The digging work of the DMRC on the Adohini-Andheriamore stretch requi

Choose a value for the total work that will be nicely divisible by the work done by the men (\(24*R_m*16\)) and the work done by the women (\(32*R_w*24\)), where \(R_m\) is the rate of 1 man, and \(R_w\) is the rate of 1 woman.
=> Choose the total work to be \(W_t=32*24\)

Find the rate of 1 man:
=> \(24*R_m*16=W_t\)
=> \(R_m=\frac{32*24}{24*16}\)
=> \(R_m=2\)

Find the rate of 1 woman:
=> \(32*R_w*24=W_t\)
=> \(R_w=\frac{32*24}{32*24}\)
=> \(R_w=1\)

Find the work done by 16 men and 16 women in 12 days:
=> \(W_{12 days}=16*R_m*12+16*R_w*12\)
=> \(W_{12 days}=(16*2+16*1)*12\)
=> \(W_{12 days}=48*12\) ... or \(24*24\)

The amount of work that remains to be done is then:
=> \(W_{remaining}=W_t-W_{12 days}=32*24-24*24\)
=> \(W_{remaining}=8*24\)

This work will be done by the 16 men and 16 women who were already working, plus an additional number of men (let's call them x), over 2 days.

This work will be done by \(16+x\) men and \(16\) women. Since \(R_m=2\), we get:
=> \(W_{remaining}=Rate*Time\)
=> \(8*24=[(16+x)*R_m+16*R_w]*2\)
=> \(4*24=(16+x)*2+16*1\)
=> \(x=24\)

Answer C.