Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 1)
1.
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
Answer: Option
Explanation:
A's 1 day's work = | 1 | ; |
15 |
B's 1 day's work = | 1 | ; |
20 |
(A + B)'s 1 day's work = | ![]() |
1 | + | 1 | ![]() |
= | 7 | . |
15 | 20 | 60 |
(A + B)'s 4 day's work = | ![]() |
7 | x 4 | ![]() |
= | 7 | . |
60 | 15 |
Therefore, Remaining work = | ![]() |
1 - | 7 | ![]() |
= | 8 | . |
15 | 15 |
Discussion:
344 comments Page 6 of 35.
Deepa said:
1 decade ago
Can you explain how 7/60 came?
(2)
Yusuf said:
4 years ago
A+b 1 days work =7/60.
(2)
Sundar said:
1 decade ago
@Poorni:
The total work = 1;
Work done by A and B in 4 days = 7/15.
Therefore, the pending work = 1 - 7/15 = 0.533333333... (or) 8/15.
Hope this help you. Have a nice day!
The total work = 1;
Work done by A and B in 4 days = 7/15.
Therefore, the pending work = 1 - 7/15 = 0.533333333... (or) 8/15.
Hope this help you. Have a nice day!
(1)
Appu said:
1 decade ago
@Deepa
(1/50+1/20) = (20+15)/300 //This get by cross multiplication.
= 35/300
= 7/60.
I think your doubt is clear now.
(1/50+1/20) = (20+15)/300 //This get by cross multiplication.
= 35/300
= 7/60.
I think your doubt is clear now.
(1)
Beer ibrahim said:
1 decade ago
Why (1-7/15) is applied ?
(1)
Sundar said:
1 decade ago
@Beer Ibrahim
Q: Why (1-7/15) is applied ?
Ans: To calculate the work remaining. Work remaining = 8/15.
Note: From the above answer, we can say 15/8 days required to
complete the remaining work by A and B.
Hope this will help you. Have nice day!
Q: Why (1-7/15) is applied ?
Ans: To calculate the work remaining. Work remaining = 8/15.
Note: From the above answer, we can say 15/8 days required to
complete the remaining work by A and B.
Hope this will help you. Have nice day!
(1)
PRAKASH said:
1 decade ago
For A:-
15 days for 1 job
in 1 day 1/15 (part of the one job)
for B:-
20 days for 1 job
in 1 day 1/20 (part of the one job)
FOR BOTH IN ONE DAY:-
1/15+1/20=7/60
FOR BOTH IN 4 DAYS:-
(7/60)*4=7/15(THEY HAVE DONE)
THE REST PART OF THE JOB IS:-
1-(7/15)=8/15 (ANS)
15 days for 1 job
in 1 day 1/15 (part of the one job)
for B:-
20 days for 1 job
in 1 day 1/20 (part of the one job)
FOR BOTH IN ONE DAY:-
1/15+1/20=7/60
FOR BOTH IN 4 DAYS:-
(7/60)*4=7/15(THEY HAVE DONE)
THE REST PART OF THE JOB IS:-
1-(7/15)=8/15 (ANS)
(1)
Swetha said:
1 decade ago
We can solve this problem by another way also..let us see..
A can do a work in 15 days
B can do a work in 20 days
Take LCM for 15 & 20 i.e Total work = 60
Then,
A's capacity = 60/15 = 4
B's capacity = 60/20 = 3
They work together for 4 days,
Then, A's capacity + B's capacity = 4 + 3 =7
AB's one day capacity = 7
since they work for 4 days, they have done 4x7 =28 work
Work left = Total work - work done by AB
= 60 - 28 = 32
Remaining work / total work = 32 / 60 = 8 / 15
This method will take less time to compute guys...please try it.
A can do a work in 15 days
B can do a work in 20 days
Take LCM for 15 & 20 i.e Total work = 60
Then,
A's capacity = 60/15 = 4
B's capacity = 60/20 = 3
They work together for 4 days,
Then, A's capacity + B's capacity = 4 + 3 =7
AB's one day capacity = 7
since they work for 4 days, they have done 4x7 =28 work
Work left = Total work - work done by AB
= 60 - 28 = 32
Remaining work / total work = 32 / 60 = 8 / 15
This method will take less time to compute guys...please try it.
(1)
Haphyz said:
1 decade ago
When dealing with addition or subtraction of fractions you consider the denominators i.e
15&20 and then you find the L.C.M which is 60 or multiply them together.
15*20 = 300 so,
(1/15 + 1/20)/300 = (20 + 15)/300.
=25/300 reduce to the lowest term and you get 7/60.
This 7/60 is the amount of work A & B will complete in one day.
Hence for 4 days we have;
7/60 * 4 = 7/15 of the total work.
Since we don't know the real value of the total work we then assume total work to be done to be 1.
Therefore,
Remaining work left will be,
1- (work done)
1-7/15 ; the denominator here is 15 & 1 (since 1 =1/1).
Do the math and you get our final answer to be 8/15.
NOTE !
If you are subtracting a fraction from a whole number, just multiply the denominator by the whole number and then subtract from the numerator. The same rule applies for addition.
15&20 and then you find the L.C.M which is 60 or multiply them together.
15*20 = 300 so,
(1/15 + 1/20)/300 = (20 + 15)/300.
=25/300 reduce to the lowest term and you get 7/60.
This 7/60 is the amount of work A & B will complete in one day.
Hence for 4 days we have;
7/60 * 4 = 7/15 of the total work.
Since we don't know the real value of the total work we then assume total work to be done to be 1.
Therefore,
Remaining work left will be,
1- (work done)
1-7/15 ; the denominator here is 15 & 1 (since 1 =1/1).
Do the math and you get our final answer to be 8/15.
NOTE !
If you are subtracting a fraction from a whole number, just multiply the denominator by the whole number and then subtract from the numerator. The same rule applies for addition.
(1)
Palak said:
4 years ago
Thanks all.
(1)
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