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 5 of 35.
Kamal said:
1 decade ago
I didn't undestand last step.
Akanksha said:
1 decade ago
(1/50+1/20) = (20+15)/300 //This get by cross multiplication.
= 35/300
= 7/60.
How did 300 came?
= 35/300
= 7/60.
How did 300 came?
Neha garg said:
1 decade ago
Let total work is 1
1/15+1/20=x/4
7/60*4=x
that means x=7/15
1-7/15=8/15
1/15+1/20=x/4
7/60*4=x
that means x=7/15
1-7/15=8/15
Fouzia said:
1 decade ago
Fouzia.
How that 7 came I can't got it?
How that 7 came I can't got it?
SRI HARSHA.P said:
1 decade ago
The way said by swetha is very long process.
that wastes time in the exam hall.
thats not the good way to do the problem.
always prepare for ashort method.
Short method takes time for the first time by practice it gives u TIME MANAGEMENT. So better to practice SM.
that wastes time in the exam hall.
thats not the good way to do the problem.
always prepare for ashort method.
Short method takes time for the first time by practice it gives u TIME MANAGEMENT. So better to practice SM.
Ramkumar said:
1 decade ago
Nice work swetha.
Swetha said:
1 decade ago
@Sri harsha.p
It is not very long process..it is short method only.i explained every steps.,so only it looks like long process.
It is not very long process..it is short method only.i explained every steps.,so only it looks like long process.
Rahul said:
1 decade ago
Ok.
I guess the easier way to explain this is to subtract(20days-14days)= 5 days,
We have calculated for 4 days, therfore remaining is 1;)
I guess the easier way to explain this is to subtract(20days-14days)= 5 days,
We have calculated for 4 days, therfore remaining is 1;)
James said:
1 decade ago
We can solve this problem more easily by taking efficiency.
Since work=efficiency *time
To take efficiency take lcm of A&B.it is 60
60/15=4 60/20=3
A'efficiency=4 & B's=3
Work=4*15 or 3*20 =60
Work=60
Since a b works together for 4 days
a+b effi=4+3=7
7*4=28
Total work-work done
60-28=32(work remaining)
32/60=8/15.
Since work=efficiency *time
To take efficiency take lcm of A&B.it is 60
60/15=4 60/20=3
A'efficiency=4 & B's=3
Work=4*15 or 3*20 =60
Work=60
Since a b works together for 4 days
a+b effi=4+3=7
7*4=28
Total work-work done
60-28=32(work remaining)
32/60=8/15.
Ghhhhgh said:
1 decade ago
Here it is given that A can do a piece of work in 15 days.
So now we can find the A's 1 day work whicvh is =1/15.
So now we can find the A's 1 day work whicvh is =1/15.
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