Time and Work problems
- 1. Krishna can do a work in 10 days while Mohan can do the same work in 20 days. They started work together. After 3 days Krishna left the work and Mohan completed it. For how many days Mohan worked alone more than the No. of days required when both worked together?
Options- A. 41/3
- B. 31/4
- C. 23/5
- D. 32/3 Discuss
Correct Answer: 41/3
Explanation:
Krishna's efficiency = 10%
Mohan's efficiency = 5%
Work done by Krishana and mohan together in 3 days = 15 x 3 = 45%
Now, number of days in which B completed rest (55%) work alone = 55/5 = 11
Total No. of days in which B worked = 3 + 11 = 14
Now No. of days required by B, when A and B both worked together = 100/15 = 62/3
? Required difference in No. of days = (11) - (62/3)
= 13/3 = 4 1/3 days
- 2. Kareena can do a piece of work in 9 days and Krishna can do the same work in 18 days. They started the work. After 3 days Shahid joined them, who can complete alone the same whole work in 3 days. What is the total No. of days in which they had completed the work?
Options- A. 12
- B. 8
- C. 4
- D. 6 Discuss
Correct Answer: 4
Explanation:
Efficiency of Kareena and Krishna = 11.11 + 5.55 = 16.66%
Work done in 3 days = 3 x 16.66 = 50%
Rest work done by Kareena, Krishna and Shahid = 50/50 = 1 day
(Since efficiency of Shahid = 33%)
Thus in ( 3 + 1) days they have completed the work = 4 days.
- 3. kavita, babita and samita started a work. 5 days later Samita left the work and Babita left the work after working 8 days. In how many more days Kavita would have completed the rest work if they take 20, 60 and 30 days individually to finish a work?
Options- A. 4
- B. 5
- C. 6
- D. 8 Discuss
Correct Answer: 6
Explanation:
Efficiency of Kavita = 5%
Efficiency of Babita = 1.66%
Efficiency of Samita = 3.33%
Work done in 5 days by K + B + S = 5 x 10 = 50%
Work done in 3 days by K + B = 3 x 6.66 = 20%
Remaning work (30%) done by Kavita alone = 30/5 = 6 days
- 4. In a school, MID-Day meal food is sufficient for 250 students for 33 days. if each students is given 125 g meals. 80 more students joined the school. If same amount of meal is given to each student, then the food will last for
Options- A. 20 days
- B. 40 days
- C. 30 days
- D. 25 days Discuss
Correct Answer: 25 days
Explanation:
M1 = 250, D 1 = 33 days
per day meal W1 = W2 = 125 g
M2 = (250 + 80) =330 and D2=?
According to the formula
M1D1W2 = M2D2W1
250 x 33 x 125 = 330 x D2x 125
? D2 = (250 x 33)/330
? D2 = 25 days
- 5. A certain number of men can do a piece of work in 80 days. If there were 10 men less, it could be finished in 20 days more. How many men are there in the starting?
Options- A. 45
- B. 50
- C. 40
- D. 60 Discuss
Correct Answer: 50
Explanation:
Let the original number of men = N
Time taken by N = 80 days
Now, (N - 10) men can finish the work in the (80 + 20) = 100 days
Here, M1 = N, M2 = (N - 10), D1 = 80 and D2 = 100
According to the formula.
M1D1 = M2D2
? N x 80 = 100 x (N - 10)
? 8N = 10N - 100
? 10N - 8N = 100
? N = 50
- 6. A can do a piece of work in 8 days, B can do it in 10 days and C can do it in 20 days. In how many days can A, B and C together complete the work?
Options- A. 37/11 days
- B. 35/11 days
- C. 32 /11 days
- D. 39/11 days Discuss
Correct Answer: 37/11 days
Explanation:
A's 1 days 's work = 1/8
B ' s 1 days ' s = 1/10
C ' s day 's work = 1/20
(A + B + C)'s 1 days 's work = 1/8 + 1/10 + 1/20 = (5 + 4 + 2)/40 = 11/40
? (A + B + C ) can finish the work in 40/11 days or 3 7/11 days.
- 7. 15 men would finish a piece of work in 210 days. But at the end of every 10 days. 15 additional men are employed. In how many days will it be finished?
Options- A. 30 days
- B. 70 days
- C. 35 days
- D. 60 days Discuss
Correct Answer: 60 days
Explanation:
10 day's work by 15 men = 10/210 = 1/21
At the end of every 10 days 15 additional men are employed i.e., for the next 10 days we have 15 + 15 = 30 men
? Next 10 day's work by 30 men = 2/21
Hence in 20 days only (1/21 + 2/21) = 3/21 work is completed
To complete the whole work we have to reach the value of (21/21) work.
Now, (1/21) + (2/21) + (3/21) + ......... (6/21) = 21/21 = 1
Hence total time to complete the whole wok 10 + 10 + 10 + 10 + 10 + 10 = 60 days
- 8. Ramesh can finish a job in 20 days. He worked for 10 days alone and completed the remaining job working with Dinesh, in 2 days. How many days would both Dinesh and Ramesh together can complete the whole work?
Options- A. 4
- B. 5
- C. 10
- D. 12 Discuss
Correct Answer: 4
Explanation:
Ramesh alone finish 1/2of the work in 10 days.
Remaining 1/2 of the job was finished by Ramesh and Dinesh together in 2 days.
Therefore, they both together can finish the complete job in 4 days.
- 9. A and B can together finish a work in 30 days. They worked for it for 20 days and then B left. the remaining work was done by A alone in 20 more days. A alone can finish the work in?
Options- A. 48 days
- B. 50 days
- C. 54 days
- D. 60 days Discuss
Correct Answer: 60 days
Explanation:
(A + B)'s 20 day's work = (20 x 1/30) = 2/3
Remaining work = (1 - 2/3) = 1/3
1/3 work is done by A in 20 days
Whole work can be done by A in (3 x 20) days = 60 days.
- 10. Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1 / 3 as efficiently as he actually did the work, then they would have completed the work in 3 days. Find the time for A to complete the job alone?
Options- A. 61/2 days
- B. 61/4 days
- C. 63/4 days
- D. 121/2 days Discuss
Correct Answer: 61/4 days
Explanation:
Efficiency is proportional to work done per day. Work done per day N number of days
= Amount of work done. Considering efficience of A and B initially as 1.
Let A alone can do the work in M days and B alone can do the same work in N days.
Then, 5/M + 5/N = Total work done = 1
Since efficiency of A and B are 2 and 1/3 respectively
? 6/M + 1/N = 1 ....(i)
and 1/M + 1/N = 5 .....(ii)
Now, subtracting equation (ii) from equation (i), we have M = 25/4 = 61/4 days.
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