Solution: (b)

Two women alone can complete a piece of work in 16 days.

Therefore, Four women can complete the same work in 8 days.

12 children can complete the work in = \(\displaystyle \frac{{4\times 8}}{{8-4}}=\frac{{4\times 8}}{4}=8days\)

Four children can complete the work in = \(\displaystyle \frac{{12\times 8}}{4}=24days\)

Solution: (e)

\(\displaystyle {{M}_{1}}{{D}_{1}}={{M}_{2}}{{D}_{2}}\)

\(\displaystyle \Rightarrow \)

\(\displaystyle \Rightarrow \) \(\displaystyle 9\times 19=18\times {{D}_{2}}\)

\(\displaystyle \Rightarrow \) \(\displaystyle {{D}_{2}}\) = \(\displaystyle \frac{{9\times 19}}{{18}}=9.5days\)

Solution: (b)

12 men complete the work in 6 days.

Therefore, 1 man’s 1 day’s work = \(\displaystyle \frac{1}{{72}}\)

Therefore, 10 men’s 3 day’s work = \(\displaystyle \frac{{10\times 3}}{{72}}=\frac{5}{{12}}work\)

Remaining work = \(\displaystyle 1-\frac{5}{{12}}=\frac{7}{{12}}\)

Therefore, 21 women do \(\displaystyle \frac{7}{{12}}\) work in 3 days.

Therefore, By \(\displaystyle \frac{{{{M}_{1}}{{D}_{1}}}}{{{{W}_{1}}}}=\frac{{{{M}_{2}}{{D}_{2}}}}{{{{W}_{2}}}}\)

\(\displaystyle \frac{{21\times 3}}{{\frac{7}{{12}}}}=\frac{{12\times {{D}_{1}}}}{1}\Rightarrow \frac{{21\times 3\times 12}}{{7\times 12}}={{D}_{2}}\)

\(\displaystyle {{D}_{2}}=9days\)

Solution: (c)

Let the A can do the task in x days

\(\displaystyle \frac{1}{x}+\frac{1}{{10}}=\frac{1}{8}\)

\(\displaystyle \frac{1}{x}=\frac{1}{8}-\frac{1}{{10}}=\frac{{10-8}}{{80}}=\frac{2}{{80}}=\frac{1}{{40}}\) 

 X = 40

Alternate Method

If total work = 80 units

Then in 1 day (A+B) will do 80/10 = 10 unit of work

Therefore, A does 2 unit of work each day

Hence, A requires 80/2 = 40 days to complete work

Solution: (e)

A’s 1 day’s work = \(\displaystyle \frac{1}{6}\)\(\displaystyle \frac{1}{6}\)

B’s 1 day’s work = \(\displaystyle \frac{1}{{12}}\)

Therefore, (A + B)’s 1 day’s work = \(\displaystyle \frac{1}{6}+\frac{1}{{12}}=\frac{{2+1}}{{12}}=\frac{1}{4}\)

Hence, A and B together will make 100 baskets in 4 days.

Alternate method

A alone can make 100 baskets in 6 days and B alone can make in 12 days.

Therefore, Rate of efficiency of A=100/6 baskets per day

And rate of efficiency of B=100/12 baskets per day

⇒ Baskets made by A and B together in 1 day = \(\displaystyle \frac{{100}}{6}+\frac{{100}}{{12}}=\frac{{300}}{{12}}=25\)= baskets per day

Therefore, Time taken by A and B together to make 100 baskets= \(\displaystyle \frac{{100}}{{25}}\)=4 days

Solution: (a)

1M = 2W

(8M + 4W) × (6 days – 2 days) = (4M + 8W) × x days

(8 × 2W + 4W) × (6 – 2) days = (4 × 2W + 8W) × x days

(16 + 4)W × 4 days = 16W × x days

X = \(\displaystyle \frac{{20\times 4}}{{16}}=5days[{{M}_{1}}{{D}_{1}}={{M}_{2}}{{D}_{2}}]\)

Answer for this Time and Work Problem is (d)

Part of the tank filled by the three pipes working simultaneously in one hour is

= \(\displaystyle \frac{1}{5}+\frac{1}{6}-\frac{1}{{12}}=\frac{{17}}{{60}}\)

i.e. it takes \(\displaystyle \frac{{60}}{{17}}\) hours to fill up the tank completely.

Now,  of the tank is filled with all the pipes open, simultaneously together in

\(\displaystyle \frac{{60}}{{17}}\times \frac{1}{2}=1\frac{{13}}{{17}}hours\)

Answer for this Time and Work Problem is (b)

A and B can complete work in 6 days

A can complete in 18 days

Let B can complete in x days

\ One day work be equal to \(\displaystyle \frac{1}{6}=\frac{1}{{18}}+\frac{1}{x}\)

\(\displaystyle \frac{1}{6}=\frac{{x+18}}{{18x}}\)

3x = x + 18

x = 9 days