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

Question

Albert, Bane and Christy working independently can paint a room in 6, 10 and 12 hours, respectively. They start working together at 10:00 am and work till the job is completed. If Bane took a break from 11 am to 12 noon, Christy took a break 12 noon to 1 pm, what fraction of work was done by Albert?

A. 1/2
B. 71/126
C. 72/125
D. 80/21
E. 81/125

A. 1/2
B. 71/126
C. 72/125
D. 80/21
E. 81/125

nick1816 · Accepted Answer

Assume total work= LCM(6,10,12) = 60

A can do 10 units of work per hour.
B can do 6 units of work per hour.
C can do 5 units of work per hour.

Assume they finish the total work in 't' hours

10t+6(t-1)+5(t-1)=60
21t=71

t= 71/21

Amount of work done by A=

fraction of work done by A=  /60 =71/ 126

ScottTargetTestPrep · Answer

We see that the work rates of Albert, Bane, and Christy are 1/6, 1/10, and 1/12, respectively. If we let x = the number of hours Albert works till the job is completed, then x - 1 = the number of hours Bane and Christy each works (since each took a 1-hour break). We can create the equation:

x/6 + (x - 1)/10 + (x - 1)/12 = 1

Multiplying the equation by 60, we have:

10x + 6x - 6 + 5x - 5 = 60

21x = 71

x = 71/21

Albert worked for 71/21 hours. Since he can paint an entire room in 6 hours by himself, then the fraction of the entire job done by Albert is (71/21)/6 = 71/126.

Answer: B

fireagablast · Answer

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

crackGMAT760 · Answer

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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AbhaGanu · Answer

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

AbhaGanu · Answer

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

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

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