Consider 3 friends A,B and C work at different speed. When the slowest

Hi,

w_s=slowest work rate, w_m=middle work rate, w_h=highest work rate. We are given:

(1) 1/(w_s+w_m)=m -> w_m=(1/m)-w_s
(2) 1/(w_m+w_h)=n -> w_m=(1/n)-w_h
(3) 3w_m=w_s+w_m+w_h -> 2w_m=w_s+w_h

(3) is w_m and not w_s or w_h, because if we assume that the highest work rate would need thrice the time than all three together, that wouldn't work (test algebraically and logically). Taking the slowest work rate, that also wouldn't work, here is more detailed why:

3w_s=w_s+w_m+w_h
3=(w_s+w_m+w_h)/w_s
3=(w_s/w_s)+(w_m/w_s)+(w_h/w_s)
Since w_s<w_m<w_h we get 1 for the first term, something greater than one for the second, and something greater than one for the third term, implying
3=(something greater than 3), and that doesn't work.

Set (1)=(2):

(4) w_h-w_s=(1/n)-(1/m)

Add (3) and (4):
2w_h=(1/n)-(1/m)+w_m
(5) w_h=(1/2n)-(1/2m)+w_m

We add can get w_s from (1) just by rearranging, and adding (5) and rearranged (1) we can eliminate the w_m term in (5), so in total we get:
w_h+w_s=(1/2n)-(1/2m)+(1/m)
(6) w_h+w_s=(m+n)/(2mn)

But from (2) we know that (w_h+w_s)/2=w_m, so adding up (6) and what we know from (2) we get:

w_s+w_m+w_h=[(m+n)/(4mn)]+[(m+n)/(2mn)]
=(3m+n)/(4mn)

That's the combined work rate, to get the time, we we have 1/WR=(4mn)/(3m+n)

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