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

max mark-90

-: Select Topics :-

Section-A

Section-B

Section-C

Section-D

### 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 x2 + y2 =
(a) 121
(b) 73
(c) 48
(d) none of these

x + y = 11, xy = 24,
We’know that, (x + y)2 = x2 + y2 + 2xy
x2 + y2 = (x + y)2 – 2xy
= (11)2 – 2 × 24 = 121 – 48 = 73 (b)

Question 2.
Factors of a3 – 64a are
(a) a(a – 8)2
(b) (a + 8) (a – 8)
(c) a(a + 8) (a – 8)
(d) a(a + 8)2

a3 – 64a = a[a2 – (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

3(y – 3) = 5(2y + 1)
⇒ 3y – 9 = 10y + 5
⇒ 3y – 10y = 5 + 9
⇒ -7y = 14
⇒ y = $\frac{14}{-7}$ = -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°

One angle of a parallelogram = 30 less then the other
n – y = 30°
But x + y= 180°
∴ Adding, 2x = 210° ⇒ x = $\frac{210}{2}$ = 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

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 cm2, then the base is
(a) 10 cm
(b) 20 cm
(c) 5 cm
(d) 25 cm

Base of the triangle is twice its height.
Let height = x cm
Then base = 2x cm
∴ Area = $\frac{1}{2}$ × base × height
⇒ $\frac{1}{2}$ × 2x × x = x2
⇒ x2 = 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

Number of possible outcomes = 52
One card is drawn, then
Probability of the card is a blackjack = $\frac{2}{52}=\frac{1}{26}$ (b)

Question 8.
Amongst the following digits, the one having exactly one line of symmetry is
(a) 0
(b) 3
(c) 6
(d) 8

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.

(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 x2y2(a4 + b4) + a2b2(x4 + y4).

x2y2(a4 + b4) + a2b2(x4 + y4)
= x2y2a4 + x2y2b4 + a2b2x4 + a2b2y4
= x2y2a4 + a2b2x4 + x2y2b4 + a2b2y4
= x2a2(y2a2 + b2x2) + y2b2(x2b2 + a2y2)
= x2a2(b2x2 + a2y2) + y2b2(b2x2 + a22y2)
= (b2x2 + a2y2) (a2x2 + b2y2)

Question 11.
Sum of three consecutive odd numbers is 153. Find the numbers.

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 = $\frac{144}{6}$ = 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.

Each interior angle of a regular polygon = 108°
Let n be the number of sides, then
$\frac{2 n-4}{n}$ × 90° = 108°
⇒ $\frac{2 n-4}{n}=\frac{108}{90}=\frac{12}{10}$
⇒ 20n – 40 = 12n
⇒ 20n – 12n = 40
8n = 40
⇒ n= $\frac{40}{8}$ = 5
∴ Number of sides = 5

Question 13.
Verify Euler’s formula for decagonal pyramid.

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?

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 = $\frac{24 \times 30 \times 36}{6 \times 6 \times 6}$ (∵ $\frac{\text { Volume of cuboid }}{\text { Volume of one cube }}$)
= 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 – 3x2 + 2y2) × (5y2 – 8xy + 3x2) + 9x3y – 8y4 + 12x4

(5xy – 3x2 + 2y2) × (5y2 – 8xy + 3x2) + 9x3y – 8y4 + 12x4
= 5xy(5y2 – 8xy + 3x2) – 3x2(5y2 – 8xy + 3x2) + 2y2(5y2 – 8xy + 3x2)
+ 9x3y – 8y4 + 12x4
= 25xy3 – 40x2y2 + 15x3y – 15x2y2 + 24x3y – 9x4 + 10y4 – 16xy3
+ 6x2y2 + 9x3y – 8y4 + 12x4
= 3x4 + 48x3y – 49x2y2 + 9xy3 + 2y4
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

15(2x – 3y)2 – 4(4x – 6y) – 16
15(2x – 3y)2 – 4 × 2(2x – 3y) – 16
Let 2x – 3y = p, then
15p2 – 8p – 16
15p2 – 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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