ICSE Class 10 Mathematics Solved Paper 2023

(a) 1, −2

Explanation:

\begin{bmatrix}2&0\\0&4\end{bmatrix}\begin{bmatrix}\mathrm{x}\\\mathrm{y}\end{bmatrix}=\begin{bmatrix}2\\-8\end{bmatrix}

⇒ \begin{bmatrix}2\mathrm{x}\\4\mathrm{y}\end{bmatrix}=\begin{bmatrix}2\\-8\end{bmatrix}

⇒ 2x = 2 and 4y = −8.

∴ x = 1 and y = −2.

(c) −2

Explanation:

x – 2 is a factor of x³ – kx – 12

⇒ x – 2 = 0

⇒ x = 2

By the factor theorem, f(2) = 0.

⇒ 2³ − 2k − 12 = 0

⇒ 8 − 2k − 12 = 0

∴ −2k = 4, therefore k = −2.

(d) 90°

Explanation:

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

Therefore, ∠PQS = ∠PST = 30°.

In triangle PQS:

∠PQS + ∠QPS + ∠PSQ = 180°

30° + 60° + ∠PSQ = 180°

Therefore, ∠PSQ = 90°.

(b) \frac{5}{26}

Explanation:

There are 5 vowels—A, E, I, O and U—among 26 English letters.

Probability = \frac{\mathrm{Number\ of\ vowels}}{\mathrm{Total\ number\ of\ letters}}=\frac{5}{26}.

(a) 4

Explanation:

3 is a root of the quadratic equation x² – px + 3 = 0

∴ 3² − 3p + 3 = 0

⇒ 9 − 3p + 3 = 0

⇒ 12 − 3p = 0

⇒ 3p = 12

Therefore, p = 4.

(c) 3 : 2

Explanation:

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB // CD.

From figure,

PA = PC + AC = 6 + 4 = 10 cm.

In △ PAB and △ PCD,

⇒ ∠PAB = ∠PCD (Both equal to 70°)

⇒ ∠BPA = ∠DPC (Both are equal)

∴ △ PAB ~ △ PCD (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional to each other.

\frac{\mathrm{PB}}{\mathrm{PD}}=\frac{\mathrm{PA}}{\mathrm{PC}}=\frac{10}{6}=\frac{5}{3}

Let PB = 5x and PD = 3x. Then DB = PB − PD = 2x.

Therefore, PD : DB = 3 : 2.

(b) ₹308

Explanation:

GST charged = GST rate × marked price

= \frac{10}{100}\times3080

= ₹308.

(d) cos² A

Explanation:

(1 + sin A)(1 − sin A)

= 1 − sin² A

= cos² A.

(a) (2, 2)

Explanation:

Centroid = \left(\frac{\mathrm{x_1+x_2+x_3}}{3},\frac{\mathrm{y_1+y_2+y_3}}{3}\right)

= \left(\frac{-4+6+4}{3},\frac{-2+2+6}{3}\right)

= (2, 2).

(c) 25

Explanation:

Given, a_n = 2n + 5.

a₁₀ = 2(10) + 5 = 20 + 5 = 25.

(b) 6

Explanation:

Let the mean proportional be x.

Then, 4 : x = x : 9

⇒ \frac{4}{x}=\frac{x}{9}

⇒ x² = 4 × 9

⇒ x² = 36

⇒ x = √36 = 6

Therefore, x = 6.

(d) Mean

Explanation:

Median and quartiles can be determined from an ogive, while mode can be estimated from a histogram.

Mean requires numerical calculation using the class values and frequencies; it cannot be determined graphically for a grouped frequency distribution.

(c) 4 cm

Explanation:

Let the radius of the cylinder be r cm.

Given, h = 3 cm and volume = 48π cm³.

⇒ Volume of cylinder = πr²h

⇒ 3πr² = 48π

⇒ 3r² = 48

⇒ r² = \frac{48}{3} =16

⇒ r = √16 = 4

Therefore, the radius is 4 cm.

(a) ₹84

Explanation:

Sum deposited = ₹800 × 6

Sum deposited = ₹4800

Interest earned = Maturity value − Sum deposited

Interest earned = ₹4884 − ₹4800

Therefore, interest earned = ₹84.

(b) {0, 1, 2, 3, 4, 5}

Explanation:

Given, 2x + 4 ≤ 14

2x ≤ 14 − 4

2x ≤ 10

x\leq\frac{10}{2}

x ≤ 5

Since x ∈ W, x can be 0, 1, 2, 3, 4 or 5.

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

Let f(x) = 3x³ + x² − ax − 81.

By the Factor Theorem, if x − a is a factor of f(x), then f(a) = 0.

Since x − a = 0, x = a.

Substituting x = a:

⇒ 3a³ + a² − a(a) − 81 = 0

⇒ 3a³ + a² − a² − 81 = 0

⇒ 3a³ − 81 = 0

⇒ 3a³ = 81

⇒ a³ = \frac{81}{3}

⇒ a³ = 27

⇒ a³ = 3³

Therefore, a = 3.

(a) Monthly deposit = ₹1000

Time = 2 years = 24 months

Total deposit = ₹1000 × 24

Total deposit = ₹24,000

Total interest = Maturity value − Total deposit

Total interest = ₹26,000 − ₹24,000

Therefore, total interest = ₹2000.

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

Interest = \frac{P\times n(n+1)}{2\times12}\times\frac{r}{100}

Substituting P = 1000, n = 24 and interest = 2000:

⇒ 2000 = \frac{1000\times24\times25}{2\times12}\times\frac{r}{100}

⇒ 2000 = 1000\times25\times\frac{r}{100}

⇒ 2000 = 250r

⇒ r = \frac{2000}{250} = 8

Therefore, the rate of interest is 8% per annum.

(a) Since BD is a diameter, the angle in a semicircle is a right angle.

Therefore, ∠BDA = 90°.

(b) In △BAD:

∠BDA + ∠BAD + ∠ABD = 180°

90° + ∠BAD + 26° = 180°

∠BAD + 116° = 180°

∠BAD = 180° − 116°

Therefore, ∠BAD = 64°.

(c) CE is tangent at A, so CE is perpendicular to radius OA.

∠CAD + ∠BAD = 90°

∠CAD + 64° = 90°

∠CAD = 90° − 64°

Therefore, ∠CAD = 26°.

(d) Join OD.

OD = OB, since both are radii.

Therefore, ∠ODB = ∠OBD.

Since O lies on AB, ∠OBD = ∠ABD = 26°.

Therefore, ∠ODB = 26°.

Comparing x² + 4x − 8 = 0 with ax² + bx + c = 0:

a = 1, b = 4 and c = −8.

Using the quadratic formula:

⇒ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

⇒ x = \frac{-4\pm\sqrt{4^2-4(1)(-8)}}{2(1)}

⇒ x = \frac{-4\pm\sqrt{16+32}}{2}

⇒ x = \frac{-4\pm\sqrt{48}}{2}

⇒ x = \frac{-4\pm4\sqrt{3}}{2}

⇒ x = -2\pm2\sqrt{3}

Using √3 = 1.732:

x = −2 ± 2(1.732)

x = −2 ± 3.464

x = −2 + 3.464 or x = −2 − 3.464

x = 1.464 or x = −5.464

Correct to one decimal place, x = 1.5 or x = −5.5.

Solving L.H.S. of the given equation :

⇒ (sin² θ – 1)(tan² θ + 1) + 1

⇒ (1 – cos² θ – 1).sec² θ + 1

⇒ -cos² θ. sec² θ + 1

\mathrm{LHS}=-\cos^2\theta\times\frac{1}{\cos^2\theta}+1

LHS = −1 + 1

LHS = 0

Since LHS = RHS, (sin² θ − 1)(tan² θ + 1) + 1 = 0 is proved.

(a) Plot A(0, 4), B(1, 1) and C(4, 0) using the given scale.

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

Therefore, E, the reflection of A(0, 4), is E(0, −4).

D, the reflection of B(1, 1), is D(1, −1).

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

Therefore, F, the reflection of B(1, 1), is F(−1, −1).

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

Therefore, H, the reflection of B(1, 1), is H(−1, 1).

G, the reflection of C(4, 0), is G(−4, 0).

(e) On joining A, B, C, D, E, F, G, H and A in order, the closed figure formed is a star.

[Completed graph space]

First, add B and C:

B+C=\begin{bmatrix}1&2\\2&4\end{bmatrix}+\begin{bmatrix}4&1\\1&5\end{bmatrix}

B+C=\begin{bmatrix}5&3\\3&9\end{bmatrix}

Now,

A(B+C)-14I=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}5&3\\3&9\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}

= \begin{bmatrix}1(5)+3(3)&1(3)+3(9)\\2(5)+4(3)&2(3)+4(9)\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}

= \begin{bmatrix}5+9&3+27\\10+12&6+36\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}

= \begin{bmatrix}14&30\\22&42\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}

= \begin{bmatrix}0&30\\22&28\end{bmatrix}

Therefore, A(B + C) − 14I = \begin{bmatrix}0&30\\22&28\end{bmatrix}.

(a) D is the midpoint of B(0, 4) and C(−6, 4).

Using the midpoint formula:

D = \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

D = \left(\frac{0+(-6)}{2},\frac{4+4}{2}\right)

D = \left(\frac{-6}{2},\frac{8}{2}\right)

Therefore, D = (−3, 4).

(b) The median AD passes through A(1, −1) and D(−3, 4).

⇒ y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

⇒ y-(-1)=\frac{4-(-1)}{-3-1}(x-1)

⇒ y+1=\frac{5}{-4}(x-1)

⇒ −4(y + 1) = 5(x − 1)

⇒ −4y − 4 = 5x − 5

⇒ 5x + 4y = 1

Therefore, the equation of median AD is 5x + 4y = 1.

(a) PAB is a secant and PT is a tangent from P.

By the tangent-secant theorem:

PT² = PA × PB

PA = PB + BA = 16 + 9 = 25 cm

PT² = 25 × 16

PT² = 400

PT = √400

Therefore, PT = 20 cm.

(b) By the alternate segment theorem, the angle between tangent PT and chord TB equals the angle subtended by chord TB in the alternate segment.

Therefore, ∠BAT = ∠PTB = 50°.

(c) The angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference.

∠BOT = 2 × ∠BAT

∠BOT = 2 × 50°

Therefore, ∠BOT = 100°.

(d) OB = OT, since both are radii.

Let ∠OBT = ∠OTB = x.

In △BOT:

x + x + 100° = 180°

2x = 80°

x = 40°

Thus, ∠OBT = 40°.

∠ABT = ∠ABO + ∠OBT

∠ABT = 45° + 40°

Therefore, ∠ABT = 85°.

(a) Taxable value of hair oil = ₹1200 − ₹100

Taxable value of hair oil = ₹1100

GST on hair oil = 1100\times\frac{18}{100}

GST on hair oil = ₹198

Taxable value of cashew nuts = ₹600

GST on cashew nuts = 600\times\frac{12}{100}

GST on cashew nuts = ₹72

Total GST = ₹198 + ₹72

Therefore, total GST paid = ₹270.

(b) Total taxable value = ₹1100 + ₹600

Total taxable value = ₹1700

Total bill amount including GST = ₹1700 + ₹270

Therefore, the total bill amount including GST is ₹1970.

Given:

−5(x − 9) ≥ 17 − 9x > x + 2

First solve −5(x − 9) ≥ 17 − 9x:

⇒ −5x + 45 ≥ 17 − 9x

⇒ −5x + 9x ≥ 17 − 45

⇒ 4x ≥ −28

⇒ x\geq\frac{-28}{4}

⇒  x ≥ −7 … (1)

Now solve 17 − 9x > x + 2:

⇒ 17 − 2 > x + 9x

⇒ 15 > 10x

⇒ 10x < 15

⇒ x < \frac{15}{10}

⇒ x < \frac{3}{2}

x < 1.5 … (2)

Combining (1) and (2):

−7 ≤ x < 1.5

Therefore, the solution set is {x : −7 ≤ x < 1.5}.

On the number line, place a closed point at −7, an open point at 1.5, and shade the interval between them.

(a) Since DE ∥ BF:

⇒ ∠DEG = ∠GBF (alternate interior angles)

⇒ ∠EDG = ∠GFB (alternate interior angles)

Therefore, △GED ∼ △GBF by the AA criterion.

(b) Corresponding sides of similar triangles are proportional.

⇒  \frac{DE}{BF}=\frac{EG}{GB}

⇒ \frac{DE}{30}=\frac{8}{16}

⇒ DE = \frac{30\times8}{16}

DE = 15 cm.

(c) In △BAC and △BDE:

∠BAC = ∠BDE (corresponding angles, since AC ∥ DE)

⇒ ∠ABC = ∠DBE (common angle)

Therefore, △BAC ∼ △BDE by the AA criterion.

Corresponding sides are proportional:

⇒ \frac{BD}{AB}=\frac{DE}{AC}

⇒ \frac{BD}{AB}=\frac{15}{24}

⇒ \frac{BD}{AB}=\frac{5}{8}

Therefore, DB : AB = 5 : 8.

Hence, required mode = ₹ 530.

Let the first term be a and the common difference be d.

The nth term of an A.P. is:

a_n = a + (n − 1)d

Given, the 5th term is 4:

⇒ a + (5 − 1)d = 4

⇒ a + 4d = 4 … (1)

Given, the 9th term is −12:

⇒ a + (9 − 1)d = −12

⇒ a + 8d = −12 … (2)

Subtracting (1) from (2):

⇒ (a + 8d) − (a + 4d) = −12 − 4

⇒ a + 8d − a − 4d = −16

⇒ 4d = −16

⇒ d = −4

Substituting d = −4 in (1):

⇒ a + 4(−4) = 4

⇒ a − 16 = 4

⇒ a = 20

(a) Therefore, the first term is 20.

(b) The common difference is −4.

(c) Sum of n terms: S_n = \frac{n}{2}[2a+(n-1)d]

⇒ S₁₆ = \frac{16}{2}[2(20)+(16-1)(-4)]

⇒ S₁₆ = 8[40 − 60]

⇒ S₁₆ = 8(−20)

Therefore, S₁₆ = −160.

(a) From the figure:

A = (4, 0) and B = (0, 4).

(b) P divides AB internally in the ratio AP : PB = 3 : 1.

Using the section formula:

P=\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)

P=\left(\frac{3(0)+1(4)}{3+1},\frac{3(4)+1(0)}{3+1}\right)

P=\left(\frac{4}{4},\frac{12}{4}\right)

Therefore, P = (1, 3).

(c) Slope of AB:

m_{AB}=\frac{4-0}{0-4}

m_{AB}=\frac{4}{-4}=-1

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

(−1) × m = −1

Therefore, m = 1.

The required line passes through P(1, 3).

Using y − y₁ = m(x − x₁):

y − 3 = 1(x − 1)

y − 3 = x − 1

Therefore, y = x + 2.

Total number of possible outcomes = 25.

(a) Multiples of 5 from 1 to 25 are 5, 10, 15, 20 and 25.

Number of favourable outcomes = 5.

⇒ P(multiple of 5) = \frac{5}{25}

⇒ P(multiple of 5) = \frac{1}{5}

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

Number of favourable outcomes = 5.

⇒ P(perfect square) = \frac{5}{25}

⇒ P(perfect square) = \frac{1}{5}

(c) Prime numbers from 1 to 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Number of favourable outcomes = 9.

⇒ P(prime number) = \frac{9}{25}

In the first case:

Distance = 100 km

Speed = x km/hr

⇒ Time taken = \frac{100}{x} hours

In the second case:

Speed = (x + 5) km/hr

⇒ Time taken = \frac{100}{x+5} hours

The faster journey takes 1 hour less.

⇒ \frac{100}{x}-\frac{100}{x+5}=1

⇒ \frac{100(x+5)-100x}{x(x+5)}=1

⇒ \frac{500}{x²+5x}=1

⇒ x² + 5x = 500

⇒ x² + 5x − 500 = 0

⇒ x² + 25x − 20x − 500 = 0

⇒ x(x + 25) − 20(x + 25) = 0

⇒ (x − 20)(x + 25) = 0

⇒ x = 20 or x = −25

Speed cannot be negative.

Therefore, the original speed is 20 km/hr.

Let the rise in water level be x cm.

Volume of displaced water = Volume of hemisphere + Volume of cone.

⇒ π(14)²x = \frac{2}{3} × π × 7³ + \frac{1}{3} × π × 7² × 4

⇒ 196πx = \frac{2}{3} × π × 343 + \frac{1}{3} × π × 196

Cancelling π:

⇒ 196x = \frac{686+196}{3}

⇒ 196x = \frac{882}{3}

⇒ x = \frac{882}{3×196}

⇒ x = \frac{882}{588}

⇒ x = 1.5

Therefore, the rise in the water level is 1.5 cm.

Take the assumed mean A = 25.

⇒ Σf = 40 and Σfd = 90

⇒ Mean = A + \frac{Σfd}{Σf}

⇒ Mean = 25 + \frac{90}{40}

⇒ Mean = 25 + 2.25

Therefore, the mean is 27.25.

Let x be added to each number.

The resulting numbers are 4 + x, 6 + x, 8 + x and 11 + x.

Since they are in proportion:

⇒ \frac{4+x}{6+x}=\frac{8+x}{11+x}

⇒ (4 + x)(11 + x) = (8 + x)(6 + x)

⇒ 44 + 4x + 11x + x² = 48 + 8x + 6x + x²

⇒ 44 + 15x + x² = 48 + 14x + x²

⇒ 15x − 14x = 48 − 44

⇒ x = 4

Therefore, the required number is 4.

Steps of construction:

1. Draw AB = 6 cm.

2. At A, construct ray AP such that ∠BAP = 120°.

3. With A as centre and radius 5 cm, cut ray AP at C. Thus, AC = 5 cm.

4. Join B and C to obtain △ABC.

5. Draw the perpendicular bisectors of AB and AC. Let them intersect at O.

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

On measurement, the radius of the circumcircle is approximately 5 cm.

Given:

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

Applying componendo and dividendo:

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

⇒ \frac{2\sqrt{2x+2}}{2\sqrt{2x-1}}=\frac{4}{2}

⇒ \frac{\sqrt{2x+2}}{\sqrt{2x-1}}=2

⇒ √(2x + 2) = 2√(2x − 1)

Squaring both sides:

⇒ 2x + 2 = 4(2x − 1)

⇒ 2x + 2 = 8x − 4

⇒ 8x − 2x = 2 + 4

⇒ 6x = 6

⇒ x = 1

Therefore, x = 1.

Given A.P.: 15, 30, 45, 60, …

First term, a = 15

Common difference, d = 30 − 15 = 15

Let the nth term be 300.

⇒ aₙ = a + (n − 1)d

⇒ 300 = 15 + (n − 1)15

⇒ 300 = 15 + 15n − 15

⇒ 300 = 15n

⇒ n = \frac{300}{15}

⇒ n = 20

Therefore, 300 is the 20th term.

Now,

⇒ Sₙ = \frac{n}{2}(a + aₙ)

⇒ S₂₀ = \frac{20}{2}(15 + 300)

⇒ S₂₀ = 10 × 315

⇒ S₂₀ = 3150

Therefore, the sum of all terms up to 300 is 3150.

Let CD be the tower, with C as its foot and D as its top.

CD = 100 m.

The angles of elevation from A and B equal the corresponding angles of depression.

Therefore, ∠A = 45° and ∠B = 38°.

In △ACD:

⇒ tan 45° = \frac{CD}{AC}

⇒ 1 = \frac{100}{AC}

⇒ AC = 100 m

In △BCD:

⇒ tan 38° = \frac{CD}{BC}

⇒ 0.7813 = \frac{100}{BC}

⇒ BC = \frac{100}{0.7813}

⇒ BC ≈ 127.99 m

Since the ships are on opposite sides:

⇒ AB = AC + BC

⇒ AB = 100 + 127.99

⇒ AB = 227.99 m

Therefore, the distance between the ships, to the nearest metre, is 228 m.

Let f(x) = 2x³ − x² − 13x − 6.

Substituting x = −2:

⇒ f(−2) = 2(−2)³ − (−2)² − 13(−2) − 6

⇒ f(−2) = 2(−8) − 4 + 26 − 6

⇒ f(−2) = −16 − 4 + 26 − 6

⇒ f(−2) = 0

Therefore, x + 2 is a factor.

Dividing, 2x³ – x² – 13x – 6 by x + 2, we get :

Hence:

⇒ 2x³ − x² − 13x − 6 = (x + 2)(2x² − 5x − 3)

Splitting the middle term:

⇒ 2x² − 5x − 3 = 2x² − 6x + x − 3

⇒ 2x(x − 3) + 1(x − 3)

⇒ (2x + 1)(x − 3)

Therefore, 2x³ − x² − 13x − 6 = (x + 2)(2x + 1)(x − 3).

First calculate the less-than cumulative frequencies:

Plot (28, 0), (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58) and (44, 60), then join them with a smooth curve.

(a) Total students, n = 60.

⇒ Median position = \frac{60}{2} = 30th observation.

From the ogive, the median ≈ 36.2 kg.

(b) Upper-quartile position = \frac{3×60}{4} = 45th observation.

From the ogive, Q₃ ≈ 38.2 kg.

(c) From the ogive, approximately 38 students weigh 37 kg or less.

⇒ Number above 37 kg = 60 − 38

⇒ Number above 37 kg = 22

Therefore: median ≈ 36.2 kg, upper quartile ≈ 38.2 kg, and 22 students weigh above 37 kg.