Latest Elitmus Aptitude Question SOLUTION: A,B and C can do a piece of work in 12,18, and 24 days respectively. they work at it together . A stops the work after 4 days and B is called off 2 days before the work is done.In

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(#M40033026) ELITMUS Numerical Ability question Time and Work Keep an EYE

A,B and C can do a piece of work in 12,18, and 24 days respectively. they work at it together . A stops the work after 4 days and B is called off 2 days before the work is done.In what time was the work finished?

Asked In Elitmus Ankit Boae (10 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Snehasis  (10 years ago)

If A can do a piece of work in 12 days then in one day he can do 1/12 work.

Similarly, B's one day work is 1/18

and C's one day work is 1/24.

Here it tells about the no of days A and B worked but its not telling anything about C.

So, lets assume C work for x days.

then, B work for x-2 days(as it given that he left the job 2 days before completion)

and, A work for 4 days(as given in the question)

Therefore, A's work + B's work + C's Work = 1

4/12 + x-2/18 + x/24 = 1

After solving,

x = 8 days.

Therefore, we assumed that C worked for x days means C is the only one who worked till the last day. Therefore x=8 is the required answer.

Therefore, the total work was finished in 8 days.

Is this solution Helpfull? Yes (62) | No (3)  

AKASH KUMAR  (10 years ago)

percentage of work done by A=(1/12)*100=8.33%
similarly B=5.55%
C=4.16%
WORK DONE IN 4 DAYS=4(8.33+5.55+4.16)=72.16% (APPROX)
WORK DONE BY C ALONE IN LAST 2 DAYS=2*4.16=8.32%
WORK LEFT TO BE DONE=100-72.16-8.32=19.52%
THIS IS TO BE DONE BY BOTH B AND C SO NO OF DAYS REQUIRED=
19.52/(5.55+4.16)=2 DAYS(APPROX)
HENCE TOTAL DAYS= 4(DONE BY ALL)+2(DONE BY B AND C)+2(C ALONE)=8DAYS

Is this solution Helpfull? Yes (12) | No (1)  

Sagar Laxmikantrao Gomte  (9 years ago)

ans:-8 days

A's work - 1/12
B's work - 1/18
C's work - 1/24

all of them worked at it for 4 days so total work:-
4(1/12+1/18+1/24)=26/36

now,
remaining work is 1-(26/36)=5/18

Now suppose that both B and C take 'x' days to complete a work of 5/18
(It is given that B is called off 2 days before the work is done)

thus,
(x-2)/18+(x/24)=5/18
solving we get x=4
So,
(B and C worked at it for 2 days followed by 2 days work of C alone)
to these 4 days we add previous 4 days of working of all of them together...

Thus
4 days (A+B+C) + 4 days((2 days B+C) + 2 days of C alone) = 8 days..

8 days

Is this solution Helpfull? Yes (10) | No   

ashish bharti  (10 years ago)

A work 4/12 in 4 days=1/3
so left work =2/3.
now let work is completed in x days then,
(x-2)/18+x/24=2/3.
x=8.

Is this solution Helpfull? Yes (7) | No   

raaj  (10 years ago)

8 days(4+2+2)

Is this solution Helpfull? Yes (3) | No (4)  

prachur srivastava  (9 years ago)

A---12
B---18
C---24
LCM of these 3 is 72
units of work done by A----6
units of work done by A----4
units of work done by A----3

NOW, 6+4+3=13

they all worked together 4 days in the starting so 13*4=52

so work remaining after 4 days=72-52=20
means 20units of work is remaining

now B left before 2 days so total unit of work remaing will become (4*2)+20=28

so no. of days=28/(4+3)=4 days

and the days they worked together=4 days

so total days=4+4=8 days

Is this solution Helpfull? Yes (2) | No   

Chirag Ranjan Lahiri  (9 years ago)

4 days (A+B+C) + 4 days((2 days B+C) + 2 days of C alone) = 8 days.

Is this solution Helpfull? Yes (1) | No   

Sudhir Kumar  (2 years ago)

8 days
Total work-72
A-12(6)
B-18(4)
C-(3)
A+B+C=(6+4+3) *4=52(for 4 days)
Then remaining work 72-52=20
B left the work 2 days before done
That means C works alone 2 days
2*3=6
Then remain work=20-6=14
14 work done by B+C=14/2=7
That is total days =4+2+2=8

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