ICSE Class 10 Mathematics Solved Paper 2025

(d) No real roots.

Explanation:

Comparing 3x² + √7x + 2 = 0 with ax² + bx + c = 0:

a = 3, b = √7 and c = 2

Discriminant, D = b² − 4ac

⇒ D = (√7)² − 4 × 3 × 2

⇒ D = 7 − 24

⇒ D = −17

Since D < 0, the equation has no real roots.

(a) ₹(500 × 18 + 570)

Explanation:

Monthly deposit = ₹500

Number of months = 18

Total deposit = ₹500 × 18

Interest earned = ₹570

Maturity amount = Total deposit + Interest

⇒ Maturity amount = ₹(500 × 18 + 570)

⇒ Maturity amount = ₹9,570

(a) \frac{5}{4}

Explanation:

The probability of any event always lies between 0 and 1, inclusive.

\frac{5}{4} = 1.25

Since 1.25 is greater than 1, it cannot be the probability of an event.

(d) 3x + 4y = 0

Explanation:

Parallel lines have the same coefficients of x and y.

A line parallel to 3x + 4y + 7 = 0 has the form:

3x + 4y + c = 0

It passes through the origin (0, 0).

Substituting x = 0 and y = 0:

3(0) + 4(0) + c = 0

⇒ c = 0

Therefore, the required equation is 3x + 4y = 0.

(c) \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}

Explanation:

A² = A × A

A² =  \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}

A² = \begin{bmatrix}(0\times0)+(1\times1) & (0\times1)+(1\times0) \\ (1\times0)+(0\times1) & (1\times1)+(0\times0)\end{bmatrix}

A² = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}

(d) 120°

Explanation:

∠ACD and ∠ACB form a linear pair because BD is a straight line.

∠ACD + ∠ACB = 180°

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 60°

In △ABC, AC = BC.

Therefore, ∠ABC = ∠BAC.

Let ∠ABC = ∠BAC = x.

x + x + 60° = 180°

⇒ 2x = 120°

⇒ x = 60°

Since ABCE is a cyclic quadrilateral, its opposite angles are supplementary.

∠ABC + ∠AEC = 180°

⇒ 60° + ∠AEC = 180°

⇒ ∠AEC = 120°

(c) (x − 2)

Explanation:

Factorizing the first polynomial:

x² − 4 = (x − 2)(x + 2)

Factorizing the second polynomial by grouping:

x³ − x² − 4x + 4

= x²(x − 1) − 4(x − 1)

= (x − 1)(x² − 4)

= (x − 1)(x − 2)(x + 2)

Thus, both (x − 2) and (x + 2) are common factors. Among the given options, the common factor is (x − 2).

(c) Above par

Explanation:

Assume the face value of one share is ₹100.

Dividend on one share = 12% of ₹100

⇒ Dividend = ₹12

Return% = \frac{\text{Dividend}}{\text{Market Value}} × 100

10 = \frac{12}{\text{Market Value}} × 100

⇒ Market Value = \frac{12\times100}{10}

⇒ Market Value = ₹120

Since the market value ₹120 is greater than the face value ₹100, the shares are above par.

(a) Both statements are true.

Explanation:

The circumcentre is the point of intersection of the perpendicular bisectors of a triangle. It is equidistant from all three vertices. Therefore, Statement 1 is true.

The incentre is the point of intersection of the three internal angle bisectors of a triangle. Therefore, Statement 2 is also true.

(c) Both A and R are true, and R is the correct reason for A.

Explanation:

Given: sin² A + sin A = 1

⇒ sin A = 1 − sin² A

Using 1 − sin² A = cos² A:

sin A = cos² A

Squaring both sides:

sin² A = cos⁴ A … (1)

Now: cos⁴ A + cos² A

= sin² A + cos² A, using (1)

= 1

Thus, the assertion is true. The reason is also true and is used directly to prove the assertion.

(a) 15 cm

Explanation:

Given, △ABC ∼ △EFG.

Therefore, corresponding sides are proportional.

\frac{AB}{EF}=\frac{BC}{FG}

From the diagram, AB = 15 cm, EF = 75 cm and BC = 3 cm.

\frac{15}{75}=\frac{3}{FG}

⇒ \frac{1}{5}=\frac{3}{FG}

⇒ FG = 3 × 5

⇒ FG = 15 cm

(a) 3 : 4

Explanation:

The volume of a sphere is proportional to the cube of its radius.

Let the radii be r and R.

\frac{r^3}{R^3}=\frac{27}{64}

Taking the cube root:

\frac{r}{R}=\sqrt[3]{\frac{27}{64}}

⇒ r/R = 3/4

Therefore, r : R = 3 : 4.

(d) ₹1,485

Explanation:

CGST rate = 4%

For an intra-state sale, SGST rate = CGST rate = 4%.

Total GST rate = 4% + 4%

⇒ Total GST rate = 8%

GST amount = \frac{8}{100} × ₹1,375

⇒ GST amount = ₹110

Price including GST = ₹1,375 + ₹110

⇒ Price including GST = ₹1,485

(a) {−5, −4, −3, −2, −1}

Explanation:

Given:

0 < – \text{x}\over3 < 2

⇒ 0 × 3 < – \text{x}\over3 × 3 < 2 × 3

⇒ 0 < – x < 6

⇒ – 1 × 0 > (- 1 )(- x) > – 1 × 6

⇒ −6 < x < 0

Since x ∈ Z, the solution set is {−5, −4, −3, −2, −1}.

(b) A is false, R is true.

Explanation:

The first 9 natural numbers are:

1, 2, 3, 4, 5, 6, 7, 8, 9

∑x = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Number of observations = 9

Mean = ∑\text{x} \over 9 = 45\over9

⇒ Mean = 5

Therefore, the assertion that the mean is 4.5 is false. The stated formula for the mean is true.

Comparing 2x² − 5x − 4 = 0 with ax² + bx + c = 0:

a = 2, b = −5 and c = −4.

Using the quadratic formula:

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

x = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-4)}}{2(2)}

⇒ x = \frac{5\pm\sqrt{25+32}}{4}

⇒ x = \frac{5\pm\sqrt{57}}{4}

Since √57 ≈ 7.54983:

x = (5 + 7.54983)/4 or x = (5 − 7.54983)/4

⇒ x ≈ 3.13746 or x ≈ −0.637458

Correct to three significant figures:

x = 3.14 or x = −0.637

(a) Monthly deposit, P = ₹250

Time = 3 years = 36 months

Total deposit = ₹250 × 36 = ₹9,000

Interest = Maturity value − Total deposit

⇒ Interest = ₹10,110 − ₹9,000

⇒ Interest = ₹1,110

For a recurring deposit:

I = \mathrm{P}\times\frac{\mathrm{n}(\mathrm{n}+1)}{2}\times\frac{\mathrm{r}}{12\times100}

1110=250\times\frac{36\times37}{2}\times\frac{\mathrm{r}}{1200}

⇒ 1110=\frac{333000\mathrm{r}}{2400}

⇒ r = \frac{1110\times2400}{333000}

⇒ r = 8%

(b) New monthly deposit = ₹250 + ₹50 = ₹300

\mathrm{I}=300\times\frac{36\times37}{2}\times\frac{8}{1200}

⇒ I = ₹1,332

Additional interest = ₹1,332 − ₹1,110

⇒ Additional interest = ₹222

Therefore, the rate is 8% and the additional interest is ₹222.

(a) In △ABC and △AED:

∠ABC = ∠AED = 90°

∠BAC = ∠DAE, as it is the common angle at A.

Therefore, △ABC ∼ △AED by the AA criterion.

(b) In right-angled △ABC:

AC² = AB² + BC²

⇒ 25² = 20² + BC²

⇒ 625 = 400 + BC²

⇒ BC² = 225

⇒ BC = 15 cm

Corresponding sides of similar triangles are proportional:

\frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{AD}}

\frac{20}{\mathrm{AE}}=\frac{15}{3}

⇒ \frac{20}{\mathrm{AE}}=5

⇒ AE = 4 cm

\frac{15}{3}=\frac{25}{\mathrm{AD}}

⇒ 5=\frac{25}{\mathrm{AD}}

⇒ AD = 5 cm

(c) Area of △ABC = \frac{1}{2} × 20 × 15 = 150 cm²

Area of △AED = \frac{1}{2} × 4 × 3 = 6 cm²

Area of BCED on the map = 150 − 6 = 144 cm²

Actual area = 576 m² = 5,760,000 cm²

If the linear scale factor is k:

k² = \frac{144}{5760000}=\frac{1}{40000}

⇒ k = \frac{1}{200}

Therefore, the scale factor is 1 : 200.

Given, AB = 6 cm, AC = 4.5 cm and ∠BAC = 120°

Steps of construction,

1. Draw AB = 6 cm.

2. At A, construct ∠BAY = 120°.

3. With A as centre and radius 4.5 cm, draw an arc cutting AY at C.

4. Join B to C to obtain △ABC.

5. Construct the perpendicular bisectors of AB and AC.

6. Mark their point of intersection as O. This is the circumcentre.

7. With O as centre and OA as radius, draw a circle through A, B and C.

8. Measure OA or OC.

The measured radius is approximately 5.2 cm.

(a) \mathrm{A}+\mathrm{C}=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}+\begin{bmatrix}-5 & 1 \\ 7 & -4\end{bmatrix}

=\begin{bmatrix}1-5 & 2+1 \\ 3+7 & 4-4\end{bmatrix}=\begin{bmatrix}-4 & 3 \\ 10 & 0\end{bmatrix}

(b) \mathrm{B}(\mathrm{A}+\mathrm{C})=\begin{bmatrix}2 & 1 \\ 4 & 2\end{bmatrix}\begin{bmatrix}-4 & 3 \\ 10 & 0\end{bmatrix}

=\begin{bmatrix}(2\times-4)+(1\times10) & (2\times3)+(1\times0) \\ (4\times-4)+(2\times10) & (4\times3)+(2\times0)\end{bmatrix}

=\begin{bmatrix}2 & 6 \\ 4 & 12\end{bmatrix}

(c) 5\mathrm{B}=5\begin{bmatrix}2 & 1 \\ 4 & 2\end{bmatrix}=\begin{bmatrix}10 & 5 \\ 20 & 10\end{bmatrix}

(d) \mathrm{B}(\mathrm{A}+\mathrm{C})-5\mathrm{B}=\begin{bmatrix}2 & 6 \\ 4 & 12\end{bmatrix}-\begin{bmatrix}10 & 5 \\ 20 & 10\end{bmatrix}

=\begin{bmatrix}-8 & 1 \\ -16 & 2\end{bmatrix}

(a) From the graph:

A(3, 3), B(0, −2), C(−4, −2) and D(−1, 3).

(b) Diagonals of a parallelogram bisect each other.

P is the midpoint of AC.

\mathrm{P}=\left(\frac{3+(-4)}{2},\frac{3+(-2)}{2}\right)

⇒ P = (−1/2, 1/2)

⇒ P = (−0.5, 0.5)

(c) Slope of CB = \frac{-2-(-2)}{0-(-4)}=\frac{0}{4}=0

Slope of DA = \frac{3-3}{3-(-1)}=\frac{0}{4}=0

Since the slopes are equal, CB ∥ DA.

(d) Slope of AC = \frac{-2-3}{-4-3}=\frac{-5}{-7}=\frac{5}{7}

Using point A(3, 3):

\mathrm{y}-3=\frac{5}{7}(\mathrm{x}-3)

⇒ 7y − 21 = 5x − 15

⇒ 5x − 7y + 6 = 0

Therefore, the equation of AC is 5x − 7y + 6 = 0.

Solving the left-hand part:

2x − \frac{5}{3} < \frac{3\mathrm{x}}{5} + 10

\frac{6\mathrm{x}-5}{3}<\frac{3\mathrm{x}+50}{5}

Multiplying by 15:

5(6x − 5) < 3(3x + 50)

⇒ 30x − 25 < 9x + 150

⇒ 21x < 175

⇒ x < \frac{25}{3} … (1)

Solving the right-hand part:

\frac{3\mathrm{x}}{5} + 10 ≤ \frac{4\mathrm{x}}{5} + 11

⇒ 3x + 50 ≤ 4x + 55

⇒ x ≥ −5 … (2)

Combining (1) and (2):

−5 ≤ x < \frac{25}{3}

Therefore, the solution set is [−5, \frac{25}{3}).

On the number line, use a closed point at −5, an open point at \frac{25}{3}, and shade between them.

First term, a = 5

Last term, l = 50

Sum, Sₙ = 440

(a) \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})

440 = \frac{\mathrm{n}}{2}(5 + 50)

⇒ 440=\frac{55\mathrm{n}}{2}

⇒ 880 = 55n

⇒ n = 16

(b) l = a + (n − 1)d

50 = 5 + (16 − 1)d

⇒ 50 = 5 + 15d

⇒ 45 = 15d

⇒ d = 3

Therefore, the number of terms is 16 and the common difference is 3.

Starting with the left-hand side:

\frac{(\cot \mathrm{A}+\tan \mathrm{A}-1)(\sin \mathrm{A}+\cos \mathrm{A})}{\sin^3 \mathrm{A}+\cos^3 \mathrm{A}}

=\frac{\left(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}+\frac{\sin \mathrm{A}}{\cos \mathrm{A}}-1\right)(\sin \mathrm{A}+\cos \mathrm{A})}{(\sin \mathrm{A}+\cos \mathrm{A})(\sin^2 \mathrm{A}-\sin \mathrm{A}\cos \mathrm{A}+\cos^2 \mathrm{A})}

Using sin² A + cos² A = 1:

=\frac{\left(\frac{\cos^2 \mathrm{A}+\sin^2 \mathrm{A}}{\sin \mathrm{A}\cos \mathrm{A}}-1\right)(\sin \mathrm{A}+\cos \mathrm{A})}{(\sin \mathrm{A}+\cos \mathrm{A})(1-\sin \mathrm{A}\cos \mathrm{A})}

=\frac{\left(\frac{1-\sin \mathrm{A}\cos \mathrm{A}}{\sin \mathrm{A}\cos \mathrm{A}}\right)(\sin \mathrm{A}+\cos \mathrm{A})}{(\sin \mathrm{A}+\cos \mathrm{A})(1-\sin \mathrm{A}\cos \mathrm{A})}

Cancelling common factors:

=\frac{1}{\sin \mathrm{A}\cos \mathrm{A}}

=\frac{1}{\cos \mathrm{A}}\times\frac{1}{\sin \mathrm{A}}

= sec A × cosec A

Therefore, LHS = RHS. Hence proved.

Given: \frac{6\mathrm{x}^2+3\mathrm{x}-5}{3\mathrm{x}-5}=\frac{9\mathrm{x}^2+2\mathrm{x}+5}{2\mathrm{x}+5}

Applying componendo and dividendo:

\frac{(6\mathrm{x}^2+3\mathrm{x}-5)+(3\mathrm{x}-5)}{(6\mathrm{x}^2+3\mathrm{x}-5)-(3\mathrm{x}-5)}=\frac{(9\mathrm{x}^2+2\mathrm{x}+5)+(2\mathrm{x}+5)}{(9\mathrm{x}^2+2\mathrm{x}+5)-(2\mathrm{x}+5)}

\frac{6\mathrm{x}^2+6\mathrm{x}-10}{6\mathrm{x}^2}=\frac{9\mathrm{x}^2+4\mathrm{x}+10}{9\mathrm{x}^2}

Since x ≠ 0:

\frac{6\mathrm{x}^2+6\mathrm{x}-10}{6}=\frac{9\mathrm{x}^2+4\mathrm{x}+10}{9}

⇒ 9(6x² + 6x − 10) = 6(9x² + 4x + 10)

⇒ 54x² + 54x − 90 = 54x² + 24x + 60

⇒ 30x = 150

⇒ x = 5

(a) Since (x − 2) is a factor, substituting x = 2 gives remainder zero.

2(2)³ − 7(2)² + 2k − 2 = 0

⇒ 16 − 28 + 2k − 2 = 0

⇒ −14 + 2k = 0

⇒ 2k = 14

⇒ k = 7

(b) The polynomial becomes:

2x³ − 7x² + 7x − 2

Dividing by (x − 2) gives 2x² − 3x + 1.

Therefore:

2x³ − 7x² + 7x − 2 = (x − 2)(2x² − 3x + 1)

= (x − 2)(2x² − 2x − x + 1)

= (x − 2)[2x(x − 1) − 1(x − 1)]

= (x − 2)(2x − 1)(x − 1)

Radius, r = 3.5 cm

Height of cylinder, h = 14 cm

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

Total surface area of one hemisphere = 3πr²

Sum of the total surface areas of all three parts:

= 2πr(h + r) + 2(3πr²)

= 2πrh + 2πr² + 6πr²

= 2πrh + 8πr²

= 2πr(h + 4r)

Substituting the values:

= 2 × \frac{22}{7} × 3.5 × (14 + 4 × 3.5)

= 22 × (14 + 14)

= 22 × 28

= 616 cm²

Therefore, the required surface area is 616 cm².

1. Plot A(1, 3), B(1, 2) and C(3, 0).

2. Reflection in the x-axis changes (x, y) to (x, −y).

Therefore, E = (1, −3) and D = (1, −2).

3. Reflection through the origin changes (x, y) to (−x, −y).

Therefore, F = (−1, −3) and G = (−1, −2).

4. Reflection in the y-axis changes (x, y) to (−x, y).

Therefore, J = (−1, 3), I = (−1, 2) and H = (−3, 0).

5. Join A, B, C, D, E, F, G, H, I and J in order, and finally join J to A.

The closed figure has ten sides. Therefore, it is a decagon.

Let the height of building CD be H m and the height of tower AB be h m.

BD = 100 m and E is the midpoint of CD.

Therefore, ED = \frac{\mathrm{H}}{2}.

Through A, draw AP parallel to BD. Then AP = 100 m and PD = AB = h.

In △ACP:

tan 35° = CP/AP

⇒ 0.70 = (H − h)/100

⇒ H − h = 70 … (1)

In △AEP:

tan 14° = EP/AP

⇒ 0.25 = (\frac{\mathrm{H}}{2} − h)/100

⇒ \frac{\mathrm{H}}{2} − h = 25 … (2)

Subtracting (2) from (1):

H − h − (\frac{\mathrm{H}}{2} − h) = 70 − 25

⇒ \frac{\mathrm{H}}{2} = 45

⇒ H = 90 m

Substituting in (1):

90 − h = 70

⇒ h = 20 m

Therefore, the tower is 20 m high and the building is 90 m high.

From graph,

The Mode = L = 64 marks (approx.)

Hence, mode = 64.

(a) From the graph:

P = (5, 0) and Q = (0, −3).

(b) Slope of PQ = \frac{-3-0}{0-5}=\frac{-3}{-5}=\frac{3}{5}

Since RS is perpendicular to PQ:

Slope of RS × \frac{3}{5} = −1

⇒ Slope of RS = -\frac{5}{3}

RS passes through the origin (0, 0).

Using y − y₁ = m(x − x₁):

y − 0 = -\frac{5}{3}(x − 0)

⇒ y = -\frac{5\mathrm{x}}{3}

⇒ 3y = −5x

⇒ 5x + 3y = 0

Therefore, the equation of RS is 5x + 3y = 0.

For Article A:

Total marked price = ₹190 × 6 = ₹1,140

GST = 12% of ₹1,140

⇒ GST = \frac{12}{100} × ₹1,140

⇒ GST = ₹136.80

Total for A = ₹1,140 + ₹136.80 = ₹1,276.80

For Article B:

Total marked price = ₹50 × 12 = ₹600

Since 12 units are purchased, discount = 5% of ₹600.

⇒ Discount = ₹30

Taxable value = ₹600 − ₹30 = ₹570

GST = 18% of ₹570

⇒ GST = \frac{18}{100} × ₹570

⇒ GST = ₹102.60

Total for B = ₹570 + ₹102.60 = ₹672.60

Total bill = ₹1,276.80 + ₹672.60

⇒ Total bill = ₹1,949.40

Rounded to the nearest ₹10, the bill is ₹1,950.

The customer paid ₹2,000.

Extra amount paid = ₹2,000 − ₹1,950 = ₹50

Therefore, the payment was not correct; the customer paid ₹50 extra.

Face value of one share = ₹200

Premium = 25% of ₹200

⇒ Premium = \frac{25}{100} × ₹200

⇒ Premium = ₹50

(a) Market value = Face value + Premium

⇒ Market value = ₹200 + ₹50

⇒ Market value = ₹250

(b) Return on one share = 5% of ₹250

⇒ Return = \frac{5}{100} × ₹250

⇒ Return = ₹12.50

Let the dividend rate be r%.

₹12.50 = r% of ₹200

⇒ 12.50 = \frac{\mathrm{r}}{100} × 200

⇒ r = 6.25%

(c) Dividend on one share = ₹12.50

Number of shares = \frac{₹1,000}{₹12.50}

⇒ Number of shares = 80

8(ii)

Prepare the calculation table:

Σf = 24 and Σfx = 319.

(a) Mean = \frac{\Sigma \mathrm{fx}}{\Sigma \mathrm{f}}

⇒ Mean = \frac{319}{24}

⇒ Mean = 13.291…

To the nearest whole number, mean = 13.

(b) N = 24, so the middle observations are the 12th and 13th.

From the cumulative frequencies, both the 12th and 13th observations have x = 13.

Median = \frac{13+13}{2}

⇒ Median = 13

Let the original speed be x km/h.

Distance = 350 km

Original time = \frac{350}{\mathrm{x}} hours

Reduced speed = (x − 20) km/h

Time at reduced speed = \frac{350}{\mathrm{x}-20} hours

The journey time increased by 2 hours:

⇒ \frac{350}{\mathrm{x}-20}-\frac{350}{\mathrm{x}}=2

⇒ \frac{350x-350(\mathrm{x}-20)}{\mathrm{x}(\mathrm{x}-20)}=2

⇒ \frac{7000}{\mathrm{x}(\mathrm{x}-20)}=2

⇒ x(x − 20) = 3,500

⇒ x² − 20x − 3,500 = 0

⇒ x² + 50x − 70x − 3,500 = 0

⇒ x(x + 50) − 70(x + 50) = 0

⇒ (x + 50)(x − 70) = 0

⇒ x = −50 or x = 70

Speed cannot be negative, so the original speed was 70 km/h.

Reduced speed = 70 − 20 = 50 km/h

Time at reduced speed = \frac{350}{50}

⇒ Time = 7 hours

External radius of the hollow sphere, R = 5 cm

Internal radius of the hollow sphere, r = 3 cm

Height of the cone, h = 8 cm

Volume of metal in the hollow sphere = \frac{4}{3}π(R³ − r³)

= \frac{4}{3}π(5³ − 3³)

= \frac{4}{3}π(125 − 27)

= \frac{392}{3}π cm³

Let the radius of the cone be m cm.

Volume of the cone = \frac{1}{3}πm²h

= \frac{1}{3}πm² × 8

= \frac{8}{3}πm²

Since the sphere is melted to form the cone, their volumes are equal.

\frac{392}{3}π = \frac{8}{3}πm²

⇒ 392 = 8m²

⇒ m² = 49

⇒ m = 7 cm

Therefore, the radius of the cone is 7 cm.

Total number of possible outcomes = 30

(a) Perfect squares from 1 to 30 are 1, 4, 9, 16 and 25.

Number of favourable outcomes = 5

Probability of winning a wall clock = \frac{5}{30} = \frac{1}{6}

(b) Even numbers that are also multiples of 3 are 6, 12, 18, 24 and 30.

Number of favourable outcomes = 5

Probability of winning a water bottle = \frac{5}{30} = \frac{1}{6}

(c) Prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 10

Probability of winning a purse = \frac{10}{30} = \frac{1}{3}

(a) OX ⟂ BX because the radius is perpendicular to the tangent at the point of contact.

Therefore, ∠OXB = 90°.

In △BOX:

∠BOX + ∠OBX + ∠OXB = 180°

⇒ ∠BOX + 32° + 90° = 180°

⇒ ∠BOX = 58°

(b) From the figure, ∠COX = ∠BOX = 58°.

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

⇒ ∠CYX = 58° ÷ 2 = 29°

(c) By the alternate segment theorem:

∠ZYX = ∠AXZ = 66°

(d) Since ZY = XY, △ZXY is isosceles.

Therefore, ∠ZXY = ∠XZY.

In △ZXY:

∠ZYX + ∠ZXY + ∠XZY = 180°

⇒ 66° + 2∠XZY = 180°

⇒ 2∠XZY = 114°

⇒ ∠XZY = 57°

By the alternate segment theorem, ∠YXB = ∠XZY = 57°.

⇒ ∠OXY = ∠OXB − ∠YXB

⇒ ∠OXY = 90° − 57° = 33°

Therefore, ∠BOX = 58°, ∠CYX = 29°, ∠ZYX = 66° and ∠OXY = 33°.

First term, a = 7

Common ratio, r = 21 ÷ 7 = 3

(a) The nth term of a G.P. is Tₙ = arⁿ⁻¹.

⇒ 1701 = 7 × 3ⁿ⁻¹

⇒ 3ⁿ⁻¹ = 1701 ÷ 7

⇒ 3ⁿ⁻¹ = 243

⇒ 3ⁿ⁻¹ = 3⁵

⇒ n − 1 = 5

⇒ n = 6

(b) Sum of n terms of a G.P. = \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}(\mathrm{r}^{\mathrm{n}}-1)}{\mathrm{r}-1}

\mathrm{S}_6=\frac{7(3^6-1)}{3-1}

= \frac{7(729-1)}{2}

= \frac{7\times728}{2}

= 7 × 364

= 2548

Therefore, n = 6 and the sum of the first six terms is 2548.

(a) By the alternate segment theorem:

∠PTR = ∠PST

Also, ∠RPT = ∠TPS, as these are the same angle at P.

Therefore, △PTR ∼ △PST by the AA criterion.

(b) Corresponding sides of similar triangles are proportional.

\frac{\mathrm{PT}}{\mathrm{PS}}=\frac{\mathrm{PR}}{\mathrm{PT}}

⇒ PT² = PR × PS

Hence proved.

(c) PR = 4 cm and PS = 16 cm

PT² = PR × PS

⇒ PT² = 4 × 16

⇒ PT² = 64

⇒ PT = 8 cm

Therefore, the length of the tangent PT is 8 cm.

From the graph, the cumulative frequencies are:

(a) The cumulative frequency at ₹12,000 is 75.

Therefore, the total number of workers is 75.

(b) Median position = \frac{\mathrm{N}+1}{2}

= \frac{75+1}{2} = 38th observation

The 38th observation lies in the class ₹6,000–₹8,000.

Therefore, the median class is ₹6,000–₹8,000.

(c) Lower-quartile position = \frac{\mathrm{N}+1}{4}

= \frac{75+1}{4} = 19th observation

The 19th observation lies in the class ₹2,000–₹4,000.

Therefore, the lower-quartile class is ₹2,000–₹4,000.

(d) Number of workers with salary less than ₹10,000 = 70

Number of workers with salary less than ₹6,000 = 35

Required number of workers = 70 − 35 = 35