i) Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
Composite numbers = 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 21, 22, 24, 25, 26, 27, 28, 30
ii) Odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31
Even numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28
iii) Multiplies of 6 = 6, 12, 18, 24, 30,
Multiplies of 4 = 4, 8, 12, 16, 20, 24, 28
Common multiples = 12, 24
LCM = 12
iv) Monday falls on = 7, 14, 21, 28

i) 2 cm² = 2 \(\times\)100 = 200 mm²
ii) 18 m² =18 \(\times\)10000 = 1,80,000 cm²
iii) 5 km² = 5 \(\times\) 1000000 = 50,00,000 m²

a = 3 cm
b = 4 cm
c = 5 cm
P = (a + b + c) units
= 3 + 4 + 5 = 12 cm
Perimeter of the triangle is 12 cm.

i) 27, 15, 14, 11, 0, −9, −14, −17
ii) 400, 78, 65, −46, −99, −120, −600
iii) 777, 555, 333, 111, −222, −444, −666, −888

The quantity of milk left over = \(5{1\over2}-3{1\over4}\)
Here, note that 5 > 3 and \({1\over2}>{1\over4}\)
The whole numbers 5 and 3 and the fractional numbers \(1\over2\) and \(1\over4\)can be subtracted separately
So, \(5{1\over2}-3{1\over4}=(5-3)+\left({1\over 2}-{1\over4}\right)\)
\(=2+\left({2\over 4}-{1\over4}\right)\)
\(=2+{1\over4}={1\over4}\)litres

Ravi’s way
Common Multiple of 2 and 4 = 4.
Equivalent Fractions of
\({1\over2}={1\times2\over 2\times2}={2\over4}\)
Now, add \({1\over 3}+{1\over 4}={2\over4}+{1\over 4}\)
\(={2+1\over 4+4}={3\over8}\)
Therefore Venkat has bought \(3\over 8\) litres of milk.
Arun’s way
Common Multiple of 2 and 4 = 4.
Equivalent Fractions of
\({1\over 2}={1\times2\over 2\times2}={2\over4}\)
Now, add  
\({1\over 3}+{1\over 4}={2\over4}+{1\over 4}\)\({1\over 3}+{1\over 4}={2\over4}+{1\over 4}\)
\(={2+1\over 4}={3\over4}\)
Therefore Venkat has bought \(3\over4\) litres of milk.

Shape	Equilateral triangle	Square	Regular pentagon	Regular hexagon	Regular octagon
Number of lines of symmetry	3	4	5	6	8
Order of rotational symmetry	3	4	5	6	8

(i) A regular polygon of 10 sides will have 10 lines of symmetry.
(ii) If a regular polygon has, 10 lines of symmetry, then its order ofrotational symmetry is 10
(iii) A regular polygon of 'n' sides has n, n lines of symmetry and the order of rotational symmetry is n.

From the figure,
Complete squares = 9
half squares = \(\cfrac { 13 }{ 2 } \)
\(\therefore\) Approximate area = Complet esquares + half squares
= \(9+\cfrac { 13 }{ 2 } \)
= \(\cfrac { 18+13 }{ 2 } =\cfrac { 31 }{ 2 } \)
= 15.5 sq.units

length = 40 cm
breadth = 20 cm
Side of the square = 10 cm
Area of the rectangle = lb
= ( 40 \(\times\) 20)cm²
= 800 cm²
(side) s = 10cm
Area of a square (S2) = 102 = 100 cm²
No. of square = \(\cfrac { 800 }{ 100 } =8\)

C3: The opposite to the second negative integer.
C5: The additive identity in whole numbers.
C6: The successor of the integer in C2.
C8: The predecessor of the integer in C7.
C9: The opposite to the integer in C5.

(i) \(\cfrac { 3 }{ 7 } \div 4\)
= \(\cfrac { 3 }{ 7 } \times \cfrac { 1 }{ 4 } =\cfrac { 3\times 1 }{ 7\times 4 } =\cfrac { 3 }{ 28 } \)
(ii) \(\cfrac { 4 }{ 3 } \div \cfrac { 5 }{ 9 } \)
= \(\cfrac { 4 }{ 3 } \times \cfrac { 9 }{ 5 } \)
= \(\cfrac { 4\times 3 }{ 5 } =\cfrac { 12 }{ 5 } =2\cfrac { 2 }{ 5 } \)
(iii) \(4\cfrac { 1 }{ 5 } \div 3\cfrac { 3 }{ 4 } \)
= \(\cfrac { 21 }{ 5 } \div \cfrac { 15 }{ 4 } \)

= \(\cfrac { 21 }{ 5 } \times \cfrac { 4 }{ 15 } =\cfrac { 7\times 4 }{ 5\times 5 } =\cfrac { 28 }{ 25 } =1\cfrac { 3 }{ 25 } \)

(iv) \(9\cfrac { 2 }{ 3 } \div 1\cfrac { 2 }{ 3 } \)
= \(\cfrac { 29 }{ 3 } \div \cfrac { 5 }{ 3 } \)
= \(\cfrac { 29 }{ 3 } \times \cfrac { 3 }{ 5 } =\cfrac { 29 }{ 5 } 5\cfrac { 4 }{ 5 } \) 

Candidate’s EMIS No 

1. Name of the candidate in Capital Letters followed by initial leaving one box blank. (Do not write Miss/Master)

3. Father’s name in Capital Letters followed by initial leaving one box blank. (Do not write Mr./Dr./Prof)

4. Mother’s Name in Capital Letters followed by initial leaving one box blank. (Do not write Mrs./Dr./Prof)

Area of a rectangle = 820 sq. cm
Width (b) = 20 cm
Length (l) =?
Perimeter =?
\(I=\cfrac { A }{ b } \)
= \(\cfrac { 820 }{ 20 } =41\\ \)
= =2(41+20)
= 2 (61) = 122 cm
\(\therefore\) I = 41 cm , P = 122 cm

15 km east = + 15 km
\(\therefore\) 15 km west = - 15 km

Length of the staircase = \(5\cfrac { 1 }{ 2 } \) m
To one step = \(\cfrac { 1 }{ 4 } \) m
No. of the steps will be there = \(\cfrac { 5\frac { 1 }{ 2 } }{ \frac { 1 }{ 4 } } \)
= \(\cfrac { \frac { 11 }{ 2 } }{ \frac { 1 }{ 4 } } \Rightarrow \cfrac { 11 }{ 2 } \times \cfrac { 4 }{ 1 } =22\)
22 steps will be there in the staircase.