Bunuel wrote:
Official Solution:
Mac can finish the job in \(M\) days: the rate of Mac is \(\frac{1}{M}\) job/day;
Jack can finish the job in \(J\) days: the rate of Jack is \(\frac{1}{J}\) job/day;
After working together for \(T\) days, Mac left and Jack alone worked to complete the remaining work in \(R\) days, hence Mac worked for \(T\) days only and did \(\frac{T}{M}\) part of the job while Jack worked for \(T+R\) days and did \(\frac{T+R}{J}\) part of the job;
Since Mac and Jack completed an equal amount of work (so half of the job each) then \(\frac{T}{M}=\frac{1}{2}\) and \(\frac{T+R}{J}=\frac{1}{2}\). So, \(T=\frac{M}{2}\) and \(T=\frac{J}{2}-R\), therefore \(\frac{M}{2}=\frac{J}{2}-R\), which gives \(J=M+2R\).
(1) \(M = 20\) days. Not sufficient, we still need the value of \(R\).
(2) \(R = 10\) days. Not sufficient, we still need the value of \(M\).
(1)+(2) \(J=M+2R=20+2*10=40\). Sufficient.
Answer: C
You made this so easy.
Can you please suggest,what was your line of thinking ..I took a longer approach and got confused at the end...
Here is my approach,
Mac's Work =1/M
Jacks' = 1/J
When both work for T days work done = ((1/M) + (1/J))*T
Remaining work = 1 - {(M+J)/MJ}*T
Now work done by jack= T+R days..
So I would make acomplex equution equation work done by Jack and Mac and this would really confuse me spotting R and T
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