\(\because\) 60 =30 x 2
30 = 10 x 3 and 10 = 5 x 2

\(\therefore\)Hence, missing numbers are 2, 3, 5 and 2.

Factors of 8 and 20 are as follows
8 = 1 x 8; 8 = 2 x 4 and 20 = 1 x 20; 20 = 2 x 10; 20 = 4 x 5
Now, all factors of 8 = 1, 2, 4, 8 and all factors of 20 = 1, 2, 4, 5, 10, 20
\(\therefore\) Common factors = 1, 2, 4
Hence, common factors of 8 and 20 are 1, 2 and 4.

(a) Ray, which contains point A is \(\vec {PA}\).
(b) Ray, which contains point B is \(\vec {QB}\).

2943

The measure of given angle = 110°.

12159 is not divisible by 4.
12159 is not divisible by 8.

On the number line, starting from -2 and moving 4 points towards right (each step being equal to 1 unit), we will reach at 2.

62 + 2

Here, angles are \(\angle AOB,\angle BOC\) and \(\angle AOC\).

Here, we have 32
32 = 29 +3

We have, 18
18 = 1 x 18; 18 = 2 x 9; 18 = 3 x 6
\(\therefore\)Factors of 18 are 1, 2, 3, 6, 9 and 18

We have, 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + (769 x 5) x 5 x 218  [\(\because\)25 = 5 x 5]
                                                              = 3845 x 5 x 782 + 3845 x 5 x 218
                                                              = 3845 x 5 x (782 + 218) [taking 3845 x 5 as common terms]
                                                              = 3845 x 5 X 1000 = 19225 x 1000 = 19225000.

Firstly, draw a line AC of 8 cm. Take a point B on AC such that AB = 5 cm.

Now, distance of BC = AC - AB = 8 - 5 = 3 cm
Thus, AB + BC = AC
So, the point B lies between points A and C.

The line segments are\(\bar { AB}\) and \(\bar { AC}\).
Yes, A is the end point of each line segment.

To find 4 + 5
Let us start from 4. Since, we have to add 5 to this number, we make 5 jumps to the right of 4. Each jump being equal to 1 unit. After five jumps, we reach at 9.

\(\therefore\)  4 + 5 = 9.

No, all whole numbers are not natural numbers because 0 is a whole number but it is not a natural number.

Given, number is 97621.
Hence, place value of 7 is 7000 and face value of 7 is 7.

Both 97632 and 96543 have same number of digits i.e. five. Now, compare them by looking at the digits at ten thousand place, which is 9 for both the numbers.
Next, compare the second digit from the left. Here, it is 7 in 97632 and 6 in 96543.
Since, 7> 6
So, 97632> 96543

For the HCF of 58, 70.

\(\therefore\) HCF of 58 and 70 is 2.

(i) The interior points of \(\angle AOB\) are S and Q.
(ii) The exterior points of \(\angle AOB\) are P and R.
(iii) The points, which lie on \(\angle AOB\) are A. O, B, T and N.

54 \(\times\) 105 = 54 \(\times\) (100 + 5)
= 5400 + 54 \(\times\) 5 = 5400 + 270 = 5670

Rs. 16570

We know that, a ray is a part of a line which starts at one point and goes endlessly in a direction.
(a) In the given figure, there are four rays namely \(\vec {TA}\), \(\vec {TN}\), \(\vec {NB}\) and \(\vec {TB}\).
(b) No, T is not a starting point of each of these rays. T is a starting point of \(\vec {TA}\), \(\vec {TB}\) and \(\vec {TN}\) only but not of \(\vec {NB}\).

Digits in ascending order are 0,1,2,3,4,5,6,7,8,9 and digits in descending order are 9,8,7,6,5,4,3,2,1, 0. Also, we know that 0 can never be taken at the leftmost place.
Keeping the digit 9 at hundreds place, we can fill thousands, tens and ones place by remaining digits.
Greatest number

[Keeping the greatest digit i.e. 8, 7 and 6 at thousands, tens and ones place]
Smallest number 

[Keeping the smallest digit l.e. 1, 0 and 2 at thousands, tens and ones place]

Firstly, draw a number line.

Let +1 be an integer and its additive inverse is -1. From the number line, we see that +1 is 2 more than its additive inverse. So, we reject this integer.
Again, let +2 be an integer, its additive inverse is -2. From the number line, we see that +2 is 4 more than its additive inverse.
Hence, the required integer is 2.

There are 10 angles,i.e. \(\angle EOD,\angle DOC,\angle COB,\angle BOA,\angle EOC,\angle DOB,\angle COA,\angle EOB,\angle DOA,\angle EOA\)

Given, van can carry a weight of 800 kg
= 800 \(\times\) 1000 g = 800000 g
According to the question,
Weight of one packet = 4 kg 500 g
= 4 \(\times\) 1000 + 500

= 4000 + 500
= 4500 g
∴ Number of packets that can be loaded in the van\(=\frac { Total\quad weight\quad strength\quad of\quad the\quad van }{ Weight\quad of\quad one\quad packet } \)
\(=\frac { 800000 }{ 4500 } =\frac { 8000 }{ 45 } \)
Hence, only 177 boxes can be loaded in the van.

The integers lying between -2 and 2 are -1, 0 and 1. So, there are 3 integers.

We know that, the product of two whole numbers is always a whole number.
e.g. 0 x 0 = 0 and 1 x 2 = 2.

Let the whole number be x.
Now, according to the question,
x + 25 + 25 - x = 50.