Ratio of income to expenditure = 7 : 5
Let the income and expenditure per month be Rs 7x and 5x respectively.
Then, saving per month = Rs 7x - Rs 5x = Rs 2x
According to the question,
2x= 2000
⇒ x =\(\frac{2000}{2}\) =1000
⇒ 7x = 7 x 1000 = 7000
Hence, his income is Rs 7000 per month.

Steps of construction are as follows:
Step I Draw a line segment \(\bar{AB}\) of length 5 cm.
Step II Take a point P and place the pointer of compasses at P draw an arc such that it intersects \(\bar{AB}\) at X and Y.
                       
Step III With X and Y as centres and radius greater than \(\bar{PX}\), draw two arcs such that they intersect at C.
Step IV Join CP Thus, \(\bar{CP}\bot\bar{AB}\)

First of all, find the HCF of 24 and 36

\(\therefore\)HCF of 24 and 36 is 12.
Now find the HCF of 12 and 40

\(\therefore\)HCF of 12 and 40 is 4.
Hence, HCF of 24, 36 and 40 is 4.

To construct a line segment of length 5 cm using a ruler and compass, we use the following steps:
                   
Step I Draw AB of length 5 cm.
Step II Draw a line I and mark a point C on it.
                   
Step III Open the compasses equal to \(\bar{AB}\) .
Step IV Keeping the same opening. Place the pointer on C and mark a point D on I. This \(\bar{CD}\) is a equal to \(\bar{AB}\) .

Rs.325

Let breadth of rectangular field = x m
Then. length of rectangular field = 2x m
Distance covered in one round = Perimeter
Distance covered in four rounds = 4 \(\times\) Perimeter
⇒ 4\(\times\)2 (l + b) = 6 km [∵ perimeter = 2 (l + b)]
⇒ 8 (l +b)= 6000 m [∵ 1 km =1000 km]
8 (2x + x)= 6000 m
x=\(\frac{6000}{24}m\)=250 m
∴ Length of the field = 500 m

r = 4 is a solution of equation r - 4 = 0.

8

5-x

Total number of students in school = 1261
Contribution of money by each student =Rs. 520
∴ Total money =1261 \(\times\) 520 =Rs. 655720

4 x 3 = 12, 5 - 3 = 2

We have, 8325 - 491
\(\because \begin{matrix} 8325\rightarrow 8300 \\ 491\rightarrow 500 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴ Rough estimate = 8325 - 491 = 8300 - 500 = 7800
\(Again\begin{matrix} 8325\rightarrow 8330 \\ 491\rightarrow 490 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 8325 - 491 = 8330 - 490 = 7840

Total length of cloth = 40 m = 40 \(\times\) 100 cm [∵1 m =100 cm]
= 4000 cm
Cloth needed for one shirt = 2 m 15 cm
= (2\(\times\)100) cm +15 cm

= (200+15) cm = 215 cm 
∴ Number of shirts stitched out of total cloth
\(=\frac { Total\quad length\quad of\quad cloth }{ Cloth\quad needed\quad for\quad stitched\quad shirt } \)
\(=\frac { 4000 }{ 215 } =18\frac { 130 }{ 215 } \)
Hence, 18 shirts can be stitched and cloth left over is 130 cm i.e. 1 m 30 cm.

The smallest 8-digit number = 10000000
∴ 1 st next number = 10000000 + 1 = 10000001
2nd next number = 10000000 + 2 = 10000002
3rd next number =10000000 + 3 = 10000003
4th next number =10000000 + 4 = 10000004
5th next number = 10000000 + 5 = 10000005
Now, ascending order of these numbers is as follows:
10000001 < 10000002 < 10000003 < 10000004 < 10000005.

The greatest 6-digit number = 999999
1st previous number =999999 -1 = 999998
2nd previous number = 999999 - 2 = 999997
3rd previous number = 999999 - 3 = 999996
4th previous number =999999 - 4 = 999995
5th previous number = 999999 - 5 = 999994
Now, descending order of these numbers is as follows:
999998 > 999997 > 999996 > 999995 > 999994.

Given digits are 3, 8 and 7.
Here, descending order of the given digits is 8, 8, 7, 3. [∵ greatest digit 8 is taken twice]
∴ Greatest 4-digit number = 8873
and ascending order of the given digits is 3, 3, 7, 8. [∵ smallest digit 3 is taken twice]
∴ Smallest 4-digit number = 3378

We have, 25286, 25245, 25270, 25210
Here, the three leftmost digits 252 are same in each number. So, we compare the tens place digit.

∴The greatest number is 25286 and the smallest number is 25210.