(a) Factors of 8 are: 1,2,4 and 8
Factors of 12 are: 1,2,3,4,6 and 12
Factors of 20 are: 1,2,4,5, 10 and 20
Hence, common factors of 8, 12 and 20 are 1, 2 and 4.
(b) Factors of9 are 1,3 and 9
Factors of 15 are 1, 3, 5 and 15
Factors of 21 are 1,3, 7 and 21
Hence, common factors of 9, 15 and 21 are 1 and 3.

 32,75,829 3  2  7  5  8  2  9
Thirty two lakh seventy five thousand eight hundred twenty nine - 3 x 10,00,000 + 2 x 1,00,000 + 7 x 10,000 + 5 x 1000 + 8 x 100 + 2 x 10 + 9.

If  \(\vec {PQ}\) is a ray, then
(a)its starting point is P.
(b)point Q lies on the way of ray \(\vec {PQ}\)
(c)No, Q is not the starting point of this ray.

We have, 136 

\(\therefore\)136=1x2x2x2x17

\(\frac { 3 }{ 4 } \) revolution

According to the given information, ΔPQR has medians \(\overline{ PS } ,\overline { QT } \) and \(\overline { RN } \). Yes, these medians passing through one point 'O'.

(a) Acute \(\angle AOB\)

(b) 
Obtuse \(\angle POQ\)

(c) Reflex \(\angle ROS\)

Here, we can write the name of angles with the help of given figure:
(a) \(\angle 1=\angle AOB\)
(b)\(\angle 2=\angle BOC\)
(c)\(\angle 3=\angle COD\)

29000 (nearest thousand)

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

In quadrilateral PQRS,

(a) Opposite sides are QR and PS, PQ and SR.
(b) Adjacent sides are \(\bar {PQ}\ and\ \bar {QR},\bar {QR}\ and\ \bar {RS},\bar {RS}\ and\ \bar{SP},\bar {SP}\ and\ \bar {PQ}\).

We have, 439 + 334 + 4317
\(\therefore \begin{matrix} 439\rightarrow 400 \\ 334\rightarrow 300 \\ 4317\rightarrow 4300 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴ Rough estimate = 439 + 334 + 4317
= 400 + 300 + 4300 = 5000
\(Again\begin{matrix} 439\rightarrow 440 \\ 334\rightarrow 330 \\ 4317\rightarrow 4320 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 439 +334 +4317
= 440 + 330 + 4320 = 5090

We have, 297144
(i) Divisibility by 2
\(\because\)Unit digit of number = 4, so 297144 is divisible by2.
(ii) Divisibility by 3
Sum of digits of given number = 2 + 9 + 7 + 1+ 4 + 4 = 27
\(\because\) 27 is divisible by 3, so 297144 is also divisible by 3.
Now, we see that 297144 is divisible by 2 and 3 both.
Hence, it is divisible by 6.

(a) In given figure, point A lies between the sides OD and OE.
So, it is in the interior of\(\angle DOE\).
(b) In given figure, points C, A and D are in the exterior of \(\angle EOF\).
(c) In given figure, point E, B, O and F lie on the sides OE and OFof \(\angle EOF\), so these points are on \(\angle EOF\).

A curve, which cut itself called closed curve.

Given, weight of one tablet = 20 mg
Total weight of 200000 tablets in the box = 200000 \(\times\) 20 mg
=4000000 mg
Now, weight in grams =\(\frac{4000000}{1000}g=4000g\) \(​​\left[ \because \quad 1mg=\frac { 1 }{ 1000 } g \right] \)
Now, weight in kilograms =\(\frac{4000}{1000}kg=4kg\) \(​​\left[ \because \quad 1g=\frac { 1 }{ 1000 } kg \right] \).

Yes, if the product of two whole numbers is zero, then one or both of them will be zero. Since, we know that the product of any whole number with zero is always zero i.e. a x 0 = 0, where a is any whole number and 0 x 0 = 0. Some examples are as follows:

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 \(\vec {NB}\).

From the given figure,
(i) Total distance covered by bus from A to D
= AB + BC + CD =4170 +3410 + 2160 =9740 km
(ii) Total distance covered by bus from D to G
= DE + EF + FG = 8140 + 4830 + 2550 = 15520 km
(iii) Total distance covered by bus = Distance from A to D + Distance from D to G + Distance from G to A
=9740+15520+1290 [from (i) and (ii) parts)
=26550 km
(iv) Distance from C to D = 2160 km
Distance from D to E = 8140 km
∴ Difference of distances from C to D and D to E
=8140 - 2160 = 5980 km
(v) To find out the time for individual places, we will use the formula of time i.e.
\(Time=\frac { Distance }{ Speed } \)
(a) Time taken by bus to reach from A to B
\(=\frac { Distance\quad from\quad AtoB }{ Speed } =\frac { 4170 }{ 60 } =\frac { 139 }{ 2 } =69\frac { 1 }{ 2 } h\)
(b) Time taken by bus to reach from C to D
\(=\frac { Distance\quad from\quad CtoD }{ Speed } =\frac { 2160 }{ 60 } =36h\)
(c) Time taken by bus to reach from E to G
\(=\frac { Distance\quad from\quad EtoG }{ Speed } =\frac { 4830+2550 }{ 60 } =\frac { 7380 }{ 60 } =123h\)
(d) Time taken by bus for total journey
\(=\frac { Total\quad distance\quad covered }{ Speed } =\frac { 26550 }{ 60 } =\frac { 885 }{ 2 } =442\frac { 1 }{ 2 } h\).

A rough figure in each of the cases is given below:
(a) Point P lies on \(\bar {AB}\).

(b)\(\overleftrightarrow {XY}\) and \(\overleftrightarrow {PO}\) intersect at M.

(c) Line I contains E and F but not D.

(d) \(\overleftrightarrow {OP}\) and \(\overleftrightarrow {OQ}\) meet at O.

Given digits are 4, 5, 6, 0, 7 and 8.
We can make many numbers by using these digits, five of them are as follows:
876540, 867540, 876450, 876045 and 867405
(a) After putting commas, numbers are as follows:
(i) 8,76,540 (ii) 8,67,540
(iii) 8,76,450 (iv) 8,76,045
(v) 8,67,405
(b) Ascending order of numbers is as follows:
867405 < 867540 < 876045 < 876450 < 876540
Descending order of numbers is as follows:
876540 > 876450 > 876045 > 867540 >  867405.