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nick1816
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Consider 3 friends A,B and C work at different speed. When the slowest 09 May 2019, 11:18
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Consider 3 friends A,B and C work at different speed. When the slowest 2 work together, they take m days to finish the task and when 2 quickest of them work together, they take n days to finish the same task. One of them if work alone would take thrice as much time as it would take when all three work together. How much time it will take when all three work together?

1. \(\frac{3mn}{2(m+n)}\)
2. \(\frac{2mn}{(m+n)}\)
3. \(\frac{4mn}{3(m+n)}\)
4. \(\frac{5mn}{3(m+n)}\)
5. \(\frac{8mn}{3(m+n)}\)
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Consider 3 friends A,B and C work at different speed. When the slowest 10 May 2019, 07:26
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Consider 3 friends A,B and C work at different speed. When the slowest 2 work together, they take m days to finish the task and when 2 quickest of them work together, they take n days to finish the same task. One of them if work alone would take thrice as much time as it would take when all three work together. How much time it will take when all three work together?

1. \(\frac{3mn}{2(m+n)}\)
2. \(\frac{2mn}{(m+n)}\)
3. \(\frac{4mn}{3(m+n)}\)
4. \(\frac{5mn}{3(m+n)}\)
5. \(\frac{8mn}{3(m+n)}\)
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Let them take A, B and C days with A>B>C..
So (1/A)+(1/B)=1/m...(I)
And (1/B)+(1/C)=1/n...(II)
The average days has to be the middle value so B..
Therefore, 1/B =(1/3)(1/A+1/B+1/C)..

Add I and II, so
(1/A)+(1/B)+ (1/B)+(1/C)=1/n+1/m.
(1/A)+(1/B)+ (1/B)+(1/C)-(1/B)=(1/n+1/m)-(1/B)...
(1/A)+(1/B)+(1/C)=(m+n)/mn-(1/3)(1/A+1/B+1/C)..
So =(4/3)(1/A+1/B+1/C)=(m+n)/mn...
1/A+1/B+1/C=(3/4)(m+n)/MN=3(m+n)/4mn..
Since this is one day work, time taken by all three = 4mn/3(m+n)
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Consider 3 friends A,B and C work at different speed. When the slowest 10 May 2019, 07:41
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Consider 3 friends A,B and C work at different speed. When the slowest 2 work together, they take m days to finish the task and when 2 quickest of them work together, they take n days to finish the same task. One of them if work alone would take thrice as much time as it would take when all three work together. How much time it will take when all three work together?

1. \(\frac{3mn}{2(m+n)}\)
2. \(\frac{2mn}{(m+n)}\)
3. \(\frac{4mn}{3(m+n)}\)
4. \(\frac{5mn}{3(m+n)}\)
5. \(\frac{8mn}{3(m+n)}\)
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Assume values since mn/(m+n) is same in all options. Only multiplication factor is different.
Say their speeds are a, b and c in increasing order. Say a and b together finish in 2 days (= m) and b and c together finish in 1 day (= n)

a + b = 1/2

b + c = 1

a + 2b + c = 3/2

a + b + c = 3/2 - b

Since one of them takes 3 times as much time, he is 1/3rd as slow as all 3 together. So he represents the average of the 3 speeds and hence must be the middle guy, b

3/2 - b = 3b
b = 3/8

a + b + c = 3/2 - 3/8 = 9/8

So all 3 will take 8/9 days. Put m = 2 and n = 1 in options to see that option (C) works.
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Consider 3 friends A,B and C work at different speed. When the slowest 23 Dec 2021, 03:44
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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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Consider 3 friends A,B and C work at different speed. When the slowest 13 Oct 2024, 00:05
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Rate A+ Rate B=\(\frac{1}{n}\)
Rate C+ Rate B= \(\frac{1}{m}\)
Rate A+ Rate B + Rate C = \(\frac{1}{x}\)
Rate B = \(\frac{1}{3x}\)
Therefore
\(\frac{1}{n}+\frac{1}{m}=\frac{1}{x}+\frac{1}{3x}\)
\(\frac{(n+m)}{mn}=\frac{4}{3x}\)
\(\frac{ 3(n+m)}{4mn }=\frac{1}{x}\)
x=\(\frac{4mn}{ 3(n+m) }\)
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Consider 3 friends A,B and C work at different speed. When the slowest 13 Oct 2024, 00:05
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Rate A+ Rate B=\(\frac{1}{n}\)
Rate C+ Rate B= \(\frac{1}{m}\)
Rate A+ Rate B + Rate C = \(\frac{1}{x}\)
Rate B = \(\frac{1}{3x}\)
Therefore
\(\frac{1}{n}+\frac{1}{m}=\frac{1}{x}+\frac{1}{3x}\)
\(\frac{(n+m)}{mn}=\frac{4}{3x}\)
\(\frac{ 3(n+m)}{4mn }=\frac{1}{x}\)
x=\(\frac{4mn}{ 3(n+m) }\)
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Consider 3 friends A,B and C work at different speed. When the slowest 28 Nov 2024, 12:36
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chetan2u
The average days has to be the middle value so B..
Therefore, 1/B =(1/3)(1/A+1/B+1/C)..
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I don't understand why B has to be the middle. It only says either A, B, or C would take 3 times as long as A, B, and C combined
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Consider 3 friends A,B and C work at different speed. When the slowest 07 Dec 2024, 14:08
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chetan2u
The average days has to be the middle value so B..
Therefore, 1/B =(1/3)(1/A+1/B+1/C)..
I don't understand why B has to be the middle. It only says either A, B, or C would take 3 times as long as A, B, and C combined
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Let's assume that all three of them can finish work in 1 day,
Therefore, (1/A + 1/B + 1/C) = 1

Now suppose one of them takes 3 times as long, which means 3 days to complete full work and hypothetically let's say it's not B
1/A = 1/3
so, 1/B + 1/C = 2/3

If we divide above fraction in 2 non-equal parts then 1/B < 1/3 and 1/C > 1/3, which results into 1/A becoming the middle fraction of these three values.

Hence, because of our initial assumption (A>B>C) that A takes the maximum time and C takes the least time, 1/B = (1/3)(1/A + 1/B + 1/C) must be true as the central fraction is the only one which can take thrice the time.
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