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- Question
- In a factory, there are equal number of women and children. Women work for 6 h a days and children for 4 h a day. During festival time, the work load goes up by 50%. The government rule does not allow children to work for more then 6 h a day. If they are equally efficient and the extra work is done by women, then extra hours of work put in by women everyday are?
Options- A. 5
- B. 3
- C. 4
- D. 9
- Correct Answer
- 3
ExplanationLet extra hours a day are N.
According to the formula,
(M1D1T1) / W1 = (M2D2T2) / W2
? [1 x 1 x (6 + 4)] / 1 = [1 x 1 x (6 + 6 + N)] / 11/2
? (3/2) x 10 = 12 + N
? 15 = 12 + N
? N = 15 - 12 = 3
Extra hours of works everyday are 3. - 1. If 14 men and 12 boys can finish work in 4 days, while 8 men and 16 boys can finish the same work in 5 days. Compare the 1 days work of 1 man and 1 boy.
Options- A. 2
- B. 11/2
- C. 1/2
- D. 3 Discuss
- 2. A contract is to be complete in 50 days and 105 men were set to work, each working 8 h a day. After 25 days, 2/5th of the work is finished. How many additional men be employed, so that the work may be completed on time, each man now working 9 h a day?
Options- A. 34
- B. 36
- C. 35
- D. 37 Discuss
- 3. 5 men started working to complete a work in 15 days. After 5 days, 10 women are accompanied by to them to complete the work in next 5 days. If the work is to be done by women only, when could the work be over, if 10 women have started it?
Options- A. 10 days
- B. 18 days
- C. 15 days
- D. 12 days Discuss
- 4. 10 men and 8 women together can complete a work in 5 days. Work done by one women in day is equal to half the work done by man in 1 day. How many days will it take for 4 men and 6 women to complete that work?
Options- A. 12 days
- B. 10 days
- C. 82/3
- D. 43/4 days Discuss
- 5. A takes twice as much time as B, and C take thrice as much time as B to finish a work. Working together, they can finish the work in 12 days. Find the number of days needed for A to do the work alone.
Options- A. 20
- B. 22
- C. 33
- D. 44 Discuss
- 6. 3 women and 18 children together take 2 days to complete a piece of work. How many days will 9 children alone take to complete the piece of work, if 6 women alone can complete the piece of work in 3 days?
Options- A. 9
- B. 7
- C. 5
- D. 6 Discuss
- 7. A contractor undertook to finish a certain work in 124 days and employed 120 men. After 64 days, he found that he had already done 2/3 of the work. How many men can be discharge now, so that the work finish in time?
Options- A. 40
- B. 50
- C. 48
- D. 56 Discuss
- 8. A contract is to be complete in 92 days and 234 men were set to work, each working 16 h a days. After 66 days, 4/7 of the work is completed. How many addition men may be employed, so that the work may be completed in time, each man now working 18 h a day?
Options- A. 162
- B. 234
- C. 262
- D. 81 Discuss
- 9. A, B and C can do a piece of work in 16, 32 and 48 days, respectively. They all begin together. A work continuously till it is finished, C leaves the work 2 days before its completion and B leaves the work in 1 days before its complete. In what time is the work finished?
Options- A. 10 days
- B. 20 days
- C. 15 days
- D. 30 days Discuss
- 10. Rashmi can do a piece of work in 16 days. Ravina can do the same work in 12 4/ 5 days, while Gitika can do it in 32 days. All of them started to work together but Rashmi leaves after 4 days. Ravina leaves the job 3 days before the completion of the work. How long would the work last?
Options- A. 9 days
- B. 6 days
- C. 18 days
- D. 5 days Discuss
Time and Work problems
Search Results
Correct Answer: 2
Explanation:
Here, a1 = 14, b1 = 12, p = 4, a2 = 8, b2 = 16 and q = 5
One days work of 1 man/One days of 1 boy
= (qb2 - pb1) / ( pa1 - qa2)
= [5 x 16 - 4 x 12] / [ 4 x 14 - 5 x 8]
= (80 - 48) / (56 - 40)
= 32/16
= 2
Correct Answer: 35
Explanation:
According to the formula
M1D1T1/W1= M2D2T2/W2
Given M1 = 105 D1 = 25 T1 = 8 W1 = 2/5
Now let the additional men be N
Then, M2 = 105 + N, T2 = 9, D2 = 25 and W2 =1 - (2/5) = 3/5
On putting these value in the above formula
(105 x 25 x 8) / (2/5) = [(105 + N) x 25 x 9] / (3/5)
? (105 x 8) / 2 = (105 + N) x (9/3)
? 105 x 4 = (105 + N) x 3
? 105 x 4 = 105 x 3 +3N
? 3N = 105
? N = 35 men
Correct Answer: 15 days
Explanation:
According to the question,
5 men's 1 day's work = 1/15 .....(i)
? 5 men's 5 day's work = (1/15) x 5 = 1/3
? Remaining work = 1 - (1/3) = 2/3
5 men + 10 women can do 2/3 of the work in 5 days.
? 5 men + 10 women can do the whole of the work in 15/2 days.
? (5 men + 10 women's) 1 day's work = 2/15 .......(ii)
10 women 1 days work = (2/15) - 5 men's 1 day's work
= (2/15) - (1/15)
= 1/15
? 10 women can finish the work in 15 days
Correct Answer: 10 days
Explanation:
According to the question,
1 man = 2 women
? 10 men + 8 women = (20 + 8) women = 28 women
4 men + 6 women = (8 + 6) women = 14 women
Given, M1 = 28, M2 = 14, D1 = 5, D2 = ?
and W1 = W2 = 1
According to the formula, M1D1W2= M2D2W1
? 28 x 5 x 1 = 14 x D2 x 1
? D2 = (28 x 5)/14 = 10 days
Correct Answer: 44
Explanation:
Let time taken by B = N
Then, time taken by A = 2N
and time taken by C = 3N
According to the question,
(1/N) + (1/2N) + (1/3N) = 1/12
? 1 + (1/2) + (1/3) = N/12
? (6 + 3 + 2)/6 = N/12
? 11 = N/2
? N = 22
? Required number of days = 2N = 2 x 22 = 44 days
Correct Answer: 6
Explanation:
Since, 3 women + 18 children complete work in 2 days. Therefore, (3 x 2) women + (18 x 2) children complete work in 1 days i.e, 6 women + 36 children complete work in 1 day.
Work of 36 children for 1 day = 1 - (1/3) = 2/3
[? work of 6 women for 1 day = 1/3]
? 36 children do 2/3 part of the work in 1 day.
36 children can do the work in 3/2 days.
9 children can do the the work in (3/2) x 4 = 6 days
Correct Answer: 56
Explanation:
Given, M1 = 120, M2 = 120 - n, D1 = 64, D2 = 60
W1= 2/3 and W2 = 1/3
According to the formula,
(M1D1) / W1 = (M2D2) / W2
? [120 x (64/2)] / 3 = [(120 - n) x 60] / (1/3)
? (120 x 64) / (2 x 60) = (120 - n)
? (120 - n) = 64
? n = 120 - 64 = 56
Now, 56 men can be discharged to finish the work in time.
Correct Answer: 162
Explanation:
Remaining work = 1 - (4/7) = 3/7
Remaining period = (92 - 66) = 26 days
Let the number of addition men = N
Given, M1 = 234, D1 = 66, T1 = 16, W1= 4/7,
M2 = (234 + x), D2 = 26, T2 = 18, W2 = 3/7
According to the question,
M1W2T1D1= M2W1T2D2
? 234 x (3/7) x 16 x 66 = (234 x N) x (4/7) x 18 x 26
? 234 + N = (3 x 66 x 16 x 234) / (4 x 26 x 18)
? 234 + N = 36 x 11 = 396
? N = 396 - 234 = 162
Additional men to be employed = 162
Correct Answer:
Explanation:
Let the work be finished in N days.
Then, A's N day's work + B's (N - 1) days + work + C's (N - 2) day's work = 1
? N/16 + (N - 1/32) + (N - 2/48) = 1
? N/1 + N - 1/2 + N - 2/3 = 16
? (6N + 3N - 3 + 2N - 4)/6 = 16
? 11N - 7 = 96
? 11N = 103
? N = 103/11 = 94/11 days
Correct Answer: 9 days
Explanation:
Let the work lasted for N days,
Then, Rashmi's 4 day's work + Ravina (N - 3) day's + Gitika's N day's work = 1
? (4/16) + (N - 3) / (124/5) + N/32 = 1
? 5(N - 3)/64 + N/32 = 1 - 1/4
? [5(N - 3) + 2N] / 64 = 3/4
? 7N - 15 = 48
? N= (48 + 15)/7 = 63/7 = 9 days
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