SOLUTION: A and B can do a work in 12 days,B and C in 16 days. A started work and afte 7 days he stopd work and B started the work,after 5days B stopd work and remaining work was completed b

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A and B can do a work in 12 days,B and C in 16 days. A started work and afte 7 days he stopd work and B started the work,after 5days B stopd work and remaining work was completed b      Log On


   

Question 1039403: A and B can do a work in 12 days,B and C in 16 days. A started work and afte 7 days he stopd work and B started the work,after 5days B stopd work and remaining work was completed by C in 13days. In how many days A ,B,C alone can finish the work.
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
a, b, c, the times in days for A, B, and C to do "one whole work". Use the work rates as 1%2FT in unit of JOB%2FDAYS.

system%281%2Fa%2B1%2Fb=1%2F12%2C1%2Fb%2B1%2Fc=1%2F16%29


A sequence of workers has done one whole job.
%281%2Fa%297%2B%281%2Fb%295%2B%281%2Fc%2913=1


Goal is to solve for a, b, and c.


Use the first two equations to solve each in terms of b.
-
1%2Fa=1%2F12-1%2Fb
1%2Fa=%28b-12%29%2F12b
highlight_green%28a=12b%2F%28b-12%29%29
-
1%2Fc=1%2F16-1%2Fb
1%2Fc=%28b-16%29%2F16b
highlight_green%28c=16b%2F%28b-16%29%29
-
Substitute these into the Finished In Thirteen Days equation.
%28%28b-12%29%2F%2812b%29%297%2B%28%28b-16%29%2F%2816b%29%2913=1

Simplify.
7%28b-12%29%2F%2812b%29%2B13%28b-16%29%2F%2816b%29=1, and simplest denominator IS 2*2*2*2*3*b;

%283%2A2%5E4%2Ab%29%287%28b-12%29%2F%2812b%29%2B13%28b-16%29%2F%2816b%29%29=3%2A2%5E4%2Ab

7%2A3%2A2%5E4%2Ab%28b-12%29%2F%2812b%29%2B13%2A3%2A2%5E4%2Ab%28b-16%29%2F%2816b%29=3%2A16b

7%2A4%28b-12%29%2B13%2A3%28b-16%29=54b

28b-336%2B39b-624=48b

28b%2B39b-48b=336%2B624

19b=960

b=960%2F19
b=50.53------------Seems strange, maybe correct, but needs to be rechecked ... Maybe a mistake was made. If you can understand the process up to here, your work MIGHT be better and without any mistake I made. You could then continue to go back to the first two equations to solve for a and c.