\(\frac { 4 }{ 9 } +\frac { 2 }{ 7 } =\frac { 46 }{ 63 } \)

This figure has four lines of symmetry.

According to International system of numeration 7,869,420

Negative integers greater than -20 on the number line are shown as below:

When we move on the number line to the right, then the value of integer increases.
So, it is clear that from the figure that when we move right of -20, we will get integers greater than -20. So, four negative integers greater than -20 are -19, -18, -17 and -16.

We have, 4275 x 125 = 4275 x (100 + 20 + 5)
                               = 4275 x 100 + 4275 x 20 + 4275 x 5 [by distributive property of multiplication over addition]
                              = 427500 + 85500 + 21375
                              = 513000 + 21375 = 534375.

We have, 36
36 = 1x 36; 36 = 2 x 18; 36 = 3 x 12;
36=4x9;36=6x6
\(\therefore\)Factors of 36 are 1, 2, 3, 4, 6, 9,12,18 and 36.

We have, 4 x 166 x 25 = (4 x 25) x 166
                                 = 100 x 166 = 16600.

Area = 5 x 3 = 15 sq cm

The additive inverse of 14 is -14.

We know that,
1 million = 1000000 (in digits)
= 10 \(\times\)100000 = ten \(\times\) one lakhs (in number name)
∴1 million = ten lakhs

Here, 270.24 + 12.52+ 1.5 can be written as

270.24 + 12.52 + 1.50 = 284.26

The required circles, are given below.

(a) 6 : 5
(b) 12 : 11

Here, the letter S has no line of symmetry.

(a) Area of Karia district = 6000 sq km
(b) Raigarh and Jashpur have the same area.
(c) Four districts have area more than 5000 sq km.

Given, area of the rectangular garden = 300 sq m
and length of the rectangular garden = 50 m
We know that,
Area of the rectangular garden = Length \(\times\) Breadth
⇒ 300 sq m = 50 m \(\times\) Breadth
∴ Breadth =\(\frac { 300\quad sqm }{ 50m } =6m\)
Hence, the breadth or width of the garden is 6 m.

We have, 9 + (- 6)

Firstly,draw a number line, then, wemove 9 steps to the right of 0 and reach at 9. Now,wemove6 steps to thelen of9 [∴  -6 is a negative integer] and reach at 3, as shown in the above figure. Hence, 9 + (-6) = 3

We have, 968 x 73 + 968 x 27
       = 968 (73 + 27) [taking 968 as common term]
       = 968 x 100 = 96800.

Using the digits 6, 2, 7, 4 and 3, we get
Greatest number = 76,432 and least number = 23,467
∴ Difference = Greatest number - Least number
=76,432 - 23,467 = 52,965.

Given, number of tickets sold on the first day = 1094
Number of tickets sold on the second day = 1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the fourth day = 2751
∴ Total number of tickets sold on all the four days
= 1094 + 1812 + 2050 + 2751 = 7707
Hence, there are 7707 tickets sold on all the four days.

In the given figure,
Number of triangles = 3; Number of squares = 2
Number of circles = 2
\(\therefore\) Total number of figures = 3 + 2 + 2 = 7
(a) Required ratio of number of triangles to the number of circles \(=\frac{Number\ of\ triangles}{Number\ of\ circles}=\frac{3}{2}=3:2\)
(b) Required ratio of number of squares to all the figure \(=\frac{Number\ of\ squares}{Total\ number\ of\ figures}=\frac{2}{7}=2:7\)
(c) Required ratio of number of circles to all the figure \(=\frac{Number\ of\ circles}{Total\ number\ of\ figures}=\frac{2}{7}=2:7\)

The table with tally marks is shown below:

Marks	Tally marks	Number of students
1	II	2
2	III	3
3	III	3
4		7
5		6
6		7
7		5
8	IIII	4
9	III	3
Total		40

(a) Here, we have to find out the students, who obtained marks equal to 7 or more than 7. So, we have to add the number of students who obtained marks equal to 7 or more than 7 i.e. 7, 8 and 9.
\(\therefore\) Number of those students = 5 + 4 + 3 = 12
Hence, 12 students obtained marks equal to 7 or more than 7.
(b) Here, we have to find out the students who obtained marks below 4.
So, we have to add the number of students who obtained marks below 4 i.e. 1, 2 and 3.
\(\therefore\) Number of those students = 3 + 3 + 2 = 8
Hence, 8 students obtained marks below 4.

The given decimals can be written as
(a) 0.29=\(0+\frac { 2 }{ 10 } +\frac { 9 }{ 100 } \)
(b) 2.08=\(2+\frac { 0 }{ 10 } +\frac { 8 }{ 100 } \)
(c) 19.60=10+9+\(\frac { 6 }{ 10 } +\frac { 0 }{ 100 } \)
(d) 14832=100+40+8+\(\frac { 3 }{ 10 } +\frac { 2 }{ 100 } \)
(e) 200.812=200+00+0+\(\frac { 8 }{ 10 } +\frac { 1 }{ 100 } +\frac { 2 }{ 1000 } \)
Now, the place value table is given below:

\(\frac { 20 }{ 30 } =\frac { 2 }{ 3 } \)= 2:3