ICSE Class 10 Mathematics — All Questions

Given,

Total GST paid = ₹ 600

Rate of GST = 12%

Let the price of the article be ₹ $x$.

By formula,

GST = Rate of GST × Price of the article

$$\begin{aligned}\Rightarrow 600 = \frac{12}{100} \times x \\[1em] \Rightarrow x = \frac{600 \times 100}{12} \\[1em] \Rightarrow x = \frac{60000}{12} \\[1em] \Rightarrow x = 5000.\end{aligned}$$

Thus, the price of the article = ₹ 5,000.

Hence, option 3 is the correct option.

Given,

Deposit per month (P) = ₹ 1000

Time (n) = 1 year = 12 months

Maturity value = ₹ 13,500

Total amount deposited = P × n

= 1000 × 12

= ₹ 12,000

By formula,

Interest earned = Maturity value − Total amount deposited

= 13500 − 12000

= ₹ 1,500

So assertion (A) is true.

We know that,

Maturity value = Total amount deposited + Interest earned

⇒ Interest earned = Maturity value − Total amount deposited

So reason (R) is true and it correctly explains the assertion (A).

Thus, Both (A) and (R) are true, and (R) is the correct explanation for (A).

Hence, option 3 is the correct option.

Given,

Number of shares bought = 50

Nominal value (N.V.) of each share = ₹ 100

Market value (M.V.) of each share = ₹ 125

Rate of dividend = 6%

We know that,

Dividend is always paid by the company on the nominal value (face value) of the shares and not on their market value.

Total nominal value of the shares = No. of shares × N.V. of 1 share

= 50 × 100

= ₹ 5,000

Thus, the company will pay the total dividend on ₹ 5,000.

Hence, option 3 is the correct option.

Given,

The roots of the quadratic equation $x^2 - 8x + p^2$ = 0 are equal.

Comparing the given equation with $ax^2 + bx + c$ = 0, we get :

$a = 1, b = -8 \text{ and } c = p^2$

We know that,

For equal roots, discriminant (D) = 0.

$$\begin{aligned}\Rightarrow b^2 - 4ac = 0 \\[1em] \Rightarrow (-8)^2 - 4 \times 1 \times p^2 = 0 \\[1em] \Rightarrow 64 - 4p^2 = 0 \\[1em] \Rightarrow 4p^2 = 64 \\[1em] \Rightarrow p^2 = 16 \\[1em] \Rightarrow p = \pm 4\end{aligned}$$

Since, the positive value of $p$ is required.

∴ $p = 4$.

Hence, option 3 is the correct option.

Given,

12, 18 and $x$ are in continued proportion.

We know that,

If $a, b$ and $c$ are in continued proportion, then $b^2 = ac$.

$$\begin{aligned}\Rightarrow 18^2 = 12 \times x \\[1em] \Rightarrow 324 = 12x \\[1em] \Rightarrow x = \frac{324}{12} \\[1em] \Rightarrow x = 27.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$(x + 2)$ and $(x + 3)$ are the factors of $x^3 + 4x^2 + x - 6$.

⇒ $(x + 2)(x + 3) = x^2 + 5x + 6$

Thus, $x^2 + 5x + 6$ is also the factor of $x^3 + 4x^2 + x - 6$.

$$\begin{array}{l} \phantom{x^2 + 5x + 6\big)\quad }\phantom{x^3 +} x - 1 \\ x^2 + 5x + 6\overline{\smash{\big)}\quad x^3 + 4x^2 + x - 6} \\ \phantom{+++++()}\underline{\underset{-}{+}x^3 \underset{-}{+}5x^2 \underset{-}{+}6x} \\ \phantom{+++++())}\phantom{x^3 +} -x^2 - 5x - 6 \\ \phantom{+++++())}\phantom{x^3 +} \underline{\underset{+}{-}x^2 \underset{+}{-}5x \underset{+}{-}6} \\ \phantom{x^2 + 5x + 6\big)\quad }\phantom{x^3 + 4x^2 +} \times \end{array}$$

Thus,

$$\begin{aligned}x^3 + 4x^2 + x - 6 = (x^2 + 5x + 6)(x - 1) \\[1em] = (x + 2)(x + 3)(x - 1).\end{aligned}$$

Thus, the third factor of the given polynomial is $(x - 1)$.

Hence, option 1 is the correct option.

Given,

Matrix A is of order 1 × 2 and matrix B is of order 2 × 1.

We know that,

If matrix A is of order $m \times n$ and matrix B is of order $n \times p$, then the order of the product AB is $m \times p$.

∴ Order of AB = 1 × 1

Calculating,

$$\Rightarrow AB = \begin{bmatrix*}[r] -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 3 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 3 + 2 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 + 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \end{bmatrix*}.$$

Hence, option 3 is the correct option.

Given,

$$\begin{aligned}n^{th} \text{ term of the G.P.} = 5 \times 2^n \\[1em] \therefore t_n = 5 \times 2^n\end{aligned}$$

Replacing $n$ by $(n + 1)$, we get :

⇒ $t_{n + 1} = 5 \times 2^{n + 1}$

By formula,

$$\begin{aligned}\text{Common ratio } (r) = \frac{t_{n + 1}}{t_n} \\[1em] = \frac{5 \times 2^{n + 1}}{5 \times 2^n} \\[1em] = 2^{n + 1 - n} \\[1em] = 2.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

Equation of the line is $y = 2x$.

A point lies on the line only if its coordinates satisfy the equation of the line.

Substituting the coordinates (1, 2) in $y = 2x$, we get :

⇒ 2 = 2 × 1

⇒ 2 = 2

Since, L.H.S. = R.H.S., the point (1, 2) lies on the line.

Substituting the coordinates (−1, −2) in $y = 2x$, we get :

⇒ −2 = 2 × (−1)

⇒ −2 = −2

Since, L.H.S. = R.H.S., the point (−1, −2) lies on the line.

Thus, the line $y = 2x$ passes through the points (1, 2) and (−1, −2).

Hence, option 3 is the correct option.

Given,

Scale of the map = 1 : 50,000

Length of the plot on the map = 4 cm

Breadth of the plot on the map = 3 cm

Actual length of the plot = 4 × 50000 = 2,00,000 cm

= $\frac{200000}{100000}$ = 2 km [∵ 1 km = 1,00,000 cm]

Actual breadth of the plot = 3 × 50000 = 1,50,000 cm

= $\frac{150000}{100000}$ = 1.5 km

Actual area of the plot = Actual length × Actual breadth

= 2 × 1.5

= 3 km².

Hence, option 1 is the correct option.

Join OA and OP.

Given,

Radius of circle (OB) = 3 cm

PB and AP are at right angles.

We know that,

Radius and tangent at point of contact are perpendicular to each other.

OA ⊥ AP

From figure,

⇒ OA = OB = 3 cm [Radius of circle]

⇒ OB is perpendicular to PB

⇒ ∠OBP = 90°

This shows APBO is square.

OA = PB = AP = OB = 3 cm.

In right angled triangle OBP,

⇒ OP² = OB² + PB²

⇒ OP² = 3² + 3²

⇒ OP² = 9 + 9

⇒ OP² = 18

⇒ OP = $\sqrt{18}$

⇒ OP = $3\sqrt{2}$ cm.

Hence, option 3 is the correct option.

Given,

Capacity (volume) of the conical tent = 616 m³

Base area of the conical tent = 154 m²

By formula,

Volume of cone = $\frac{1}{3} \times \text{Base area} \times \text{Height}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow 616 = \frac{1}{3} \times 154 \times h \\[1em] \Rightarrow 154h = 616 \times 3 \\[1em] \Rightarrow h = \frac{1848}{154} \\[1em] \Rightarrow h = 12 \text{ m}.\end{aligned}$$

Thus, the height of the tent = 12 m.

Hence, option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sec\theta + \tan\theta = a \\[1em] \Rightarrow \sec\theta - \tan\theta = b \\[1em] \Rightarrow ab = (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) \\[1em] \Rightarrow ab = \sec^2\theta - \tan^2\theta \\[1em] \Rightarrow ab = 1\end{aligned}$$

So assertion (A) is true.

We know that,

⇒ $\sec^2\theta - \tan^2\theta = 1$

This is a fundamental trigonometric identity.

Thus, Both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 3 is the correct option.

Given,

The distribution is 30, 34, 35, 36, 37, 38, 39, 40.

The data is already arranged in ascending order.

Here, $n = 8$, which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{th} \text{term} + \Big(\frac{n}{2} + 1\Big)^{th} \text{term}\Big] \\[1em] = \frac{1}{2}[4^{th} \text{ term} + 5^{th} \text{ term}] \\[1em] = \frac{36 + 37}{2} \\[1em] = \frac{73}{2} \\[1em] = 36.5\end{aligned}$$

On removing the value 35, the distribution becomes 30, 34, 36, 37, 38, 39, 40.

Here, $n = 7$, which is odd.

By formula,

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{th} \text{ term} \\[1em] = 4^{th} \text{ term} \\[1em] = 37\end{aligned}$$

Change in median = 37 − 36.5 = 0.5.

Hence, option 1 is the correct option.

Given,

The arrow can come to rest pointing to one of the numbers 1, 2, 3, ......, 12.

Total number of possible outcomes = 12

Favourable outcomes = Arrow pointing to an odd number

{1, 3, 5, 7, 9, 11}

Number of favourable outcomes = 6

By formula,

Probability = $\frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$

= $\frac{6}{12}$

= $\frac{1}{2}$.

Hence, option 3 is the correct option.

Given,

For the cone formed by A,

Radius (R) = 7 cm

Height (H) = 28 cm

Since, all the three students are given an equal amount of Play-Dough, the volumes of the cone, the sphere and the cylinder are equal.

(a) Volume of sphere = Volume of cone

$$\begin{aligned}\Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 H \\[1em] \Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3} \times \pi \times 7^2 \times 28 \\[1em] \Rightarrow 4r^3 = 7^2 \times 28 \\[1em] \Rightarrow 4r^3 = 49 \times 28 \\[1em] \Rightarrow r^3 = \frac{1372}{4} \\[1em] \Rightarrow r^3 = 343 \\[1em] \Rightarrow r^3 = 7^3 \\[1em] \Rightarrow r = 7 \text{ cm}.\end{aligned}$$

Hence, radius of the sphere = 7 cm.

(b) Given,

Radius of the cylinder (R') = 14 cm

Let height of cylinder be H'.

Volume of cylinder = Volume of sphere

$$\begin{aligned}\Rightarrow \pi R'^2 H' = \frac{4}{3}\pi r^3 \\[1em] \Rightarrow 14^2 \times H' = \frac{4}{3} \times 7^3 \\[1em] \Rightarrow 196 \times H' = \frac{4 \times 343}{3} \\[1em] \Rightarrow H' = \frac{1372}{3 \times 196} \\[1em] \Rightarrow H' = \frac{7}{3} \\[1em] \Rightarrow H' = 2.33 \text{ cm} \approx 2 \text{ cm}.\end{aligned}$$

Hence, height of the cylinder = 2 cm.

(c) By formula,

Volume of the dough given to each of them = Volume of cone

$$\begin{aligned}= \frac{1}{3}\pi R^2 H \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 28 \\[1em] = \frac{1}{3} \times 22 \times 7 \times 28 \\[1em] = \frac{4312}{3}\\[1em] = 1437.33 \text{ cm}^3 ≈ 1437 \text{ cm}^3.\end{aligned}$$

Hence, volume of the dough given to each of them = 1437 cm³.

Given,

$$\begin{aligned}\Rightarrow (x - 2)^2 + 11x - 7 = 0 \\[1em] \Rightarrow x^2 - 4x + 4 + 11x - 7 = 0 \\[1em] \Rightarrow x^2 + 7x - 3 = 0\end{aligned}$$

Comparing the above equation with $ax^2 + bx + c$ = 0, we get :

$a = 1, b = 7 \text{ and } c = -3$

By quadratic formula,

$$\begin{aligned}\Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-3)}}{2 \times 1} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{49 + 12}}{2} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{61}}{2} \\[1em] \Rightarrow x = \frac{-7 \pm 7.8102}{2}\end{aligned}$$

Solving further,

$$\begin{aligned}\Rightarrow x = \frac{-7 + 7.8102}{2} \text{ or } x = \frac{-7 - 7.8102}{2} \\[1em] \Rightarrow x = \frac{0.8102}{2} \text{ or } x = \frac{-14.8102}{2} \\[1em] \Rightarrow x = 0.4051 \text{ or } x = -7.4051\end{aligned}$$

Correct to 3 significant figures,

$x = 0.405 \text{ or } x = -7.41$.

Hence, $\bm{x}$ = 0.405 or $\bm{x}$ = −7.41.

(a) Since, R lies on the x-axis, its ordinate is zero.

Let the point R$(x, 0)$ divide the line segment joining P(−4, 1) and Q(11, −2) in the ratio $m : n$.

By section formula,

$y = \frac{my_2 + ny_1}{m + n}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow 0 = \frac{m(-2) + n(1)}{m + n} \\[1em] \Rightarrow -2m + n = 0 \\[1em] \Rightarrow n = 2m \\[1em] \Rightarrow \frac{m}{n} = \frac{1}{2}.\end{aligned}$$

Hence, PR : RQ = 1 : 2.

(b) From part (a),

P(−4, 1), Q(11, −2) and $m : n$ = 1 : 2

By section formula,

$x = \frac{mx_2 + nx_1}{m + n}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow x = \frac{1(11) + 2(-4)}{1 + 2} \\[1em] \Rightarrow x = \frac{11 - 8}{3} \\[1em] \Rightarrow x = \frac{3}{3} \\[1em] \Rightarrow x = 1.\end{aligned}$$

R = $(x, 0)$ = (1, 0).

Hence, the coordinates of the point R are (1, 0).

(c) By formula,

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow m_{PQ} = \frac{-2 - 1}{11 - (-4)} \\[1em] \Rightarrow m_{PQ} = \frac{-3}{15} \\[1em] \Rightarrow m_{PQ} = -\frac{1}{5}\end{aligned}$$

We know that,

The product of the slopes of two perpendicular lines is −1.

∴ Slope of the line perpendicular to PQ = $\frac{-1}{-\frac{1}{5}}$ = 5

By point-slope form,

$y - y_1 = m(x - x_1)$

Equation of the line perpendicular to PQ and passing through R(1, 0) is :

$$\begin{aligned}\Rightarrow y - 0 = 5(x - 1) \\[1em] \Rightarrow y = 5x - 5 \\[1em] \Rightarrow 5x - y = 5.\end{aligned}$$

Hence, the equation of the line perpendicular to PQ at R is $\bm{5x − y = 5}$.

Given,

Marked price of the article = ₹ 2000

Rate of GST = 18%

Amount available with the customer = ₹ 1888 (including GST)

Let the reduced price of the article be ₹ $x$.

⇒ $x + 18% \text{ of } x = 1888$

$$\begin{aligned}\Rightarrow x + \frac{18}{100} \times x = 1888 \\[1em] \Rightarrow x\Big(1 + \frac{18}{100}\Big) = 1888 \\[1em] \Rightarrow x \times \frac{118}{100} = 1888 \\[1em] \Rightarrow x = \frac{1888 \times 100}{118} \\[1em] \Rightarrow x = 1600.\end{aligned}$$

Reduced price of the article = ₹ 1,600

Reduction in price = ₹ 2000 − ₹ 1600 = ₹ 400

By formula,

Discount % = $\frac{\text{Reduction in price}}{\text{Marked price}} \times 100$

= $\frac{400}{2000} \times 100$

= 20%.

Hence, the shopkeeper must give a discount of 20%.

(a) Given,

The A.P. consists of all whole numbers which are divisible by 3 and 5.

A number which is divisible by both 3 and 5 is divisible by their L.C.M.

L.C.M. of 3 and 5 = 15

Thus, the A.P. is 15, 30, 45, 60, ......

First term (a) = 15

Common difference (d) = 30 − 15 = 15.

Hence, first term = 15 and common difference = 15.

(b) By formula,

t_n = a + (n - 1)d

Substituting values, we get :

⇒ t₁₀ = 15 + (10 − 1) × 15

⇒ t₁₀ = 15 + 9 × 15

⇒ t₁₀ = 15 + 135

⇒ t₁₀ = 150.

Hence, the 10^th term of the A.P. = 150.

(c) By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

⇒ S₁₀ = $\frac{10}{2}[2 \times 15 + (10 - 1) \times 15]$

⇒ S₁₀ = 5[30 + 9 × 15]

⇒ S₁₀ = 5[30 + 135]

⇒ S₁₀ = 5 × 165

⇒ S₁₀ = 825.

Hence, the sum of the first 10 terms of the A.P. = 825.

(a) From the graph,

The coordinates of the points are A(0, 3), B(1, 2), C(2, 2), D(2, 1) and E(3, 0).

(b) We know that,

The image of a point $(x, y)$ under reflection on the y-axis is $(-x, y)$.

∴ Image of A(0, 3) on reflection in y-axis is (0, 3).

Since, A lies on the y-axis, it remains invariant under reflection on the y-axis.

Hence, the image of A under reflection on the y-axis is A(0, 3).

(c) We know that,

The image of a point $(x, y)$ under reflection through the origin is $(-x, -y)$.

∴ Image of C(2, 2) on reflection in origin is (−2, −2).

From the graph, the point (−2, −2) is the point K.

Hence, the image of C under reflection through the origin is K(−2, −2).

(d) We know that,

The image of a point $(x, y)$ under reflection on the y-axis is $(-x, y)$.

So, on reflecting the polygon ABCDEFGHI on the y-axis, each vertex is mapped to the point lying at the same distance from the y-axis but on the opposite side of it.

Since, the vertices A and I lie on the y-axis, they remain invariant under reflection on the y-axis.

From the graph, the images of the vertices of the polygon under reflection in y-axis are :

A → A, B → P, C → O, D → N, E → M, F → L, G → K, H → J and I → I

Thus, on joining these image points in order, we get the polygon APONMLKJI, which is the mirror image of the polygon ABCDEFGHI in the y-axis.

Hence, the image of the polygon ABCDEFGHI under reflection on the y-axis is the polygon APONMLKJI.

(e) From the graph,

N = (−2, 1) and D = (2, 1)

Since, both the points have the same ordinate (y-coordinate) equal to 1, the line joining them is parallel to the x-axis.

Hence, the equation of the line joining the points N and D is $\bm{y}$ = 1.

Given,

Deposit per month (P) = ₹ 1000

Time (n) = 2 years = 24 months

Maturity value = 26 × ₹ 1000 = ₹ 26,000

(a) Total amount deposited = P × n

= 1000 × 24

= ₹ 24,000

By formula,

Interest = Maturity value − Total amount deposited

= ₹ 26,000 − ₹ 24,000

= ₹ 2,000.

Hence, interest paid by the bank = ₹ 2,000.

(b) Let the rate of interest be r% per annum.

By formula,

$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow 2000 = 1000 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times \frac{24 \times 25}{24} \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times 25 \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 250r \\[1em] \Rightarrow r = \frac{2000}{250} \\[1em] \Rightarrow r = 8.\end{aligned}$$

Hence, rate of interest (r%) = 8% per annum.

(a) Given,

Matrix A is of order 2 × 2 and matrix B is of order 2 × 1.

We know that,

If matrix A is of order $m \times n$ and matrix B is of order $n \times p$, then the order of the product AB is $m \times p$.

∴ Order of AB = 2 × 1

Since, AB is a null matrix, the order of the null matrix is the same as the order of AB.

Hence, the order of the null matrix is 2 × 1.

(b) Calculating AB,

$$\Rightarrow AB = \begin{bmatrix*}[r] x & 1 \\ 2 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ x - 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x \times x + 1 \times (x - 2) \\ 2 \times x + 2 \times (x - 2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x^2 + x - 2 \\ 2x + 2x - 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x^2 + x - 2 \\ 4x - 4 \end{bmatrix*}.$$

Since, AB is a null matrix,

$$\Rightarrow \begin{bmatrix*}[r] x^2 + x - 2 \\ 4x - 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ 0 \end{bmatrix*}$$

Comparing the corresponding elements, we get :

$$\begin{aligned}\Rightarrow 4x - 4 = 0 \\[1em] \Rightarrow 4x = 4 \\[1em] \Rightarrow x = 1\end{aligned}$$

Also,

$$\begin{aligned}\Rightarrow x^2 + x - 2 = 0 \\[1em] \Rightarrow x^2 + 2x - x - 2 = 0 \\[1em] \Rightarrow x(x + 2) - 1(x + 2) = 0 \\[1em] \Rightarrow (x + 2)(x - 1) = 0 \\[1em] \Rightarrow (x + 2) = 0 \text{ or } (x - 1) = 0 \qquad [\text{Using zero product rule}] \\[1em] \Rightarrow x = -2 \text{ or } x = 1\end{aligned}$$

The value of $x$ which satisfies both the equations and is greater than 0 is 1.

Hence, $\bm{x}$ = 1.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow \frac{\cos^2\theta}{1 - \tan\theta} + \frac{\sin^2\theta}{1 - \cot\theta} \\[1em] \Rightarrow \frac{\cos^2\theta}{1 - \frac{\sin\theta}{\cos\theta}} + \frac{\sin^2\theta}{1 - \frac{\cos\theta}{\sin\theta}} \\[1em] \Rightarrow \frac{\cos^2\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} + \frac{\sin^2\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta}} \\[1em] \Rightarrow \frac{\cos^3\theta}{\cos\theta - \sin\theta} + \frac{\sin^3\theta}{\sin\theta - \cos\theta} \\[1em] \Rightarrow \frac{\cos^3\theta}{\cos\theta - \sin\theta} - \frac{\sin^3\theta}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \frac{\cos^3\theta - \sin^3\theta}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \frac{(\cos\theta - \sin\theta)(\cos^2\theta + \cos\theta,\sin\theta + \sin^2\theta)}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \cos^2\theta + \sin^2\theta + \cos\theta,\sin\theta \\[1em] \Rightarrow 1 + \cos\theta,\sin\theta \qquad [\because \sin^2\theta + \cos^2\theta = 1]\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\mathbf{\frac{\cos^2\theta}{1 - \tan\theta} + \frac{\sin^2\theta}{1 - \cot\theta} = 1 + \cos\theta,\sin\theta}$.

Given,

For initial investment,

Investment = ₹ 8,000

Face Value = ₹ 100

Market Value = ₹ 80

Dividend Rate = 6%

By formula,

Number of shares = $\frac{\text{Investment}}{\text{Market value of each share}}$

= $\frac{8000}{80}$

= 100

By formula,

Dividend for the first year = No. of shares × Rate of div. × N.V. of 1 share

= 100 × $\frac{6}{100}$ × 100

= ₹ 600

Hence, dividend for first year = ₹ 600.

(b) Given,

Number of shares sold = 100

Selling price per share = ₹ 75

Proceeds from sale = Number of shares × selling price

= 100 × 75

= ₹ 7,500

Total proceeds = Proceeds from sale + Dividend received = 7500 + 600 = ₹ 8,100

Total investment = ₹ 8,100

Market value per share = ₹ 27

Number of new shares = $\frac{\text{Investment}}{\text{Market value of each share}}$

= $\frac{8100}{27}$

= 300.

Hence, number of shares purchased by Ms. Kaur = 300.

Solving L.H.S. of the inequation, we get :

$$\begin{aligned}\Rightarrow -6 + \frac{12x}{5} \le \frac{7x}{5} - 3 \\[1em] \Rightarrow \frac{12x}{5} - \frac{7x}{5} \le 6 - 3 \\[1em] \Rightarrow \frac{12x - 7x}{5} \le 3 \\[1em] \Rightarrow \frac{5x}{5} \le 3 \\[1em] \Rightarrow x \le 3 \text{ ......(1)}\end{aligned}$$

Solving R.H.S. of the inequation, we get :

$$\begin{aligned}\Rightarrow \frac{7x}{5} - 3 \lt \frac{2}{5} + 2x \\[1em] \Rightarrow \frac{7x}{5} - 2x \lt \frac{2}{5} + 3 \\[1em] \Rightarrow \frac{7x - 10x}{5} \lt \frac{2 + 15}{5} \\[1em] \Rightarrow \frac{-3x}{5} \lt \frac{17}{5} \\[1em] \Rightarrow -3x \lt 17 \\[1em] \Rightarrow x \gt \frac{-17}{3} \\[1em] \Rightarrow x \gt -5\frac{2}{3} \text{ ......(2)}\end{aligned}$$

From equation (1) and (2),

Solution set = ${x : -5\frac{2}{3} \lt x \le 3, x \in R}$

Hence, solution set = {$\bm{x} : \mathbf{-5\frac{2}{3}} \lt \bm{x} \le 3, \bm{x}$ ∈ R}.

Given,

O is the centre of the circle, AD = AB and ∠ABC = 32°.

From the figure, BC passes through the centre O and hence BC is a diameter of the circle.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠BAC = 90°

In ∆ABC,

⇒ ∠ACB + ∠BAC + ∠ABC = 180° [Angle sum property of triangle]

$$\begin{aligned}\Rightarrow x + 90^{\circ} + 32^{\circ} = 180^{\circ} \\[1em] \Rightarrow x = 180^{\circ} - 122^{\circ} \\[1em] \Rightarrow x = 58^{\circ}.\end{aligned}$$

We know that,

The angle between a chord and a tangent at the point of contact is equal to the angle in the alternate segment.

$$\begin{aligned}\Rightarrow y = \angle ACB \\[1em] \Rightarrow y = x \\[1em] \Rightarrow y = 58^{\circ}.\end{aligned}$$

We know that,

Angles in the same segment of a circle are equal.

Here, ∠ADB and ∠ACB are the angles in the same segment subtended by the chord AB.

$$\begin{aligned}\Rightarrow z = \angle ADB = \angle ACB \\[1em] \Rightarrow z = x \\[1em] \Rightarrow z = 58°.\end{aligned}$$

In ∆ADB,

⇒ AD = AB [Given]

⇒ ∠ABD = ∠ADB [Angles opposite to equal sides are equal]

⇒ $\angle ABD = z = 58°$

By angle sum property of triangle,

$$\begin{aligned}\Rightarrow v + \angle ADB + \angle ABD = 180° \\[1em] \Rightarrow v + 58° + 58° = 180° \\[1em] \Rightarrow v = 180° - 116° \\[1em] \Rightarrow v = 64°.\end{aligned}$$

In cyclic quadrilateral ADEB,

We know that,

The sum of the opposite angles of a cyclic quadrilateral is 180°.

$$\begin{aligned}\Rightarrow v + w = 180° \\[1em] \Rightarrow 64° + w = 180° \\[1em] \Rightarrow w = 180° - 64° \\[1em] \Rightarrow w = 116°.\end{aligned}$$

Hence, $\bm{x = 58°}, \bm{y = 58°}, \bm{z = 58°}, \bm{v = 64°}$ and $\bm{w} = 116°$.

Given,

$f(x) = 2x^3 + 5x^2 - 28x - 15$

Substituting $x = 3$ in $f(x)$, we get :

$$\begin{aligned}\Rightarrow f(3) = 2(3)^3 + 5(3)^2 - 28(3) - 15 \\[1em] \Rightarrow f(3) = 2 \times 27 + 5 \times 9 - 84 - 15 \\[1em] \Rightarrow f(3) = 54 + 45 - 84 - 15 \\[1em] \Rightarrow f(3) = 99 - 99 \\[1em] \Rightarrow f(3) = 0\end{aligned}$$

Since, the remainder is zero.

∴ $(x - 3)$ is a factor of $f(x)$.

Dividing $2x^3 + 5x^2 - 28x - 15$ by $(x - 3)$, we get :

$$\begin{array}{l} \phantom{x - 3\big)\quad } 2x^2 + 11x + 5 \\ x - 3\overline{\smash{\big)}\quad 2x^3 + 5x^2 - 28x - 15} \\ \phantom{+++(}\underline{\underset{-}{+}2x^3 \underset{+}{-}6x^2} \\ \phantom{x - 3\big)\quad }\phantom{2x^3 +} 11x^2 - 28x \\ \phantom{+++( }\phantom{2x^3 +} \underline{\underset{-}{+}11x^2 \underset{+}{-}33x} \\ \phantom{++++( }\phantom{2x^3 + 5x^2 -} 5x - 15 \\ \phantom{++++) }\phantom{2x^3 + 5x^2 -} \underline{\underset{-}{+}5x \underset{+}{-}15} \\ \phantom{x - 3\big)\quad }\phantom{2x^3 + 5x^2 - 28x} \times \end{array}$$

∴ $2x^3 + 5x^2 - 28x - 15 = (x - 3)(2x^2 + 11x + 5)$

Factorising $2x^2 + 11x + 5$, we get :

$$\begin{aligned}= (x - 3)[2x^2 + 10x + x + 5] \\[1em] = (x - 3)[2x(x + 5) + 1(x + 5)] \\[1em] = (x - 3)(x + 5)(2x + 1).\end{aligned}$$

Hence, $\bm{{2x^3} + 5x^2 − 28x − 15 = (x − 3)(x + 5)(2x + 1)}$.

Given,

AD ∥ GE ∥ BC, DE = 18 cm, EC = 3 cm and AD = 35 cm

(a) In ∆ADC,

Since, GE ∥ AD, we have FE ∥ AD.

By Basic Proportionality Theorem (B.P.T.),

$$\begin{aligned}\Rightarrow \frac{AF}{FC} = \frac{DE}{EC} \\[1em] \Rightarrow \frac{AF}{FC} = \frac{18}{3} \\[1em] \Rightarrow \frac{AF}{FC} = \frac{6}{1}.\end{aligned}$$

∴ AF : FC = 6 : 1

In ∆ABC,

Since, GE ∥ BC, we have GF ∥ BC.

By Basic Proportionality Theorem (B.P.T.),

$\frac{AG}{GB} = \frac{AF}{FC} = \frac{6}{1}$

∴ AG : GB = 6 : 1.

Hence, AF : FC = 6 : 1 and AG : GB = 6 : 1.

(b) In ∆ADC and ∆FEC,

⇒ ∠ADC = ∠FEC [Corresponding angles as AD ∥ FE]

⇒ ∠ACD = ∠FCE [Common angle]

∴ ∆ADC ~ ∆FEC (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{EF}{AD} = \frac{EC}{DC} \\[1em] \Rightarrow \frac{EF}{35} = \frac{3}{18 + 3} \\[1em] \Rightarrow \frac{EF}{35} = \frac{3}{21} \\[1em] \Rightarrow EF = \frac{35 \times 3}{21} \\[1em] \Rightarrow EF = 5 \text{ cm}.\end{aligned}$$

Hence, length of EF = 5 cm.

(c) We know that,

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

⇒ $\frac{\text{Area of ∆ADC}}{\text{Area of ∆EFC}} = \frac{AD^2}{EF^2}$

From the figure,

Area of trapezium ADEF = Area of ∆ADC − Area of ∆EFC

Dividing each side by Area of ∆EFC,

$$\begin{aligned}\therefore \frac{\text{Area of trapezium ADEF}}{\text{Area of ∆EFC}} = \frac{\text{Area of ∆ADC} - \text{Area of ∆EFC}}{\text{Area of ∆EFC}} \\[1em] = \frac{\text{Area of ∆ADC}}{\text{Area of ∆EFC}} - \frac{\text{Area of ∆EFC}}{\text{Area of ∆EFC}}\\[1em] = \frac{AD^2}{EF^2} - 1 \\[1em] = \frac{35^2}{5^2} - 1 \\[1em] = \frac{1225}{25} - 1 \\[1em] = \frac{1225 - 25}{25} \\[1em] = \frac{1200}{25} \\[1em] = \frac{48}{1}.\end{aligned}$$

Hence, area (trapezium ADEF) : area (∆ EFC) = 48 : 1.

Steps of construction :

1. Draw a line segment AB of length 6 cm.

2. Draw XY, the perpendicular bisector of AB. XY is the locus of points which are equidistant from A and B.

3. With A as centre and radius 4.5 cm, draw an arc cutting XY at the point O. Thus, O is the point on the locus which is 4.5 cm from A.

4. With O as centre and radius 4.5 cm, draw a circle. This circle is the locus of all points which are 4.5 cm from O.

5. Mark P, the point of intersection of the perpendicular bisector XY and the circle, which lies on the same side of AB as O.

6. Join AP.

7. Draw AZ, the bisector of ∠PAB. AZ is the locus of points which are equidistant from AP and AB.

Given,

The bag contains identical cards numbered 1 to 50.

Total number of possible outcomes = 50

(a) Favourable outcomes = Getting a card with a number divisible by 6.

Numbers divisible by 6 = {6, 12, 18, 24, 30, 36, 42, 48}

Number of favourable outcomes = 8

By formula,

Probability = $\frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$

P(card divisible by 6) = $\frac{8}{50} = \frac{4}{25}$.

Hence, probability that the card selected is divisible by 6 = $\mathbf{\frac{4}{25}}$.

(b) We know that,

A number which is divisible by both 2 and 3 is divisible by 6.

∴ P(card divisible by 2 and 3) = P(card divisible by 6) = $\frac{4}{25}$

By formula,

P(card not divisible by 2 and 3) = 1 − P(card divisible by 2 and 3)

$$\begin{aligned}= 1 - \frac{4}{25} \\[1em] = \frac{25 - 4}{25}\\[1em] = \frac{21}{25}.\end{aligned}$$

Hence, probability that the card selected is not divisible by 2 and 3 = $\mathbf{\frac{21}{25}}$.

Join AC and BD.

(a) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

∴ AP = AS, BP = BQ, CR = CQ and DR = DS

Adding the equations, we get :

⇒ AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC ......(1)

We know that,

Opposite sides of a parallelogram are equal.

∴ AB = CD and AD = BC

Substituting the above values in equation (1), we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, proved that AB = BC.

(b) From part (a),

⇒ AB = BC

Also, AB = CD and AD = BC [Opposite sides of a parallelogram are equal]

∴ AB = BC = CD = AD

Since, all the four sides of the parallelogram ABCD are equal, ABCD is a rhombus.

We know that,

The diagonals of a rhombus bisect each other at right angles.

Hence, proved that the diagonals AC and BD bisect each other at right angles.

Given,

$\frac{a + 2b + x}{a - 2b + x} = \frac{(a + b)^2}{(a - b)^2}$

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{(a + 2b + x) + (a - 2b + x)}{(a + 2b + x) - (a - 2b + x)} = \frac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} \\[1em] \Rightarrow \frac{2a + 2x}{4b} = \frac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow \frac{2(a + x)}{4b} = \frac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow a + x = \frac{a^2 + b^2}{a} \\[1em] \Rightarrow x = \frac{a^2 + b^2}{a} - a \\[1em] \Rightarrow x = \frac{a^2 + b^2 - a^2}{a} \\[1em] \Rightarrow x = \frac{b^2}{a}.\end{aligned}$$

Hence, $\bm{x} = \mathbf{\frac{b^2}{a}}$.

Given,

Mean = 27 and sum of all the frequencies = 50

Class	Class Mark (xi)	Frequency (fi)	fixi
0 - 10	5	8	40
10 - 20	15	x	15x
20 - 30	25	12	300
30 - 40	35	13	455
40 - 50	45	y	45y
Total		Σfi = 33 + x + y	Σfixi = 795 + 15x + 45y

Given,

Sum of all the frequencies = 50

⇒ 33 + x + y = 50

⇒ x + y = 17 ......(1)

By formula,

Mean = $\frac{∑f_ix_i}{∑f_i}$

Substituting values, we get :

⇒ 27 = $\frac{(795 + 15x + 45y)}{50}$

⇒ 795 + 15x + 45y = 1350

⇒ 15x + 45y = 555

⇒ x + 3y = 37 ......(2)

Subtracting equation (1) from equation (2), we get :

⇒ (x + 3y) − (x + y) = 37 − 17

⇒ 2y = 20

⇒ y = 10

Substituting y = 10 in equation (1), we get :

⇒ x + 10 = 17

⇒ x = 7.

Hence, the missing frequencies are x = 7 and y = 10.

Given,

Area of the shaded region = 57 m²

From the figure,

Length of the outer rectangle = 12 m

Breadth of the outer rectangle = 10 m

Area of the outer rectangle = 12 × 10 = 120 m²

Length of the inner (unshaded) rectangle = (12 − x) m

Breadth of the inner (unshaded) rectangle = (10 − x) m

Area of the inner rectangle = (12 − x)(10 − x) m²

From the figure,

Area of shaded region = Area of outer rectangle − Area of inner rectangle

⇒ 12 × 10 − (12 − x)(10 − x) = 57

⇒ 120 − (120 − 12x − 10x + x²) = 57

⇒ 120 − 120 + 22x − x² = 57

⇒ 22x − x² = 57

⇒ x² − 22x + 57 = 0

⇒ x² − 19x − 3x + 57 = 0

⇒ x(x − 19) − 3(x − 19) = 0

⇒ (x − 19)(x − 3) = 0

[Using zero product rule]

⇒ (x − 19) = 0 or (x − 3) = 0

⇒ x = 19 or x = 3

Since, the breadth of the rectangle is 10 m, the width x cannot be 19 m.

∴ x = 3 m.

Hence, the possible value of x = 3 m.

Marks	No. of students (f)	Cumulative frequency (cf)
0 - 10	8	8
10 - 20	12	20
20 - 30	16	36
30 - 40	24	60
40 - 50	29	89
50 - 60	38	127
60 - 70	42	169
70 - 80	31	200

Steps :

1. Take 2 cm = 10 marks on the x-axis.

2. Take 1 cm = 10 students on the y-axis.

3. Plot the points (0, 0), (10, 8), (20, 20), (30, 36), (40, 60), (50, 89), (60, 127), (70, 169) and (80, 200).

4. Join the points by a free hand curve.

(a) Here, $n$ (no. of students) = 200, which is even.

Median = $\frac{n}{2} = \frac{200}{2}$ = 100^th term.

Through point A = 100 on the y-axis draw a horizontal line parallel to the x-axis touching the graph at point B, through B draw a vertical line parallel to the y-axis touching the x-axis at point C = 53.

Hence, median = 53.

(b) Through point D = 75 on the x-axis draw a vertical line parallel to the y-axis touching the graph at point E, through E draw a horizontal line parallel to the x-axis touching the y-axis at point F = 186.

∴ 186 students scored 75 marks and below.

∴ Number of students who scored above 75 marks = 200 − 186 = 14.

Hence, 14 students scored above 75 marks.

(c) Maximum marks = 80

Pass marks = 35% of 80 = $\frac{35}{100} \times 80$ = 28 marks

Through point G = 28 on the x-axis draw a vertical line parallel to the y-axis touching the graph at point H, through H draw a horizontal line parallel to the x-axis touching the y-axis at point I = 33.

Hence, 33 students scored 35% and below.

Let AD be the lighthouse and B and C be the two ships on the opposite sides of the lighthouse.

Given,

Height of the lighthouse (AD) = 120 m

The angles of depression of the two ships are 32° and 58°.

From figure,

⇒ ∠FAB = ∠ABD = 32° (alternate angles are equal)

⇒ ∠EAC = ∠ACD = 58° (alternate angles are equal)

In right angled triangle ABD,

$$\begin{aligned}\Rightarrow \tan 32° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 32° = \frac{AD}{BD} \\[1em] \Rightarrow BD = \frac{AD}{\tan 32°} \\[1em] \Rightarrow BD = \frac{120}{0.6249} \\[1em] \Rightarrow BD = 192.03 \text{ m}.\end{aligned}$$

In right angled triangle ACD,

$$\begin{aligned}\Rightarrow \tan 58° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 58° = \frac{AD}{CD} \\[1em] \Rightarrow CD = \frac{AD}{\tan 58°} \\[1em] \Rightarrow CD = \frac{120}{1.6003} \\[1em] \Rightarrow CD = 74.99 \text{ m}.\end{aligned}$$

From the figure,

Distance between the two ships (BC) = BD + CD

= 192.03 + 74.99

= 267.02 m ≈ 267 m.

Hence, the distance between the two ships = 267 m.

Let the first term of the G.P. be $a$ and the common ratio be $r$.

Since, the G.P. contains 10 terms, the terms are $a, ar, ar^2, ......, ar^9$.

Given,

Sum of the first three terms = 7

$$\begin{aligned}\Rightarrow a + ar + ar^2 = 7 \\[1em] \Rightarrow a(1 + r + r^2) = 7 \text{ ......(1)}\end{aligned}$$

Sum of the last three terms = 896

$$\begin{aligned}\Rightarrow ar^7 + ar^8 + ar^9 = 896 \\[1em] \Rightarrow ar^7(1 + r + r^2) = 896 \text{ ......(2)}\end{aligned}$$

Dividing equation (2) by equation (1), we get :

$$\begin{aligned}\Rightarrow \frac{ar^7(1 + r + r^2)}{a(1 + r + r^2)} = \frac{896}{7} \\[1em] \Rightarrow r^7 = 128 \\[1em] \Rightarrow r^7 = 2^7 \\[1em] \Rightarrow r = 2.\end{aligned}$$

Substituting $r = 2$ in equation (1), we get :

$$\begin{aligned}\Rightarrow a(1 + 2 + 2^2) = 7 \\[1em] \Rightarrow a(1 + 2 + 4) = 7 \\[1em] \Rightarrow 7a = 7 \\[1em] \Rightarrow a = 1.\end{aligned}$$

Hence, first term = 1 and common ratio = 2.

(a) We know that,

The lengths of two tangents drawn from an external point to a circle are equal.

∴ PB = PA

⇒ PB = 9 cm.

Hence, length of PB = 9 cm.

(b) Join OA, OB and OP.

Figure unavailable in the source: In a circle with centre O, PA and PB are tangents drawn from an external point P. If PA = 9 cm and ∠APB = 60°, find the:. Specimen 2027, ICSE Specimen 2027 Maths Solved Question Paper.

We know that,

The tangents drawn from an external point are equally inclined to the line joining the centre of the circle and that point.

∴ OP bisects ∠APB.

⇒ ∠OPA = $\frac{1}{2} \times ∠APB = \frac{1}{2} \times 60° = 30°$

We know that,

Radius and tangent at the point of contact are perpendicular to each other.

∴ ∠OAP = 90°

In right angled triangle OAP,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{OA}{PA} \\[1em] \Rightarrow OA = PA \times \tan 30° \\[1em] \Rightarrow OA = 9 \times \frac{1}{\sqrt{3}} \\[1em] \Rightarrow OA = \frac{9 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow OA = \frac{9\sqrt{3}}{3} \\[1em] \Rightarrow OA = 3\sqrt{3} \\[1em] \Rightarrow OA = 3 \times 1.7321 \\[1em] \Rightarrow OA = 5.1963 \\[1em] \Rightarrow OA = 5.2 \text{ cm}.\end{aligned}$$

Hence, radius of the circle = 5.2 cm.

Steps :

1. Take 2 cm = 5 kg along the x-axis.

2. Take 2 cm = 5 students along the y-axis.

3. Draw the rectangles (bars) for the given class intervals, taking the corresponding frequencies as their heights.

4. The class interval 50 - 55 has the highest frequency 25, so 50 - 55 is the modal class.

5. In the highest rectangle, draw two lines AD and BC from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AD and BC.

6. From the point of intersection P, draw a perpendicular to the x-axis. The point R = 51.5, where this perpendicular meets the x-axis gives the mode.

From the histogram,

Mode = 51.5 kg.

Hence, modal weight = 51.5 kg.

Given,

(x + 3), 1, (3x − 7) and −5 are in proportion.

$$\begin{aligned}\therefore \frac{(x + 3)}{1} = \frac{(3x - 7)}{-5} \\[1em] \Rightarrow (x + 3)= \frac{(3x - 7)}{-5} \\[1em] \Rightarrow -5(x + 3) = (3x - 7) \\[1em] \Rightarrow -5x - 15 = 3x - 7 \\[1em] \Rightarrow -5x - 3x = 15 - 7 \\[1em] \Rightarrow -8x = 8 \\[1em] \Rightarrow x = -\frac{8}{8} \\[1em] \Rightarrow x = -1.\end{aligned}$$

Hence, option 1 is the correct option.

Let rate of GST be x.

Given,

Marked price of refrigerator = ₹ 12,000

GST paid = ₹ 2,160

GST amount = $\frac{\text{Rate}}{100}$ × Marked price

⇒ 2160 = $\frac{x}{100}$ × 12000

⇒ 2160 = 120x

⇒ x = $\frac{2160}{120}$

⇒ x = 18%

Hence, option 3 is the correct option.

Given,

1, 6, 9, 8, 9, 1, 7, 8, 9.

Mode = The value which occurs most frequently in the data.

The number 9 occurs 3 times, which is the highest.

Mode = 9

Hence, option 4 is the correct option.

Given,

For A,

P = ₹ 1,500

n = 12 months

r = 9%

We know that,

Amount Invested = P × n

= ₹ 1,500 × 12

= ₹ 18,000.

For B,

Amount Invested = P × n

= ₹ 1,200 × 15

= ₹ 18,000.

Both invested ₹ 18,000.

Hence, option 4 is the correct option.

A line parallel to the x-axis has :

Slope (m) = 0

y-intercept (c) = 6

Substituting the values in equation y = mx + c, we get :

y = 0x + 6

y = 6.

Hence, option 1 is the correct option.

Given,

Face Value (FV) of share = ₹ 20

Dividend = 9%

Return on investment = 12%

By formula,

Dividend % × N.V. = Return % × M.V.

$\frac{9}{100} \times 20 = \frac{12}{100}$ × M.V.

180 = 12 × M.V.

M.V. = $\frac{180}{12}$ = ₹ 15.

Hence, option 2 is the correct option.

Let the radius of the sphere (S₁) be R and hemisphere (S₂) be r.

Total surface area of sphere = 4πR²

Total surface area of hemisphere = 3πr²

Given,

Since, total surface area of sphere and hemisphere are equal,

4πR² = 3πr²

$\frac{R^2}{r^2} = \frac{3}{4}$

$\frac{R}{r} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} = \sqrt{3} : 2$.

Hence, option 3 is the correct option.

In a cyclic quadrilateral,

Exterior angle is equal to the interior opposite angle.

∠CBA = ∠CDE = 65°

We know that,

The angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle.

∠AOC = 2 × ∠ABC

x = 2 × 65°

x = 130°.

Hence, option 4 is the correct option.

Comparing 3x² − 6x − 3 = 0 with ax² + bx + c = 0, we get :

a = 3, b = -6, c = -3.

We know that,

⇒ D = b² - 4ac

⇒ D = (-6)² - 4 × (3) × (-3)

⇒ D = 36 + 36

⇒ D = 72.

∴ D > 0 and not a perfect square. Roots are real, distinct and irrational.

Hence, option 3 is the correct option.

When a die is rolled,

Total number of possible outcomes = 6

S = {1, 2, 3, 4, 5, 6}

Let E be the event of getting number greater than 6,

Number of favorable outcomes = 0

P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{0}{6} = 0$.

Assertion (A) is false.

There are six possible outcomes when a dice is rolled,

S = {1, 2, 3, 4, 5, 6}

Reason (R) is true.

(A) is false and (R) is true.

Hence, option 2 is the correct option.

Let the areas be A₁ and A₂, and the altitudes be h₁ and h₂ respectively.

Since the triangles are similar,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides (or altitudes).

$$\begin{aligned}\Rightarrow \frac{A_1}{A_2} = \Big(\frac{h_1}{h_2}\Big)^2 \\[1em] \Rightarrow \frac{9}{64} = \Big(\frac{h_1}{h_2}\Big)^2 \\[1em] \Rightarrow \frac{h_1}{h_2} = \sqrt{\frac{9}{64}} \\[1em] \Rightarrow \frac{h_1}{h_2} = \frac{3}{8} \\[1em] \Rightarrow h_1 : h_2 = 3 : 8.\end{aligned}$$

Hence, option 1 is the correct option.

Using remainder theorem,

If polynomial P(x) is divided by (x - a), the remainder is P(a).

Given,

x + 2 = 0

x = -2.

Substituting x = -2 into the polynomial x³ + 7x² + 3x + 2, we get :

⇒ P(-2) = (-2)³ + 7(-2)² + 3(-2) + 2

= -8 + 28 - 6 + 2

= 20 - 6 + 2

= 16.

For a polynomial to be completely divisible by x + 2, the remainder must be 0.

We have a remainder of 16, we need to add a number k such that the new remainder becomes zero i.e.,:

k + 16 = 0

k = -16.

Hence, option 2 is the correct option.

Given,

C is midpoint of AB.

We know that,

In a right-angled triangle, the point equidistant from all three vertices is the midpoint of the hypotenuse.

Using midpoint formula :

M = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Substituting values, we get :

$$\begin{aligned}\Rightarrow C = \Big(\frac{0 + 2x}{2}, \frac{2y + 0}{2} \Big) \\[1em] \Rightarrow C = \Big(\frac{2x}{2}, \frac{2y}{2} \Big) \\[1em] \Rightarrow C = \text{(x, y)}.\end{aligned}$$

Hence, option 1 is the correct option.

For AB to exist :

Number of columns in A = Number of rows in B

Order of matrix A = 2 × 2

Order of matrix B = 1 × 2

Columns in A = 2

Rows in B = 1

∴ Product AB is not possible.

Hence, option 4 is the correct option.

Given,

6, −12, 24, −48... is a G.P.

a = 6

r = $\frac{-12}{6}$ = -2

By formula,

a_n = ar^{n - 1}

For 9th term,

⇒ a₉ = (6).(-2)^{9 - 1}

= 6.(-2)⁸

= 6.(256)

= 1536 > 0.

∴ Assertion (A) is true.

(-2)⁸

Any negative number raised to an even power is always positive.

∴ Reason (R) is true.

Thus, both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 3 is the correct option.

(a) Given,

The fourth and seventh terms of an Arithmetic Progression (A.P.) are 60 and 114 respectively.

We know that,

a_n = a + (n - 1)d

Fourth term :

⇒ a₄ = a + (4 - 1)d

⇒ a + 3d = 60 ....(1)

Seventh term :

⇒ a₇ = a + (7 - 1)d

⇒ a + 6d = 114 ....(2)

Subtracting equation (1) from equation (2), we get :

⇒ (a + 6d) - (a + 3d) = 114 - 60

⇒ a + 6d - a - 3d = 54

⇒ 3d = 54

⇒ d = $\frac{54}{3}$ = 18.

Substituting value of d = 18 in equation (1), we get :

⇒ a + 3d = 60

⇒ a + 3(18) = 60

⇒ a + 54 = 60

⇒ a = 60 - 54

⇒ a = 6.

Hence, first term = 6 and common difference = 18.

(b) Sum of the first 10 terms :

By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

$$\begin{aligned}\Rightarrow S_{10} = \frac{10}{2}[2(6) + (10 - 1)18] \\[1em] = 5[12 + 162] \\[1em] = 5[174] \\[1em] = 870.\end{aligned}$$

Hence, sum of its first 10 terms = 870.

Given,

A = $$\begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}$$

B = $$\begin{bmatrix} -1 & a \\ 3 & -5 \end{bmatrix}$$

AB = $$\begin{bmatrix} b & 7 \\ 4 & 5 \end{bmatrix}$$

Solving,

$$\Rightarrow AB = \begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix} \times \begin{bmatrix} -1 & a \\ 3 & -5 \end{bmatrix} \\[1em] \Rightarrow AB = \begin{bmatrix} 3(-1) + 1(3) & 3(a) + 1(-5) \\ 5(-1) + 3(3) & 5(a) + 3(-5) \end{bmatrix} \\[1em] \Rightarrow AB = \begin{bmatrix} -3 + 3 & 3a - 5 \\ -5 + 9 & 5a - 15 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} b & 7 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 0 & 3a - 5 \\ 4 & 5a - 15 \end{bmatrix} \\[1em]$$

Comparing corresponding elements,

∴ b = 0

⇒ 3a - 5 = 7

⇒ 3a = 7 + 5

⇒ 3a = 12

⇒ a = $\frac{12}{3}$

⇒ a = 4

Hence, a = 4 and b = 0.

Given,

∠ADB = 32°

From figure,

∠ABC = 90° (Angle in a semicircle is a right angle)

We know that,

Angle between a tangent and a chord through point of contact is equal to an angle in the alternate segment.

∠BCA = ∠ABE = y

∠DBC = ∠BAC = x

Since, DE is a straight line, thus

⇒ ∠DBC + ∠ABC + ∠ABE = 180°

⇒ x + 90° + y = 180°

⇒ x + y = 180° - 90°

⇒ x + y = 90° ....(1)

In triangle ADB,

⇒ ∠ADB + ∠BAD + ∠ABD = 180°

⇒ 32° + x + (∠DBC + ∠ABC) = 180°

⇒ 32° + x + x + 90° = 180°

⇒ 2x + 122° = 180°

⇒ 2x = 180° - 122°

⇒ 2x = 58°

⇒ x = $\frac{58°}{2}$ = 29°.

Substituting value of x in equation (1), we get :

⇒ 29° + y = 90°

⇒ y = 90° - 29°

⇒ y = 61°.

Hence, x = 29° and y = 61°.

(a) Using remainder theorem,

If polynomial P(x) is divided by (x - a), the remainder is P(a).

x + 1 = 0

x = -1.

Given,

The polynomial kx³ + 3x² − 11x − 6 when divided by (x + 1), leaves a remainder of 6.

Thus, on substituting value x = -1 in kx³ + 3x² − 11x − 6, we get :

⇒ k(-1)³ + 3(-1)² − 11(-1) − 6 = 6

⇒ -k + 3 + 11 − 6 = 6

⇒ -k + 8 = 6

⇒ -k = 6 - 8

⇒ k = 2.

Hence, the value of k = 2.

(b) P(x) = 2x³ + 3x² − 11x − 6

Substituting value x = 2 in P(x), we get :

⇒ 2(2)³ + 3(2)² − 11(2) − 6

⇒ 2(8) + 3(4) − 22 − 6

⇒ 16 + 12 − 22 − 6

⇒ 28 − 28

⇒ 0.

Since, P(2) = 0, thus (x − 2) is a factor of P(x).

$$\begin{array}{l} \phantom{x - 2)}{2x^2 + 7x + 3} \\ x - 2\overline{\smash{\big)}2x^3 + 3x^2 − 11x − 6} \\ \phantom{x - 2}\phantom{}\underline{\underset{-}{}2x^3 \underset{+}{-}4x^2} \\ \phantom{{x - 2}x^3,.-2}7x^2 - 11x \\ \phantom{{x - 2)}x^3-2}\underline{\underset{-}{}7x^2 \underset{+}{-} 14x} \\ \phantom{{x - 2)}{x^3-2x^{2}(3-.)}}3x - 6 \\ \phantom{{x - 2)}{x^3-2x^{2}(31),}}\underline{\underset{-}{}3x \underset{+}{-} 6} \\ \phantom{{x - 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$$

∴ 2x³ + 3x² − 11x − 6 = (x - 2)(2x² + 7x + 3)

= (x - 2)(2x² + 6x + x + 3)

= (x - 2)[2x(x + 3) + 1(x + 3)]

= (x - 2)(2x + 1)(x + 3).

Hence, 2x³ + 3x² − 11x − 6 = (x - 2)(2x + 1)(x + 3).

(a) Given,

Height of the cylindrical and conical parts = 7 cm

Minimum height of the cylindrical box required to pack this bottle = Height of cylindrical part + Height of conical part = 7 + 7 = 14 cm.

Hence, minimum height of the cylindrical box required to pack this bottle = 14 cm.

(b) From figure,

Diameter of cylindrical part = 7 cm

Radius of cylindrical part (r) = $\frac{7}{2}$ = 3.5 cm

Volume of Hemisphere = $\frac{2}{3}πr^3$

= $\frac{2}{3} \times \frac{22}{7} \times (3.5)^3$

= $\frac{2}{3} \times \frac{22}{7} \times 42.875$

= 89.83 cm³

Volume of cylinder = πr²h

= $\frac{22}{7} \times (3.5)^2 \times 7$

= 22 × 12.25

= 269.5 cm³

Total volume of liquid medicine (shaded part) = Volume of cylinder - Volume of hemisphere

= 269.5 - 89.83

= 179.67 ≈ 180 cm³

Hence, volume of liquid medicine = 180 cm³.

Steps of construction :

1. Draw a line segment AB = 5 cm.

2. At point A with radius = 5 cm draw an arc.

3. At point B with radius = 5 cm draw another arc, cutting previous arc at point C.

4. Join AC and BC.

5. Construct perpendicular bisectors of AB and BC, let the bisectors meet at point O.

6. With O as centre and OA as radius draw a circumcircle.

7. Construct angle bisector of ∠ABC, mark the point as D where angle bisector intersects circumcircle.

8. Join AD and CD.

Since, points A, B, C and D lie on the circumcircle, thus ABCD is a cyclic quadrilateral.

Hence, ABCD is a cyclic quadrilateral.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\cos θ} - \cos θ\Big) \Big(\frac{1}{\sin θ} - \sin θ\Big) \\[1em] \Rightarrow \Big(\frac{1 - \cos^2 θ}{\cos θ}\Big) \Big(\frac{1 - \sin^2 θ}{\sin θ}\Big) \\[1em] \Rightarrow \Big(\frac{\sin^2 θ}{\cos θ}\Big) \Big(\frac{\cos^2 θ}{\sin θ}\Big) \\[1em] \Rightarrow \sin θ \cos θ.\end{aligned}$$

Hence, proved (sec θ − cos θ)(cosec θ − sin θ) = sin θ cos θ.

(a) Given,

Cost price of T.V. = ₹ 20,000

Marked price = ₹ 24,000

Discount = 10%

GST = 12%

Discount provided by shopkeeper = 10% of ₹ 24,000

= $\frac{10}{100} \times 24000$

= ₹ 2,400

Discounted price of the TV set = ₹ 24,000 - ₹ 2,400 = ₹ 21,600

Hence, discounted price of the TV set = ₹ 21,600.

(b) GST = 12% of 21,600

= $\frac{12}{100} \times 21600$

= ₹ 2,592

Total amount paid by customer = ₹ 21,600 + ₹ 2,592 = ₹ 24,192

Hence, amount paid by customer = ₹ 24,192.

(a) Given,

DE ∥ BC

In triangle △ADE and △ABC,

∠ADE = ∠ABC (Corresponding angles are equal)

∠AED = ∠ACB (Corresponding angles are equal)

Therefore, △ADE ∼ △ABC (By AA similarity)

Given,

AD : DB = 2 : 3

Let AD = 2x and DB = 3x.

From figure,

AB = AD + DB = 2x + 3x = 5x.

We know that,

Corresponding sides of similar triangle are in proportion.

$\frac{DE}{BC} = \frac{AD}{AB} = \frac{2x}{5x} = \frac{2}{5}$.

DE : BC = 2 : 5.

Hence, proved ΔADE ~ ΔABC and DE : BC = 2 : 5.

(b) Given,

DE ∥ BC

In triangle ΔDFE and ΔCFB,

∠DFE = ∠CFB (Vertically opposite angles are equal)

∠DEF = ∠CBF (Alternate interior angles are equal)

ΔDFE ~ ΔCFB [By AA similarity]

Hence, proved that ΔDFE ~ ΔCFB.

(c) Let area of ΔCFB be x square units.

We know that,

The areas of two similar triangles are proportional to the squares of their corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of Δ DFE}}{\text{Area of Δ CFB}} = \Big(\frac{DE}{BC}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of Δ DFE}}{\text{Area of Δ CFB}} = \Big(\frac{2}{5}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of Δ DFE}}{\text{Area of Δ CFB}} = \frac{4}{25} \\[1em] \Rightarrow \frac{16}{x} = \frac{4}{25} \\[1em] \Rightarrow x = \frac{16 \times 25}{4} \\[1em] \Rightarrow x = \frac{400}{4} \\[1em] \Rightarrow x = 100.\end{aligned}$$

Hence, area of △CFB = 100 sq.units.

(a) From graph,

Students whose height is 150 cm and above :

150 - 160 = 9 students

160 - 170 = 4 students

Total number of students whose height is 150 cm and above = 9 + 4 = 13

Hence, number of students whose height is 150 cm and above = 13.

(b) From graph,

The vertical line intersects the x-axis at 137.

Hence, the modal height = 137 cm.

(c) Total number of students = 6 + 2 + 9 + 14 + 12 + 9 + 4 = 56.

Hence, total number of students = 56.

(a) Let point at which AB intersects x-axis be P(x, 0) and P divides AB in the ratio m : n.

Using section formula,

y = $\frac{my_2 + ny_1}{m + n}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 0 = \frac{m \times 10 + n \times -2}{m + n} \\[1em] \Rightarrow 0 = \frac{10m - 2n}{m + n} \\[1em] \Rightarrow 10m - 2n = 0 \\[1em] \Rightarrow 10m = 2n \\[1em] \Rightarrow \frac{m}{n} = \frac{2}{10} \\[1em] \Rightarrow \frac{m}{n} = \frac{1}{5} \\[1em] \Rightarrow m : n = 1 : 5.\end{aligned}$$

Hence, P divides AB in the ratio 1 : 5.

(b) Using section formula,

x = $\frac{mx_2 + nx_1}{m + n}$

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{1 \times 2 + 5 \times -10}{1 + 5} \\[1em] \Rightarrow x = \frac{2 - 50}{6} \\[1em] \Rightarrow x = \frac{-48}{6} \\[1em] \Rightarrow x = -8.\end{aligned}$$

P = (x, 0) = (-8, 0).

Hence, coordinates of point P are (−8, 0).

Given,

⇒ (x − 2)² − 5x − 3 = 0

⇒ x² - 4x + 4 − 5x − 3 = 0

⇒ x² - 9x + 1 = 0

Comparing x² - 9x + 1 = 0 with ax² + bx + c = 0 , we get :

a = 1, b = -9 and c = 1

By formula,

$$\begin{aligned}\Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(1)}}{2(1)} \\[1em] \Rightarrow x = \frac{9 \pm \sqrt{81 - 4}}{2} \\[1em] \Rightarrow x = \frac{9 \pm \sqrt{77}}{2} \\[1em] \Rightarrow x = \frac{9 \pm 8.775}{2} \\[1em] \Rightarrow x = \frac{9 + 8.775}{2} \text{ or } \frac{9 - 8.775}{2} \\[1em] \Rightarrow x = \frac{17.775}{2} \text{ or } \frac{0.225}{2} \\[1em] \Rightarrow x = 8.8875 \text{ or } 0.1125 \\[1em] \Rightarrow x = 8.89 \text{ or } 0.113\end{aligned}$$

Hence, x = 8.89 or 0.113.

(a) Given,

Number of shares = 120

Nominal value = ₹ 100

Premium = ₹ 25

Market price per share = ₹100 + ₹25 = ₹125

Total money invested by Kabir = Number of shares × Market price per share

= 120 × 125 = ₹ 15,000.

Hence, the money invested by Kabir = ₹ 15,000.

(b) Given,

Total dividend = ₹ 1080

Let rate of dividend be R.

By formula,

Dividend = No. of shares × $\frac{\text{Rate of dividend}}{100}$ × N.V. of share

1080 = 120 × $\frac{R}{100}$ × 100

1080 = 120 × R

R = $\frac{1080}{120}$

R = 9%.

Hence, rate of dividend = 9%.

(c) Given,

Investment = ₹ 15,000

Rate of return = $\frac{\text{Total dividend}}{\text{Total investment}}$ x 100

= $\frac{1080}{15000} \times 100$

= 7.2%.

Hence, rate of return = 7.2%.

Here, i = class size = 8 and assumed mean (A) = 28.

Class Interval	Frequency (f)	Class mark (x)	d = x - A	t = (x - A)/i	ft
0 - 8	10	4	-24	-3	-30
8 - 16	20	12	-16	-2	-40
16 - 24	14	20	-8	-1	-14
24 - 32	16	28	0	0	0
32 - 40	18	36	8	1	18
40 - 48	22	44	16	2	44
Total	∑f = 100				∑ft = -22

By formula,

Mean = $A + \Big(\frac{\sum ft}{\sum f} \Big) \times i$

= 28 + $\Big(-\frac{22}{100}\Big) \times 8$

= 28 + 8(-0.22)

= 28 - 1.76

= 26.24

Hence, mean = 26.24

Let the two natural numbers be: x and x + 5

Given,

The sum of the reciprocals of two numbers is $\frac{3}{10}$.

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{x + 5} = \frac{3}{10} \\[1em] \Rightarrow \frac{x + 5 + x}{x(x + 5)} = \frac{3}{10} \\[1em] \Rightarrow \frac{2x + 5}{x(x + 5)} = \frac{3}{10} \\[1em] \Rightarrow 10(2x + 5) = 3[x(x + 5)] \\[1em] \Rightarrow 20x + 50 = 3x^2 + 15x \\[1em] \Rightarrow 3x^2 + 15x - 20x - 50 = 0 \\[1em] \Rightarrow 3x^2 - 5x - 50 = 0 \\[1em] \Rightarrow 3x^2 + 10x - 15x - 50 = 0 \\[1em] \Rightarrow x(3x + 10) - 5(3x + 10) = 0 \\[1em] \Rightarrow (x - 5)(3x + 10) = 0 \\[1em] \Rightarrow (x - 5) = 0 \text{ or } (3x + 10) = 0 \\[1em] \Rightarrow x = 5 \text{ or } 3x = -10 \\[1em] \Rightarrow x = 5 \text{ or } x = -\frac{10}{3}\end{aligned}$$

Since, the numbers are natural numbers, thus x cannot be negative.

Thus, x = 5 and x + 5 = 5 + 5 = 10.

Hence, numbers are 5 and 10.

Let height of building (BC) be h meters and height of flagpole (CD) be x meters.

Distance from point = 100 m

In triangle ABC,

$$\begin{aligned}\Rightarrow \tan 31° = \frac{BC}{AB} \\[1em] \Rightarrow \tan 31° = \frac{h}{100} \\[1em] \Rightarrow 0.6009 = \frac{h}{100} \\[1em] \Rightarrow h = 0.6009 \times 100 \\[1em] \Rightarrow h = 60.09 m\end{aligned}$$

In triangle ADB,

$$\begin{aligned}\Rightarrow \tan 33° = \frac{DB}{AB} \\[1em] \Rightarrow \tan 33° = \frac{h + x}{100} \\[1em] \Rightarrow h + x = 0.6494 \times 100 \\[1em] \Rightarrow h + x = 64.94 \\[1em] \Rightarrow 60.09 + x = 64.94 \\[1em] \Rightarrow x = 64.94 - 60.09 \\[1em] \Rightarrow x = 4.85 \text{ m} \approx 5 \text{ m}.\end{aligned}$$

Hence, the height of flagpole = 5 m.

The cumulative frequency table for the given continuous distribution is :

Weight (in kg)	Number of students	Cumulative frequency
35 – 40	4	4
40 – 45	6	10
45 – 50	10	20
50 – 55	24	44
55 – 60	26	70
60 – 65	17	87
65 – 70	8	95
70 – 75	5	100

(a) Steps of Construction :

1. Since, the scale on x-axis starts at 35, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 35.

2. Take 1 cm along x-axis = 5 kg (weight)

3. Take 1 cm along y-axis = 10 (students)

4. Plot the points (40, 4), (45, 10), (50, 20), (55, 44), (60, 70), (65, 87), (70, 95), (75, 100) representing upper class limits and the respective cumulative frequencies. Also plot the point representing lower limit of the first class i.e. 35 - 40.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

Here, n (no. of students) = 100.

To find the median :

Let B be the point on y-axis representing frequency = $\frac{n}{2} = \frac{100}{2} = 50$.

Through B draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents .

From graph, N = 56

Hence, the median weight = 56 kg.

(b) Percentage of students whose weight is 60 kg or more.

Let O be the point on x-axis representing weight = 60.

Through O draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents .

From graph,

C = 70

Students weighing more than 60 kg = 100 - 70 = 30.

Percentage of students weighing more than 60 kg = $\frac{30}{100} \times 100$ = 30%

Hence, students weighing more than 60 kg = 30%.

(c) The weight above which 20% of the students lie.

Total number of students = 100

20% of students = $\frac{20}{100}$ × 100 = 20.

So, 80 students are below that weight.

Let P be the point on y-axis representing number of students = 80.

Through P draw a horizontal line to meet the ogive at T. Through T, draw a vertical line to meet the x-axis at S. The ordinate of the point S represents weight above which 20% of the students lie.

From graph,

S = 63

Hence, weight above which 20% students lie = 63 kg.

(a) Time = 2 years = 24 months

r = 8%

P = ₹ 300 for Vinay

By formula,

I = $\frac{P \times n(n + 1)}{2} \times \frac{r}{12 \times 100}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow I = \frac{300 \times 24(24 + 1)}{2} \times \frac{8}{1200} \\[1em] = \frac{300 \times 24(25)}{2} \times \frac{8}{1200} \\[1em] = 300 \times 300 \times \frac{8}{1200} \\[1em] = 300 \times 2 \\[1em] = ₹ 600.\end{aligned}$$

Hence, interest earned by Vinay = ₹ 600.

(b) Rohit’s interest is ₹800 more than Vinay’s.

Interest earned by Rohit = ₹600 + ₹800 = ₹1400.

Let Rohit deposit ₹P per month.

By formula,

I = $\frac{P \times n(n + 1)}{2} \times \frac{r}{12 \times 100}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow I = \frac{P \times 24(24 + 1)}{2} \times \frac{8}{1200} \\[1em] \Rightarrow 1400 = P \times 300 \times \frac{8}{1200} \\[1em] \Rightarrow 1400 = 2P \\[1em] \Rightarrow P = \frac{1400}{2} = ₹ 700.\end{aligned}$$

Hence, Rohit deposits ₹700 per month.

(a) We know that,

a_n = ar^{n - 1}

Given,

a₄ = ar³ = 16 ....(1)

a₇ = ar⁶ = 128 .....(2)

Dividing equation (2) by equation (1), we get :

$$\begin{aligned}\Rightarrow \frac{ar^6}{ar^3} = \frac{128}{16} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} = 2.\end{aligned}$$

Hence, common ratio = 2.

(b) The first term,

Substituting r = 2 in equation (1), we get :

⇒ a(2)³ = 16

⇒ a(8) = 16

⇒ a = $\frac{16}{8}$

⇒ a = 2.

Hence, first term = 2.

(a) The point O(0, 0), A(2, 3), B(5, 3) and C(3, 0) are plotted on the graph below :

From graph, on reflecting O, A, B, C on x-axis we get,

O(0, 0) ⇒ O(0, 0)

A(2, 3) ⇒ D(2, -3)

B(5, 3) ⇒ E(5, -3)

C(3, 0) ⇒ C(3, 0)

(b) From graph, on reflecting O, A, B, C trough origin we get,

O(0, 0) ⇒ O(0, 0)

A(2, 3) ⇒ I(-2, -3)

B(5, 3) ⇒ J(-5, -3)

C(3, 0) ⇒ H(-3, 0)

(c) From graph, on reflecting O, A, B, C on y-axis we get,

O(0, 0) ⇒ O(0, 0)

A(2, 3) ⇒ F(-2, 3)

B(5, 3) ⇒ G(-5, 3)

C(3, 0) ⇒ H(-3, 0)

Given,

$-1 \lt \frac{2x - 3}{3} - \frac{x}{5} \le 1, x \in R$

Solving L.H.S of the inequation,

$$\begin{aligned}\Rightarrow -1 \lt \frac{2x - 3}{3} - \frac{x}{5} \\[1em] \Rightarrow -1 \lt \frac{10x - 15 - 3x}{15} \\[1em] \Rightarrow -1 \lt \frac{7x - 15}{15} \\[1em] \Rightarrow -15 \lt 7x - 15 \\[1em] \Rightarrow -15 + 15 \lt 7x \\[1em] \Rightarrow 0 \lt 7x \\[1em] \Rightarrow x \gt 0.\end{aligned}$$

Solving R.H.S of the inequation,

$$\begin{aligned}\Rightarrow \frac{2x - 3}{3} - \frac{x}{5} \le 1 \\[1em] \Rightarrow \frac{10x - 15 - 3x}{15} \le 1 \\[1em] \Rightarrow \frac{7x - 15}{15} \le 1 \\[1em] \Rightarrow 7x - 15 \le 15 \\[1em] \Rightarrow 7x \le 15 + 15 \\[1em] \Rightarrow 7x \le 30 \\[1em] \Rightarrow x \le \frac{30}{7}.\end{aligned}$$

∴ 0 < x ≤ $\frac{30}{7}$

Solution set = $\Big(0, \frac{30}{7}\Big]$

Hence, solution set = $\Big(0, \frac{30}{7}\Big]$.

(a) From graph,

Coordinates of :

A(4, 8), B(-1, 2), C(6, 2)

Hence, A(4, 8), B(-1, 2), C(6, 2).

(b) Midpoint of AC,

By formula,

Midpoint = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Substituting values, we get :

Mid-point of AC = $\Big(\frac{4 + 6}{2}, \frac{8 + 2}{2}\Big)$ = (5, 5).

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Thus,

Slope of AB = $\frac{8 - 2}{4 - (-1)}$

= $\frac{6}{5}$.

Since, slope of parallel lines are equal, thus slope of required line (m) = $\frac{6}{5}$.

By point-slope form,

Equation : y - y₁ = m(x - x₁)

Equation of line parallel to AB and passing through mid-point of AC is given by,

⇒ y - 5 = $\frac{6}{5}(x - 5)$

⇒ 5(y - 5) = 6(x - 5)

⇒ 5y - 25 = 6x - 30

⇒ 5y - 25 + 30 = 6x - 30

⇒ 5y - 5 = 6x

⇒ 6x - 5y - 5 = 0.

Hence, equation of required line = 6x - 5y - 5 = 0.

(a) Given,

Radius of each solid (r) = 7 cm

Height of cone = Height of cylinder = h = 24 cm

Let slant height of cone be l.

By formula,

$$\begin{aligned}l = \sqrt{r^2 + h^2} \\[1em] = \sqrt{7^2 + 24^2} \\[1em] = \sqrt{49 + 576} \\[1em] = \sqrt{625} \\[1em] = 25.\end{aligned}$$

Total surface area of sphere = 4πr²

= 4 × $\frac{22}{7}$ × 7²

= 616 cm².

Total surface area of cylinder = 2πr(h + r)

= 2 × $\frac{22}{7}$ × 7 × (24 + 7)

= 2 × 22 × 31

= 1364 cm².

Total surface area of cone = πr(r + l)

= $\frac{22}{7}$ × 7 × (7 + 25)

= 22 × 32

= 704 cm²

Total surface area of toy = Total surface area of (sphere + cylinder + cone)

= 616 + 1364 + 704

= 2684 cm².

Hence, total surface area of toy = 2684 cm².

(b) Given,

Cost of painting = ₹ 0.50 per cm².

The cost of painting the total surface = 2684 × ₹ 0.50 = ₹ 1,342

Hence, total cost of painting = ₹ 1,342.

(a) Given,

$x = \frac{5ab}{a - b}$

Multiplying $\frac{1}{a}$ on both sides of equation :

$$\begin{aligned}\Rightarrow \frac{x}{a} = \frac{1}{a} \Big(\frac{5ab}{a - b}\Big) \\[1em] \Rightarrow \frac{x}{a} = \frac{5b}{a - b}.\end{aligned}$$

Hence, $\frac{x}{a} = \frac{5b}{a - b}.$

(b) Since,

$\frac{x}{a} = \frac{5b}{a - b}$

Applying componendo and dividendo, we get :

$$\begin{aligned}\Rightarrow \frac{x + a}{x - a} = \frac{5b + (a - b)}{5b - (a - b)} \\[1em] \Rightarrow \frac{x + a}{x - a} = \frac{a + 4b}{6b - a}.\end{aligned}$$

Hence, $\frac{x + a}{x - a} = \frac{a + 4b}{6b - a}.$

(a) Total number of families = 300

Let E be the event of selecting family with one girl child,

∴ Number of favorable outcomes = 165

P(E) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{165}{300} = \frac{11}{20}.$

Hence, probability of selecting family with one girl child = $\frac{11}{20}.$

(b) Let A be the event of selecting family with one or more girl child,

∴ Number of favorable outcomes = 165 + 95 = 260

P(A) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{260}{300} = \frac{13}{15}.$

Hence, probability of selecting family with one or more girl child = $\frac{13}{15}.$

(c) Let B be the event of selecting family with no boy child.

A family will have no boy child, if there are 2 girl child in the family.

∴ Number of favorable outcomes = 95

P(B) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{95}{300} = \frac{19}{60}.$

Hence, probability of selecting family with no boy child = $\frac{19}{60}.$

Given,

∠CBQ = 40°

In a circle, the angle between a tangent and a chord through the point of contact is equal to the angle in the opposite (alternate) segment of the circle.

∠BAC = ∠CBQ = 40°

x = 40°

Since, AB = AC.

∠ABC = ∠BCA [Angles opposite to equal sides of a triangle are equal]

In triangle ABC,

∠ABC + ∠BAC + ∠BCA = 180°

2∠ABC + 40° = 180°

2∠ABC = 180° - 40°

2∠ABC = 140°

∠ABC = $\frac{140°}{2}$

∠ABC = 70°.

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∠BOC = 2∠BAC

y = 2x

y = 80°.

In triangle OBC,

OB = OC (Radii of same circle)

∠OBC = ∠OCB (Angles opposite to equal sides in a triangle are equal)

By angle sum property of triangle,

∠OBC + ∠OCB + ∠BOC = 180°

2∠OBC + 80° = 180°

2∠OBC = 100°

∠OBC = 50°

From figure,

w = ∠ABC - ∠OBC = 70° - 50° = 20°.

We know that,

Sum of opposite angles of a cyclic quadrilateral is 180°.

In cyclic quadrilateral ABCD,

∠ABC + ∠ADC = 180°

70° + z = 180°

z = 180° - 70° = 110°.

Hence, x = 40°, y = 80°, z = 110°, w = 20°.

Given,

Marked price = ₹ 18,000

Rate of GST = 18%

GST = $\frac{18}{100} \times 18000$ = ₹ 3,240

Price paid by customer = Marked price + GST

= ₹ 18,000 + ₹ 3,240

= ₹ 21,240.

Hence, option 4 is the correct option.

Given,

Ratio of heights ($h_1 : h_2$) = 2 : 1

Ratio of radii ($r_1 : r_2$) = 1 : 2

By formula,

Volume of cone = $\frac{1}{3} \pi r^2 h$

$$\begin{aligned}\therefore \frac{V_1}{V_2} = \frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2} \\[1em] = \frac{r_1^2 h_1}{r_2^2 h_2} \\[1em] = \frac{1^2 \times 2}{2^2 \times 1} \\[1em] = \frac{2}{4} \\[1em] = \frac{1}{2}.\end{aligned}$$

Ratio of volumes = 1 : 2.

Hence, option 2 is the correct option.

By formula,

Interest = $P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

In this formula, the interest is directly proportional to the rate of interest, as the principal (P) and the number of months ($n$) remain unchanged.

When the rate is revised from $r$% to $2r$%, the rate becomes double.

∴ The interest earned also becomes double.

Hence, option 4 is the correct option.

Dividend is always calculated on the face value (nominal value) of the shares.

Here, the face value of the share is ₹ 100.

∴ The person will receive an annual dividend which is 9% of the face value.

Hence, option 2 is the correct option.

Given,

Quadratic equation : $2x^2 + 5x + c = 0$

Comparing with $ax^2 + bx + c = 0$,

$a = 2, b = 5, c = c$

By formula,

Discriminant (D) = $b^2 - 4ac$

$$\begin{aligned}\Rightarrow 1 = 5^2 - 4 \times 2 \times c \\[1em] \Rightarrow 1 = 25 - 8c \\[1em] \Rightarrow 8c = 25 - 1 \\[1em] \Rightarrow 8c = 24 \\[1em] \Rightarrow c = 3.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

$(x + 3), (x + 6)$ and $(x + 10)$ are in continued proportion.

We know that,

Three quantities $a, b$ and $c$ are said to be in continued proportion if the ratio of the first to the second is equal to the ratio of the second to the third.

i.e. $a : b = b : c$

$$\begin{aligned}\Rightarrow \frac{a}{b} = \frac{b}{c} \\[1em] \Rightarrow b^2 = ac.\end{aligned}$$

So, the square of the middle term is equal to the product of the first and the last terms.

Here,

$a = (x + 3), b = (x + 6)$ and $c = (x + 10)$

∴ $(x + 6)^2 = (x + 3)(x + 10)$

Expanding both sides,

$\Rightarrow x^2 + 12x + 36 = x^2 + 13x + 30$

Cancelling $x^2$ from both sides,

$\Rightarrow 12x + 36 = 13x + 30$

Transposing the like terms,

$$\begin{aligned}\Rightarrow 36 - 30 = 13x - 12x \\[1em] \Rightarrow 6 = x \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

$(x – 1)$ is a factor of $x^3 + 2x^2 – x – k$.

∴ On putting $x = 1$, the value of the polynomial is zero.

$$\begin{aligned}\Rightarrow (1)^3 + 2(1)^2 - (1) - k = 0 \\[1em] \Rightarrow 1 + 2 - 1 - k = 0 \\[1em] \Rightarrow 2 - k = 0 \\[1em] \Rightarrow k = 2.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

$$mA = m\begin{bmatrix} -1 & 2 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} -m & 2m \\ 3m & 0 \end{bmatrix}$$

Comparing this with the given matrix $$\begin{bmatrix} 3 & -6 \\ -9 & 0 \end{bmatrix}$$,

$$\begin{aligned}\Rightarrow -m = 3\\[1em] \Rightarrow m = -3.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

The points A ($x$, 0), B (−1, −1) and C (0, $y$) are collinear.

We know that,

Collinear points lie on the same straight line, so they cannot enclose a triangle.

∴ Area of the triangle formed by them = 0.

By formula,

Area of triangle = $\frac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$

Here,

$$\begin{aligned}\Rightarrow x_1 = x, y_1 = 0 \\[1em] \Rightarrow x_2 = −1, y_2 = −1 \\[1em] \Rightarrow x_3 = 0, y_3 = y\end{aligned}$$

Substituting values we get :

$\Rightarrow \frac{1}{2}[x(-1 - y) + (-1)(y - 0) + 0(0 - (-1))] = 0$

Multiplying both sides by 2,

$$\begin{aligned}\Rightarrow x(-1 - y) + (-1)(y - 0) + 0(0 - (-1)) = 0 \\[1em] \Rightarrow -x - xy - y + 0 = 0 \\[1em] \Rightarrow -(x + y + xy) = 0 \\[1em] \Rightarrow x + y + xy = 0 \\[1em] \Rightarrow x + y = -xy.\end{aligned}$$

Since $\frac{1}{x} + \frac{1}{y}$ is required, we divide both sides by $xy$ (where $x ≠ 0$ and $y ≠ 0$),

$$\begin{aligned}\Rightarrow \frac{x + y}{xy} = \frac{-xy}{xy} \\[1em] \Rightarrow \frac{x}{xy} + \frac{y}{xy} = -1 \\[1em] \Rightarrow \frac{1}{y} + \frac{1}{x} = -1 \\[1em] \Rightarrow \frac{1}{x} + \frac{1}{y} = -1.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

PQ // BC and ar ∆APQ = ar quad BPQC.

From figure,

ar ∆ABC = ar ∆APQ + ar quad BPQC = ar ∆APQ + ar ∆APQ = 2 ar ∆APQ.

$\Rightarrow \frac{\text{ar ∆APQ}}{\text{ar ∆ABC}} = \frac{1}{2}$

In ∆APQ and ∆ABC,

⇒ ∠PAQ = ∠BAC [Common]

⇒ ∠APQ = ∠ABC [Corresponding angles as PQ // BC]

∴ ∆APQ ~ ∆ABC (By A.A. axiom)

We know that,

The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{ar ∆APQ}}{\text{ar ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{1}{2} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{AP}{AB} = \frac{1}{\sqrt{2}}.\end{aligned}$$

Let AP = $k$, then AB = $\sqrt{2}k$.

$$\begin{aligned}\Rightarrow PB = AB - AP = \sqrt{2}k - k = (\sqrt{2} - 1)k \\[1em] \Rightarrow \frac{AP}{PB} = \frac{k}{(\sqrt{2} - 1)k} = \frac{1}{\sqrt{2} - 1}.\end{aligned}$$

Rationalizing the denominator,

$\Rightarrow \frac{AP}{PB} = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \frac{\sqrt{2} + 1}{1}.$

AP : PB = $(\sqrt{2} + 1)$ : 1.

Hence, option 4 is the correct option.

Given,

PA = PB = AB.

Since all three sides of ∆PAB are equal, ∆PAB is an equilateral triangle.

∴ ∠APB = 60°.

We know that,

Radius is perpendicular to the tangent at the point of contact.

∴ ∠OAP = ∠OBP = 90°.

In quadrilateral OAPB,

⇒ ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

⇒ ∠AOB + 90° + 60° + 90° = 360°

⇒ ∠AOB = 360° - 240° = 120°.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

$\Rightarrow \angle \text{ ACB }= \frac{1}{2} \angle \text{ AOB }= \frac{1}{2} \times 120° = 60°.$

Hence, option 3 is the correct option.

When a semi-circular sheet is folded to form a right circular cone, the radius of the semi-circle becomes the slant height of the cone.

Radius of semi-circular sheet = 7 cm.

∴ Slant height of the cone = 7 cm.

Hence, option 2 is the correct option.

Solving the L.H.S. of the Assertion,

$$\begin{aligned}\Rightarrow \sec θ(1 + \sin θ)(\sec θ + \tan θ) \\[1em] \Rightarrow \sec θ(1 + \sin θ)\Big(\frac{1}{\cos θ} + \frac{\sin θ}{\cos θ}\Big) \\[1em] \Rightarrow \sec θ(1 + \sin θ) \times \frac{1 + \sin θ}{\cos θ} \\[1em] \Rightarrow \frac{1}{\cos θ} \times \frac{(1 + \sin θ)^2}{\cos θ} \\[1em] \Rightarrow \frac{(1 + \sin θ)^2}{\cos^2 θ} \\[1em] \Rightarrow \frac{(1 + \sin θ)^2}{1 - \sin^2 θ} \\[1em] \Rightarrow \frac{(1 + \sin θ)^2}{(1 - \sin θ)(1 + \sin θ)} \\[1em] \Rightarrow \frac{1 + \sin θ}{1 - \sin θ}.\end{aligned}$$

Since, $\frac{1 + \sin θ}{1 - \sin θ} \ne 1$, the Assertion (A) is false.

Now, $\sec^2 θ - \tan^2 θ = 1$ is a fundamental trigonometric identity which is true.

So, Reason (R) is true.

Thus, (A) is false, (R) is true.

Hence, option 2 is the correct option.

The Sun rising from the west is an impossible event, so its probability is 0.

∴ Assertion (A) is true.

The probability of an impossible event is always 0.

∴ Reason (R) is true.

Also, Reason (R) correctly explains why the probability of the Sun rising from the west (an impossible event) is 0.

Thus, both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 3 is the correct option.

Given,

Mean of $x_1, x_2, x_3, ……, x_n$ = $m$

$\Rightarrow \frac{x_1 + x_2 + \dots + x_n}{n} = m$

When each variate is increased by 2, the new mean is:

$$\begin{aligned}\Rightarrow \frac{(x_1 + 2) + (x_2 + 2) + \dots + (x_n + 2)}{n} \\[1em] \Rightarrow \frac{(x_1 + x_2 + \dots + x_n) + 2n}{n} \\[1em] \Rightarrow \frac{x_1 + x_2 + \dots + x_n}{n} + \frac{2n}{n} \\[1em] \Rightarrow m + 2.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

Radius of cylinder (R) = 3.5 cm

Height of cylinder (H) = 7 cm

Since the sphere completely fits into the cylinder,

Radius of sphere ($r$) = 3.5 cm

By formula,

Volume of cylinder = $\pi R^2 H$

$$\begin{aligned}= \frac{22}{7} \times (3.5)^2 \times 7 \\[1em] = \frac{22}{7} \times 12.25 \times 7 \\[1em] = 22 \times 12.25 \\[1em] = 269.5 \text{ cm}^3.\end{aligned}$$

By formula,

Volume of sphere = $\frac{4}{3} \pi r^3$

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times (3.5)^3 \\[1em] = \frac{4}{3} \times \frac{22}{7} \times 42.875 \\[1em] = \frac{4312}{24} \\[1em] = 179.67 \text{ cm}^3.\end{aligned}$$

Volume of water required = Volume of cylinder − Volume of sphere

= 269.5 − 179.67

= 89.83

≈ 90 cm³.

Hence, the volume of water required = 90 cm³.

(a) Given,

Distance covered = 400 km

Let the original speed of the car be $x$ km/h.

Since the speed is 12 km/h more in the second case,

Increased speed = $(x + 12)$ km/h.

By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

∴ Time taken with the original speed = $\frac{400}{x}$ hours

∴ Time taken with the increased speed = $\frac{400}{x + 12}$ hours

Converting the given difference of time into hours,

1 hour 40 minutes = $1\frac{40}{60}$ hours = $\frac{5}{3}$ hours.

The question says that with the increased speed the journey takes 1 hour 40 minutes less. So the time taken with the increased speed is smaller, and subtracting it from the time taken with the original speed must give $\frac{5}{3}$ hours.

∴ Time with original speed − Time with increased speed = $\frac{5}{3}$

According to the question,

$$\begin{aligned}\Rightarrow \frac{400}{x} - \frac{400}{x + 12} = \frac{5}{3} \\[1em] \Rightarrow 400\Big[\frac{(x + 12) - x}{x(x + 12)}\Big] = \frac{5}{3} \\[1em] \Rightarrow 400 \times \frac{12}{x^2 + 12x} = \frac{5}{3} \\[1em] \Rightarrow \frac{4800}{x^2 + 12x} = \frac{5}{3} \\[1em] \Rightarrow 5(x^2 + 12x) = 14400 \\[1em] \Rightarrow x^2 + 12x = 2880 \\[1em] \Rightarrow x^2 + 12x - 2880 = 0 \\[1em] \Rightarrow x^2 + 60x - 48x - 2880 = 0 \\[1em] \Rightarrow x(x + 60) - 48(x + 60) = 0 \\[1em] \Rightarrow (x - 48)(x + 60) = 0 \\[1em] \Rightarrow x = 48 \text{ or } x = -60.\end{aligned}$$

Since speed cannot be negative, $x = 48$.

Hence, the original speed of the car = 48 km/h.

(b) Increased speed = $x + 12 = 48 + 12 = 60$ km/h.

By formula,

Time taken with increased speed = $\frac{\text{Distance}}{\text{Speed}}$

$$\begin{aligned}= \frac{400}{60} \\[1em] = \frac{20}{3} \\[1em] = 6\frac{2}{3} \text{ hours}.\end{aligned}$$

= 6 hours 40 minutes.

Hence, the time taken with the increased speed = 6 hours 40 minutes.

(a) Given,

Vertices of ∆ABC are A ($a, b$), B (5, −3) and C (−5, 3).

Centroid G = $\Big(\frac{2}{3}, \frac{1}{3}\Big)$

By formula,

Centroid = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Here,

$x_1 = a, y_1 = b$

$x_2 = 5, y_2 = -3$

$x_3 = -5, y_3 = 3$

Substituting values we get :

$$\begin{aligned}\Rightarrow \Big(\frac{2}{3}, \frac{1}{3}\Big) = \Big(\frac{a + 5 + (-5)}{3}, \frac{b + (-3) + 3}{3}\Big) \\[1em] \Rightarrow \Big(\frac{2}{3}, \frac{1}{3}\Big) = \Big(\frac{a}{3}, \frac{b}{3}\Big).\end{aligned}$$

Comparing the coordinates,

$$\begin{aligned}\Rightarrow \frac{a}{3} = \frac{2}{3} \\[1em] \Rightarrow a = 2 \\[1em] \Rightarrow \frac{b}{3} = \frac{1}{3} \\[1em] \Rightarrow b = 1.\end{aligned}$$

Hence, the coordinates of vertex A are (2, 1).

(b) Let CM be the median, passing through C, here M is the mid-point of AB.

A = (2, 1) and B = (5, −3)

By formula,

Mid-point of AB = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Here,

$x_1 = 2, y_1 = 1$

$x_2 = 5, y_2 = -3$

Substituting values we get :

$$\begin{aligned}\text{ M }= \Big(\frac{2 + 5}{2}, \frac{1 + (-3)}{2}\Big) \\[1em] = \Big(\frac{7}{2}, \frac{-2}{2}\Big) \\[1em] = \Big(\frac{7}{2}, -1\Big).\end{aligned}$$

Hence, M = $\Big(\frac{7}{2}, -1\Big)$.

The median through C is the line joining C (−5, 3) and M $\Big(\frac{7}{2}, -1\Big)$.

By formula,

Slope of a line = $\frac{y_2 - y_1}{x_2 - x_1}$

Here,

$x_1 = -5, y_1 = 3$

$x_2 = \frac{7}{2}, y_2 = -1$

Substituting values we get :

$$\begin{aligned}\text{Slope (m)} = \frac{-1 - 3}{\frac{7}{2} - (-5)} \\[1em] = \frac{-4}{\frac{7 + 10}{2}} \\[1em] = \frac{-4}{\frac{17}{2}} \\[1em] = \frac{-8}{17}.\end{aligned}$$

By point-slope form,

Equation of a line : $y - y_1 = m(x - x_1)$

Substituting values we get :

$$\begin{aligned}\Rightarrow y - 3 = \frac{-8}{17}(x - (-5)) \\[1em] \Rightarrow 17(y - 3) = -8(x + 5) \\[1em] \Rightarrow 17y - 51 = -8x - 40 \\[1em] \Rightarrow 8x + 17y - 51 + 40 = 0 \\[1em] \Rightarrow 8x + 17y - 11 = 0.\end{aligned}$$

Hence, the equation of the median through vertex C is $\bm{8x + 17y - 11 = 0}$.

(a) For Anamika,

P = ₹ 300, $n$ = 3 years = 36 months, $r$ = 10%

By formula,

Interest = $P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}= 300 \times \frac{36(36 + 1)}{2 \times 12} \times \frac{10}{100} \\[1em] = 300 \times \frac{36 \times 37}{24} \times \frac{10}{100} \\[1em] = 300 \times 55.5 \times \frac{10}{100} \\[1em] = 1665.\end{aligned}$$

Hence, the interest earned by Anamika = ₹ 1,665.

(b) For Sanjeev,

P = ₹ 500, $n$ = 2 years = 24 months, $r$ = 10%

By formula,

Interest = $P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}= 500 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{10}{100} \\[1em] = 500 \times \frac{24 \times 25}{24} \times \frac{10}{100} \\[1em] = 500 \times 25 \times \frac{10}{100} \\[1em] = 1250.\end{aligned}$$

Interest earned by Anamika = ₹ 1,665

Interest earned by Sanjeev = ₹ 1,250

Difference = ₹ 1,665 − ₹ 1,250 = ₹ 415.

Hence, Anamika got more interest, by an amount of ₹ 415.

(a) Given,

$\sqrt{2}, \sqrt{8}, \sqrt{18}$ ......

Simplifying the terms,

$\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}$ ......

Second term − First term = $2\sqrt{2} - \sqrt{2} = \sqrt{2}$

Third term − Second term = $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$

Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (A.P.) with first term $a = \sqrt{2}$ and common difference $d = \sqrt{2}$.

Hence, the progression is an Arithmetic Progression.

(b) Given,

$a = \sqrt{2}, d = \sqrt{2}, n = 10$

By formula,

Sum of first $n$ terms = $\frac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$$\begin{aligned}= \frac{10}{2}[2\sqrt{2} + (10 - 1)\sqrt{2}] \\[1em] = 5[2\sqrt{2} + 9\sqrt{2}] \\[1em] = 5 \times 11\sqrt{2} \\[1em] = 55\sqrt{2}.\end{aligned}$$

Given, the sum of first 10 terms = $p\sqrt{2}$.

$$\begin{aligned}\Rightarrow p\sqrt{2} = 55\sqrt{2} \\[1em] \Rightarrow p = 55.\end{aligned}$$

Hence, the value of $\bm{p}$ = 55.

(a) The points A (2, 3) and B (4, 5) are plotted on the graph :

(b) From figure,

On reflecting points A and B in the x-axis, we get :

A′ = (2, −3) and B′ = (4, −5).

(c) From figure,

On reflecting point A' in y-axis, we get :

A″ = (−2, −3).

(d) Point A (2, 3) is mapped to A″ (−2, −3).

Since both the coordinates change their signs, the single transformation that maps A to A″ is a reflection in the origin.

Hence, the single transformation that maps A to A″ is reflection in the origin.

(a) For Component A,

Repairing cost = ₹ 5,000, Discount = 20%

Discount = $\frac{20}{100} \times 5000$ = ₹ 1,000

For Component B,

Repairing cost = ₹ 6,500, Discount = 30%

Discount = $\frac{30}{100} \times 6500$ = ₹ 1,950

Total discount = ₹ 1,000 + ₹ 1,950 = ₹ 2,950.

Hence, the total discount = ₹ 2,950.

(b) For Component A,

Cost after discount = ₹ 5,000 − ₹ 1,000 = ₹ 4,000

For Component B,

Cost after discount = ₹ 6,500 − ₹ 1,950 = ₹ 4,550

Total taxable amount = ₹ 4,000 + ₹ 4,550 = ₹ 8,550

GST = $\frac{18}{100} \times 8550$ = ₹ 1,539.

Hence, the total amount of GST = ₹ 1,539.

(a) Calculating A²,

$$A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 3 \times 3 + (-5)(-4) & 3 \times (-5) + (-5)(2) \\ (-4)(3) + 2(-4) & (-4)(-5) + 2 \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 20 & -15 - 10 \\ -12 - 8 & 20 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}.$$

Calculating 5A,

$$5A = 5\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}.$$

Hence, $$A^2 = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}$$ and $$5A = \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix}$$.

(b) Calculating A² − 5A,

$$A^2 - 5A = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} \\[1em] = \begin{bmatrix} 29 - 15 & -25 - (-25) \\ -20 - (-20) & 24 - 10 \end{bmatrix} \\[1em] = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \\[1em] = 14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\[1em] = 14 \cdot I.$$

Comparing with $A^2 - 5A = k.I$,

$\Rightarrow k = 14.$

Hence, the value of $\bm{k}$ = 14.

Solving the L.H.S.,

$$\begin{aligned}\Rightarrow \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} + \sqrt{\frac{\sec θ - 1}{\sec θ + 1}} \\[1em] \Rightarrow \frac{\sqrt{\sec θ + 1}}{\sqrt{\sec θ - 1}} + \frac{\sqrt{\sec θ - 1}}{\sqrt{\sec θ + 1}} \\[1em] \Rightarrow \frac{\sqrt{\sec θ + 1} \times \sqrt{\sec θ + 1} + \sqrt{\sec θ - 1} \times \sqrt{\sec θ - 1}}{\sqrt{\sec θ - 1} \times \sqrt{\sec θ + 1}} \\[1em] \Rightarrow \frac{(\sec θ + 1) + (\sec θ - 1)}{\sqrt{(\sec θ + 1)(\sec θ - 1)}} \\[1em] \Rightarrow \frac{\sec θ + 1 + \sec θ - 1}{\sqrt{\sec^2 θ - 1}} \\[1em] \Rightarrow \frac{2\sec θ}{\sqrt{\sec^2 θ - 1}} \\[1em] \Rightarrow \frac{2\sec θ}{\sqrt{\tan^2 θ}} \\[1em] \Rightarrow \frac{2\sec θ}{\tan θ} \\[1em] \Rightarrow \frac{\frac{2}{\cos θ}}{\frac{\sin θ}{\cos θ}} \\[1em] \Rightarrow 2 \times \frac{1}{\cos θ} \times \frac{\cos θ}{\sin θ} \\[1em] \Rightarrow \frac{2}{\sin θ} \\[1em] \Rightarrow \text{2 cosec θ}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\bm{\sqrt{\frac{\sec θ + 1}{\sec θ - 1}} + \sqrt{\frac{\sec θ - 1}{\sec θ + 1}} = 2,\text{cosec},θ}$.

(a) Given,

Investment = ₹ 7,500, Market value of each share = ₹ 150

By formula,

Number of shares = $\frac{\text{Investment}}{\text{Market value of each share}}$

$= \frac{7500}{150}$

= 50.

Hence, the number of shares purchased = 50.

(b) Given,

Face value = ₹ 100, Dividend rate = 6%, Number of shares = 50

By formula,

Annual income = Number of shares × Rate of dividend × Face value

$= 50 \times \frac{6}{100} \times 100$

= ₹ 300.

Hence, his annual income = ₹ 300.

(c) Number of shares sold = 50% of 50 = $\frac{50}{100} \times 50$ = 25 shares.

Selling price of each share = ₹ 175

Cost price of each share = ₹ 150

Gain on each share = ₹ 175 − ₹ 150 = ₹ 25

Total gain = 25 × ₹ 25 = ₹ 625.

Hence, his gain on the shares he sold = ₹ 625.

Given,

$3x + \frac{14}{3} \gt \frac{4x}{3} - 2 \geq 3x - 4, ; x \in R$

Solving the L.H.S. of the inequation, we get:

$$\begin{aligned}\Rightarrow 3x + \frac{14}{3} \gt \frac{4x}{3} - 2 \\[1em] \Rightarrow 3x - \frac{4x}{3} \gt -2 - \frac{14}{3} \\[1em] \Rightarrow \frac{9x - 4x}{3} \gt \frac{-6 - 14}{3} \\[1em] \Rightarrow \frac{5x}{3} \gt \frac{-20}{3} \\[1em] \Rightarrow 5x \gt \frac{-20}{3} \times 3 \\[1em] \Rightarrow 5x \gt -20 \\[1em] \Rightarrow x \gt \frac{-20}{5} \\[1em] \Rightarrow x \gt -4 \text{ ......(1)}\end{aligned}$$

Solving the R.H.S. of the inequation, we get:

$$\begin{aligned}\Rightarrow \frac{4x}{3} - 2 \geq 3x - 4 \\[1em] \Rightarrow \frac{4x}{3} - 3x \ge -4 + 2 \\[1em] \Rightarrow \frac{4x - 9x}{3} \ge -2 \\[1em] \Rightarrow \frac{-5x}{3} \ge -2 \\[1em] \Rightarrow \frac{-5x}{3} \times 3 \ge -2 \times 3 \\[1em] \Rightarrow -5x \ge -6 \\[1em] \Rightarrow -5x \times -1 \le -6 \times (-1) \\[1em] \Rightarrow 5x \le 6 \\[1em] \Rightarrow x \le \frac{6}{5} \text{ ......(2)}\end{aligned}$$

From equations (1) and (2),

Solution set = $\Big\lbrace x : -4 \lt x \le \frac{6}{5}, x \in R\Big\rbrace$

Representing on the real number line :

Hence, solution set = $\bm{\Big\lbrace x : -4 \lt x \le \frac{6}{5}, x \in R\Big\rbrace}$.

(a) Since ∆ABC ~ ∆DEC, their corresponding sides are proportional.

$\frac{AB}{DE} = \frac{AC}{DC} = \frac{BC}{EC}$

Hence, $\mathbf{\frac{AB}{DE} = \frac{AC}{DC} = \frac{BC}{EC}}$.

(b) Since ∆ABC ~ ∆DEC, their corresponding angles are equal.

∴ ∠ABC = ∠DEC.

Hence, ∠ABC is equal to ∠DEC.

(c) We know that,

The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{ar (∆ABC)}}{\text{ar (∆DEC)}} = \frac{AB^2}{DE^2} \\[1em] = \Big(\frac{2}{3}\Big)^2 \\[1em] = \frac{4}{9} \\[1em] = 4 : 9.\end{aligned}$$

Hence, ar (∆ABC) : ar (∆DEC) = 4 : 9.

(a) PQRS is a cyclic quadrilateral.

We know that,

The sum of opposite angles of a cyclic quadrilateral is 180°.

$$\begin{aligned}\Rightarrow \angle PQR + \angle PSR = 180° \\[1em] \Rightarrow 112° + \angle PSR = 180° \\[1em] \Rightarrow \angle PSR = 180° - 112° = 68°.\end{aligned}$$

Hence, ∠PSR = 68°.

(b) We know that,

By the alternate segment theorem, the angle between the tangent and a chord equals the angle subtended by the chord in the alternate segment.

∠BPR is the angle between tangent PB and chord PR, and its alternate segment contains Q.

∴ ∠BPR = ∠PQR = 112°.

Hence, ∠BPR = 112°.

(c) Given, PS = SR.

Equal chords subtend equal arcs.

∴ arc PS = arc SR.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended by it on the circumference.

∠PQR is the inscribed angle standing on arc PSR.

$\Rightarrow \text{arc PSR} = 2 \times \angle PQR = 2 \times 112° = 224°.$

Since arc PS = arc SR,

$\Rightarrow \text{arc PS} = \frac{224°}{2} = 112°.$

∠PTS is the inscribed angle standing on arc PS.

$\Rightarrow \angle PTS = \frac{1}{2} \times \text{arc PS} = \frac{1}{2} \times 112° = 56°.$

Hence, ∠PTS = 56°.

Marks	Class Mark (x)	d = x - A	u = d/i	Frequency (f)	fu
20 – 25	22.5	−10	−2	10	−20
25 – 30	27.5	−5	−1	7	−7
30 – 35	A = 32.5	0	0	9	0
35 – 40	37.5	5	1	13	13
40 – 45	42.5	10	2	5	10
45 – 50	47.5	15	3	6	18
Total				Σf = 50	Σfu = 14

Let assumed mean A = 32.5 and class size $i = 5$.

By formula,

Mean = A + $\frac{Σfu}{Σf} \times i$

$$\begin{aligned}= 32.5 + \frac{14}{50} \times 5 \\[1em] = 32.5 + \frac{70}{50} \\[1em] = 32.5 + 1.4 \\[1em] = 33.9 \\[1em] ≈ 34.\end{aligned}$$

Hence, the mean = 34.

(a) By the Remainder Theorem, the remainder when a polynomial is divided by $(x − a)$ is the value of the polynomial at $x = a$.

Let R₁ be the remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$. Put $x = 2$ :

$$\begin{aligned}\Rightarrow R_1 = m(2)^2 - 3(2) + 6 \\[1em] \Rightarrow R_1 = 4m - 6 + 6 \\[1em] \Rightarrow R_1 = 4m.\end{aligned}$$

Let R₂ be the remainder when $3x^2 + 5x − m$ is divided by $(x + 3)$. Put $x = −3$ :

$$\begin{aligned}\Rightarrow R_2 = 3(-3)^2 + 5(-3) - m \\[1em] \Rightarrow R_2 = 27 - 15 - m \\[1em] \Rightarrow R_2 = 12 - m.\end{aligned}$$

According to the question, R₁ = 2R₂.

$$\begin{aligned}\Rightarrow 4m = 2(12 - m) \\[1em] \Rightarrow 4m = 24 - 2m \\[1em] \Rightarrow 4m + 2m = 24 \\[1em] \Rightarrow 6m = 24 \\[1em] \Rightarrow m = 4.\end{aligned}$$

Hence, the value of $\bm{m}$ = 4.

(b) The remainder when $mx^2 − 3x + 6$ is divided by $(x − 2)$ is R₁ = $4m$.

$\Rightarrow R_1 = 4 \times 4 = 16.$

Hence, the remainder = 16.

The word 'BINOCULARS' has 10 letters, all distinct.

Sample space = {B, I, N, O, C, U, L, A, R, S}

Total number of outcomes = 10.

(a) The vowels in the word are I, O, U, A.

Number of vowels = 4.

By formula,

Probability = $\frac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$

$\Rightarrow P(\text{vowel}) = \frac{4}{10} = \frac{2}{5}.$

Hence, the probability that the letter is a vowel = $\mathbf{\frac{2}{5}}$.

(b) The first 8 letters of the English alphabet are A, B, C, D, E, F, G, H.

The letters among these which appear in 'BINOCULARS' are B, C and A.

Number of favourable outcomes = 3.

$\Rightarrow P = \frac{3}{10}.$

Hence, the required probability = $\mathbf{\frac{3}{10}}$.

Given,

$\frac{\sqrt{2x + 3} + \sqrt{2x - 3}}{\sqrt{2x + 3} - \sqrt{2x - 3}} = \frac{3}{1}$

Applying componendo and dividendo,

$$\begin{aligned}\Rightarrow \frac{(\sqrt{2x + 3} + \sqrt{2x - 3}) + (\sqrt{2x + 3} - \sqrt{2x - 3})}{(\sqrt{2x + 3} + \sqrt{2x - 3}) - (\sqrt{2x + 3} - \sqrt{2x - 3})} = \frac{3 + 1}{3 - 1} \\[1em] \Rightarrow \frac{2\sqrt{2x + 3}}{2\sqrt{2x - 3}} = \frac{4}{2} \\[1em] \Rightarrow \frac{\sqrt{2x + 3}}{\sqrt{2x - 3}} = 2.\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\Rightarrow \frac{2x + 3}{2x - 3} = 4 \\[1em] \Rightarrow 2x + 3 = 4(2x - 3) \\[1em] \Rightarrow 2x + 3 = 8x - 12 \\[1em] \Rightarrow 3 + 12 = 8x - 2x \\[1em] \Rightarrow 15 = 6x \\[1em] \Rightarrow x = \frac{15}{6} = \frac{5}{2}.\end{aligned}$$

Hence, $\bm{x = \frac{5}{2}}$.

(a) In ∆PTR and ∆QTP,

⇒ ∠PTR = ∠QTP [Common angle]

⇒ ∠TPR = ∠TQP [Angle between tangent and chord is equal to the angle in the alternate segment]

∴ ∆PTR ~ ∆QTP (By A.A. axiom)

Hence, proved that ∆PTR ~ ∆QTP.

(b) Since, ∆PTR ~ ∆QTP, their corresponding sides are proportional.

$$\begin{aligned}\Rightarrow \frac{PT}{QT} = \frac{TR}{TP} \\[1em] \Rightarrow PT^2 = QT \times TR.\end{aligned}$$

From figure,

QT = QR + RT = 9 + 16 = 25 cm.

Substituting the values,

$$\begin{aligned}\Rightarrow PT^2 = 25 \times 16 \\[1em] \Rightarrow PT^2 = 400 \\[1em] \Rightarrow PT = \sqrt{400} \\[1em] \Rightarrow PT = 20 \text{ cm}.\end{aligned}$$

Hence, the length of PT = 20 cm.

Steps of construction :

1. Take 2 cm = 10 units along the x-axis and 2 cm = 2 units along the y-axis.

2. Draw the rectangles (bars) for each class with heights equal to their frequencies.

3. The modal class is the class with the highest frequency, i.e. 50 – 60 (frequency 18).

4. In the highest rectangle, draw two lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

5. From the point of intersection of AC and BD, draw a vertical line to meet the x-axis. The point where it meets the x-axis gives the mode.

From the graph, the mode = 54.

Hence, the mode = 54.

Given,

$x^2 - 6x - 8 = 0$

Comparing with $ax^2 + bx + c = 0$,

$a = 1, b = -6, c = -8$

By formula,

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-8)}}{2(1)} \\[1em] \Rightarrow x = \frac{6 \pm \sqrt{36 + 32}}{2} \\[1em] \Rightarrow x = \frac{6 \pm \sqrt{68}}{2} \\[1em] \Rightarrow x = \frac{6 \pm 8.2462}{2}.\end{aligned}$$

Case 1:

$\Rightarrow x = \frac{6 + 8.2462}{2} = \frac{14.2462}{2} = 7.1231 ≈ 7.12.$

Case 2:

$\Rightarrow x = \frac{6 - 8.2462}{2} = \frac{-2.2462}{2} = -1.1231 ≈ -1.12.$

Hence, $\bm{x}$ = 7.12 or $\bm{x}$ = −1.12.

(a) Given,

Surface area of the solid metallic sphere = 616 cm²

Let the radius of the sphere be $r$ cm.

By formula,

Surface area of sphere = $4\pi r^2$

Equating this to the given surface area and substituting $\pi = \frac{22}{7}$, we get :

$$\begin{aligned}\Rightarrow 4 \times \frac{22}{7} \times r^2 = 616 \\[1em] \Rightarrow \frac{88}{7} \times r^2 = 616 \\[1em] \Rightarrow r^2 = \frac{616 \times 7}{88} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = 7 \text{ cm}.\end{aligned}$$

Hence, the radius of the sphere = 7 cm.

(b) We know that,

When a solid is melted and recast into a number of smaller solids, the material only changes its shape and no material is lost.

∴ Volume of the sphere = Number of cones × Volume of one cone

$\Rightarrow \text{Number of cones} = \frac{\text{Volume of sphere}}{\text{Volume of one cone}}$

So, we first find the volume of the sphere and the volume of one cone.

Volume of the sphere :

From part (a), radius of the sphere ($r$) = 7 cm.

By formula,

Volume of sphere = $\frac{4}{3} \pi r^3$

Substituting values we get :

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 7^3 \\[1em] = \frac{4}{3} \times \frac{22}{7} \times 343 \\[1em] = \frac{4}{3} \times 22 \times 49 \\[1em] = \frac{4312}{3} \text{ cm}^3.\end{aligned}$$

Volume of one cone :

Given,

Diameter of cone = 3.5 cm

$\Rightarrow \text{Radius of cone (r)} = \frac{\text{Diameter}}{2} = \frac{3.5}{2} = \frac{7}{4} \text{ cm}$

Height of cone ($h$) = 3.5 cm = $\frac{7}{2}$ cm

By formula,

Volume of cone = $\frac{1}{3} \pi r^2 h$

Substituting values we get :

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times \Big(\frac{7}{4}\Big)^2 \times \frac{7}{2} \\[1em] = \frac{1}{3} \times \frac{22}{7} \times \frac{49}{16} \times \frac{7}{2} \\[1em] = \frac{22 \times 49 \times 7}{3 \times 7 \times 16 \times 2} \\[1em] = \frac{539}{48} \text{ cm}^3.\end{aligned}$$

Number of cones :

$$\begin{aligned}\text{Number of cones} = \frac{\text{Volume of sphere}}{\text{Volume of one cone}} \\[1em] = \frac{\frac{4312}{3}}{\frac{539}{48}} \\[1em] = \frac{4312}{3} \times \frac{48}{539} \\[1em] = \frac{4312}{539} \times \frac{48}{3} \\[1em] = 8 \times 16 \\[1em] = 128.\end{aligned}$$

Hence, the number of cones formed = 128.

(a) Let T be the top of the lighthouse and L be its foot. Let A and B be the two positions of the ship, where the angles of depression are 35° and 55° respectively.

(b) From figure,

∠TAL = ∠ATX = 35° (alternate angles are equal)

∠TBL = ∠BTX = 55° (alternate angles are equal)

Height of lighthouse, TL = 80 m.

In right-angled triangle TLA,

$$\begin{aligned}\Rightarrow \tan 35° = \frac{TL}{LA} \\[1em] \Rightarrow LA = \frac{TL}{\tan 35°} \\[1em] \Rightarrow LA = \frac{80}{0.7002} \\[1em] \Rightarrow LA = 114.25 \text{ m}.\end{aligned}$$

In right-angled triangle TLB,

$$\begin{aligned}\Rightarrow \tan 55° = \frac{TL}{LB} \\[1em] \Rightarrow LB = \frac{TL}{\tan 55°} \\[1em] \Rightarrow LB = \frac{80}{1.4281} \\[1em] \Rightarrow LB = 56.02 \text{ m}.\end{aligned}$$

Distance travelled by the ship (AB) = LA − LB

= 114.25 − 56.02

= 58.23 m

≈ 58 m.

Hence, the distance travelled by the ship = 58 m.

Forming the cumulative frequency table :

Height (in cm)	No. of students	Cumulative frequency (cf)
150 – 155	8	8
155 – 160	15	23
160 – 165	24	47
165 – 170	17	64
170 – 175	11	75
175 – 180	5	80

Steps :

1. Take 2 cm = 10 students on the y-axis and 2 cm = 5 cm of height on the x-axis.

2. Plot the points (150, 0), (155, 8), (160, 23), (165, 47), (170, 64), (175, 75) and (180, 80).

3. Join the points with a free hand smooth curve to obtain the ogive.

(a) Here, $n$ = 80.

Median = $\frac{n}{2} = \frac{80}{2}$ = 40^th term.

Through the point A = 40 on the y-axis, draw a horizontal line to meet the curve at B, then a vertical line down to meet the x-axis at point C = 163.5.

Hence, the median height = 163.5 cm.

(b) Lower quartile = $\frac{n}{4} = \frac{80}{4}$ = 20^th term.

Through the point D = 20 on the y-axis, draw a horizontal line to meet the curve at E, then a vertical line down to meet the x-axis at point F = 159.

Hence, the lower-quartile height = 159 cm.

(c) Through the point G = 173 on the x-axis, draw a vertical line to meet the curve at H, then a horizontal line to meet the y-axis at point I = 71.

Thus, 71 students have a height less than or equal to 173 cm..

∴ Number of students with height more than 173 cm = 80 − 71 = 9.

Hence, the number of students whose height is more than 173 cm = 9.

Let the first three terms of the G.P. be $\frac{a}{r}, a$ and $ar$.

(a) Given,

Product of the terms = 27.

$$\begin{aligned}\Rightarrow \frac{a}{r} \times a \times ar = 27 \\[1em] \Rightarrow a^3 = 27 \\[1em] \Rightarrow a = 3.\end{aligned}$$

Given,

Sum of the terms = $\frac{21}{2}$.

$$\begin{aligned}\Rightarrow \frac{a}{r} + a + ar = \frac{21}{2} \\[1em] \Rightarrow a\Big(\frac{1}{r} + 1 + r\Big) = \frac{21}{2} \\[1em] \Rightarrow 3\Big(\frac{1}{r} + 1 + r\Big) = \frac{21}{2} \\[1em] \Rightarrow \frac{1}{r} + 1 + r = \frac{7}{2} \\[1em] \Rightarrow \frac{1}{r} + r = \frac{7}{2} - 1 \\[1em] \Rightarrow \frac{1}{r} + r = \frac{5}{2}.\end{aligned}$$

Multiplying both sides by $2r$,

$$\begin{aligned}\Rightarrow 2 + 2r^2 = 5r \\[1em] \Rightarrow 2r^2 - 5r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 4r - r + 2 = 0 \\[1em] \Rightarrow 2r(r - 2) - 1(r - 2) = 0 \\[1em] \Rightarrow (2r - 1)(r - 2) = 0 \\[1em] \Rightarrow r = \frac{1}{2} \text{ or } r = 2.\end{aligned}$$

Since $r \lt 1, r = \frac{1}{2}$.

Hence, the common ratio = $\mathbf{\frac{1}{2}}$.

(b) With $a = 3$ and $r = \frac{1}{2}$,

First term = $\frac{a}{r} = \frac{3}{\frac{1}{2}} = 6$

Second term = $a = 3$

Third term = $ar = 3 \times \frac{1}{2} = \frac{3}{2}$.

Hence, the first three terms are 6, 3 and $\mathbf{\frac{3}{2}}$.

(a) Given,

$3x - 4y + 7 = 0$

Writing in the form $y = mx + c$,

$$\begin{aligned}\Rightarrow 4y = 3x + 7 \\[1em] \Rightarrow y = \frac{3}{4}x + \frac{7}{4}.\end{aligned}$$

Comparing with $y = mx + c$, slope $m = \frac{3}{4}$.

Hence, the slope of the line = $\mathbf{\frac{3}{4}}$.

(b) We know that,

The product of the slopes of two perpendicular lines is −1.

Slope of the required line = $-\frac{1}{m} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$.

Given, y-intercept $c = 2$.

By slope-intercept form, $y = mx + c$,

$$\begin{aligned}\Rightarrow y = -\frac{4}{3}x + 2 \\[1em] \Rightarrow 3y = -4x + 6 \\[1em] \Rightarrow 4x + 3y - 6 = 0.\end{aligned}$$

Hence, the equation of the required line is $\bm{4x + 3y - 6 = 0}$.

(a) Steps of construction :

1. Draw a line segment AB = 5 cm.

2. At B, construct ∠ABX = 60°.

3. With A as centre and radius 6.3 cm, draw an arc cutting the ray BX at C.

4. Join AC. Thus, ∆ABC is constructed.

5. In ∆ABC, draw the perpendicular bisectors of any two sides (say AC and BC). Let them intersect at O.

6. With O as centre and radius OA, draw a circle. This is the circumcircle of ∆ABC passing through A, B and C.

7. In ∆ABC, draw the bisector of ∠ABC. This bisector is the locus of points equidistant from AB and BC.

Hence, the required construction is completed, where the bisector of ∠ABC is the locus of points equidistant from AB and BC.

Given,

(x - 2) and (x + 2) are the factors of x³ + x² - 4x - 4.

⇒ (x - 2)(x + 2) = x² - 4

Thus, x² - 4 is also the factor of x³ + x² - 4x - 4.

$$\begin{array}{l} \phantom{x - 3x)}{\quad x + 1} \\ x^2 - 4\overline{\smash{\big)}\quad x^3 + x^2 - 4x - 4 } \\ \phantom{x^2 + n}\phantom(\underline{\underset{-}{+}x^3 \underset{+}{-}4x} \\ \phantom{x^2 + 3x -])7 + o)()} x^2 - 4 \\ \phantom{x^2 + 3x - ((5) + )}\underline{\underset{-}{+}x^2 \underset{+}{-}4} \\ \phantom{x^2 + 3x - 5) + 24x )}\times \end{array}$$

Thus, x³ + x² - 4x - 4 = (x² - 4)(x + 1)

= (x + 2)(x - 2)(x + 1).

Hence, option 3 is the correct option.

Given,

P = ₹ 400

n = 18 months

By formula,

Qualifying Sum = $P \times \frac{n(n + 1)}{2}$

= $400 \times \frac{18(18 + 1)}{2}$

= 400 × 9 × 19

= 400 × 171

= ₹ 68,400

Hence, option 3 is the correct option.

Given,

We know that,

Angle in a semi-circle is a right angle.

∴ ∠ABC = 90°.

In ∆APQ and ∆ABC,

⇒ ∠APQ = ∠ABC [Both equal to 90°]

⇒ ∠BAC = ∠PAQ [Common angles]

∴ ∆APQ ∼ ∆ABC (By A.A. axiom)

We know that,

The ratio of area of similar triangles is equal to the ratio of the square of the corresponding sides.

$$\begin{aligned}\Rightarrow \frac{\text{Area of ∆APQ}}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{18}{\text{Area of ∆ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC}= \frac{(4 + 3)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of ∆ABC} = \frac{49}{9} \times 18 = 98 \text{ cm}^2.\end{aligned}$$

From figure,

Area QPBC = Area of ∆ABC − Area of ∆APQ

= 98 - 18 = 80 cm².

Hence, option 3 is the correct option.

Join OA and OP.

Given,

Radius of circle (OB) = 3 cm

PB and AP are at right angles.

We know that,

Radius and tangent at point of contact are perpendicular to each other.

OA ⊥ AP

From figure,

⇒ OA = OB = 3 cm [Radius of circle]

⇒ OB is perpendicular to PB

⇒ ∠OBP = 90°

This shows APBO is square.

OA = PB = AP = OB = 3 cm.

In right angled triangle OBP,

⇒ OP² = OB² + PB²

⇒ OP² = 3² + 3²

⇒ OP² = 9 + 9

⇒ OP² = 18

⇒ OP = $\sqrt{18}$

⇒ OP = $3\sqrt{2}$ cm.

Hence, option 3 is the correct option.

Given,

⇒ sec θ + tan θ = a

⇒ sec θ − tan θ = b

⇒ ab = (sec θ + tan θ)(sec θ - tan θ)

⇒ ab = sec² θ - tan² θ

⇒ ab = 1

So assertion (A) is true.

We know that,

⇒ sec² θ - tan² θ = 1

This is a fundamental trigonometric identity.

Thus, Both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 3 is the correct option.

By formula,

Volume of sphere = $\frac{4}{3} \pi r^3$

Given,

A solid sphere is cut into two identical hemispheres.

Volume of hemisphere = $\frac{2}{3} \pi r^3$

Volume of two identical hemispheres = $2 \times \frac{2}{3} \pi r^3$

= $\frac{4}{3} \pi r^3$

Thus, volume of a sphere = volume of two identical hemispheres.

So assertion (A) is true.

We know that,

Surface area of sphere = 4πr²

When a sphere is cut into two hemispheres, two new flat circular surfaces are created,

Surface area of a single hemisphere = Curved surface area + Area of its flat circular face

= 2πr² + πr²

= 3πr²

Total surface area of two hemispheres = 2 × 3πr² = 6πr².

Thus, surface area of a single hemisphere ≠ surface area of two hemispheres.

So reason (R) is false.

(A) is true, (R) is false.

Hence, option 1 is the correct option.

Given,

N = 7

Sum of the squares of the first seven natural numbers = 140.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

= $\frac{140}{7}$

= 20.

Hence, option 1 is the correct option.

Given,

Number of red marbles = 3

Number of blue marbles = 2

The total number of marbles in the bag is 3 + 2 = 5.

Number of black marbles in the bag = 0

Probability = $\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

= $\frac{0}{5}$

= 0.

Hence, option 1 is the correct option.

Calculating,

$$\Rightarrow AB = \begin{bmatrix*}[r] -1 & 2 \\ \end{bmatrix*}\begin{bmatrix*}[r] 3 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 3 + 2 \times 4 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 + 8 \\ \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \end{bmatrix*}.$$

Hence, option 3 is the correct option.

Given,

2 parts of red pigments is mixed with 5 parts of the base.

$$\begin{aligned}\therefore \frac{4}{a} = \frac{2}{5} \\[1em] \Rightarrow a = \frac{4 \times 5}{2} = \frac{20}{2} = 10. \\[1em] \therefore \frac{b}{12.5} = \frac{2}{5} \\[1em] \Rightarrow b = \frac{2}{5} \times 12.5 = 5. \\[1em] \therefore \frac{6}{c} = \frac{2}{5} \\[1em] \Rightarrow c = \frac{6 \times 5}{2} = 3 \times 5 = 15.\end{aligned}$$

Hence, option 4 is the correct option.

Market price = ₹ 1,200

Discount = 20% = $\frac{20}{100} \times 1200$ = ₹ 240

Price after discount = ₹ 1200 - ₹ 240 = ₹ 960.

GST = 18%

SGST = $\frac{GST}{2} = \frac{18}{2} = 9%$

= $\frac{9}{100} \times 960$

= 9 × 9.6

= ₹ 86.40

Hence, option 4 is the correct option.

M.V of share = ₹ 38.50

No of shares bought = 50

Money required = ₹ 38.50 × 50 = ₹ 1925.

Hence, option 3 is the correct option.

⇒ x² – 1 = 0

⇒ x² – 1² = 0

⇒ (x + 1)(x - 1) = 0

⇒ (x + 1) = 0 or (x - 1) = 0 [Using zero product rule]

⇒ x = -1 or x = 1.

Hence, option 4 is the correct option.

Equation:

⇒ x - y = 7

⇒ y = x - 7

Comparing above equation with y = mx + c, we get :

m = 1.

A line is equally inclined to the axes if slope = 1.

Hence, option 2 is the correct option.

Solving the given equation :

$$\begin{aligned}\Rightarrow x + 2 \le \frac{x}{3} + 3 \\[1em] \Rightarrow x - \frac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \frac{3x - x}{3} \le 1 \\[1em] \Rightarrow \frac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \frac{3}{2} \\[1em] \Rightarrow x \le 1.5\end{aligned}$$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, option 1 is the correct option.

Given,

⇒ 2x + 1 = 0

⇒ 2x = -1

⇒ x = $-\frac{1}{2}$

Substituting x = $-\frac{1}{2}$ in 2x³ + 7x² + 2x – 3, we get:

$$\begin{aligned}\Rightarrow 2 \times \Big(-\frac{1}{2}\Big)^3 + 7 \times \Big(-\frac{1}{2}\Big)^2 + 2 \times \Big(-\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times -\frac{1}{8} + 7 \times \frac{1}{4} + (-1) - 3 \\[1em] \Rightarrow -\frac{1}{4} + \frac{7}{4} - 4 \\[1em] \Rightarrow \frac{-1 + 7}{4} - 4 \\[1em] \Rightarrow \frac{6}{4} - 4 \\[1em] \Rightarrow \frac{6 - 16}{4} \\[1em] \Rightarrow \frac{-10}{4} \\[1em] \Rightarrow \frac{-5}{2}.\end{aligned}$$

Since, remainder is not equal to zero.

Hence, (2x + 1) is not a factor of the given polynomial.

Substituting x = $\frac{1}{2}$ in 2x³ + 7x² + 2x – 3, we get:

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{1}{2}\Big)^3 + 7 \times \Big(\frac{1}{2}\Big)^2 + 2 \times \Big(\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times \frac{1}{8} + 7 \times \frac{1}{4} + 1 - 3 \\[1em] \Rightarrow \frac{1}{4} + \frac{7}{4} - 2 \\[1em] \Rightarrow \frac{1 + 7}{4} - 2 \\[1em] \Rightarrow \frac{8}{4} - 2 \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.\end{aligned}$$

Since, remainder is equal to zero. Thus, x - $\frac{1}{2}$ is the factor of the polynomial.

$$\begin{aligned}\Rightarrow x - \frac{1}{2} = 0 \\[1em] \Rightarrow x = \frac{1}{2} \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow 2x - 1.\end{aligned}$$

2x - 1 is factor of polynomial.

Dividing 2x³ + 7x² + 2x – 3 by 2x - 1, we get:

$$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x – 3} \\ \phantom{x^2 +o 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{x^2 + 3x - 7)[][]} 8x^2 + 2x \\ \phantom{x^2 + 3x - 5))}\underline{\underset{-}{+}8x^2 \underset{+}{-}4x} \\ \phantom{x^2 + 3x - 5) + (24)}6x - 3 \\ \phantom{{x^2 + 3x - 5) + 24}}\underline{\underset{-}{+}6x \underset{+}{-}3} \\ \phantom{{x^2 + 3x - 54)} + 2x ,k}\times\ \end{array}$$

2x³ + 7x² + 2x – 3 by 2x - 1 = (2x - 1)(x² + 4x + 3)

= (2x - 1)[x² + 3x + x + 3]

= (2x - 1)[x(x + 3) + 1(x + 3)]

= (2x - 1)(x + 1)(x + 3).

Hence, 2x³ + 7x² + 2x – 3 = (2x - 1)(x + 1)(x + 3).

(a) Let the point P(x, 0) divide the line segment joining A(-6, 2) and B(9, -4) in the ratio m : n.

By section formula,

$y = \frac{my_2 + ny_1}{m + n}$

Substituting the values we get :

$$\begin{aligned}\Rightarrow 0 = \frac{m(-4) + n(2)}{m+n} \\[1em] \Rightarrow 0 = -4m + 2n \\[1em] \Rightarrow 4m = 2n \\[1em] \Rightarrow \frac{m}{n} = \frac{2}{4} \\[1em] \Rightarrow \frac{m}{n} = \frac{1}{2}.\end{aligned}$$

Hence, the ratio in which P divides the line segment AB is 1 : 2.

(b) From part (a),

A (-6, 2) and B (9, - 4)

m : n = 1 : 2

By section-formula,

$x = \frac{mx_2 + nx_1}{m + n}$

$$\begin{aligned}\Rightarrow x = \frac{1(9) + 2(-6)}{1+2} \\[1em] \Rightarrow x = \frac{9 - 12}{3} \\[1em] \Rightarrow x = \frac{-3}{3} \\[1em] \Rightarrow x = -1.\end{aligned}$$

P = (x, 0) = (-1, 0).

Hence, the coordinates of the point P are (-1, 0).

(c) By formula,

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

$$\begin{aligned}\Rightarrow m_{AB} = \frac{-4 - 2}{9 - (-6)} \\[1em] \Rightarrow m_{AB} = \frac{-6}{15} \\[1em] \Rightarrow m_{AB} = \frac{-2}{5}\end{aligned}$$

Line parallel to AB will have the same slope, so the slope of the new line is also m = $\frac{-2}{5}$.

By point-slope formula,

y - y₁ = m(x - x₁)

Line passing through (-3, -2) and parallel to AB is :

$$\begin{aligned}\Rightarrow y - (-2) = \frac{-2}{5}(x - (-3)) \\[1em] \Rightarrow y + 2 = \frac{-2}{5}(x + 3) \\[1em] \Rightarrow 5(y + 2) = -2(x + 3) \\[1em] \Rightarrow 5y + 10 = -2x - 6 \\[1em] \Rightarrow 5y + 10 + 2x + 6 = 0 \\[1em] \Rightarrow 2x + 5y + 16 = 0 \\[1em]\end{aligned}$$

Hence, the equation of the line is 2x + 5y + 16 = 0.

(a) Join AE.

We know that,

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

∴ ∠AOB = 2∠AEB

⇒ 80° = 2∠AEB

⇒ ∠AEB = $\frac{80^\circ}{2} = 40°$

We know that,

Angle in a semi-circle is a right angle.

⇒ ∠AEC = 90°

From figure,

⇒ ∠BEC = ∠AEC - ∠AEB = 90° - 40° = 50°.

Hence, ∠BEC = 50°.

(b) From figure,

⇒ ∠ECD = ∠CEB = 50° (Alternate angles are equal)

We know that,

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

⇒ ∠AOB = 2∠BCA

⇒ 80° = 2∠BCA

⇒ ∠BCA = $\frac{80^\circ}{2}$ = 40°.

From figure,

⇒ ∠BCD = ∠BCA + ∠ACE + ∠ECD = 40° + 20° + 50° = 110°.

Hence, ∠BCD = 110°.

(c) As sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠BED + ∠BCD = 180°

⇒ ∠BED = 180° - ∠BCD = 180° - 110° = 70°.

From figure,

⇒ ∠BED = ∠BEC + ∠CED

⇒ 70° = 50° + ∠CED

⇒ ∠CED = 70° - 50° = 20°.

Hence, ∠CED = 20°.

(a) Given,

-11, -7, -3, ......,49, 53

Second term - First term : −7 − (−11) = 4

Third term - Second term : −3 − (−7) = 4

Last term - Second last term : 53 − 49 = 4

Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (AP) with first term (a) = -11 and common difference (d) = 4.

Hence, the progression is an Arithmetic Progression.

(b) Let 53 be the nth term.

By formula,

⇒ t_n = a + (n - 1)d

⇒ 53 = -11 + (n - 1)4

⇒ 53 = -11 + 4n - 4

⇒ 53 = -15 + 4n

⇒ 4n = 53 + 15

⇒ 4n = 68

⇒ n = $\frac{68}{4}$

⇒ n = 17.

Hence, there are a total of 17 terms in the progression.

(c) Since, n = 17, is odd.

By formula,

Middle term = $\frac{n + 1}{2}$

= $\frac{17 + 1}{2}$

= $\frac{18}{2}$

= 9.

Thus, 9th term is middle term.

By formula,

⇒ t₉ = a + (n - 1)d

= -11 + (9 - 1) × 4

= -11 + 8 × 4

= -11 + 32

= 21.

Hence, the value of the middle most term is 21.

When vertically placed,

Height of liquid (h) = $\frac{1 + 6}{2} = \frac{7}{2}$ cm

Diameter of base = 7 cm

Radius (r) = $\frac{7}{2}$ cm

Area of wet surface = πr² + 2πrh

= πr(r + 2h)

= $\frac{22}{7} \times \frac{7}{2} \Big(\frac{7}{2} + 2 \times \frac{7}{2}\Big)$

= 11 × (3.5 + 7)

= 11 × 10.5

= 115.5 cm².

Hence, area of wet surface = 115.5 cm².

Steps of construction :

1. With O as center draw a circle of radius 4.5 cm.

2. Take a point A on the circumference with A as center cut an arc of radius 6.5 cm, intersecting circumference at point B.

3. Construct ∠CAB = 75°, where C lies on the circumference of the circle.

4. Draw XY, the perpendicular bisector of AB.

5. Draw AZ, the angular bisector of angle A.

6. Mark point P as the intersection of AZ and XY.

7. Measure CP.

Hence, the length of CP = 5.2 cm.

Given,

For initial investment,

Investment = ₹ 8,000

Face Value = ₹ 100

Market Value = ₹ 80

Dividend Rate = 6%

By formula,

Number of shares = $\frac{\text{Investment}}{\text{Market value of each share}}$

= $\frac{8000}{80}$

= 100

By formula,

Dividend for the first year = No. of shares × Rate of div. × N.V. of 1 share

= 100 × $\frac{6}{100}$ × 100

= ₹ 600

Hence, dividend for first year = ₹ 600.

(b) Given,

Number of shares sold = 100

Selling price per share = ₹ 75

Proceeds from sale = Number of shares × selling price

= 100 × 75

= ₹ 7,500

Total proceeds = Proceeds from sale + Dividend received = 7500 + 600 = ₹ 8,100

Total investment = ₹ 8,100

Market value per share = ₹ 27

Number of new shares = $\frac{\text{Investment}}{\text{Market value of each share}}$

= $\frac{8100}{27}$

= 300.

Hence, number of shares purchased by Ms. Kaur = 300.

Solving L.H.S of inequation : 5x - 21 < $\frac{5x}{7}$ - 6 ≤ $-3\frac{3}{7}$ + x, we get:

$$\begin{aligned}\Rightarrow 5x - 21 \lt \frac{5x}{7} - 6 \\[1em] \Rightarrow 5x - \frac{5x}{7} \lt 21 - 6 \\[1em] \Rightarrow \frac{35x - 5x}{7} \lt 15 \\[1em] \Rightarrow \frac{30x}{7} \lt 15 \\[1em] \Rightarrow x \lt \frac{7 \times 15}{30} \\[1em] \Rightarrow x \lt \frac{7}{2}\text{ ......(1)}\end{aligned}$$

Solving R.H.S of inequation 5x - 21 < $\frac{5x}{7}$ - 6 ≤ $-3\frac{3}{7}$ + x, we get:

$$\begin{aligned}\Rightarrow \frac{5x}{7} - 6 \le -3\frac{3}{7} + x \\[1em] \Rightarrow \frac{5x}{7} - 6 \le -\frac{24}{7} + x \\[1em] \Rightarrow x - \frac{5x}{7} \ge \frac{24}{7} - 6 \\[1em] \Rightarrow \frac{7x - 5x}{7} \ge \frac{24 - 42}{7} \\[1em] \Rightarrow \frac{2x}{7} \ge \frac{-18}{7} \\[1em] \Rightarrow 2x \ge -18 \\[1em] \Rightarrow x \ge \frac{-18}{2} \\[1em] \Rightarrow x \ge -9 \text{ .....(2)}\end{aligned}$$

From equation (1) and (2),

Solution set = {x : -9 ≤ x < $\frac{7}{2}$ , x ∈ R}

Hence, solution set = {x : -9 ≤ x < $\frac{7}{2}$ , x ∈ R}.

Solving,

(sin θ + cos θ)(cosec θ - sec θ)

$$\begin{aligned}\Rightarrow (\sin θ + \cos θ) \times \Big(\frac{1}{\sin θ} - \frac{1}{\cos θ}\Big) \\[1em] \Rightarrow (\sin θ + \cos θ) \times \Big(\frac{\cos θ - \sin θ }{\sin θ \cos θ}\Big) \\[1em] \Rightarrow \frac{\cos^2 θ - \sin^2 θ }{\sin θ \cos θ} \\[1em] \Rightarrow \frac{1- \sin^2 θ - \sin^2 θ}{\sin θ \cos θ} \\[1em] \Rightarrow \frac{1 - 2\sin^2 θ }{\sin θ \cos θ} \\[1em] \Rightarrow \frac{1}{\sin θ. \cos θ} - \frac{2 \sin^2 θ}{\sin θ.\cos θ} \\[1em] \Rightarrow \cosec θ.\sec θ - \frac{2 \sin θ}{\cos θ} \\[1em] \Rightarrow \cosec θ.\sec θ - 2\tan θ \\[1em]\end{aligned}$$

Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.

(a) In ∆APB and ∆CPD,

⇒ ∠APB = ∠CPD (Vertically opposite angles are equal)

⇒ ∠BAP = ∠DCP (Angles in same segment are equal)

∴ ∆APB ~ ∆CPD (By A.A. axiom)

Hence, proved that ∆APB ~ ∆CPD.

(b) We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{CD}{AB} = \frac{PD}{PB} \\[1em] \Rightarrow \frac{CD}{9} = \frac{2}{3} \\[1em] \Rightarrow CD = \frac{2}{3} \times 9 \\[1em] \Rightarrow CD = 6 \text{ cm}.\end{aligned}$$

Hence, CD = 6 cm.

(c) We know that,

Ratio of area of similar triangles is equal to the ratio of square of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of ∆APB}}{\text{Area of ∆CDP}} = \frac{PB^2}{PD^2} \\[1em] = \frac{3^2}{2^2} \\[1em] = \frac{9}{4} \\[1em] = 9 : 4.\end{aligned}$$

Hence, area ∆APB : area ∆CPD = 9 : 4.

Let rate of interest be r%.

Given,

P = ₹ 600/month

n = 2 years or 24 months

M.V. = ₹ 15,600

By formula,

M.V = $P \times n + P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$\Rightarrow 15600 = 600 \times 24 + 600 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{r}{100}$

⇒ 15600 = 14400 + 6 × 25 × r

⇒ 15600 − 14400 = 6 × 25 × r

⇒ 150r = 1200

⇒ r = $\frac{1200}{150}$ = 8%.

Hence, rate of interest = 8%.

In the given table i is the class interval which is equal to 15.

Class	Class Mark (x)	d = (x - A)	u = d/i	Frequency (f)	fu
0 - 15	7.5	-45	-3	3	-9
15 - 30	22.5	-30	-2	4	-8
30 - 45	37.5	-15	-1	7	-7
45 - 60	A = 52.5	0	0	6	0
60 - 75	67.5	15	1	8	8
75 - 90	82.5	30	2	2	4
Total				Σf = 30	Σfu = -12

By formula,

Mean = A + $\frac{Σfu}{Σf} \times i$

= 52.5 + $\frac{-12}{30}$ × 15

= = 52.5 - $\frac{180}{30}$

= 52.5 - 6

= 46.50

Hence, mean = 46.50.

By formula,

Centroid of triangle = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get:

⇒ Centroid of △ ABC = $\Big(\frac{-1 + 3 + 0}{3}, \frac{3 + (-1) + 0}{3}\Big)$

⇒ P = $\Big(\frac{2}{3}, \frac{2}{3}\Big)$

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

$$\begin{aligned}\text{Slope of AB} = \frac{-1 - 3}{3 - (-1)} \\[1em] = \frac{-4}{4} \\[1em] = -1.\end{aligned}$$

We know that,

Slope of parallel lines are equal.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁ )

Substituting values we get :

Equation of line passing through P and parallel to AB :

$$\begin{aligned}\Rightarrow y - \frac{2}{3} = -1 \Big(x - \frac{2}{3}\Big) \\[1em] \Rightarrow \frac{3y - 2}{3} = -1 \times \Big(\frac{3x - 2}{3}\Big) \\[1em] \Rightarrow 3y - 2 = -1 \times (3x - 2) \\[1em] \Rightarrow 3y - 2 = -3x + 2 \\[1em] \Rightarrow 3x + 3y = 2 + 2 \\[1em] \Rightarrow 3x + 3y = 4.\end{aligned}$$

Hence, equation of a line passing through P and parallel to AB is 3x + 3y = 4.

(a) We know that,

If two tangents are drawn to a circle from an exterior point the tangents are equal length.

AP = AS, BP = BQ, CQ = CR, DR = DS

From figure,

⇒ AB + CD = (AP + BP) + (DR + CR)

⇒ AB + CD = (AS + BQ) + (DS + CQ)

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC .......(1)

We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD and AD = BC

Substituting above values in equation (1), we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, proved that AB = BC.

(b) Since, AB = BC, AB = CD and AD = BC.

∴ AB = BC = CD = AD.

A parallelogram with all four sides equal is a rhombus.

Hence, ABCD is a rhombus.

For Dry fruits :

M.P. = ₹ 1,200

Discount = ₹ 100

Discounted price = ₹ 1200 - ₹ 100 = ₹ 1,100

GST = 12%

Tax = $\frac{12}{100} \times 1100$ = ₹ 132

Price including tax = ₹ 1,100 + ₹ 132 = ₹ 1,232.

For Wheat flour :

M.P. = ₹ 286

Discount = Nil

GST = 5%

Tax = $\frac{5}{100} \times 286$ = ₹ 14.30

Price including tax = ₹ 286 + ₹ 14.30 = 300.30

For bakery products :

M.P. = ₹ 500

Discount = 10% = $\frac{10}{100} \times 50$ = ₹ 50.

Discounted price = ₹ 500 - ₹ 50 = ₹ 450

GST = 12%

Tax = $\frac{12}{100} \times 450$ = ₹ 54

Price including tax = ₹ 450 + ₹ 54 = ₹ 504.

Total price = ₹ 1,232 + ₹ 300.30 + ₹ 504 = ₹ 2,036.30

Hence, total amount to be paid = ₹ 2,036.30.

From figure,

Let the top of the lighthouse be T, the base of the lighthouse be L, and the two ships be A and B.

Given,

The height of the lighthouse, TL, is 80 m.

The angles of depression are 30° and 50°.

These are equal to the alternate interior angles of elevation at the ships.

∠TBL = 50° and ∠TAL = 30°

In triangle TLA,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 30° = \frac{TL}{LA} \\[1em] \Rightarrow LA = \frac{TL}{\tan 30°} \\[1em] \Rightarrow LA = \frac{80}{0.577} \\[1em] \Rightarrow LA = 138.65 \text{ m}.\end{aligned}$$

In triangle TLB,

$$\begin{aligned}\Rightarrow \tan 50° = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 50° = \frac{TL}{LB} \\[1em] \Rightarrow LB = \frac{TL}{\tan 50°} \\[1em] \Rightarrow LB = \frac{80}{1.192} \\[1em] \Rightarrow LB = 67.11 \text{ m}.\end{aligned}$$

Distance between ships = LA + LB

= 138.65 m + 67.11 m = 205.76 m.

Hence, the distance between the two ships is 206 m.

Marks %	No.of students	cf
0 - 10	5	5
10 - 20	7	12
20 - 30	11	23
30 - 40	20	43
40 - 50	40	83
50 - 60	52	135
60 - 70	36	171
70 - 80	15	186
80 - 90	9	195
90 - 100	5	200

Steps :

1. Take 1 cm = 10 marks on x-axis.

2. Take 1 cm = 20 students on y-axis.

3. Plot the points (10, 5), (20, 12), (30, 23), (40, 43), (50, 83), (60, 135), (70, 171), (80, 186), (90, 195) and (100, 200).

4. Join the points by free hand curve.

(a) Here, n (no. of students) = 200, which is even.

Median = $\frac{n}{2} = \frac{200}{2}$ = 100^th term.

Through point L = 100 draw a horizontal line parallel to x-axis touching the graph at point M, through M draw a vertical line parallel to y-axis touching x-axis at point N = 53.

Hence, median = 53.

(b) Total marks = 100

65% of 100 = 65

Through point O = 65 draw a vertical line parallel to y-axis touching the graph at point P, through P draw a horizontal line parallel to x-axis touching y-axis at point Q = 154.

∴ 154 students score less than or equal to 65%.

∴ 46 (200 - 154) students score more than 65%.

Hence, 46 students score more than 65%.

(c) Total marks = 100

35% of 100 = 35

Through point R = 35 draw a vertical line parallel to y-axis touching the graph at point S, through S draw a horizontal line parallel to x-axis touching y-axis at point T = 33.

Hence, 33 students did not pass the exam.

The cards are numbered from 0 to 100.

Total number of cards = 101 (as there is no. zero card also)

(a) Favorable outcome = Getting a card having a number which is divisible by 6.

{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

Probability $=\frac{\text{No of favorable outcomes}}{\text{Total number of outcomes}} = \frac{17}{101}$.

Hence, probability of getting a card which is divisible by 6 = $\frac{17}{101}$.

(b) Favorable outcome = Getting a card having a number which is not divisible by 6.

P(getting a card not divisible by 6) = 1 - P(getting a card divisible by 6)

= $1 - \frac{17}{101}$

= $\frac{101 - 17}{101}$

= $\frac{84}{101}$.

Hence, probability of getting a card which is not divisible by 6 = $\frac{84}{101}$.

Given,

x, y and z are in continued proportion.

$\therefore \frac{x}{y} = \frac{y}{z}$

⇒ y² = xz

To prove :

$\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$

Solving L.H.S,

$$\begin{aligned}\Rightarrow \frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.y^2.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^2.xz.z^2} \\[1em] \Rightarrow \frac{x^3 + y^3 + z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{x^3}{x^3.z^3} + \frac{y^3}{x^3.z^3} + \frac{z^3}{x^3.z^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(xz)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{(y^2)^3} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{y^3}{y^6} + \frac{1}{x^3} \\[1em] \Rightarrow \frac{1}{z^3} + \frac{1}{y^3} + \frac{1}{x^3}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{x}{y^2.z^2} + \frac{y}{z^2.x^2} + \frac{z}{x^2.y^2} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$.

(a) Given,

Radius of ball bearings = 7 mm

Volume of box = 2156 cm³ = 2156 × 10³ mm³

Number of ball bearings that each box can have (N)

= $\frac{\text{Volume of box}}{\text{Volume of each ball}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow N = \frac{2156 \times 10^3}{\frac{4}{3} \times \frac{22}{7} \times 7^3} \\[1em] = \frac{2156 \times 10^3}{\frac{88}{3} \times 7^2} \\[1em] = \frac{2156 \times 3 \times 1000}{88 \times 49} \\[1em] = \frac{6468000}{4312} \\[1em] = 1500.\end{aligned}$$

Hence, maximum no. of ball bearings in a box = 1500.

(b) Mass of each box = No. of balls × Mass of each ball

= 1500 × 4 gm

= 6000 gm

= $\frac{6000}{1000}$

= 6 kg.

Hence, mass of each box = 6 kg.

(a) From histogram,

The number of batsmen who scored between 500 and 600 runs is 3.

The number of batsmen who scored between 600 and 700 runs is 2.

Thus, the total number of batsmen who scored 500 to 700 runs is 3 + 2 = 5.

Hence, number of batsmen who scored 500 to 700 runs = 5.

(b) The modal class is the class with the highest frequency.

Hence, the modal class interval is 400 - 500.

(c) From graph,

Mode = 430.

Hence, mode = 430.

Let common difference be d.

First term (a) = 3

By formula,

Sum of first n terms of an A.P. = $\frac{n}{2}[2a + (n - 1)d]$

Given,

The sum of the first 8 terms is twice the sum of the first 5 terms.

⇒ S₈ = 2 × S₅

$\therefore \frac{8}{2}[2 \times a + (8 - 1) \times d] = \frac{5}{2}[2 \times a + (5 - 1) \times d]$

⇒ 4[2a + 7d] = 5[2a + 4d]

⇒ 4[2 × 3 + 7d] = 5[2 × 3 + 4d]

⇒ 4[6 + 7d] = 5[6 + 4d]

⇒ 24 + 28d = 30 + 20d

⇒ 28d - 20d = 30 - 24

⇒ 8d = 6

⇒ d = $\frac{6}{8} = \frac{3}{4}$.

Hence, common difference = $\frac{3}{4}$.

Given,

The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (r - p)² - 4 × (q - r) × (p - q) = 0

⇒ r² + p² - 2pr - 4(qp - q² - rp + qr) = 0

⇒ r² + p² - 2pr - 4qp + 4q² + 4rp - 4qr = 0

⇒ r² + p² + 2pr - 4qp - 4qr + 4q² = 0

⇒ (r + p)² - 4q(p + r) + 4q² = 0

Let p + r = y,

⇒ y² - 4qy + 4q² = 0

⇒ (y - 2q)² = 0

⇒ y - 2q = 0

⇒ y = 2q

⇒ p + r = 2q.

Hence, proved that p + r = 2q.

Let the three consecutive even numbers be represented as x, x + 2, and x + 4.

Given,

The sum of their squares is 596.

⇒ x² + (x + 2)² + (x + 4)² = 596

⇒ x² + x² + 4x + 4 + x² + 8x + 16 = 596

⇒ 3x² + 12x + 20 = 596

⇒ 3x² + 12x + 20 - 596 = 0

⇒ 3x² + 12x - 576 = 0

⇒ 3(x² + 4x − 192) = 0

⇒ x² + 4x − 192 = 0

⇒ x² + 16x - 12x − 192 = 0

⇒ x(x + 16) - 12(x + 16) = 0

⇒ (x - 12)(x + 16) = 0

⇒ (x - 12) = 0 or (x + 16) = 0 [Using zero product rule]

⇒ x = 12 or x = -16.

Case 1 : x = 12

The three consecutive even numbers are:

⇒ x = 12

⇒ x + 2 = 12 + 2 = 14

⇒ x + 4 = 12 + 4 = 16

Case 2: x = −16

The three consecutive even numbers are:

⇒ x = −16

⇒ x + 2 = −16 + 2 = −14

⇒ x + 4 = −16 + 4 = −12

Hence, the two possible sets of three consecutive even numbers are 12, 14, 16 and -16, -14, -12.

Given,

X² = 4X + 5I

Solving for L.H.S.,

$$X^2 = \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1 \times 3 \\ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 + 8 & 1 + 3 \\ 8 + 24 & 8 + 9 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.$$

Solving for R.H.S.,

$$4X + 5I = 4\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 & 4 \\ 32 & 12 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \\ 0 & 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 4 + 5 & 4 + 0 \\ 32 + 0 & 12 + 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 9 & 4 \\ 32 & 17 \end{bmatrix*}.$$

Since, L.H.S. = R.H.S.

Hence, proved that X² = 4X + 5I.

Steps of Construction:

1. Plot the points : A(4, 0), B(4, 2), C(2, 2), D(2,4) and E(0, 4).

2. On reflecting the points A, B, C and D on the y-axis, they become F, G, H and I respectively.

3. Join the points ABCDEIHGF.

4. Reflect the figure ABCDEIHGF in x-axis.

5. On reflecting the points A and F in x-axis, they remain same as they lie on x-axis.

6. On reflecting points B, C, D, E, I, H and G on the x-axis, they become M, N, P, Q, R, S and T respectively.

7. Join the points A, E, F and Q.

In figure AEFQ,

There are four congruent isosceles right angle triangles.

Thus, all the sides AE, EF, FQ and QA are equal.

Also each interior angle is equal to 90°.

Hence, the figure AEFQ is square.

Comparing the given quadratic equation with ax² + bx + c = 0, we get :

a = 3, b = $\sqrt{7}$ and c = 2

Discriminant = b² - 4ac

= $(\sqrt{7})^2$ - 4 x 3 x 2

= 7 - 24

= -17.

Since, D < 0, there are no real roots.

Hence, option 4 is the correct option.

Given,

Monthly deposit = ₹500

Number of months = 18

Interest earned = ₹570

By formula,

Matured amount = Total deposit + Interest

= Monthly deposit x number of months + Interest

= ₹(500 x 18 + 570)

Hence, option 1 is the correct option.

Probability of any event is always between 0 and 1.

$\frac{5}{4}$ = 1.25, which is greater than 1, which is not possible.

Hence, option 1 is the correct option.

Given,

⇒ 3x + 4y + 7 = 0

⇒ 4y = -3x - 7

⇒ y = $-\frac{3}{4}x - \frac{7}{4}$

Comparing above equation with y = mx + c, we get :

m = $-\frac{3}{4}$.

By point-slope formula,

Equation of line : y - y₁ = m(x - x₁)

Thus, equation of line parallel to 3x + 4y + 7 = 0 and passing through origin (0, 0) is :

⇒ y - 0 = $-\frac{3}{4}$ (x - 0)

⇒ y = $-\frac{3x}{4}$

⇒ 4y = -3x

⇒ 4y + 3x = 0.

Hence, option 4 is the correct option.

Given,

$$A = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}$$

Substituting values we get :

$$A^2 = \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*} . \begin{bmatrix*}[r] 0 & 1 \\ 1 & 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{bmatrix*}\\[1em] = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}.$$

Hence, option 3 is the correct option.

From figure,

∠ACD and ∠ACB forms linear pairs [BD is a straight line].

⇒ ∠ACD + ∠ACB = 180°

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180° - 120°

⇒ ∠ACB = 60°.

In ΔABC,

⇒ BC = AC [Given]

⇒ ∠ABC = ∠BAC = x° (let) [Angles opposite to equal sides in triangle are equal]

According to angle sum property in ΔABC,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 60° + x° + x° = 180°

⇒ 2x° = 180° - 60°

⇒ 2x° = 120°

⇒ x° = $\frac{120°}{2}$ = 60°.

From figure,

ABCE is cyclic quadrilateral.

We know that,

Opposite angles of cyclic quadrilateral are supplementary.

⇒ ∠ABC + ∠AEC = 180°

⇒ 60° + ∠AEC = 180°

⇒ ∠AEC = 180° - 60°

⇒ ∠AEC = 120°.

Hence, option 4 is the correct option.

Factorize :

⇒ x² - 4

⇒ (x - 2)(x + 2)

Factorize :

⇒ x³ - x² - 4x + 4

⇒ (x³ - x²) - (4x - 4)

⇒ x²(x - 1) - 4(x - 1)

⇒ (x - 1)(x² - 4)

⇒ (x - 1)(x² - 2²)

⇒ (x - 1)(x - 2)(x + 2).

Common factors between x² - 4 and x³ - x² - 4x + 4 = (x - 2) and (x + 2).

Hence, option 3 is the correct option.

Given,

Dividend rate = 12% of face value.

So, if face value = ₹100, dividend = ₹12.

Return = 10%

By formula,

Return % = $\frac{\text{Dividend on one share}}{\text{Investment on one share}} \times 100%$

Investment on one share equals to the market value of the share.

Substituting values we get :

$$\begin{aligned}\Rightarrow 10 = \frac{\text{12}}{\text{Market value}} \times 100 \\[1em] \Rightarrow \text{Market value} = \frac{12}{10} \times 100 \\[1em] \Rightarrow \text{Market value} = ₹120.\end{aligned}$$

Since market value > face value, the shares are said to be above par.

Hence, option 3 is the correct option.

The circumcenter of a triangle is defined as the point where the perpendicular bisectors of the sides intersect. It is equidistant from the three vertices (D, E, and F) of the triangle.

∴ Statement 1 is true.

The incenter of a triangle is defined as the point of concurrency of the three angle bisectors of the triangle. This point is also the center of the inscribed circle and is equidistant from the three sides of the triangle.

∴ Statement 2 is true.

∴ Both statements are true.

Hence, option 1 is the correct option.