Solution:

The cost of 12 m of clothes = Rs 600
The cost od 1 m of clothes = Rs \(\frac{600}{12}\)  
= Rs 50
The cost of 8 m of clothes = 8 × Rs 50
= Rs 400 
So, the required cost of 8 m of clothes is Rs 400.

Solution:

Here,
the length of the floor (l) = 14 m
the breadth of the floor (b) = 8.5m
∴ Area of the floor = l × b
= 14m × 8.5m
= 119 m²
Now, 
the cost of carpeting  119 m² is 119 × Rs 75
= Rs  8925
So, the required cost of carpeting the floor is Rs 8925.

Solution:

Here, 
Rs 2 is the tax for the income of Rs 100 
Re 1 is the tax for the income of Rs \(\frac{100}{2}\) 
= Rs 50
Rs 525 is the tax for the income of Rs 525 × 50 
= Rs 26250
So, the required income of the man is Rs 26250

Solution:

Here, 
the cost of \(\frac{3}{4}\) parts of a land = Rs 15,000
The cost of 1 (whole) land = Rs \(\frac{15000}{\frac{3}{4}}\) 
= Rs \(\frac{15000 × 4}{3}\) 
= Rs 20,000
The cost of \(\frac{2}{5}\) parts of land = Rs 20,000 × \(\frac{2}{5}\) 
= Rs 8,000
so, the required cost is Rs 8,000.

Solution:

Here, 
In 20 days, 7 workers build a wall
In 1 days, 20 × 7 workers build the wall
In 14 days, \(\frac{20 × 7}{14}\) workers build the wall = 10 workers build the wall 
∴ The required number of workers to be added = 10 - 7 
= 3 workers

Solution:

In 10 days, (X + y) can finish 1 work 
In 1 days, (X = y) can finish \(\frac{1}{10}\) parts of the work
Also,
In 15 days, X can finish 1 work 
In 1  days , X can finish \(\frac{1}{15}\) parts of the work.
Now, 
In 1 days (X + Y - X) can finish (\(\frac{1}{10}\) - \(\frac{1}{15}\)) parts of the work
In 1 days Y can finish (\(\frac{3 - 2}{30}\)) parts of the work = \(\frac{1}{30}\) parts of the work.
∴ Y can finish \(\frac{1}{30}\) parts of the work in 1 day.
Y can finish 1 work in \(\frac{1}{\frac{1}{30}}\) days = 1 × \(\frac{30}{1}\) = 30 days.
So, Y alone can finish the work in 30 days

Solution:

Here, 
In 8 days, Hari can do 1 work 
In 1 day, Hari can do \(\frac{1}{8}\) parts of the work.
In 6 days, Hari can do \(\frac{6}{8}\) parts of the work = \(\frac{3}{4}\) parts of the work.
Now,
Remaining parts of work = (1 - \(\frac{3}{4}\)) 
= \(\frac{1}{4}\) parts  of the work
So, Gopal finished \(\frac{1}{4}\) part of the work

Solution:

In 6 days, A can do 1 work 
In 1 day, A can do \(\frac{1}{6}\) parts of the work
Also,
In 12 days, B can do 1 work 
In 1 day, B can do \(\frac{1}{12}\) parts of the work
Now,
In 1 day, (A + b) can do (\(\frac{1}{6}\) + \(\frac{1}{12}\)) partss of the work
= (\(\frac{2 + 1}{12}\)) parts of the work = \(\frac{3}{12}\) = \(\frac{1}{4}\) parts of the work
∴ (A + B) can do \(\frac{1}{4}\) parts of the work = 1 
(A + B) can do 1 work in \(\frac{1}{\frac{1}{4}}\) day = 1 × \(\frac{1}{4}\) = 4 days
So, A and B finish the work in 4 days working together.

Solution:

Here, 
after joining 10 more people, total number of the people in the group = 0 + 10 = 50 
40 people had provisions for 60 days 
1 people had provisions for 40 × 60 days
50 people had provisions for \(\frac{40 × 60}{50}\) days = 48 days
So, the provisions would long last for 48 days.

Solution:

Here, Cost of 5 pens = Rs. 100

 or, Cost of 1 pen = \(\frac{Rs.100}{5}\) = Rs. 20

Then, The cost of 2 pens = 2 × Rs. 20 = Rs. 40 ans.

Solution:

6 men finish a piece of work in 5 days.

Then, 1 man finishes the same piece of work in 6 × 5 days = 30 days.

Now, 3 men finish the same piece of work in \(\frac{30}{3}\) = 10 days ans.

Solution:

Here, The cost of 12 litres of milk = Rs. 432

or, The cost of 1 litres of milk = Rs. \(\frac{432}{12}\) = Rs. 36

The cost of 5 litres of milk = 5 × Rs. 36 = Rs. 180

So the required cost of 5 litres of milk is Rs. 180.