+288 To finish painting a house, "A" working alone needs 2 times the time that "B" and "C" work together. "B" working alone needs 3 times the time that "A" and "C" work together. "C" working alone needs x times the time that "A" and "B" work together. Find x.
+129907 Let the part of the job that "A" performs in one hour = 1/A
So "A" takes A hours to finish the job alone
Let the part of the job that "B" performs in one hour = 1/B
"B" takes B hours to finish the job alone
Let the part of the job that "C" performs in one hour = 1/C
So.....it takes "C" C hours to finish the job alone
Time for B,C to finish the job working together = 1/B + 1/C = (B + C) /BC
Take the reciprocal of this = BC /(B + C)
Similarly....the time for A,C to finish the job working together = AC / (A + C)
We have this system
2 *(BC) / (B + C) = A (1)
3 *(AC) / (A + C) = B (2)
x (AB)/ (A + B) = C (3)
Rearranging the first two equations we get that
2BC = AB + AC (4)
3AC = AB + BC subtract these
3AC - 2BC = BC - AC rearrange
4AC = 3BC
4A = 3B
A = (3/4)B (5)
Sub (5) into (4)
2BC = (3/4)B * B + (3/4)B*C divide through by B
2C = (3/4)B + (3/4)C simplify
(5/4)C = (3/4)B
5C = 3B
C = (3/5)B (6)
Rearrange (3) as
x (AB) = AC + BC sub (5) , (6) into this
x [ (3/4)B * B ] = (3/4)B* (3/5)B + B * (3/5)B
x (3/4) B^2 + (9/20)B^2 + (3/5)B^2 divide through by B^2
x (3/4) = (9/20) + (3/5)
(9/20) + (3/5) 7
x = ____________ = ____ = 1 . 4
(3/4) 5
