Aptitude - Time and Work - Discussion

Good question. Reasoning is difficult for this one.

Hi friends,

(A+C-A)=C
(1/2-1/4)=1/4

so C's 1 hour work will be 1/4.

Similarly

(B+C-C)=B
(1/3-1/4)=1/12

Therefore B's 1 hour work is 1/12.

A's 1 hour work is 1/4
(A+C)'s 1 hour work is 1/2
(B+C)'s 1 hour work is 1/3
so, A+C=1/2 ->X and B+C=1/3 ->Y
subtract X-Y
we will get A-B=1/6
now, substitute A's value in d equation u vl get B's value as 1/12.
which is 1 day,s work of B...
hence B alone require 12 days...

I HOPE IT VL B HELPFULL...

A=1/4 (given)
(B+C)=1/3 (given)
(A+B+C)=(1/4+1/3) then take L.C.M
3+4/12=7/12
FIND FOR B
A+B+C=7/12
B=7/12-(A+C) given(A+C=1/2)
B=7/12-(1/2)=1/12
ANS:12

Hi.

A = 1/4....(1).
B+C = 1/3....(2).
A+C = 1/2....(3).

Now solve equation (1) & (3).

1/4+C = 1/2.
C = 1/4....(4).

Put C value in equation (2).

B+1/4 = 1/3.
B = 1/3-1/4.

B = 1/12 that's the answer.

A = 4 (ie A completes the work in 4 hours)
B + C in 3.
A + C in 2.

By taking LCM 4, 3, 2 = Total work or LCM is 12hrs work.

A 1-hour work is 3/ 3*4 = 12 (ie takes 4 hours to complete total work 12 hours already given in question follow same with B + C and A + C).

B + C 1 hour work is 4/4*3 = 12.
A + C 1 hour work is 6/6*2 = 12.

A = 3
B + C = 4
A + C = 6

3A + 3C
1B + 3C

A = 3
B = 1
C = 3
So, B takes 1hr x12 = 12 hour to complete the work.

a = 1/4;
b+c = 1/3;
a+c = 1/2.

Therefore find the c value first:- a+c = 1/2.

1/4+c = 1/2.
c = 1/4.

Now substitute in b+c = 1/3.

b+1/4 = 1/3.
b = 1/12 hence it takes 12 hours.

A's 1 day work is 1/4.

A and B 1 day work is 1\3.

So 1/(A+B)-1\A = 1\B.

1\3 - 1\4 = 1\12.

(A + B) - B = B + C (A and B, C works equal).

1/10 - B = B + C.

1/2B = 1/10 - 1/50.

B = 25.

Following method focuses on avoiding fractions as much as possible, because calculating with fractions takes longer amount of time than calculating with integers.

In most problems on work-time-rate, fractions can be avoided altogether, or at least the role of fractions can be minimized.

Solution:

Assume total work = LCM (4, 3, 2) units = 12 units.

A's time = 4 hr.

==> A's rate = 12 units/4 hr = 3 units/hr.....(1).

(B+C)'s time = 3 hr.

==> B's rate + C's rate = 12 units/3 hr = 4 units/hr.....(2).

(A+C)'s time = 2 hr.

==> A's rate + C's rate = 12 units/2 hr = 6 units/hr.....(3).

From (1) and (3),

C's rate = 6-3 = 3 units/hr.....(4).

From (2) and (4),

B's rate = 4-3 = 1 unit/hr.

Hence B's time = Total work/B's rate = 12/1 = 12 hours.

Note: If you want to avoid spending time calculating the LCM, you can do so by saying that total amount of work = 4x3x2 = 24 units. Further calculations will change accordingly, but final answer will still be 12 hours.

The trick is "Don't define the total work to be simply 1 unit". Whenever the total work is defined to be 1 unit, fractions are introduced in the solution almost inevitably. Instead, we keep the total amount of work problem specific, and choose the total number of units to be the multiplication (or LCM) of suitable numbers.

While there is no guarantee that the fractions will be avoided altogether, they will at least be reduced to a large extent. In many cases, even some complicated ratios can be avoided by this method.

I will try to post as many solutions as I can using this method.