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Useful for the students who prepare for numerical aptitude in recruitment process.

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- 1. TIME & WORK Standard level c.s.veeraragavan 9894834264 veeraa1729@gmail.com
- 2. MEN Normally it takes 5 students 1 hour to arrange chairs for a function. If one of them is absent, how much time will it take for the remaining 4 students to arrange the same number of chairs? a) 80 mins b) 75 mins c) 90 mins d) 100 mins Total units of work by 5 students = 5 * 60 = 300 units. No of units for 4 students = 300 4 =75 minutes.
- 3. MEN A group of workers can do a piece of work in 60 days. However 4 workers were absent due to illness and it took 100 days to complete the work. How many workers are there in total? a)6 b)8 c)10 d)12 Let the number of workers be x. 60x = 100(x – 4) 100x – 400 = 60x X = 10
- 4. MEN 66 men can finish a job in 42 days. If the work has to be completed 9 days earlier, how many additional men must be engaged to complete the work? a) 18 b) 5 c) 15 d) 9 No of men required to complete the work = 66 * 42 33 = 84 No of additional men required = 84 – 66 = 18
- 5. Beginning A group of men decided to do a job in 7 days. But as 10 men dropped out everyday, the job got completed at the end of tenth day. How many men were at the beginning? a) 200 b) 180 c) 175 d) 150 Let no. of men in the beginning be x. Overall work = 7x man-days. 7x = x + (x – 10) + (x – 20) +…+(x – 90). 7x = 10x – 450 3x = 450 x = 150
- 6. EFFICIENCY A is twice as good as B and together they finish a piece of work in 18 days. In how many days will B alone finish the work? a) 27 b) 35 c) 42 d) 54 Ratio of efficiency = A:B = 2:1 = 2 x : 1 x Work done by together in 1 day= 2 x + 1 x = 3 x = 1 18 X = 54. Total work done by B in = 54 days.
- 7. EFFICIENCY A can do a piece of work in 20 days. B is 25% more efficient than A. How many days will B take to complete the same work? a) 15 b) 18 c) 16 d) 14 B is 25% more efficient than A Hence B’s efficiency is 125% of A = 5 4 of A Work done by B = 4 5 * 20= 16 days.
- 8. Days A, B and C can together do a piece of work in 7.5 days. A can do alone in 40 days and B alone in 24 days. In how many days can C do it alone? a) 9 days b) 12 days c) 15 days d) None of these (A + B + C)’s 1 day work = 1 7.5 = 2 15 (A + B)’s 1 day work = 64 40 ∗ 24 = 1 15 C’s 1 day work = 1 15 C can complete the work alone in 15 days LCM OF 7.5, 40, 24 = 120 WORK DONE BY A IN 1 DAY = 3 WORK DONE BY B IN 1 DAY = 5 WORK DONE BY ALL IN 1 DAY = 16 WORK DONE BY C IN 1 DAY = 8 TOTAL NO OF DAYS FOR C = 120/8 = 15
- 9. DAYS Venkat and Prasad can do a job in 12 days. Prasad alone can finish it in 28 days. In how many days can Venkat alone finish the work? a) 21 days b) 20 days c) 15 days d)18 days Amount of work per day for Venkat = 1 12 − 1 28 = 7 −3 12∗7 = 1 21 No of days required for Venkat alone = 21 days.
- 10. Days 5 men can do a work in 20 days of 8 hrs each. In how many days of 10 hrs each can 8 men do the same work? a) 6 days b) 8 days c) 10 days d) 12 days Days = 5 8 ∗ 8 10 ∗ 20 = 10 days men hours Days 5 8 20 8 10 ?
- 11. DAYS Amar and Akbar can do a job in 20 days. Akbar alone can finish it in 30 days. In how many days can Amar alone finish the work? a)70 days b)60 days c)50 days d)54 days 1 day work of Amar = 1 20 − 1 30 = 1 60 Amar can finish in 60 days.
- 12. Days 10 horses and 9 donkeys can do a piece of work in 20 days. 6 horses and 12 donkeys can do the same work in 30 days. In how many days can the same work be done by 1 horse and 1 donkey? a)174 days b) 182 days c) 190 days d) 198 days (10 H + 9 D) * 20 = ( 6 H + 12 D) * 30 200 H + 180 D = 180 H + 360 D 20 H = 180 D H = 9 D Hence total units of work = (90 D + 9 D) * 20 = 1980 D units. Hence 1 Horse and 1 Donkey can do 10 D units in 1 day. Hence no of days to complete the work by 1 Horse and 1 donkey = 1980 10 = 198.
- 13. Days Two workers Gopal and Shyam can do a work alone in 10 and 15 days respectively. Find the amount of work done by them in 5 days. a) 4 9 parts b) 3 8 parts c) 5 6 parts d) 3 4 parts LCM of 10 and 15 is 30 units. Work done by both in 1 day = 30 10 + 30 15 = 3 + 2 = 5 units. Work done in 5 days = 25 units. Hence amount of work done = 25 30 = 5 6
- 14. Days Bimal starts working on a job and works on it for 20 days and completes 50% of the work. To help him complete the work, he employs Ritesh and together they work for another 12 days and the work gets completed. In how many days Ritesh alone can complete the work? a) 40 days b) 50 days c) 60 days d) None of these As Bimal completes 50% work in 20 days, Bimal can complete in 40 days. Let Ritesh can complete in y days. Let the entire work be x. 0.5x = 12 days * 𝑥 40 + 𝑥 𝑦 Solving y = 60 days.
- 15. TIME Three friends Sam, Stephen and Durai can complete a piece of work in 13,5 and 10 days respectively. What is the approximate difference in the number of days between the number of days when Sam alone does the work and the number of days when Stephen and Durai together do the work? a)11.3 days b)8.5 days c)9.6 days d)13 days LCM of 13,5,10 is130 units. No of units of work done by Stephen and Durai = 130 5 + 130 10 = 26 + 13 = 39 No of days = 130 39 No of units of work done by Sam alone = 13 - 130 39 = 9.66 days.
- 16. TIME Four machines can mow a large playground in 12 hours. How long will it take 6 machines to mow a playground of same size? a) 7 hrs b) 8 hrs c) 9 hrs d) 10 hrs Total units of work for 4 machines = 4 * 12 = 48 hrs Total units of work for 6 machines = 48 6 = 8 hrs
- 17. TIME Carpenters A and B can do a work in 10 hrs when working at the same time. B works twice as fast as A. How long would it take for B to complete the work if B worked alone? a) 3.33 hrs b) 30 hrs c) 20 hrs d) 15 hrs Let A do x units of work and B do 2x units of work. Total units of work = 10 * 3x = 30x Time taken by B alone = 30𝑥 2𝑥 = 15 hrs.
- 18. TIMEA software company in Noida has been doing well in the last four years. The company recruited people from ECE, CSE, IT streams. All programmers receive equal amount as salary and perform equal amount of work. Suppose 15 such programmers take 15 minutes to write 15 lines of code in total. How long will it take for 84 programmers to write 84 lines of code in total? a)15 minutes b) 84 minutes c) 30 minutes d) 42 minutes Let no of lines coded by 1 programmer in 1 minute be x. 15 programmers takes 15 minutes to write 15 lines of code. X = 15 15 ∗15 Assume 84 programmers take y minutes to write 84 lines. X = 84 84∗y Thus x = 15 15 ∗15 = 84 84 ∗ y . Solving y = 15 minutes PROGRAMMERS LINES OF CODE TIME 15 15 15 84 84 ? More programmers less time (Indirect) 15 : 84 :: x : 15 More lines more time (direct) 15 : 84 :: 15 : x Time taken = 15 84 ∗ 84 15 ∗15 = 15
- 19. TIME 7 men can produce 7 candles in 7 minutes. 1 man can produce 1 candle in how many minutes? a)49 b)5 c) 7 d) 9 Less men More minutes (inverse) 7:1 :: x : 7 Less candles less minutes (direct) 1:7 :: x : 7 7*1*7 = 1 * 7 * x (or) 7 1 ∗ 1 7 ∗ 7 = 7 X = 7 men candles Minutes 7 7 7 1 1 ?
- 20. TIME If a pipe can fill the tank within 3 hrs but due to a leak at the bottom it took 15 minutes more. Now once the tank is full, how much time will it take to get emptied through the leak? a)39 hrs b)64 hrs c)75 hrs d)None of these Difference in time taken in 1 hr = 1 3 − 1 3.25 = 0.25 3∗3.25 = 25 3∗325 = 1 39 Hence total time taken to get emptied = 39 hours.
- 21. PIPES & CISTERN Two pipes A and B can fill a tank in 16 minutes and 20 minutes respectively. Both the pipes are opened together but after 5 minutes, pipe A is turned off. What is the total time required to fill the tank? a) 8.75 minutes b) 9.45 minutes c) 8.65 minutes d) 10.15 min Work done in 5 minutes = 5 16 + 20 16 ∗ 20 = 36 64 = 9 16 Remaining work = 7 16 Time required for B = 7 16 * 20 = 35 4 = 8.75 minutes.
- 22. PIPES & CISTERN Two pipes can fill a tank in 10 and 20 hrs respectively. There is a leakage at the bottom of the tank which takes 40 hrs to empty the tank. If both the pipes are used together, then how long will it take to fill the tank? (with the leakage still being present)? a) 5hrs b) 12 hrsc) 8 hrs d) 14 hrs. Work done in one hour = 1 10 + 1 20 − 1 40 = 4 + 2 − 1 40 = 5 40 = 1 8 Total work done in = 8 hrs. Alternate Method: LCM OF 10,20,40 IS 40. PIPE 1 (4 units), Pipe2 (2) Pipe 3 (1) Capacity = 4 + 2 – 1 = 5 Total work done = 40/5 = 8 hrs.
- 23. PIPES & CISTERN A pipe can fill a tank in 3 hrs. Due to a leakage at the bottom of the tank, it takes 3.5hrs to fill the same tank. In how many hrs will the leakage empty the tank once it is full? a) 12 hrs b) 15hrs c) 18hrs d) 21 hrs. Difference in time taken to fill the tank in 1 hr = 1 3 − 1 3.5 = 1 3 − 2 7 = 1 21 Total time taken to empty = 21 hrs.
- 24. PIPES & CISTERN Three pipes A, B and C can together fill a tank in 5 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 8 hours. The number of hours taken by C alone to fill the tank is a)8 hours b)9 hours c)10 hours d)12 hours. Work done by A, B and C together in 2 hours = 2 5 3 5 of work done by A, B together in = 8 hours. Work done by A,B together in 2 hours = 3 5 * 2 8 = 3 20 Work done by C alone in 2 hours = 2 5 − 3 20 = 5 20 = 1 4 Total work done by C alone in = 8 hours.
- 25. PIPES & CISTERN Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30 more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut? a)6 hrs b)8 hrs c)10 hrs d)12 hrs Difference in time taken = 1 2 − 1 2.5 = 0.5 5 = 0.1 Hence time taken to empty the tank = 10 hrs.
- 26. Ratio 12 Women can do a work in 16 days. 16 men can complete the same work in 17 days. What is the ratio between the working capacity of a woman and a man? a)12:17 b)11:16 c) 16:11 d) 17:12 Work done by 1 woman in 1 day = 1 12 ∗ 16 Work done by 1 man in 1 day = 1 17 ∗ 16 Ratio = 1 12 ∗ 16 : 1 17 ∗ 16 = 17 : 12

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