Aptitude - Time and Work - Discussion

@Ankush

Here is answer for your question.

Take LCM of 36, 42 & 48 i.e. 1008
Now Rehman and Sandeep star work and worked for 6 days.
Now per day work of Rehman = 1008/36 = 28
And per day work of Sandeep = 1008/42=24
Together they have completed per day work = 28+24=52.
Now in 6 days, they completed 52*6 = 312 work out of 1008.
Now last 10 days again Rehman and Sandeep worked so the work completed in 10 days = 52*10=520.
So total work completed by Rehman and Sandeep = 312+520=832.
Remaining work = 1008-832= 176 which is completed by Abbas alone and Abbas per day work is 1008/48= 21.
So Abbas worked for 176/21 Days. Or 8 8/21 days.

Hi Guys,

A can complete work in 18 days
A's 1 day work = 1/18--------------(1)

B can complete work in 15 Days
B's 1 day work = 1/15--------------(2)
[but B is working 10 days only]
Remaining work is 5 days
so, remaining 1 day work is 5*(1/15) =>1/3

A's 1/3 work can complete in N days
so A's 1 day work = 1/3N--------------(3)

Equation=> (2)=(3)

1/3N = 1/18
so, N = 18/3
N = 6 Days

A's (1/3) remaining work can complete in 6 Days

N = 6 Days

It will be easy if we use chocolate method.

A takes 18 days and B takes 15 days. LCM of 18 and 15 is 90.
Let 90 be the total number of chocolates.

A takes 18 days to eat 90 chocolates. so A eats 5 chocolates per day.
B takes 15 days to eat 90 chocolates. So B eats 6 chocolates per day.

in 10 days B finishes 6*10 = 60 chocolates.
Remaining chocolates are 30.

So to eat 30 chocolates A need 30/5 = 6 days.
Ans = 6 days.

Rahman Sandeep and Abass can complete a work in 36, 42 and 48 days respectively. Rahman and Sandeep start the work after 6 days they take a leave for few days and handed the work to Abaas. Rahman and Sandeep leave ends 10 days before the work complete and Abass leaves the work on the return of Rahman and Sandeep. Find for how many days Abaas work?

Can anyone solve this time and work problem?

TAKE LCM OF 18 and 15.

LCM = 90
TOTAL WORK=90 UNITS,
A' s one day work=90/18 = 5 units,
B's one day work=90/15 = 6 units,
B's 10 days work=10 days * 6 units = 60 units.
now remaining work= total units-work done by B in 10 days.
i.e, = 90-60= 30 unit of work remaining which is done by A.
A's one-day work is 5 units.
So he completes the remaining 30 units in 6 days(30/5=6 days).

Let us suppose Total work to be W = 90 units.
(LCM of 15 &18)
According to that,
A's rate of doing work =(90/18)=5 units per day
B's rate of doing work =(90/15)=6 units per day
B worked for 10 days so work done per 10 days would be = 6*10=60 units
Remaining work= 90-60= 30 units
A completes 5 units per day so it will take 6 days to complete the remaining work.

Two men can do a piece of work in 6 hours and 4 hours. After the first has worked for 2 hours he is joined by another man. By when should the work be completed?

Can anyone please solve this question. I am facing a lot of problem in this question. This is from composite mathematics. The answer at the back is 1 hour 36 minutes and I am getting 1 hour 33 Minutes.

Hi @Kusum...........

We have to find how many days have to work to A for finishing the the remaining work. remaining work is 1/3
so we know the processes
. .
. A finishes 1/18 work in 1 day

.
. . ,, ,, 1 ,, ,, 1/(1/18) day

.
. . ,, ,, 1/3 ,, ,,1*(18/1)*(1/3)

After solving you well get 6days answer.
Understand???????????

A can do it in 18 days.
B can do it in 15 days.

Take LCM of those 18 and 15 I.e., 90.

Consider total work as 90,
ThenA's efficiency = total work/time taken by A.
=>90/18=5.
Similarly for B=>90/15=6.
So B's 10days work is 10*6=60.
Remaining work is 90-60=30.

Time taken by A to complete a work is =remaining work/efficiency of A.
=>30/5 = 6days.

Answer explained

A's one day work =1/18
B's one day work =1/15

So B worked for 10 days to do a work 1
So B's 10 day work =(1/15)*10 = 10/15 =2/3

B worked only 10 days to complete a work (i.e)1
Therefore remaining work is 1-2/3 =1/3

Now the remaining work is to be completed by A
A's 1 day work is 1/18
To complete 1/3 of work A takes= 1/18*3 = 1/6
Ans 6