ICSE Class 10 Mathematics — 2027 Specimen

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.