Albert, Bane and Christy working independently can paint a room in

The time taken by A, B and C to complete the task is given as follows:
A - 6 hours
B - 10 hours
C - 12 hours

We can let the Total Work be the LCM of 6, 10, 12 ie 60 and calculate the efficiency or unit/time for A, B, C

A=60/10=6
B=60/6=10
C-60/12=5

10 AM -11 AM,
All three of them worked, therefore total work done in that period = 6+10+5=21 units
Work Left = 60-21=39 units

11 AM - 12 PM,
A & C worked, therefore work done = 10+5 = 15 units
Work Left = 24 units

12 PM - 1 PM
A & B worked, therefore work completed = 10 + 6 = 16 units
Work left = 8 units

Now comes the tricky part, since only 8 units are left (< Total work completed by them in one hour - 21 units), we divide the remaining units in the ratio of their works. Therefore 8 units will be distributed among A,B and C in the ratio 10:6:5.

Now, for the question, fraction of work done by A = (Work done from 10 - 1 PM + Fraction of work done from 1 - 2 PM)/Total Work = (10+10+10+(8*10/21))/60 = 71/126


Ans : B

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1 = 1/6 + 1/10(t-1) + 1/12(t-1)
1 = 21/60t - 11/60
71/60 = 21/60t
t=71/21

Contribution by Albert = 1/6t = 1/6(71/21) = 71/126
Contribution by Bane = 1/10(t-1) = 1/10(71/21 - 21/21) = 1/10(50/21) = 50/210
Contribution by Christy = 1/12(t-1) = 1/12(71/21 - 21/21) = 1/12(50/21) = 50/252

Is there a mistake here? A has worked for 6 hours, and the LCM of all 3 is 60. So units worked by A shoubd be = 6/60 = 10

A=60/10=6
B=60/6=10
C-60/12=5

Hello.. why have you multiplied the equation by 60?

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