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# m14#07

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19 Jun 2010, 03:32
If working together, brothers Tom and Jack can paint a wall in 4 hours, how much time would it take Jack to paint the wall alone?

(1) Jack is painting twice as fast as Tom.
(2) If Tom painted twice as fast as he actually does, the brothers would finish the work in 3 hours.
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17 Jul 2014, 01:07
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Given : $$\frac{1}{T} + \frac{1}{J} = \frac{1}{4}$$

1) $$2\frac{1}{T} = \frac{1}{J}$$
Substitute in given equation. we can find J and T. SUFF.

2) $$2\frac{1}{T} + \frac{1}{J} = \frac{1}{3}$$
Solve given equation and new equation together. we can find J and T. SUFF.

Ans : D
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20 Jun 2010, 07:02
Guys,
Thank but I will be happy with your explanations.
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20 Jun 2010, 07:09
sorry.... just use the algebric approach:

A - if j+t=1/4 and j=2t than j=2/12 and t=1/12. - sufficient.

B - the same deal - if j+t=1/4 and 2t+y= 1/3 than t=1/12 and j=1/6=2/12 - sufficient.
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24 Jun 2011, 22:12
I have a very generic doubt regarding statement 2 ONLY.

if I write an equation in rates form I get the solution :

1/j+2/t = 1/3; when I solve this with the problem statement->
1/j+1/t=1/4; it gives T=12,J=6.perfect.

Doubt :
If I write an equation in time form I get the solution:
statement 2->If Tom painted twice as fast as he actually does, the brothers would finish the work in 3 hours.->therefore tom would take half the time than he normally does T=.5T.
J+0.5T = 3
from problem statement :
J+T=4.

SOlving these we get T=2, which is wrong. Can someone help me with this ?
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09 Feb 2013, 03:49
Bunuel/Karishma,
Could you please explain how S2 is sufficient?
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10 Feb 2013, 03:01
Sachin9 wrote:
Bunuel/Karishma,
Could you please explain how S2 is sufficient?

Sure.

If working together, brothers Tom and Jack can paint a wall in 4 hours, how much time would it take Jack to paint the wall alone?

Say the rates of Tom and Jack are T job/hour and J job/hour, respectively. Their combined rate is T+J job/hour, and we are told that it equals to 1/4 job/hour.

Thus given that T+J=1/4 job/hour.

(1) Jack is painting twice as fast as Tom --> J=2T --> T+2T=1/4 --> T=1/12 --> J=2/12=1/6 --> (time)=(reciprocal of rate)=6 hours. Sufficient.

(2) If Tom painted twice as fast as he actually does, the brothers would finish the work in 3 hours. This statement implies that if the rate of Tom were 2T instead of T, the brothers combined rate would be 1/3 job/hour, thus 2T+J=1/3. Solving T+J=1/4 and 2T+J=1/3 gives J=1/6. Sufficient.

Hope it's clear.
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16 Jul 2014, 11:49
Bunuel
I took 12 Units of wall..So
3 units of work per hour= T + J
Stmt- I- T +2T = 3..so we get J and hence suff
Stmt- II- T + T + J = 4..so T causes a difference of 1 unit and hence we can find J now..

D it is

Is my reasoning correct..
For such questions, I always assume amount of work in some convenient multiple ...Can I ever get stuck with such an approach..i.e assuming the total work in units
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16 Jul 2014, 11:55
JusTLucK04 wrote:
Bunuel
I took 12 Units of wall..So
3 units of work per hour= T + J
Stmt- I- T +2T = 3..so we get J and hence suff
Stmt- II- T + T + J = 4..so T causes a difference of 1 unit and hence we can find J now..

D it is

Is my reasoning correct..
For such questions, I always assume amount of work in some convenient multiple ...Can I ever get stuck with such an approach..i.e assuming the total work in units

That's correct approach. Generally, if the question is such that you can assume some number for a job, then this approach will work. Not sure that it would be the best approach but most of the case it'll work fine.
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# m14#07

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