silverflamein wrote:Can 3 men and 7 women finish a certain piece of work in 15 days???
1) 2men and 4 women can finish the work in 10 days
2) each of the women can do twice the work that each of the men can do in a given time.
pemdas wrote:the question tests average speed: Can it be that 1/3m (throughput of one men for one job is 1/3 of whole unit) and 1/7w (throughput of one women for one job is 1/7 of whole unit) *multiplied* by time, 15 days be enough to complete one job?
st(1) 10*(1/2m+1/4w)=1 means 10 hours is enough to complete one job with the given throughputs, and if we match 10(2+1)/4mw=1, 30=4mw we see that men and women can be only integer and not fractions
out of possible values for m*w (15=2mw). The question asked was: 15(1/3m+1/7w)=1? and can answer No since 15(7+3)=21mw and 50=7mw is not equivalent to 15=2mw, 50/15 =! 7/2 <> 3 1/3 =! 3 1/2 Sufficient
st(2) w=2m although we are not given day-people function, we can find whether 15(1/m + 1/2m)=1 to test 50=7mw; 15*3=2m and m=45/2, w=45. The question asked was 50=7mw and f we plug in m and w we get 50=! 7*(45/2)*45. We can answer No, Sufficient
dsilverflamein wrote:Can 3 men and 7 women finish a certain piece of work in 15 days???
1) 2men and 4 women can finish the work in 10 days
2) each of the women can do twice the work that each of the men can do in a given time.
You don't need to solve this question at all. you can answer this by using common sense.bpdulog wrote:I think the answer is C.
With A, you still have 2 unknowns. I think it be okay if the same amount of men and women were added, but in this case it is not.
I went for Csilverflamein wrote:Can 3 men and 7 women finish a certain piece of work in 15 days???
1) 2men and 4 women can finish the work in 10 days
2) each of the women can do twice the work that each of the men can do in a given time.
