Data Sufficiency

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.

OA - D I got the answer D. Did you get Js speed = 12 hours?

I got D too

Jacks time for me was 6 hours and Toms was 12 hours

D

By solving 1 Tom timing is 6 and that of Jack is 12. however, not able to solve with second and getting calculation error

1/T + 1/J = 1/4 (1)
1/2T + 1/J = 1/3 (2)

Solving 1 & 2
1/T - 1/2T = 1/4 -1/3 = -ive (not possible ) . Please suggest.

JeffB wrote:I got D too

Jacks time for me was 6 hours and Toms was 12 hours

Ok, solution is 6 for Jack. I am messing up somewhere. For the below statement, can you provide your equation and solution?

Jack is painting twice as fast as Tom.

For me, J = 2T (I am guessing here is where I am missing)

1/J + 1/T = 1/4

1/2T + 1/T = 1/4

3/2T = 1/4, hence T = 6, hence J = 12

Where am I going wrong?

@crackgmat007:

Both working together have a speed of 1/4 wall/hour.

If you define J and T as wall/hour, then

1/4 = J + T (not 1/J, 1/T)

From stml. 1 we know that J = 2T, hence T = J/2.

1/4 = J + J/2 => J = 1/6 wall/hour

It takes J 6 hours to complete a wall.

T = J/2 => T = 1/12

It takes T 12 hours to complete a wall.

Hope that helps.

Btw. I chose (D), too.

Got it...thanks much.