Question 427705
First, lets deal in hours and fractions of hours
Let x=amount of time it takes B to do the task
Then B works at the rate of 1/x of the task per hr
Then 2x=amount of time it takes M to do the task (twice as long as B)
M, then, works at the rate of 1/2x of the task per hr
And 2x-2=amount of time it takes K to do the task (2hrs less than M)
K, then, works at the rate of 1/(2x-2) of the task per hr
Soooo:
We are told that M+B+K can do the task in 1 1/3 hr =4/3 hr
So M+B+K works at the rate of 1/(4/3)=3/4 of the task per hr
Our equation to solve, then, is:
1/x + 1/2x + 1/(2x-2)=3/4  Multiply each term by 4x(2x-2)
4(2x-2)+2(2x-2)+4x=3x(2x-2) simplify
8x-8+4x-4+4x=6x^2-6x  and this simplifies to:
16x-12=6x^2-6x collect like terms
6x^2-22x+12=0 divide each term by 2
3x^2-11x+6=0  quadratic in standard form and it can be factored
(3x-2)(x-3)=0
x=2/3 hr 
or 
x=3 hr
If x=2/3 hr ---time it takes B to do the task
And 2x=2*(2/3)=4/3 hr time it takes M to do the task
Then 2x-2=4/3-2=4/3-6/3 -----NEGATIVE TIME FOR K, So 2/3 hr is NOT a solution 

x=3 hr---time required for B to do the task
2x=2*3=6 hr ---time required for M to do the task
2x-2=6-2=4 hr----time required for K to do the task

CK
1/3 + 1/6 + 1/4 =3/4
4/12+2/12+3/12=3/4
9/12=3/4
3/4=3/4

Hope this helps----ptaylor