Aptitude - Time and Work - Discussion

Lets assume that A completes the work in 'x' days.

B completes the same amount of work in 'x/2' days because A takes twice the time than B.

C completes the same amount of work in 'x/3' days because A takes thrice the time than C.

Now the work completed by A in one day is '1/x'.

Work completed by B in one day is '2/x'
Work completed by C in one day is '3/x'

All three together can complete the work in '2' days as given in the problem. So the work done by all three together in one day is '1/2'.

Hence the equation: '(1/x + 2/x + 3/x) = 1/2'.

Evaluating the above equation gives the value 'x=12' which is the number of days taken by A to complete the work.

As we know that number of days taken by B complete the work is 'x/2', we substitute the 'x' value.

Hence the evaluation '12/2' where 12 is the value of 'x'.

HENCE THE TIME REQUIRED BY B TO COMPLETE THE WORK IS 6 DAYS.

As the question says,

The time taken by A is twice that of B and thrice that of C, so
This means that;
Efficiency,
A=2*B=3*C

Efficiency is inversely proportional to the time taken to complete a task.

Time ratio
A/B=2/1,
A/C=3/1

So, the Efficiency ratio

A/B=1/2,
A/C=1/3
=>A/B/C=1/2/3.
A:B:C=1:2:3.

So in terms of RATIO,

Let 1-day work of A,B, C be 1unit, 2unit, 3unit.
i.e. work for A: 1 Day= 1 U---->eqn 1
work for B: 1 Day= 2 U----> eqn 2
Work for C: 1 Day= 3 U----eqn 3
Work for A+B+C: 1 Day= 6 U.
2 Day=12U----eqn iv
[Total work(12U) completes in Time =2days(given)
i.e.Total Works time taken by A+B+C, Time =2days]
Comparing (eqn ii & iv) find B can do the work alone
Work for B: 1 Day = 2 U ----> eqn 2
work for B: 3x2 Day = 6x2 U.
6 DAYS = 12 U.

B alone do the work in time = 6 days.

Best Solution,

Let A take x days to complete the work, Therefore in one day he completes 1/x of the work
Thus B will complete the work in x/2 days, Therefore in one day B will complete twice of A's work, i.e 2*(1/x)................(i)

Similarly in one day C will complete thrice of A's work, i.e 3*(1/x)

Together they take 2 days to complete the work, Therefore in one day they complete 1/2 or 50% of the work,.

Thus (work done by A + work done by B + work done by C) in one day = 1/2 or 50%
1/x + 2/x + 3/x = 1/2,
(1+2+3)/x = 1/2,
6/x = 1/2,
x=12, i.e The days required by A to complete the work.
Therefore B will take = x/2 = 12/2 = 6 days......................from eq.(i).

My thoughts:

Let the time for A to do the job be X.
Let the time for B to do the job be 2X (since A takes twice as much time as B).
Let the time for B to do the job be 3X (since A takes thrice as much time as C).

So it takes all three X+2X+3X = 1/2 (days to finish the job).
This gives 6x=1/2. Therefore x=3.

Therefore the time taken for B to do the work is 2X.
Which is 2*3=6 which is equal to 6days.

It takes 3X that is 3*3 for C to do the work.
Which is equal 9 days.

With the above, you would find out that indeed A takes twice as much time as B and thrice as much time as C.

I hope this would suffice.

There are many methods to solve a problem. As I feel Subhas 's method was good, follow it. A can do 1 unit of work in 1 day.
B can do 2 unit of work in 1 day.
C can do 3 unit of work in 1 day.
Totally A+B+C's 1 day's work = 12 units of work.
To do 12 units of work B take 6 days.

See this solution:
Suppose A's 1 day's work= x
then B's 1 day's work = 2x,(Since its efficiency is twice of A)
C's 1 day's work will be 3x.(Since its efficiency is thrice of A)

Now, their combined 1 day's work will be
x+2x+3x

their combined 2 day's work will be
2(x+2x+3x)=1
gives x=1/12.

Since B's 1 day work is 2x =2 *1/12 = 1/6
Hence total days taken by B to complete the same work will be = 6 .

A takes twice much time them B i.e. 2A=B...(i)
A takes thrice much time thn C i.e. 3A=C...(ii)

It is given dat workin together they can finish in 2 days
so A+B+C=1/2
Now put B=2A AND C=3A in this equ...
A+B+C=1/2
A+2A+3A=1/2
6A=1/2
A=1/2*6=1/12
Now put the value of A in equ.(i)
.i.e B=2A
=2*1/12
=1/6
Therefore B takes 6 days to complete the work.

As per given:

2A = B => A = B/2......(1).

3A = C......(2).

=> 3*B/2 = C......(2) (from 1st A = B/2).

(A+B+C)'s do the work in 2 days.

So (A+B+C)'s 1 day work is = 1/2.

Put the value of A and C from 1st and 2nd.

= (B/2+B+3*B/2) 's 1 day work = 1/2.

=> 6B/2, 1 day work = 1/2.

=> B's 1 day work = 1/2*2/6 = 1/6.

So B complete work in 6 days.

Can anyone please explain me that how these x, x/2, x/3 become 1/x + 2/x + 3/x.

Here I don't get is we take 1 day's work of A as 1/x and after that it's taken 2/x + 3/x as B and C's 1 day work but actually after A's 1 day work B's & C's work should be taken as 1/x/2 and 1/x/3.

Which becomes x/2 and x/3 So finally it should look like 1/x + x/2 + x/3.

Suppose C can complete work in 18 days then A and B can complete the work in 54 and 27 days respectively.
Now A's 1 day work is 1/54 and B's 1 day work is 1/27 and C's 1/18.
Together 1/54+1/27+1/18=1/9.
Work finishes by together in 9 days.
Now when work complete in 9 days B's days are 27 so when work complete in 2 days B's days=?

27*2/9=6 days.