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=> Time and work : In this type of questions , a specified number of person work to complete a task in certain days . examiner asks to find out if the same work can be completed by working for more time or employing more workers .
=> Important formula :
(S X T) / W = (sXt)/w
where S is Strength(number of workers)
T is Time (duration)
W is Work.
=> S X T = W
so, (SXT)/W = 1
=> If A can complete a piece of work in ' M ' days and B can complete a piece of work in 'N' days, then A and B together can complete the work in = (M X N) / (M+N)
=> If A can complete a piece of work in 10 days then he can complete 1/10 work in single day.
Similarly, if A does 1/10th work in a single day then he can complete the total work in 10 days.
Using above simple formula you can solve most of the problems.
See the below examples along with explanation on how to use above formula in different situations
Questions on time and work
1. 45 men can do a piece of work in 15 days. How many days are needed for 30 men to do the same work ?
Answer: (SXT)/w = (sXt)/w
(45X15)/w=(30Xt)/w
→ here work is same in both cases. so,
=> (45X15)/w=(30Xt)/w
=>675=30Xt
=>t=675/30
=>t=22.5 days
2. 20 men can do a piece of work in 15 days working for 8 hours a day. In how many days will 20 men can complete the work if they work for 10 hours a day ?
Answer:
(MXD XT)/w = (mXdXt)/w
(20X15X8)= 20X10Xd
=2400=200Xd
=12 days
3. A can do a peace of work in 20 days days and B can do it 30 days. In how many days can they finish it if they work together?
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Answer:
Short cut method : (20X30) / 50 = 12 .
12 days
Routine method:
A’s one day work = 1/20
B’s one day work = 1 / 30
→ A and B ‘s work together in 1 day = (1/20) + (1/ 30) =
=> Important formula :
(S X T) / W = (sXt)/w
where S is Strength(number of workers)
T is Time (duration)
W is Work.
=> S X T = W
so, (SXT)/W = 1
=> If A can complete a piece of work in ' M ' days and B can complete a piece of work in 'N' days, then A and B together can complete the work in = (M X N) / (M+N)
=> If A can complete a piece of work in 10 days then he can complete 1/10 work in single day.
Similarly, if A does 1/10th work in a single day then he can complete the total work in 10 days.
Using above simple formula you can solve most of the problems.
See the below examples along with explanation on how to use above formula in different situations
Questions on time and work
1. 45 men can do a piece of work in 15 days. How many days are needed for 30 men to do the same work ?
Answer: (SXT)/w = (sXt)/w(45X15)/w=(30Xt)/w
→ here work is same in both cases. so,
=> (45X15)/w=(30Xt)/w
=>675=30Xt
=>t=675/30
=>t=22.5 days
2. 20 men can do a piece of work in 15 days working for 8 hours a day. In how many days will 20 men can complete the work if they work for 10 hours a day ?
Answer:
(MXD XT)/w = (mXdXt)/w
(20X15X8)= 20X10Xd
=2400=200Xd
=12 days
3. A can do a peace of work in 20 days days and B can do it 30 days. In how many days can they finish it if they work together?
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Answer:
Short cut method : (20X30) / 50 = 12 .
12 days
Routine method:
A’s one day work = 1/20
B’s one day work = 1 / 30
→ A and B ‘s work together in 1 day = (1/20) + (1/ 30) =
→ (3+2)/60
→5/60 = 1/12
→ 1 day they together complete 1/12 th part
→ So in 12 days they complete total work .
4. India wants to build a wall along china border. 15 men build a wall of 50 meters long in 25 days working for 8 hours a day. How many days can 60 men does it to take build a similar wall of 250 meters working for 10 hours a day?
Answer:
→5/60 = 1/12
→ 1 day they together complete 1/12 th part
→ So in 12 days they complete total work .
4. India wants to build a wall along china border. 15 men build a wall of 50 meters long in 25 days working for 8 hours a day. How many days can 60 men does it to take build a similar wall of 250 meters working for 10 hours a day?
Answer:
(DXH XT)/w = (dXhXt)/w
Here work(w) is building a wall of different length
(15 X 25X 8)/ 50 = (60XdX10)/250
=>3000/50 = dX600/250
=>60=dX2.4
=>d=25 days.
5. A toy maker employed 15 laborers to make 300 toys in 5 days. After 3 days he found that only 200 days were made. How many additional men he has to employ to finish the work in target time ?
Answer:
(15 X 4) / 200 = ((15+X) X 2)/ 300
→ (60 / 200) X 300 = 30+2X
→90-30= 2X
→ 60 =2X
→X = 30.
→So. 30 additional men are to be employed to finish the work in target time .
Here work(w) is building a wall of different length
(15 X 25X 8)/ 50 = (60XdX10)/250
=>3000/50 = dX600/250
=>60=dX2.4
=>d=25 days.
5. A toy maker employed 15 laborers to make 300 toys in 5 days. After 3 days he found that only 200 days were made. How many additional men he has to employ to finish the work in target time ?
Answer:
(15 X 4) / 200 = ((15+X) X 2)/ 300
→ (60 / 200) X 300 = 30+2X
→90-30= 2X
→ 60 =2X
→X = 30.
→So. 30 additional men are to be employed to finish the work in target time .
6. A and B can compete a work in 12 days .B and C in 15 days and C and A in 20 days. In how many how many they can complete the given task if they all work together ?
Answer:
A+B= 14
B+C=17
C+A=20
So as you know “if A can do the given work in n days, then he can do 1/n th work in one day”
Same formula applies here also
A+B ‘s 1 day work = 1/(A+B) = 1/12
B+C’s 1 day work = 1/ (B+C) = 1/15
C+A’s 1 day work = 1/ (C+A) = 1/20
So together there 1 day work = 1/(A+B+B+C+C+A)= (1/12)+(1/15)+1/(20)
= ½(A+B+C)= ½(48/240)
=1/2(1/5)
=1/10
So all together can do 1/10thwork in work in day.
=>Remember the formula
“If A can do 1/nth work in one day , he can finish it in n days”
So they can complete the given work in 10 days.
7. A and B can do complete a task in 10 days. B and C can do the same job in 15 days and C and A in 20 days. In how many days can A alone will be able to finish the given task ?
Answer:
As you know A+B ‘s 1 day work = 1/10
B+C 1 day work= 1/15
C+A 1 day work= 1/20
We need only A work , so let’s subtract the pair which is not having A from pairs having A
→ (A+B)-(B+C)+(C+A)
= 2A
So 2A= (1/10)-(1/ 15)+(1/20)
= 30-20+15/300
=25/300
=1/14.
So A 1 day work = 1/14 then
Remember the formula
“If A can do 1/n work in one day , he can finish the work in n days”
Therefore, A can finish the complete work in 14 days
8. X and Y can finish the important project in 20 days .X alone can finish it in 30 days . In how many days can Y alone finish project ?
Answer:
A+B one day work= 1/20
A one day work= 1/30
So B one day work = (A+B)-A = (1/20)- (1/30)
=3-2/60
=1/60
So A can do 1/60th work in one day
A alone can complete the project the project in 60 days
9. Raju can finish a job in 18 days and Kedar can finish the same job in 9 days. If they work together how much portion of work can they complete in a single day?
Answer:
Together they finish the work in
Formula = (mXn) /(m+n)
= (18X9) / (18+9)
=162/27
=6 days
So, Raju and Kedar can finish the job in 6 days. Which means they compete 1/6th of work in a single day.
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