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The objective is to find the work done by 1 man and 1 woman

Equation 1: 4M + 3W=\(\frac{1}{6}\) (work done in 1 day) or 24M + 18W = 1


Equation 2: 5M + 7W = \(\frac{1}{4}\) or 20M + 28W = 1


Equating and solving: 24M + 18W = 20M + 28W


4M = 10W


Therefore 1M = \(\frac{10}{4}\)W = \(\frac{5}{2}\) - - (Equation 3)


Putting in equation 1: 4*\(\frac{5}{2}\)W + 3W = 13W = \(\frac{1}{6}\)


Therefore 1 Woman's 1 days work = \(\frac{1}{78}\)


Putting in Equation (3) : 1M = \(\frac{5}{2}\) = \(\frac{5}{2}\) * \(\frac{1}{78}\) = \(\frac{2.5}{78}\)


Work done by 1 man and 1 woman in 1 day = \(\frac{5}{156}\) + \(\frac{1}{78}\) = \(\frac{5+2}{156}\) = \(\frac{7}{156}\)

Therefore Time taken = Inverse of 1 days work = \(\frac{156}{7}\)

Option B

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given
(4M+3W)6=(5M+7W)4 (same work)
24M+ 18W = 20M + 28 W
4M =10 W
now 1M + 1W= 5/2W+ W= 7/2W
so 7/2WxT = (4M+3W)6 (T= no of days)
7/2 WxT= 13Wx6 ( as 4M =10W)
T= 156/7
answer B
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This approach takes me 3-4 minutes.

I basically use the substitution method and find out what M and W are.

4m+3w=1/6 and 5m+7w=1/4

1. I make m the same number for each equation by multiplying the first equation by 5 and the second by 4. This gets me: 20m+15w=5/6 and 20m+28w=1
2. I subtract the smaller equation from the larger and get w=1/78
3. Plug back into any original equation to get m. m=5/156
4. Answer the question: 5/156 + 1/78 =1/x

Answer is B: 156/7
Bunuel
4 men and 3 women can finish a piece of work in 6 days whereas if there were 5 men and 7 women, they would have taken 4 days to finish the work. How many days will 1 man and 1 woman take to finish the work?

A. 156/5
B. 156/7
C. 156/9
D. 156/11
E. 156/13
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