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Manager  Joined: 20 Mar 2015
Posts: 54
A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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7 00:00

Difficulty:   45% (medium)

Question Stats: 71% (02:32) correct 29% (02:34) wrong based on 172 sessions

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A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?

A. 11.30 am
B. 11.40 am.
C. 12.00 pm
D.11.10 am
E. 11.20am

Let's take the LCM of 4,6, and 12. LCM = 36
We know,
R* T = W
(For A) R * 4 = 36 R =9
(For B) R * 6 = 36 R =6
(For C) R * 12 = 36 R =3
A+B+C combined rate = 18
Now,
A completes 9 hrs of work in 1 hour. A and B combined completes 9+6 hrs of work in 1 hour. Total work completed is 24. Remaining work is 36-24 = 12.
Rate of A+B+C = 18.
Therefore,
R * T = W ==> 18 * t = 12 ==> t = 12/18 = 2/3hrs [2/3 * 60 = 40 min]
40 min + 1+1 hour = 2 hour 40 min.

therefore, answer is b.

Is this a sub-600 level question?
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A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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bimalr9 wrote:
A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?

A. 11.30 am
B. 11.40 am.
C. 12.00 pm
D.11.10 am
E. 11.20am

Let the total work be 12 units

Efficiency of A = 3 units/ hour
Efficiency of B = 2 units/ hour
Efficiency of C = 1 units/ hour

Quote:
A starts alone and works for 1 hr.............If they start working at 9 am

So, A completes 3 units of work by 10 am

Quote:
He is then joined by B and they work together for 1 hr.

So, A and B completes 5 units of work by 11 am

Quote:
Finally C joins them and they all work together to finish the work.

Work left is 4 units ( 12 - 3 - 5 ) and combined efficiency of A, B and C is 6 units

So, time required to complete 4 units will be 4/6 => 4/6*60 = 40 minutes...

So they will require 40 minutes after 11 am to complete the remaining work...

Hence answer will be (B) 11:40 am ! This problem doesn't require the use of the formula R* T = WSo do not involve in the concept of RTW unnecesarily

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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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A's 1 hr work= 1/4
B's 1 hr work= 1/6
C's 1hr work= 1/12

A+B+C's 1 hr work = 1/4 +1/6+1/12= 1/2

Now A worked alone for 1 hr. That means he did 1/4th of the work.
A and B worked together for 1 hr. So, they both did 1/4+1/6= 10/24 of the total work.

After 2 hrs the total work done was= 1/4 +10/24= 16/24

And, the remaining work was= 8/24= 1/3

All three of them did 1/3 work together.
If they do 1/2 work in 1 hr, they will do 1/3 work in 2/3*60= 40 mins

Total time taken to finish the job
1hr +1hr+40 mins= 2hr 40 mins

They will finish the job at 11:40 am.

IMHO, this is not a sub 600 level question.
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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bimalr9 wrote:
A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?

A. 11.30 am
B. 11.40 am.
C. 12.00 pm
D.11.10 am
E. 11.20am

Let's take the LCM of 4,6, and 12. LCM = 36
We know,
R* T = W
(For A) R * 4 = 36 R =9
(For B) R * 6 = 36 R =6
(For C) R * 12 = 36 R =3
A+B+C combined rate = 18
Now,
A completes 9 hrs of work in 1 hour. A and B combined completes 9+6 hrs of work in 1 hour. Total work completed is 24. Remaining work is 36-24 = 12.
Rate of A+B+C = 18.
Therefore,
R * T = W ==> 18 * t = 12 ==> t = 12/18 = 2/3hrs [2/3 * 60 = 40 min]
40 min + 1+1 hour = 2 hour 40 min.

therefore, answer is b.

Is this a sub-600 level question?

Say they work for t fraction of an hour after 11.

(Rate*Time) for each = (2 + t)/4 + (1 + t)/6 + t/12 = 1 = Total Work

6t/12 = 1/3

t = 2/3 hrs = 40 mins

Time at which the work gets done = 11:40
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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From the question we know :

Ra x 4 = 1 => Ra = 1/4
Rb x 6 = 1 => Rb = 1/6
Rc x 12 = 1 => Rc = 1/12

stage 1 :
Ra x 1hour = 1/4 x 1hour = 1/4 = 6/24

stage 2 :
(Ra + Rb) x 1hour = (1/4 + 1/6) x 1hour = 10/24

stage 3 :
(Ra + Rb + Rc ) x Time = remaining work

remaining work = 1 - (6/24 + 10/24) = 8/24

So, (1/4 + 1/6 + 1/12) x Time = 8/24
12/24 x Time = 8/24
Time = 2/3 = 40 min

Finally,
Stage 1 + stage 2 + stage 3 = 1h + 1h + 40min = 2h40min
9h + 2h40 = 11h40, answer choice B
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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I like putting things in a table , that helps to proceed without making a mistake.

A , B and C can do 1/4, 1/6 and 1/12 each individually and respectively.

so
Time___Person involved_____Word done_________Remaining
9:00_______A________________0________________1
10:00____A (B just joins)_______1/4_______________3/4
11:00__A,B,(C just joins)__1/4 + (1/4 +1/6) = 2/3__1 - (1/4 + (1/4 +1/6)) = 1/3

So 1/3 is done by A+B+C, who can do 1/4+1/6+1/12 = 1/2 in an hour, so would need, 2/3 hour for remaining, which is 2/3* 60min = 40 mins.
Therefore 11:40.
(Excuse me for my formatting.)
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A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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let t=total time
1/4+5/12+(t-2)(1/2)=1
t=2 2/3 hrs=2 hrs 40 min
9 am+2 hrs 40 min=11:40 am
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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bimalr9 wrote:
A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?

A. 11.30 am
B. 11.40 am.
C. 12.00 pm
D.11.10 am
E. 11.20am

We are given that A, B, and C, working alone, finish constructing a house in 4, 6, and 12 hours respectively.

Thus, we can let the rate of A = 1/4, the rate of B = 1/6, and the rate of C = 1/12.

We can let x = the number of hours the three people work together when C joins in.

Thus, the number of hours A works is 2 + x, the number of hours B works is 1 + x, and the number of hours C works is x.

We can now determine x with the following equation:

work of A + work of B + work of C = 1

(2 + x)/4 + (1 + x)/6 + x/12 = 1

Multiplying the entire equation by 12, we have:

3(2 + x) + 2(1 + x) + x = 12

6 + 3x + 2 + 2x + x = 12

6x = 4

x = 4/6 = 2/3 hour = 40 minutes

Since A started at 9 a.m., they finished constructing the house 2 hours and 40 minutes later, or 11:40 a.m.

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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re  [#permalink]

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