Aptitude - Time and Work - Discussion

A+B ----> (5/30),

B+C ----> (5/24),

C+A ----> (5/20).


Numerator (A+B+C) worked together for 10 days that means ' A ' Worked 5 days with ' B ' and 5 days with C . Similarly ' B ' Worked 5 days with ' A ' and 5 days with C. And Similarly ' C ' Worked 5 days with ' A ' and 5 days with B.

Now Denominator is their 1-day work :

(5/30) + (5/24) + (5/20) + A * (x) = 1.

Here ' A* (x) ' Means A Completed the Remaining Work (Given in Question) but We Don't Know How many days he Completed the work that is why I have taken (x). Here Represents the days ' A ' Has Worked to Complete that Job.

Now Equation Becomes :
(1/6) + (5/24) + (1/4) + (x * A) = 1.

(5/8) + (x * A) = 1 = > x * A = 1 - (5/8).

=> x * A = (3/8).

To Find ' A's ' 1 Day Work = Total Work Done By all of them - ( Eliminate 2 Variables to Get Value ) (i.e B and C )

So Equation Become,
A's 1 Day Work = (A + B + C)'s 1 day's work - (B + C)'s 1 day's work.
A's 1 Day Work = (1/16) - (1/24) = (1/48).

Now we Got A's 1 Day Work Substitute Above i.e in (x * A).

=> (x/ 48) = (3/8).

Cross Multiplying we get;

=> 8x = 48 * 3.
=> x = 18 days.

Hope You Understood.

Hi friends, let me explain one simple method to solve this problem.

It is lengthy process, but you will understand easily.

(a+b)'s 1 day work = 1/30.
(b+c)'s 1 day work = 1/24
(c+a)'s 1 day work = 1/20.

solving that equations we will get
a =1/48 , b = 7/240 , c = 1/80

now,
(a+b+c)'s one day work = 1/48 + 7/240 + 1/80.
= 15/240.
= 1/16.

i.e,
(a+b+c)'s 10 days work = 10/16
=5/8

remaining work = (1- 5/8)
= 3/8

3/8 portion of work is completed by a alone....

now,

a completes 1/48 portion of work in 1 day....
a completes 3/8 portion of work in _ days.....?

1/48------> 1
3/8-------> x

by cross multyplying,we will get

x /48 = 3/8

x = 18..

So, A alone can complete the work in 18 days.

A+B = 30 days.
B+C = 24 days.
C+A = 20 days.
--------------------

divide by taking lcm (30,24,20) = 120.
A+B = 4 units.
B+C = 5 units.
C+A = 6 units.
-------------------

2( A+B+C) = 15 units.
A+B+C =7.5 units.

They work together 10 days, means in 10 days they have completed 75 units out of 120 units. now,

Remaining units are (120-75) = 45.

A+B+C = 7.5 (put B+C= 5 as given),

A = 2.5.

So for completing 45 units, A need (45/2.5) = 18 days.

From the very first statement, we can calculate,

A's one day work as 1/60 i.e. (1/30) *(1/2). But they have done some confused calculations and saying that A's 1-day work is 1/48 this is not possible.

The ANSWER IS 22 + 1/2 days.

You all just check it by putting (10days A + B + C work ) + (22 + 1/2 days A's work alone) = 1. Then, you will get 1 which is the total work done

LCM of 30, 24, 20 is 120.

A+B B+C C+A A+B+C.

4 5 6 2 (a+b+c) = 15.

=> a+b+c = 15/2.

A+B+C do for 10 means they work together 10*15/2 = 75 units of work.

Work remains 120-75 = 45 unit.

A 's one day work = (A+B+C)-(B+C) = 15/2-5 = 5/2.

Hence A can finish the remaining work = 45*2/5 = 18.

GIVEN: A+B B+C A+C.
FORMULA: 2(A+B+C) = (A+B)(B+C)(A+C)
=2(1/8)
=1/16

GET A:
A+B+C=1/16, B+C=1/24.

(A+B+C) - (B+C) =1/16-1/24.

By solving we get A as 1/48.

10/16 + x/48=1 [ABC worked for 10days]
x=18.

@Tanvir,

A's one-day work = Total 1-day work of (A+B+C) - One day work of (B+C)

Total 1-day work of (A+B+C) is found by us while solving.
One day work of (B+C) is given in the question.

Therefore,
A's one-day work = (1/16) - (1/24).
A's one-day work = 1/48.

2(a+b+c) 1 days work =1/8 then a+b+c 1 day work=1/16 then a's 1 day work=1/16-1/24=1/48 then in 10 days a+b+c can do 10/16 work now remaining work=1-10/16=3/8 now in 48 days a can do 1 work then 3/8 work a can do in 48*3/8=18.

Nice work guys. I really didn't understand how we calculated A's one days work 1/48 but after reading discussions I got the answer, nice work by all of you.

A+B=1/30 AND B+C=1/24 AND C+A=1/20.

As per question all work together means : a+b+b+c+c+a=2a+2b+2c;

2 is common factor so we can write it as 2(a+b+c).