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- Question
4 men can repair a road in 7 hours. How many men are required to repair the road in 2 hours ?
Options- A. 17 men
- B. 14 men
- C. 13 men
- D. 16 men
- Correct Answer
- 14 men
ExplanationM x T / W = Constant
where, M= Men (no. of men)
T= Time taken
W= Work load
So, here we apply
M1 x T1/ W1 = M2 x T2 / W2
Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?
Note that here, W1 = W2 = 1 road, ie. equal work load.
Clearly, substituting in the above equation we get, M2 = 14 men. - 1. When A, B and C are deployed for a task , A and B together do 70% of the work and B and C together do 50% of the work. who is most efficient?
Options- A. A
- B. B
- C. C
- D. can't be determined Discuss
- 2. 16 women can do a piece of work in 10 days while 15 men can do the same work in 12 days. All men and all women work alternately starting with all men, then find the time taken by them to complete the whole work?
Options- A. 10 5/2 days
- B. 11 days
- C. 11 2/5 days
- D. 12 1/2 days Discuss
- 3. 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:
Options- A. 1/24 days
- B. 7/24 days
- C. 24/7 days
- D. 4 days Discuss
- 4. There are three boats B1, B2 and B3 working together they carry 60 people in each trip. One day an early morning B1 carried 50 people in few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on an average 300 people cross the river using only one of the 3 boats B1, B2 and B3. How many trips it would take to B1, to carry 150 passengers alone?
Options- A. 15
- B. 30
- C. 25
- D. 10 Discuss
- 5. A does half as much work as Band C does half as much work as A and B together. If C alone can finish the work in 40 days, then together ,all will finish the work in ?
Options- A. 17 + 4/7 days
- B. 13 + 1/3 days
- C. 15 + 3/2 days
- D. 16 days Discuss
- 6. A and B together can do a piece of work in 40 days. A having worked for 20 days, B finishes the remaining work alone in 60 days. In How many days shall B finish the whole work alone?
Options- A. 60
- B. 70
- C. 80
- D. 90 Discuss
- 7. If 2 men and 3 women can do a piece of work in 8 days and 3 men and 2 women in 7 days. In how many days can the work be done by 5 men and 4 women working together?
Options- A. 3 days
- B. 6 days
- C. 4 days
- D. 2 days Discuss
- 8. 10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
Options- A. 215 days
- B. 225 days
- C. 235 days
- D. 240 days Discuss
- 9. A can finish a work in 18 days and B can do the same work in half the time taken by A. then, working together, what part of the same work they can finish in a day ?
Options- A. Total work
- B. One-fourth work
- C. Half work
- D. Two-third work Discuss
- 10. A can do a piece of work in 30 days. He works at it for 6 days and then B finishes it in 18 days. In what time can A and B together it ?
Options- A. 14 1/2 days
- B. 11 days
- C. 13 1/4 days
- D. 12 6/7 days Discuss
Time and Work problems
Search Results
Correct Answer: A
Explanation:
A + B= 70%
B + C =50%
=> B= 20% A= 50% and C=30%
Hence A is most efficient
Correct Answer: 10 5/2 days
Explanation:
Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively
15×12×m = 10×16×w
? m/w = 8/9
Total work = 15 × 12 × 9 = 1620 units
In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units
So, in 10 days work done will be = 263 × 5 = 1315 units
Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days
Total days = 10 5/2 days.
Correct Answer: 24/7 days
Explanation:
(A+B+C)'s 1 day's work = (1/24 + 1/6 + 1/12) = 7/24
so, A,B and C together will complete the work in 24/7 days.
Correct Answer: 15
Explanation:
Combined efficiency of all the three boats = 60 passenger/trip
Now, consider option(a)
15 trips and 150 passengers means efficiency of B1 = 10 passenger/trip
which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.
Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip
Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.
Correct Answer: 13 + 1/3 days
Explanation:
C alone can finish the work in 40 days.
As given C does half as much work as A and B together
=> (A + B) can do it in 20 days
(A + B)s 1 days wok = 1/20.
A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)
A's 1 day?s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]
B's 1 days work = (1/20) x (2/3) = 1/30
(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40
All the three together will finish it in 40/3 = 13 and 1/3 days.
Correct Answer: 80
Explanation:
Let A's 1 day's work=x and B's 1 day's work=y
Then x+y = 1/40 and 20x+60y=1
Solving these two equations , we get : x= 1/80 and y= 1/80
Therefore B's 1 day work = 1/80
Hence,B alone shall finish the whole work in 80 days
Correct Answer: 4 days
Explanation:
From the given data,
=> (2 M + 3W) 8 = (3M + 2W)7
=> 16M + 24W = 21M + 14 W
=> 10W = 5M
=> 2W = M
=> 14W × ? = 7W × 8
? = 4 days
Correct Answer: 225 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
Correct Answer: Half work
Explanation:
A can do the work = 18 days
B can do the work = 18/2 = 9 days
(A + B)'s 1 day work = 1/18 + 1/9 = 1/6
=> In 3 days = 3x1/6 = 1/2 work is completed.
Correct Answer: 12 6/7 days
Explanation:
Let 'B' alone can do the work in 'x' days
6/30 + 18/x = 1
=> x = 22.5
1/30 + 1/22.5 = 7/90
=> 90/7 = 12 6/7 days
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