# Model Questions Paper-6 Class-8 ML Aggarwal Solutions

Model Questions Paper-6 Class-8 ML Aggarwal ICSE Mathematics Solutions. APC Understanding Mathematics for ICSE Class-8 Model Question Paper-6 Solutions of Section-A , B, C and D based on Chapter-10 to 19. Visit official Website **CISCE **for detail information about ICSE Board Class-8 Mathematics.

## Model Questions Paper-6 Class-8 ML Aggarwal ICSE Mathematics Solutions.

( based on chapter-10 to 19)

Time Allowed-2 and 1/2 hour

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### Section-A, Paper-6 Class-8 ML Aggarwal

**Questions 1 to 8 are of 1 mark each.**

Question 1.

If x + y = 11 and xy = 24, then x^{2} + y^{2} =

(a) 121

(b) 73

(c) 48

(d) none of these

#### Answer

x + y = 11, xy = 24,

We’know that, (x + y)^{2} = x^{2} + y^{2} + 2xy

x^{2} + y^{2} = (x + y)^{2} – 2xy

= (11)^{2} – 2 × 24 = 121 – 48 = 73 (b)

Question 2.

Factors of a^{3} – 64a are

(a) a(a – 8)^{2}

(b) (a + 8) (a – 8)

(c) a(a + 8) (a – 8)

(d) a(a + 8)^{2}

#### Answer

a^{3} – 64a = a[a^{2} – (8)^{2}]

= a(a + 8)(a – 8) (c)

Question 3.

If 3(y – 3) = 5(2y + 1) then y is equal to

(a) 3

(b) -3

(c) 2

(d) -2

#### Answer

3(y – 3) = 5(2y + 1)

⇒ 3y – 9 = 10y + 5

⇒ 3y – 10y = 5 + 9

⇒ -7y = 14

⇒ y = = -2 (d)

Question 4.

If one angle of a parallelogram is 30° less than other angle, then the larger angle is

(a) 150°

(b) 75°

(c) 105°

(d) 210°

#### Answer

One angle of a parallelogram = 30 less then the other

n – y = 30°

But x + y= 180°

∴ Adding, 2x = 210° ⇒ x = = 105°

∴ One angle = 105°

and other angle = 180° – 105° = 75°

Larger angle = 105° (c)

Question 5.

A polyhedron has 7 faces, 10 vertices and 15 edges. Name of the polyhedron is

(a) hexagonal prism

(b) pentagonal prism

(c) hexagonal pyramid

(d) none of these

#### Answer

A polyhedron has 7 faces, 10 vertices and 15 edges,

It is a pentagonal prism. (b)

Question 6.

The base of a triangle is twice its height. If the area of the triangle is 25 cm^{2}, then the base is

(a) 10 cm

(b) 20 cm

(c) 5 cm

(d) 25 cm

#### Answer

Base of the triangle is twice its height.

Let height = x cm

Then base = 2x cm

∴ Area = × base × height

⇒ × 2x × x = x^{2}

⇒ x^{2} = 25 = (25)^{2}

⇒ x = 5

∴ Base = 2x = 2 × 5 = 10 cm (a)

Question 7.

From a well-shuffled deck of 52 playing cards, one card is drawn at random then the probability that card drawn is a blackjack

#### Answer

Number of possible outcomes = 52

One card is drawn, then

Probability of the card is a blackjack = (b)

Question 8.

Amongst the following digits, the one having exactly one line of symmetry is

(a) 0

(b) 3

(c) 6

(d) 8

#### Answer

One line of symmetry can be drawn of the digit 3. (b)

**Model Questions Paper-6 Class-8 ML Aggarwal Section-B**

**Questions 9 to 14 are of 2 marks each.**

Question 9.

Using suitable identity, evaluate (9.7)^{2}.

#### Answer

(9.7)^{2} = (10 – 0.3)^{2}

= (10)^{2} – 2 × 10 × 0.3 + (0.3)^{2}

= 100 – 6 + 0.09

= 94.09

Question 10.

Factorize x^{2}y^{2}(a^{4} + b^{4}) + a^{2}b^{2}(x^{4} + y^{4}).

#### Answer

x^{2}y^{2}(a^{4} + b^{4}) + a^{2}b^{2}(x^{4} + y^{4})

= x^{2}y^{2}a^{4} + x^{2}y^{2}b^{4} + a^{2}b^{2}x^{4} + a^{2}b^{2}y^{4}

= x^{2}y^{2}a^{4} + a^{2}b^{2}x^{4} + x^{2}y^{2}b^{4} + a^{2}b^{2}y^{4}

= x^{2}a^{2}(y^{2}a^{2} + b^{2}x^{2}) + y^{2}b^{2}(x^{2}b^{2} + a^{2}y^{2})

= x^{2}a^{2}(b^{2}x^{2} + a^{2}y^{2}) + y^{2}b^{2}(b^{2}x^{2} + a2^{2}y^{2})

= (b^{2}x^{2} + a^{2}y^{2}) (a^{2}x^{2} + b^{2}y^{2})

Question 11.

Sum of three consecutive odd numbers is 153. Find the numbers.

#### Answer

Let first odd number = 2x + 1

Second odd number = 2x + 3

and third odd number = 2x + 5

According to the condition,

2x + 1 + 2x + 3 + 2x + 5 = 153

⇒ 6x + 9 = 153

⇒ 6x = 153 – 9 = 144

⇒ x = = 24

∴ First odd number = 2 × 24 + 1 = 49

Second number = 49 + 2 = 51

and third number = 51 + 2 = 53

Hence numbers are 49, 51, 53

Question 12.

If each interior angle of a regular polygon is 108°, then find the number of sides of the polygon.

#### Answer

Each interior angle of a regular polygon = 108°

Let n be the number of sides, then

× 90° = 108°

⇒

⇒ 20n – 40 = 12n

⇒ 20n – 12n = 40

8n = 40

⇒ n= = 5

∴ Number of sides = 5

Question 13.

Verify Euler’s formula for decagonal pyramid.

#### Answer

In a decagonal pyramid.

Number of faces = 11

Number of vertices = 11

Number of edges = 20

Using Euler’s formula

F + V = E + 2

⇒ 11 + 11 = 20 + 2

⇒ 22 = 22

Hence proved.

Question 14.

A cuboid is of dimensions 24 cm × 30 cm × 36 cm. How many small cubes with side 6 cm can be placed in the cuboid?

#### Answer

Dimensions of a cuboid = 24 cm × 30 cm × 36 cm

Edge (side) of a small cube = 6 cm

∴ Number of small cubes which will be kept in the cuboid = (∵ )

= 4 × 5 × 6 = 120

**Section-C Model Questions Paper-6 Class-8 ML Aggarwal **

**Questions 15 to 24 are of 4 marks each.**

**
**Question 15.

Simplify the following expressions and evaluate it when x = 1, y = -1

(5xy – 3x

^{2}+ 2y

^{2}) × (5y

^{2}– 8xy + 3x

^{2}) + 9x

^{3}y – 8y

^{4}+ 12x

^{4}

#### Answer

(5xy – 3x^{2} + 2y^{2}) × (5y^{2} – 8xy + 3x^{2}) + 9x^{3}y – 8y^{4} + 12x^{4}

= 5xy(5y^{2} – 8xy + 3x^{2}) – 3x^{2}(5y^{2} – 8xy + 3x^{2}) + 2y^{2}(5y^{2} – 8xy + 3x^{2})

+ 9x^{3}y – 8y^{4} + 12x^{4}

= 25xy^{3} – 40x^{2}y^{2} + 15x^{3}y – 15x^{2}y^{2} + 24x^{3}y – 9x^{4} + 10y^{4} – 16xy^{3}

+ 6x^{2}y^{2} + 9x^{3}y – 8y^{4} + 12x^{4}

= 3x^{4} + 48x^{3}y – 49x^{2}y^{2} + 9xy^{3} + 2y^{4}

When x = 1, y = -1, then

= 3(1)^{4} + 48(1)^{3}(-1) – 49(1)^{2}(-1 )^{2} + 9(1)(-1)^{3} + 2(-1)^{4}

= 3 – 48 – 49 – 9 + 2

= -106 + 5 = -101

Question 16.

Factorize: 15(2x – 3y)^{2} – 4(4x – 6y) – 16

#### Answer

15(2x – 3y)^{2} – 4(4x – 6y) – 16

15(2x – 3y)^{2} – 4 × 2(2x – 3y) – 16

Let 2x – 3y = p, then

15p^{2} – 8p – 16

15p^{2} – 20p + 12p – 16

{∵15 × (-16) = -240

-240}

= 5p(3p – 4) + 4(3p – 4)

= (3p – 4) (5p + 4)

= [3(2x – 3y) – 4] [5(2x – 3y) + 4]

= (6x – 9y – 4) (10x – 15y + 4)

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