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.

Fill the 9 litre can. Remove 5 litres from it using the 5 litre can. Empty the 5 litre can. Pour four litres remaining in the 9 litre can to the 5 litre can.
Fill the 9 litre can again. Fill the remaing 5 litre can from the water in it. This leaves 8 litres in the 9 litre can. Empty the five litre can. Fill it from the 9 litre can. You now have 3 litres left in the 9 litre can.

The man takes an empty vessel other than these.
With the help of 3 litre can he takes out 9 litres of milk from the 10 litre can and pours it in the extra can. So, 1 litre milk remains in the 10 litre can. With the help of 7 litre can he takes out 7 litres of milk from the extra can, and pours it in the 10 litre can. The 10 litre can now has 1 + 7 = 8 litres of milk.
With the help of 3 litre can he takes out 3 litres milk from the 10 litre can. The 10 litre can now has 8 - 3 = 5 litres of milk, which he gives to the customer.

15 = 3 x 5, 21= 3 x 7,
36 = 2 x 2 x 3 x 3,
42 = 2 x 3 x 7,
48 = 2 x 2 x 2 x 2 x 3
∴ HCF=3
Hence, the largest length of the rod is 3 m.

(a) Diameter of circle is \(\bar{AB}\) .
(b) Radius of the circle is \(\bar{OA}\), \(\bar{OB}\) and \(\bar{OC}\) .
(c) The chord of the circle are \(\bar{EF}\) and \(\bar{AB}\) .
(d) Centre of circle is 'O'.

8 cm

CD \(\bot \) AB

\(\angle\)APB = 90^o

To draw two circles such that each one of them passes through the centre of the other, we use the following steps:
Step I Firstly, mark two points A and B on the paper.
Step II Take the distance between A and B as radius and draw a circle with centre A.
Step III Now, take B as centre and draw a circle with radius AB.
Thus, we get two circles which passes through the centres of each other. Let these circles intersect each other at C and D. Join C and D, which intersect AB at O. Then, we observe that, the \(\angle\)AOC and \(\angle\)COB are equal to 90°.
                         
Hence, \(\bar{AB}\bot\bar{CD}\)
So, we can say that \(\bar{AB}\) and \(\bar{CD}\) are at right angles.

Firstly, draw a circle with O as centre and of any radius.Then, draw any two diameters, say AOB and COD.
                           
Now, join DA, AC, CB and BD.1t is clear from the given figure that, DACB is a rectangle.
When the diameters AOB and DOC are perpendicular to each other, then figure obtained by joining AC, CB, BD and DA is a square ADBC.
To check our answer, we can compare lengths of sides by using divider.
                      

∵ 5-1=9-5=13-9
=17-13=21-17=4
∴ Next number in the sequence
= 21+4=25

Divider.

Diameter

Square.

FD = 3 cm.

Yes.

(i) By paper folding
(ii) using the ruler and compasses
(iii) Using ruler and a set-square.

3.8 cm.

PQ = 2.1 cm.

C lies on the circle.

Steps of construction are as follows:
Step I Draw a line segment \(\bar{PQ}\) = 9 cm.
                    
Step II With P as centre and a convenient radius (more than \(\frac{1}{2}\bar{PQ}\)) draw arc.
Step III With 0 as centre and same radius, draw another arc such that it intersects the previous arc at A and B.
               
Step IV Join A and B.
Thus, \(\bar{AB}\) is perpendicular bisector of PO.
i.e. OP = OQ = 4.5 cm.
(a) Length of each part is 4.5 ft.
(b) The value depicted here is gender equality.

To construct an angle of 15^o, steps of construction are given below:
(i) Firstly, construct an angle of 60°.
(ii) Bisect this angle to obtain an angle of 30°.
(iii) Finally,bisect the angle of 30° to obtain an angle of 15°.
Steps of construction
Step I Draw a line l and mark a point O on it.
                    
Step II Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at a point say A.
                      
Step III Without disturbing the radius on the compasses draw an arc with A as centre which cuts the first arc at B.
                       
Step IV Join OB. We get \(\angle\)BOA, whose measure is 60°.
                      
Step V Now, bisect this angle. For this, take distance more than half of length AB as radius and A as centre draw an arc.
                    
Step VI Take B as centre and radius same as in Step V. Draw another arc which intersects the arc drawn in Step V at C.
                     
Step VII Join OC by dotted line which intersects the arc AB at I. Then, \(\angle\)COA= 30°. Now, again we bisect this angle.
                      
Step VIII Take A and I as centres and radius more than \(\frac{1}{2}\) AI, draw two arcs, respectively such that both intersect each other at point D.
                     
Step IX Join OD. Thus, \(\angle\)DOA = 15°.
                             

To make \(\bar{PQ}\), we use the following steps:
Step I First of all, draw a line segment \(\bar{AB}\), whose length is not known.
                    
Step II Fix the pointer of compasses on A and the pencils end on B. The opening of the instrument now gives the length of \(\bar{AB}\).
Step III Draw any line I. Choose a point P on l, without changing the compasses setting, place the pointer on P, swing an arc that cuts I at point R.
                    
Step IV Now, place the pointer on R and  Without changing the compasses setting, swing another arc that cuts I at a point Q.
Step V Thus, \(\bar{PQ}\) is the required line segment who length is twice that of \(\bar{AB}\). Hence, \(\bar{PQ}\) = 2\(\bar{AB}\).

Given, \(\bar{AB}\) = 7.3 cm and \(\bar{CD}\) = 3.4 cm
Now, to construct required line segment \(\bar{XY}\) , we use the following steps:
Step I Firstly, draw \(\bar{AB}\) = 7.3 cm and \(\bar{CD}\) = 3.4 cm.
                 
Step II Now, place the pointer of compasses on C of pencil on D. The opening of the instrument gives the length of \(\bar{CD}\) i.e. 3.4 cm.
Step III Without changing the opening of the compasses place the pointer on A and swing an arc to cut \(\bar{AB}\) at R.
                   
Step IV Thus, \(\bar{AR}\) = 3.4 ern and RE is the difference between the length of \(\bar{AB}\) and \(\bar{CD}\).
Step V Now, draw a line 1 and mark a point X on it.
Step VI Place the pointer of compasses on R and of pencil on B. The opening of the compasses gives the length of \(\bar{RB}\) .
Step VII Without changing the opening of the compasses, place the pointer on X and swing arc to cut l at Y.
                       
Thus, \(\bar{XY}\) is a line segment whose length is equal to the difference between the lengths of \(\bar{AB}\) and \(\bar{CD}\).
Verification By actual measurement, we have \(\bar{XY}\)= 3.9 ern
Now, \(\bar{AB}\) - \(\bar{CD}\) = 7.3 cm- 3.4 ern = 3.9 cm
\(\Rightarrow\) \(\bar{XY}\) = \(\bar{AB}\) - \(\bar{CD}\)
i.e. Length of \(\bar{XY}\) = Difference of lengths \(\bar{AB}\) and \(\bar{CD}\).

Given, \(\bar{AB}\) =3.9 cm. Now, to construct required line segment by using compasses, we use the following steps :
Step I Firstly, draw \(\bar{AB}\) = 3.9 cm.
                
Step II Now, to draw an another line I, mark a point P on it.
Step III Place the pointer of compasses at the zero mark of the ruler. Open it to the place of the pencil point up to 3.9 cm mark.
Step IV Without changing the opening of the compasses, place the pointer on P and swing an arc to cut 1 at X.
                  
Step V Measure \(\bar{PX}\), we get \(\bar{PX}\) = 3.9 cm = \(\bar{AB}\).
Step VI Again, without changing the opening of the compasses, place the pointer on X and swing an arc to cut 1at Q.
                      
Step VII Now, measure \(\bar{XQ}\), we get \(\bar{XQ}\) = 3.9 cm = \(\bar{AB}\)
Step VIII \(\bar{PQ}\) = \(\bar{PX}\) + \(\bar{XQ}\) = (3.9 + 3.9) cm
                    = \(\bar{AB}\) + \(\bar{AB}\) = 2\(\bar{AB}\)
Hence, \(\bar{PQ}\) is twice that of \(\bar{AB}\).
Verification: On measuring the length of \(\bar{PQ}\) and \(\bar{AB}\).
We get, \(\bar{PQ}\) = 7.8 ern, \(\bar{AB}\) = 3.9 cm and \(\bar{PQ}\) = 2(\(\bar{AB}\)) =7.8 cm
Thus, twice of \(\bar{AB}\) is equal to \(\bar{PQ}\).