+0  
 
0
41
1
avatar+150 

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.

 Mar 9, 2020
 #1
avatar+109490 
+4

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

 

 

 

cool cool cool

 Mar 9, 2020

28 Online Users

avatar
avatar
avatar
avatar
avatar