## ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 4

**Choose the correct answer from the given four options (1-2):**

Question 1.

\(\frac{103^{2}-97^{2}}{200}\) is equal to

(a) 3

(b) 4

(c) 5

(d) 6

Solution:

Question 2.

If the’ sum of three consecutive even integers is 36, then the largest integer is

(a) 10

(b) 12

(c) 14

(d) 16

Solution:

Let first number = x

Second number = x + 2

and third = x + 4 x + x + 2 + x + 4 = 36

⇒ 3x + 6 = 36 ⇒ 3x = 36 – 6 = 30

⇒ x= \(\frac{30}{3}\) = 10

∴ Number are : 10, 10 + 2, 10 + 4

= 10, 12, 14

Largest integer =14 (c)

Question 3.

Find the area of rectangle whose length and breadth are respectively (4x^{2} – 3x + 7) and (3 – 2x + 3x^{2}).

Solution:

Length of a rectangle (l) = 4x^{2} – 3x + 7

and breadth of a rectangle (b) = 3 – 2x + 3x^{2}

Area of a rectangle

= (4x^{2} – 3x + 7) (3 – 2x + 3x^{2})

= 4x^{2}(3 – 2x + 3x^{2}) – 3x(3 – 2x + 3x^{2}) + 7(3 – 2x + 3x^{2})

= 12x^{2} – 8x^{3} + 12x^{42} – 9x + 6x^{2} – 9x^{3} + 21 – 14x + 21x^{2}

= 12x^{4} – 17x^{3} + 39x^{2} – 23x + 21

Question 4.

Factorize: a^{2} – c^{2} – 2ab + b^{2}.

Solution:

a^{2} – 2ab + b^{2} – c^{2}

= (a – b)^{2} – (c)^{2}

= (a – b + c) (a – b – c)

Question 5.

The ages of A and B are in the ratio 3 : 4. Five years later the sum of their ages will be 31 years. What are their present ages?

Solution:

Ratio in the present ages of A and B = 3 : 4

Let A’s age = 3x, then B’s age = 4x

5 years later,

A’s age = 3x + 5 and B’s age = 4x + 5

According to the condition,

3x + 5 + 4x + 5 = 31 7x + 10 = 31

⇒ 7x = 31 – 10 = 21 ⇒ x = \(\frac{21}{7}\) = 3

∴ A’s present age = 3 × 3 = 9 years

and B’s age = 4 × 3 = 12 years

Question 6.

The sum of the digits of a two digit number is 13. If the number obtained by reversing the digits is 45 more than the original number. Find the original number.

Solution:

Sum of two digits of a 2-digit number = 13

Let unit’s digit = x

Then ten’s digit = 13 – x

and number = x + 10(13 – x)

= x + 130 – 10x

= 130 – 9x …(i)

By reversing the digits,

Unit’s digit =13 – x

and ten’s digits = x

∴ Number = 13 – x + 10x = 13 + 9x

According to the condition,

13 + 9x= 130 – 9x + 45

9x + 9x = 130 – 13 + 45

⇒ 18x= 162

⇒ x= \(\frac{162}{18}\) = 9

∴ Original number = 130 – 9x

= 130 – 9 × 9 = 130 – 81 =49

Question 7.

The ratio between an exterior angle and interior angle of a regular polygon is 1 : 5, find:

(i) the measure of each exterior angle,

(ii) the measure of each interior angle.

(iii) the number of sides of the polygon.

Solution:

In a regular polygon,

Ratio in exterior angle and interior angle = 1 : 5

Let exterior angle = x

Then interior angle = 5x

but sum of interior angle and exterior angle = 180° (Linear pair)

∴ x + 5x – 180°

⇒ 6x = 30°

∴ x = \(\frac{180^{\circ}}{6}\) = 30°

(i) ∴ Measure of exterior angle = 30°

(ii) Measure of interior angle = 30° × 5 = 150°

(iii) Number of sides

Question 8.

Solve the inequality: 3 – \(\frac{x}{2}\) > 2 – \(\frac{x}{3}\) , x ϵ W.

Also represent its solution set on the number line.

Solution:

Question 9.

Factorise: x^{2} + \(\frac{1}{x^{2}}-7\left(x-\frac{1}{x}\right)\) + 8.

Solution:

Question 10.

In the given figure, ABCD is a parallelogram. Find x, y, z and w.

Solution:

In ||gm ABCD

DL ⊥ AB and DM ⊥ BC

∠ADL = 20°

In ∆ADL, ∠L = 90°

∴ ∠A + ∠L + ∠ADL = 180° (Sum of angles of a triangle)

x + 90° + 20° = 180°

⇒ x + 110° = 180°

x= 180°- 110° = 70°

∠C = ∠A (Opposite angles of a ||gm)

∠C = 70°

Similarly in ∆CDM

∠C + ∠CDM + ∠M= 180°

70° + y + 90° = 180°

⇒ y + 160° = 180°

⇒ y= 180°- 160° = 20°

In ||gm ABCD,

∠A + ∠B = 180° (Co-interior angles)

70° + y = m° ⇒ z = 180°- 70°= 110°

∠ADC = ∠B (Opposite angles of a ||gm)

∠ADC =110°

20 ° + w + y = z

20° + w + 20° = 110°

w + 40° = 110°

w= 110° -40° = 70°

Hence, x = 70°, y = 20°, y = 110° and w = 70°