complete a work in 18 days. 6 days after they had started working, 4 men join them. Howmany more days will all of them take to complete the remaining work?- a)10 days
- b)12 days
- c)15 days
- d)9 days
Correct answer is option 'D'. Can you explain this answer?
12 men complete a work in 18 days. 6 days after they had started working, 4 men join them. Howmany more days will all of them take to complete the remaining work?
a)
10 days
b)
12 days
c)
15 days
d)
9 days
|
|
Pallavi Shah answered |
12 × 18 = 12 × 6 + 16 × t Æ t = 9
Three diggers dug a ditch of 324 m deep in six days working simultaneously. During one shift, the third digger digs as many metres more than the second as the second digs more than the first. The third digger’s work in 10 days is equal to the first digger’s work in 14 days. How many metres does the first digger dig per shift?- a)15 m
- b)18 m
- c)21 m
- d)27 m
Correct answer is option 'A'. Can you explain this answer?
Three diggers dug a ditch of 324 m deep in six days working simultaneously. During one shift, the third digger digs as many metres more than the second as the second digs more than the first. The third digger’s work in 10 days is equal to the first digger’s work in 14 days. How many metres does the first digger dig per shift?
a)
15 m
b)
18 m
c)
21 m
d)
27 m
|
Priya Rane answered |
The per day digging of all three combined is 54 meters. Hence, their average should be 18. This
means that the first should be 18 – x, the second, 18 & the third 18 + x.
The required conditions are met if we take the values as 15,18,21 meters for the first, second and
third diggers respectively. Hence, (a) is the correct answer.
means that the first should be 18 – x, the second, 18 & the third 18 + x.
The required conditions are met if we take the values as 15,18,21 meters for the first, second and
third diggers respectively. Hence, (a) is the correct answer.
Ajay and Vijay can do a piece of work in 28 days. With the help of Manoj, they can finish it in 21days. How long will Manoj take to finish the work all alone?- a)84 days
- b)80 days
- c)75 days
- d)70 days
Correct answer is option 'A'. Can you explain this answer?
Ajay and Vijay can do a piece of work in 28 days. With the help of Manoj, they can finish it in 21days. How long will Manoj take to finish the work all alone?
a)
84 days
b)
80 days
c)
75 days
d)
70 days
|
|
Devanshi Sarkar answered |
Ajay + Vijay = 1/28 and Ajay + Vijay + Manoj = 1/21.
Hence, Manoj = 1/21 – 1/28 = 1/84.
Hence, Manoj will take 84 days to do the work.
Hence, Manoj = 1/21 – 1/28 = 1/84.
Hence, Manoj will take 84 days to do the work.
Ajit can do as much work in 2 days as Baljit can do in 3 days and Baljit can do as much in 4 days as Diljit in 5 days. A piece of work takes 20 days if all work together. How long would Baljittake to do all the work by himself?- a)82 days
- b)44 days
- c)66 days
- d)50 days
Correct answer is option 'C'. Can you explain this answer?
Ajit can do as much work in 2 days as Baljit can do in 3 days and Baljit can do as much in 4 days as Diljit in 5 days. A piece of work takes 20 days if all work together. How long would Baljittake to do all the work by himself?
a)
82 days
b)
44 days
c)
66 days
d)
50 days
|
Sameer Rane answered |
6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?- a)6
- b)8
- c)9
- d)15
Correct answer is option 'A'. Can you explain this answer?
6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?
a)
6
b)
8
c)
9
d)
15
|
|
Bhaskar Das answered |
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
∴ 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
∴ 10 children can do the work in 6 days.
If 20 engineers and 20 workers can together construct a 20 km road in 20 days. 40 engineers and 40 workers together will construct 40 km road in how many days?- a)10
- b)20
- c)40
- d)15
Correct answer is option 'B'. Can you explain this answer?
If 20 engineers and 20 workers can together construct a 20 km road in 20 days. 40 engineers and 40 workers together will construct 40 km road in how many days?
a)
10
b)
20
c)
40
d)
15
|
|
Akash Sharma answered |
Equate the man-days
For 20 km road, 20 x 20 = 400 man-days are required
∴ For 40 km road 800 man-days are required So, 800 = 40 * x
► ⇒ x = 20
For 20 km road, 20 x 20 = 400 man-days are required
∴ For 40 km road 800 man-days are required So, 800 = 40 * x
► ⇒ x = 20
X can do a piece of work in 20 days. He worked at it for 5 days and then Y finished it in 15 days. In how many days can X and Y together finish the work?- a)12 days
- b)15 days
- c)10 days
- d)5 days
Correct answer is option 'C'. Can you explain this answer?
X can do a piece of work in 20 days. He worked at it for 5 days and then Y finished it in 15 days. In how many days can X and Y together finish the work?
a)
12 days
b)
15 days
c)
10 days
d)
5 days
|
|
Rahul Mehta answered |
- X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
- This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.
- X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.
There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone.
Find the percentage of work done by tap A itself?- a)72%
- b)70%
- c)71%
- d)65%
Correct answer is option 'B'. Can you explain this answer?
There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone.
Find the percentage of work done by tap A itself?
Find the percentage of work done by tap A itself?
a)
72%
b)
70%
c)
71%
d)
65%
|
|
Akshita Kaur answered |
Given Information:
- Tap A can fill the tank in 25 hours.
- Tap B can fill the tank in 20 hours.
- Tap C can fill the tank in 10 hours.
- All taps are opened simultaneously for the first hour.
- After the first hour, tap C is closed and only taps A and B are kept running.
- After the fourth hour, tap B is also closed.
- The remaining work is done by tap A alone.
Calculation:
Let's assume the capacity of the tank is 100 units.
Work done by each tap in 1 hour:
- Tap A can fill 100/25 = 4 units of the tank in 1 hour.
- Tap B can fill 100/20 = 5 units of the tank in 1 hour.
- Tap C can fill 100/10 = 10 units of the tank in 1 hour.
Work done in the first hour:
- Since all taps are opened simultaneously, the total work done in the first hour is 4 + 5 + 10 = 19 units.
Work done in the second and third hour:
- After the first hour, only taps A and B are running.
- So, the work done in the second and third hour is (4 + 5) * 2 = 18 units.
Work done in the fourth hour:
- After the fourth hour, tap B is closed and only tap A is running.
- So, the work done in the fourth hour is 4 units.
Remaining work:
- The total capacity of the tank is 100 units.
- The work done in the first four hours is 19 + 18 + 4 = 41 units.
- Therefore, the remaining work is 100 - 41 = 59 units.
Work done by tap A alone:
- The remaining work of 59 units is done by tap A alone.
- Since tap A can fill 4 units of the tank in 1 hour, it will take 59/4 = 14.75 hours for tap A to complete the remaining work.
Percentage of work done by tap A:
- The total time taken to complete the remaining work is 14.75 hours.
- The total time taken to fill the tank by all taps is 4 hours (first four hours) + 14.75 hours (remaining work by tap A) = 18.75 hours.
- The percentage of work done by tap A is (14.75 / 18.75) * 100 = 78.67%.
Therefore, the percentage of work done by tap A itself is approximately 78.67%, which is closest to option 'b' (70%).
- Tap A can fill the tank in 25 hours.
- Tap B can fill the tank in 20 hours.
- Tap C can fill the tank in 10 hours.
- All taps are opened simultaneously for the first hour.
- After the first hour, tap C is closed and only taps A and B are kept running.
- After the fourth hour, tap B is also closed.
- The remaining work is done by tap A alone.
Calculation:
Let's assume the capacity of the tank is 100 units.
Work done by each tap in 1 hour:
- Tap A can fill 100/25 = 4 units of the tank in 1 hour.
- Tap B can fill 100/20 = 5 units of the tank in 1 hour.
- Tap C can fill 100/10 = 10 units of the tank in 1 hour.
Work done in the first hour:
- Since all taps are opened simultaneously, the total work done in the first hour is 4 + 5 + 10 = 19 units.
Work done in the second and third hour:
- After the first hour, only taps A and B are running.
- So, the work done in the second and third hour is (4 + 5) * 2 = 18 units.
Work done in the fourth hour:
- After the fourth hour, tap B is closed and only tap A is running.
- So, the work done in the fourth hour is 4 units.
Remaining work:
- The total capacity of the tank is 100 units.
- The work done in the first four hours is 19 + 18 + 4 = 41 units.
- Therefore, the remaining work is 100 - 41 = 59 units.
Work done by tap A alone:
- The remaining work of 59 units is done by tap A alone.
- Since tap A can fill 4 units of the tank in 1 hour, it will take 59/4 = 14.75 hours for tap A to complete the remaining work.
Percentage of work done by tap A:
- The total time taken to complete the remaining work is 14.75 hours.
- The total time taken to fill the tank by all taps is 4 hours (first four hours) + 14.75 hours (remaining work by tap A) = 18.75 hours.
- The percentage of work done by tap A is (14.75 / 18.75) * 100 = 78.67%.
Therefore, the percentage of work done by tap A itself is approximately 78.67%, which is closest to option 'b' (70%).
Sagar is 20% more efficient than Pranesh. If Pranesh can complete a piece of work in 30 days, then in how many days can both Sagar and Pranesh complete the work?- a)(11+5/11) days
- b)(13+ 7/11) days
- c)(15+ 5/11) days
- d)(17+ 7/11) days
Correct answer is option 'B'. Can you explain this answer?
Sagar is 20% more efficient than Pranesh. If Pranesh can complete a piece of work in 30 days, then in how many days can both Sagar and Pranesh complete the work?
a)
(11+5/11) days
b)
(13+ 7/11) days
c)
(15+ 5/11) days
d)
(17+ 7/11) days
|
|
Arshiya Bose answered |
In a military camp there is sufficient food supply for 200 soldiers for 40 days. After 10 days, certain number of men join and everyone eats 50% more than earlier. Now the food lasts for another 10 days. How many soldiers joined the camp after 10 days?- a)150
- b)300
- c)250
- d)200
Correct answer is option 'D'. Can you explain this answer?
In a military camp there is sufficient food supply for 200 soldiers for 40 days. After 10 days, certain number of men join and everyone eats 50% more than earlier. Now the food lasts for another 10 days. How many soldiers joined the camp after 10 days?
a)
150
b)
300
c)
250
d)
200
|
EduRev CLAT answered |
Let eating capacity of each soldier initially = 1 unit/day.
Initial food supply = 200 × 1 × 40 = 8000 units.
Supply left after 10 days = 200 × 1 × 30 units.
Let ‘s’ numbers of soldiers joined the camp.
New capacity of each soldier = 1.5 units/day
∴ The remaining supply is consumed by (200 + s) solders in another 10 days.
∴ 200 × 1 × 30 = (200 + s) × 1.5 × 10
⇒ 400 = 200 + s
⇒ s = 200
Hence, option D.
Two typists of varying skills can do a job in 6 minutes if they work together. If the first typisttyped alone for 4 minutes and then the second typist typed alone for 6 minutes, they would be left with 1/5 of the whole work. How many minutes would it take the slower typist to complete the typing job working alone?
- a) 10 minutes
- b) 15 minutes
- c) 12 minutes
- d) 20 minutes
Correct answer is option 'B'. Can you explain this answer?
Two typists of varying skills can do a job in 6 minutes if they work together. If the first typisttyped alone for 4 minutes and then the second typist typed alone for 6 minutes, they would be left with 1/5 of the whole work. How many minutes would it take the slower typist to complete the typing job working alone?
a)
10 minutesb)
15 minutesc)
12 minutesd)
20 minutes|
|
Ishani Rane answered |
Working efficiency of both typist together,
= 100/6 = 16.66% per minute
Now, let work efficiency of first typist be x and then second typist will be (16.66 - x)
First typist typed alone for 4 minutes and second typed alone for 6 minutes and they left with 1/5 (i.e 20%) of job, means they have completed 80% job.
Now,
First Typist typed in 4 minute + Second typed in 6 minutes = 80%
4 *x + 6 *(16.66 - x) = 80%
4x + 100% - 6x = 80%
x = 10%
First Typist typed 10% per minutes. Then second typed (16.66 - 10) = 6.66% per minute
Then, Second typist complete the whole job in 100/ 6.66 = 15.01 = 15 minutes.
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?
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
|
|
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'.
- 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'.
Efficiency of Asha is 25% more than Usha and Usha takes 25 days to complete a piece of work, Asha started a work alone and then Usha joined her 5 days before actual completion of the work. For how many days Asha worked alone?- a)9
- b)11
- c)10
- d)15
Correct answer is option 'B'. Can you explain this answer?
Efficiency of Asha is 25% more than Usha and Usha takes 25 days to complete a piece of work, Asha started a work alone and then Usha joined her 5 days before actual completion of the work. For how many days Asha worked alone?
a)
9
b)
11
c)
10
d)
15
|
|
Priya Sharma answered |
∴ Number of days required by Asha to finish the work alone = 20
► ∴ (4 x = 4 x 5)
(Alternatively, from percentage change graphic, number of days taken by Asha will be 20% less than Usha, if efficiency of Asha is 25% more than Usha)
Now, since Asha and Usha did work together for last 5 days = 5 x 9 = 45%
(Since efficiency of Asha = 5% and Usha's efficiency = 4%)
It means Asha completed 55% work alone
∴ Number of days taken by Asha to complete 55% work = 55 / 5 = 11
A tank can be filled in 30 minutes by 20 pumps. If five pumps go out of order, what time will be taken by the remaining pumps?- a)40 mins
- b)38 mins
- c)32 mins
- d)30 mins
Correct answer is option 'A'. Can you explain this answer?
A tank can be filled in 30 minutes by 20 pumps. If five pumps go out of order, what time will be taken by the remaining pumps?
a)
40 mins
b)
38 mins
c)
32 mins
d)
30 mins
|
Glance Learning Institute answered |
Answer: Option A
Explanation :We know,
Time ∝ 1/no. of pumps
∴ Time × (Number of pumps) = constant.
⇒ 20 × 30 = 15 × x
⇒ x = 20 × 30/15 = 40 mins
Hence, option (a).
Directions for question : A can do a work in 15 days and B can do it in 18 days. with the help of C work is completed in 6 days.
Q. A,B and C received total ₹ 27,000 for the whole work. What is the share of B, if the money is distributed in the ratio of amount of work done, individually?
- a)7200
- b)9000
- c)14400
- d)6300
Correct answer is option 'B'. Can you explain this answer?
Directions for question : A can do a work in 15 days and B can do it in 18 days. with the help of C work is completed in 6 days.
Q. A,B and C received total ₹ 27,000 for the whole work. What is the share of B, if the money is distributed in the ratio of amount of work done, individually?
a)
7200
b)
9000
c)
14400
d)
6300
|
|
Asha Reddy answered |
Ratio of efficiency of A,B and C = 6 : 5 : 4
∴ Share of C =
∴ Share of C =
Raju is twice as good as Vijay. Together, they finish the work in 14 days. In how many days canVijay alone do the same work?- a)16 days
- b)21 days
- c)32 days
- d)42 days
Correct answer is option 'D'. Can you explain this answer?
Raju is twice as good as Vijay. Together, they finish the work in 14 days. In how many days canVijay alone do the same work?
a)
16 days
b)
21 days
c)
32 days
d)
42 days
|
|
Devanshi Sarkar answered |
Raju being twice as good a workman as Vijay, you can solve the following equation to get the
required answer:
1/R + 1/2R = 1/14.
Solving will give you that Vijay takes 42 days.
required answer:
1/R + 1/2R = 1/14.
Solving will give you that Vijay takes 42 days.
Two pipes, when working one at a time, can fill a cistern in 3 hours and 4 hours, respectively while a third pipe can drain the cistern empty in 8 hours. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?- a)2 hours
- b)1 hour 45 minutes
- c)2 hour 11 minutes
- d)2 hour 10 minutes
Correct answer is option 'A'. Can you explain this answer?
Two pipes, when working one at a time, can fill a cistern in 3 hours and 4 hours, respectively while a third pipe can drain the cistern empty in 8 hours. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?
a)
2 hours
b)
1 hour 45 minutes
c)
2 hour 11 minutes
d)
2 hour 10 minutes
|
|
EduRev CLAT answered |
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 11/12th of the work = 11/12 × 24/ 11 = 2 hours
∴ The correct answer is 2 hours.
Dev and Tukku can do a piece of work in 45 and 40 days respectively. They began the work together, but Dev leaves after some days and Tukku finished the remaining work in 23 days. After how many days did Dev leave- a)7 days
- b)8 days
- c)9 days
- d)11 days
Correct answer is option 'C'. Can you explain this answer?
Dev and Tukku can do a piece of work in 45 and 40 days respectively. They began the work together, but Dev leaves after some days and Tukku finished the remaining work in 23 days. After how many days did Dev leave
a)
7 days
b)
8 days
c)
9 days
d)
11 days
|
|
Pallavi Shah answered |
n(1/45 + 1/40) + 23/40 = 1 Æ n = 9.
Chapter doubts & questions for Work & Time - General Aptitude for GATE 2025 is part of Mechanical Engineering exam preparation. The chapters have been prepared according to the Mechanical Engineering exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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