Model Question Paper 4

Using the identity,

a² - b² = (a + b)(a - b)

Solving,

$$\begin{aligned}\Rightarrow \frac{103^2 - 97^2}{200}\\[1em] \Rightarrow \frac{(103 + 97)(103 - 97)}{200}\\[1em] \Rightarrow \frac{200 \times 6}{200}\\[1em] \Rightarrow 6\end{aligned}$$

Hence, option 4 is the correct option.

It is given that the sum of three consecutive even integers is 36.

Let the three consecutive even integers be x, x + 2 and x + 4.

⇒ x + (x + 2) + (x + 4) = 36

⇒ 3x + 6 = 36

⇒ 3x = 36 - 6

⇒ 3x = 30

⇒ x = $\frac{30}{3}$

⇒ x = 10

⇒ x + 2 = 10 + 2 = 12

⇒ x + 4 = 10 + 4 = 14

Thus, the three consecutive even integers are 10, 12 and 14.

So, the largest integer is 14.

Hence, option 3 is the correct option.

Given,

Length = (4x² - 3x + 7)

Breadth = (3 - 2x + 3x²) or (3x² - 2x + 3)

By formula,

Area of a rectangle = length x breadth

Area = (4x² - 3x + 7)(3x² - 2x + 3)

= 4x²(3x² - 2x + 3) - 3x(3x² - 2x + 3) + 7(3x² - 2x + 3)

= 12x⁴ - 8x³ + 12x² - 9x³ + 6x² - 9x + 21x² - 14x + 21

= 12x⁴ - 8x³ - 9x³ + 12x² + 6x² + 21x² - 9x - 14x + 21

= 12x⁴ - 17x³ + 39x² - 23x + 21

Hence, the area of the rectangle is 12x⁴ - 17x³ + 39x² - 23x + 21.

⇒ a² - c² - 2ab + b²

Rearranging the terms,

⇒ a² - 2ab + b² - c²

Using the identity a² - 2ab + b² = (a - b)²,

⇒ (a - b)² - c²

Using the identity p² - q² = (p + q)(p - q),

⇒ (a - b + c)(a - b - c)

Hence, a² - c² - 2ab + b² = (a - b + c)(a - b - c).

It is given that the ages of A and B are in the ratio 3 : 4 and five years later the sum of their ages will be 31 years.

Let the common factor be a. So, the present age of A is 3a years and the present age of B is 4a years.

Five years later,

Age of A = (3a + 5) years and Age of B = (4a + 5) years

⇒ (3a + 5) + (4a + 5) = 31

⇒ 7a + 10 = 31

⇒ 7a = 31 - 10

⇒ 7a = 21

⇒ a = $\frac{21}{7}$

⇒ a = 3

Thus,

Present age of A = 3a = 3 x 3 = 9 years

Present age of B = 4a = 4 x 3 = 12 years

Hence, the present ages of A and B are 9 years and 12 years respectively.

Let the digit at the tens place be a and the digit at the units place be b.

So, the original number is 10a + b and the number obtained by reversing the digits is 10b + a.

As the sum of the digits is 13,

a + b = 13 ...............(1)

Given,

Number obtained by reversing the digits is 45 more than the original number.

⇒ 10b + a = (10a + b) + 45

⇒ 10b + a - 10a - b = 45

⇒ 9b - 9a = 45

⇒ 9(b - a) = 45

⇒ b - a = $\frac{45}{9}$

⇒ b - a = 5 ...............(2)

Adding equations (1) and (2), we get:

⇒ a + b + (b - a) = 13 + 5

⇒ 2b = 18

⇒ b = $\frac{18}{2}$

⇒ b = 9

Substituting b = 9 in equation (1), we get:

⇒ a + 9 = 13

⇒ a = 13 - 9

⇒ a = 4

Thus, the original number = 10a + b = 10 x 4 + 9 = 49

Hence, the original number is 49.

It is given that the ratio between an exterior angle and interior angle of a regular polygon is 1 : 5.

Let the common factor be a. So, the exterior angle is a and the interior angle is 5a.

As the sum of an interior angle and its adjacent exterior angle is 180°,

⇒ a + 5a = 180°

⇒ 6a = 180°

⇒ a = $\frac{180^{\circ}}{6}$

⇒ a = 30°

(i) Each exterior angle = a = 30°

Hence, the measure of each exterior angle is 30°.

(ii) Each interior angle = 5a = 5 x 30° = 150°

Hence, the measure of each interior angle is 150°.

(iii) According to the properties of a polygon,

If each exterior angle of a regular polygon is a°, then the number of sides in the regular polygon is $\frac{360^{\circ}}{a^{\circ}}$.

Number of sides = $\frac{360^{\circ}}{30^{\circ}}$ = 12

Hence, the number of sides of the polygon is 12.

Given,

Inequation : 3 - $\frac{x}{2} \gt 2 - \frac{x}{3}$

L.C.M. of 2 and 3 = 6

Multiplying both sides of the inequality by 6, we get :

$\Rightarrow \Big(3 - \frac{x}{2}\Big) \times 6 \gt \Big(2 - \frac{x}{3}\Big) \times 6$

⇒ 6 x 3 - 6 x $\frac{x}{2}$ > 6 x 2 - 6 x $\frac{x}{3}$

⇒ 18 - 3x > 12 - 2x

⇒ 18 - 12 > 3x - 2x

⇒ 6 > x

⇒ x < 6

As x ∈ W, the values of x are 0, 1, 2, 3, 4 and 5.

So, the solution set is {0, 1, 2, 3, 4, 5}.

The solution set is represented on the number line as shown below:

Hence, the solution set is {0, 1, 2, 3, 4, 5}.

Given,

$x^2 + \frac{1}{x^2} - 7\Big(x - \frac{1}{x}\Big) + 8$

Let $x - \frac{1}{x} = y$

Squaring both the sides,

$$\begin{aligned}\Rightarrow \left(x - \frac{1}{x}\right)^2 = y^2\\[1em] \Rightarrow x^2 - 2 + \frac{1}{x^2} = y^2\\[1em] \Rightarrow x^2 + \frac{1}{x^2} = y^2 + 2\end{aligned}$$

Substituting these values in the given expression,

$$\begin{aligned}\Rightarrow y^2 + 2 - 7y + 8\\[1em] \Rightarrow y^2 - 7y + 10 \\[1em] \Rightarrow y^2 - 5y - 2y + 10\\[1em] \Rightarrow y(y - 5) - 2(y - 5)\\[1em] \Rightarrow (y - 5)(y - 2) \\[1em] \Rightarrow \left(x - \frac{1}{x} - 5\right)\left(x - \frac{1}{x} - 2\right)\end{aligned}$$

Hence, $\bm{x^2 + \frac{1}{x^2} - 7\Big(x - \frac{1}{x}\Big) + 8 = \left(x - \frac{1}{x} - 5\right)\left(x - \frac{1}{x} - 2\right)}$.

It is given that ABCD is a parallelogram in which the perpendicular from D meets AB at M and BC at N, and ∠ADM = 20°, ∠MDN = w, ∠NDC = y, ∠DAB = x and ∠ABC = z.

In ΔADM, ∠AMD = 90° as DM ⊥ AB.

As the sum of all the angles of a triangle is 180°,

⇒ x + ∠ADM + ∠AMD = 180°

⇒ x + 20° + 90° = 180°

⇒ x + 110° = 180°

⇒ x = 180° - 110°

⇒ x = 70°

As the adjacent angles of a parallelogram are supplementary,

⇒ z + x = 180°

⇒ z + 70° = 180°

⇒ z = 180° - 70°

⇒ z = 110°

As the opposite angles of a parallelogram are of equal measure,

∠BCD = ∠DAB = x = 70°

From figure,

∠DCN = ∠BCD = 70°

In ΔDNC, ∠DNC = 90° as DN ⊥ BC.

As the sum of all the angles of a triangle is 180°,

⇒ y + ∠DCN + ∠DNC = 180°

⇒ y + 70° + 90° = 180°

⇒ y + 160° = 180°

⇒ y = 180° - 160°

⇒ y = 20°

As DC ∥ AB and DM ⊥ AB, we have DM ⊥ DC.

So, ∠MDC = 90°

From the figure,

⇒ w + y = ∠MDC

⇒ w + 20° = 90°

⇒ w = 90° - 20°

⇒ w = 70°

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