\begin{align} & \left( {20 \times 16} \right){\text{women}}\, {\text{can}}\, {\text{complete}}\, {\text{the}}\, {\text{work}}\, {\text{in}}\, {\text{1}}\, {\text{day}} \cr & \therefore {\text{1}}\, {\text{woman's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{320}} \cr & \left( {16 \times 15} \right)\, {\text{men}}\, {\text{can}}\, {\text{complete}}\, {\text{the}}\, {\text{work}}\, {\text{in}}\, {\text{1}}\, {\text{day}} \cr & \therefore {\text{1}}\, {\text{man's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{240}} \cr & {\text{So, }}\, {\text{required}}\, {\text{ratio}} \cr & = \frac{1}{{240}}:\frac{1}{{320}} \cr & = \frac{1}{3}:\frac{1}{4} \cr & = 4:3\, \left( {{\text{cross}}\, {\text{multiplied}}} \right) \cr\end{align}

Work rate of A = ¹⁰⁰/₉ = 11.11% work per day.

Work rate of B = ¹⁰⁰/₁₅ = 6.66% work per day.

They together can do (A + B) = 11.11 + 6.66 = 18% work per day.

They are working in alternate day, so we take 2 days = 1 unit of day.

Therefore, in one unit of day they can complete 18% of work.

(A +B) can complete 90% of work in 5 units of days. i.e. (5 * 18).

And the rest 10% work will be completed by A in Next day.

So, total number of day,

= 5 Unit of days + 1 day of A

= 2*5 + 1 = 11 days.

Variation Method:

In 100 days only 2.5 km rod i.e. 16.66 % of work has been completed.

Men Days Road (km)

45 100 ↓ 2.5

X ↑ 200 12.5↑

Arrows show the directions of variation of quantity with respect to each other.

^x/₄₅ = ^100*12.5/_200*2.5;

x = 113 men;

Required men to be increased,

= 113 - 45 = 68.

Total efficiency of two persons = 50% [As they complete work in 2 days. ]

First Person completes work = ¹/₃ = 33.33% [In 2 days]

Rest work will be completed by Second man = ²/₃ = 66.66% [In 2 days]

So, efficiency of second person is greater.

Efficiency of second person = ^66.66/₂ = 33.33% per day

Then, Second person will complete whole work in,

= ¹⁰⁰/_33.33 = 3 days.

Let pipe C alone can empty the cistern in x min.

A fills cistern in 1 min = ¹/₁₀;

B fills cistern in 1 min = ¹/₁₅;

A and B together fill in 1 min = ^(10*15)/_(10+15) = 16 min

Since, waste pipe was left open for 6 min then,

6 min, ⁶/_x part of cistern will be emptied by C.

Now,

⁶/_x part of the cistern would be filled by A and B in 2 min.

Hence,

cistern will be filled in ³/_x min. And

^x/₃ = 6

x = 18 min.