\begin{align} & {\text{Let}}\, {\text{1}}\, {\text{man's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = x\, {\text{and}} \cr & {\text{1}}\, {\text{woman's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = y \cr & {\text{Then}}, \, 4x + 6y = \frac{1}{8}\, {\text{and}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 3x + 7y = \frac{1}{{10}} \cr & {\text{Solving}}\, {\text{the}}\, {\text{two}}\, {\text{equations, }}\, {\text{we}}\, {\text{get}} \cr & x = \frac{{11}}{{400}}\, \, , \, \, \, y = \frac{1}{{400}} \cr & \therefore {\text{1}}\, {\text{woman's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} = \frac{1}{{400}} \cr & \Rightarrow {\text{10}}\, {\text{women's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} \cr & = \left( {\frac{1}{{400}} \times 10} \right) = \frac{1}{{40}} \cr & {\text{Hence, }}\, {\text{10}}\, {\text{women}}\, {\text{will}}\, {\text{complete}}\, {\text{thw}}\, {\text{work}}\, {\text{in}}\, {\text{40}}\, {\text{days}} \cr\end{align}

1^st Method:

Whenever, workers are working alternatively on one work, we take 2 as 1 unit.

In this case, we take 2 hours as 1 unit.

Part of the field moved by Ganga and Saraswati in 2 hours (1 unit) = ¹/₈ + ¹/₁₂ = ⁵/₂₄;

Time taken to complete the work,

= ¹/_(5/24) = ²⁴/₅ unit of time;

Then actual time taken by them to complete the work,

=2 * ²⁴/₅ = 9.6 hours.

The work starts at 9 a.m. then it will complete at 6:36 pm.

2^nd Method:

% of the field moved by Ganga and Saraswati in 2 hours (1 unit),

= [(¹⁰⁰/₈)% + (¹⁰⁰/₁₂)%]

= 12.5 + 8.33

= 20.83%; Time taken to move the whole field,

= ^100%/_20.83%

= 4.8 unit of time;

Hence, actual time = 2*4.8 = 9.6 hours.

Let x be the required number of days.

Given,

12 men and 12 women can complete a work separately in 14 days and 21 days respectively.

Then,

12 man's 1 day work = ¹/₁₄

And,

12 women's 1 day work = ¹/₂₁

Then,

12 women's 3 days work = ³/₂₁ = ¹/₇

The remaining work = 1 - ¹/₇ = ⁶/₇

Man's group leaves 3 days before the completion of work.

That is, they were working together for x-3 days.

Thus, we have 1/7 work left to be done in last 3 days by the women's group. This also means 6/7 th of work has been done by both the groups (before men left).

Now, (12 men + 12 women)'s 1 day work = ¹/₁₄ + ¹/₂₁ = ⁵/₄₂.

i.e., ⁵/₄₂ work is done by 2 groups in 1 day.

So, ⁶/₇ of work is done by 2 groups together in (⁴²/₅)(⁶/₇) = ³⁶/₅ days.

As we know, women's group will work to

\begin{align} & \left( {{\text{A's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}}} \right){\text{:}}\left( {{\text{B's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}}} \right) \cr & = \frac{7}{4}:1 = 7:4 \cr & {\text{Let}}\, {\text{A's}}\, {\text{and}}\, {\text{B's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}}\, {\text{be}} \cr & 7x\, {\text{and}}\, 4x\, {\text{respectively}} \cr & {\text{Then}}, \, 7x + 4x = \frac{1}{7} \cr & \Rightarrow 11x = \frac{1}{7} \cr & \Rightarrow x = \frac{1}{{77}} \cr & \therefore {\text{A's}}\, {\text{1}}\, {\text{day's}}\, {\text{work}} \cr & = \left( {\frac{1}{{77}} \times {\text{7}}} \right) = \frac{1}{{11}} \cr\end{align}

Let E₁ and E₂ are the working efficiency of each student of IIT and NIT per hour respectively.

Using work equivalence method,

4 *10 *60 *E₁ = 5 *8 *80 *E₂

^E₁/_E2 = ⁴/₃

As, 3E₁ = 4E

% less efficient of NITians = ^(1*100)/₄ = 25%

Thus, each engineer from NIT is 25% efficient than that of IIT.