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 12 of 35.
PriyA said:
1 decade ago
But still I can't understand the logic behind the subtraction of 1. I have read @Mangesh answer. But why do we assume "one"? They did't mention anything right?
Virat said:
5 years ago
A=15 Days.
B=20 Days.
LCM 60.
A's 1 Day work=4.
B's 1 Day work=3.
Both together 1 Day work= 7/60.
Both together 4 Days work= 28/60.
Rest work= 32/60 = 8/15.
B=20 Days.
LCM 60.
A's 1 Day work=4.
B's 1 Day work=3.
Both together 1 Day work= 7/60.
Both together 4 Days work= 28/60.
Rest work= 32/60 = 8/15.
Meena said:
9 years ago
Fraction of work done in 4 days = 7/15.
(It means that out of 15 parts, 7 parts has been done. So 8 parts left)
Therefore, the fraction of work left = 8/15.
(It means that out of 15 parts, 7 parts has been done. So 8 parts left)
Therefore, the fraction of work left = 8/15.
Prabhu said:
8 years ago
Can anyone solve this?
Problem: 'A' can do a piece of in 6days, 'B' in 9days and 'C' in 12 days. In how many days will all of the together finish the work?
Problem: 'A' can do a piece of in 6days, 'B' in 9days and 'C' in 12 days. In how many days will all of the together finish the work?
VisHnu RaM said:
3 years ago
Thank you for giving the clear explanation @Swetha.
And 60 will come, No of days LCM
15 days to 15 * 4 = 60.
20 days to 20 * 3 = 60.
So, the LCM is 60.
And 60 will come, No of days LCM
15 days to 15 * 4 = 60.
20 days to 20 * 3 = 60.
So, the LCM is 60.
(13)
Radhika said:
3 years ago
@All.
As 1 is the whole work, so when we subtract 7/15 from 1.
Then,
LCM of 1 and 15 will be 15 ( in 1 the denominator will be 1 ).
So, 15-7/15 is 8/15.
As 1 is the whole work, so when we subtract 7/15 from 1.
Then,
LCM of 1 and 15 will be 15 ( in 1 the denominator will be 1 ).
So, 15-7/15 is 8/15.
(18)
Sai Kumar reddy said:
7 years ago
The common Multiple for 15 and 20 is 5.
A's one day work =15*5=75 units,
B's one day work =20/5=4 units,
Both working together for 4 days=7*4=28 units.
A's one day work =15*5=75 units,
B's one day work =20/5=4 units,
Both working together for 4 days=7*4=28 units.
Niruban said:
1 decade ago
A does the work for a day and B does it for a same day. If we add both then it should be work done for 2 days but why is it considered as one day work?
K.Nirdesh said:
1 decade ago
1/50+1/20) = (20+15)/300 //This get by cross multiplication.
= 35/300
= 7/60.
why we should Division do for (20+15)/300, how did the 300 is come?
= 35/300
= 7/60.
why we should Division do for (20+15)/300, how did the 300 is come?
Laxman said:
7 years ago
A's 1 day's work = 1/15.
B's 1 day's work = 1/20.
(A+B)'s 1 day's work= (1/15+1/20),
L.C.M of 15,20 is 60 then,
15*4=60, 20*3=60,
4 + 3/60=7/60.
B's 1 day's work = 1/20.
(A+B)'s 1 day's work= (1/15+1/20),
L.C.M of 15,20 is 60 then,
15*4=60, 20*3=60,
4 + 3/60=7/60.
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