ICSE Class 10 Mathematics 2023 Question Paper with Solutions

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Last Updated on November 13, 2026 by sanjjeett

ICSE Class 10 Mathematics PYQ

Practice the complete ICSE Class 10 Mathematics 2023 Question Paper with Solutions, including MCQs, algebra, geometry, trigonometry, statistics, graphs and constructions. Use this solved paper to revise important concepts, practise board-style questions and strengthen your exam preparation.

Year
2023
Maximum Marks
80
Time Allowed
Two hours and a half
📄
Paper Instructions
ICSE Class 10 Mathematics 2023

Attempt all questions from Section A and any four questions from Section B.

All working, including rough work, must be clearly shown, and must be done on the same sheet as the rest of the answer.

Omission of essential working will result in loss of marks.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables and graph papers are provided.

✓
Section A

ICSE Mathematics 2023 — Section A

40 marks • Attempt all questions from this Section.
40
Marks
1
Question 1
Choose the correct answer from the given options.
i

If \(\left[\begin{array}{ll}2 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ -8\end{array}\right]\), the value of \(\boldsymbol{x}\) and \(\boldsymbol{y}\) respectively are: [1]

A1, -2
B-2, 1
C1, 2
D-2, -1
Answer & Solution

Option (a) is correct.

Explanation: \(\left[\begin{array}{ll}2 & 0 \\ 0 & 4\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ -8\end{array}\right]\)

\[ \begin{aligned} {\left[\begin{array}{l} 2 x \\ 4 y \end{array}\right] } & =\left[\begin{array}{c} 2 \\ -8 \end{array}\right] \\ 2 x & =2, x=1 \\ 4 y & =-8, y=-2 \end{aligned} \]
ii

If \(\boldsymbol{x}-\mathbf{2}\) is a factor of \(\boldsymbol{x}^{\mathbf{3}}-\boldsymbol{k} \boldsymbol{x}-\mathbf{1 2}\), then the value of \(\boldsymbol{k}\) is: [1]

A3
B2
C-2
D-3
Answer & Solution

Option (c) is correct.

Explanation: \((x-2)\) is a factor of \(f(x)\) ∴

\[ \begin{aligned} \text { Remainder } & =f(2)=0 \\ f(x) & =x^{3}-k x-12 \\ f(2) & =(2)^{3}-k(2)-12 \\ 0 & =8-2 k-12 \\ k & =-2 \end{aligned} \]
iii

In the given diagram RT is a tangent touching the circle at \(\mathbf{S}\). If \(\angle \mathbf{P S T}=\mathbf{3 0}^{\boldsymbol{\circ}}\) and \(\angle \mathbf{S P Q}=\mathbf{6 0}^{\boldsymbol{\circ}}\) then \(\angle \mathrm{PSQ}\) is equal to: [1]

ICSE Class 10 Mathematics 2023 Question 1(iii) figure
A40°
B30°
C60°
D90°
Answer & Solution

Option (d) is correct.

Explanation:

ICSE Class 10 Mathematics 2023 Question 1(iii) solution figure
\[ \angle \mathrm{PST}=\angle \mathrm{PQS}=30^{\circ} \]

(alternate segment angle) In \(\triangle \mathrm{PSQ}\),

\[ \angle \mathrm{PSQ}+\angle \mathrm{PQS}+\angle \mathrm{SPQ}=180^{\circ} \]

(Sum of all angles of a triangle)

\[ \begin{aligned} \angle \mathrm{PSQ}+30^{\circ}+60^{\circ} & =180^{\circ} \\ \angle \mathrm{PSQ} & =90^{\circ} \end{aligned} \]
iv

A letter is chosen at random from all the letters of the English alphabets. The probability that the letter chosen is a vowel, is: [1]

A\(\frac{4}{26}\)
B\(\frac{5}{26}\)
C\(\frac{21}{26}\)
D\(\frac{5}{24}\)
Answer & Solution

Option (b) is correct.

Explanation: Total no. of letters in English alphabets, \(n(\mathrm{~S})=26\) Total no. of vowels in English alphabets, \(n(\mathrm{E})=05\) ∴ Required probability \(\mathrm{P}(\mathrm{E})=\frac{5}{26}\)

v

If 3 is a root of the quadratic equation \(\boldsymbol{x}^{\mathbf{2}}-\boldsymbol{p} \boldsymbol{x}+\) \(\boldsymbol{3}=\mathbf{0}\) then \(\boldsymbol{p}\) is equal to: [1]

A4
B3
C5
D2
Answer & Solution

Option (a) is correct.

Explanation: \(f(x)=x^{2}-p x+3=0\) 3 is a root of \(f(x)\) ∴

\[ \begin{array}{r} f(3)=0 \\ (3)^{2}-3 p+3=0 \end{array} \]

∴

\[ p=4 \]
vi

In the given figure \(\angle \mathbf{B A P}=\angle \mathbf{D C P}=\mathbf{7 0}{ }^{\circ}, \mathbf{P C}=\) \(\mathbf{6 c m}\) and \(\mathbf{C A}=\mathbf{4 c m}\), then PD : DB is: [1]

ICSE Class 10 Mathematics 2023 Question 1(vi) figure
A5 : 3
B3 : 5
C3 : 2
D2 : 3
Answer & Solution

Option (c) is correct.

Explanation:

ICSE Class 10 Mathematics 2023 Question 1(vi) solution figure
\[ \angle A=\angle C=70^{\circ} \]

(Given) These are corresponding angles

\(\therefore D C \| B A\)

∴

\[ \frac{P D}{D B}=\frac{P C}{A C} \]

(Basic proportional Theorem)

\[ \frac{P D}{D B}=\frac{6}{4}=\frac{3}{2} \]
vii

The printed price of an article is ₹3080. If the rate of GST is 10% then the GST charged is: [1]

A₹154
B₹308
C₹30.80
D₹15.40
Answer & Solution

Option (b) is correct.

Explanation: G.S.T. \(=\%\) Rate × printed price

\[ \begin{aligned} & =\frac{10}{100} \times 3080 \\ & =\text{₹} 308 \end{aligned} \]

Therefore, the GST charged on the article is (rupees symbol) 308.

viii

\((\mathbf{1}+\boldsymbol{\operatorname { s i n }} \mathbf{A})(\mathbf{1}-\sin \mathbf{A})\) is equal to: [1]

A\(\operatorname{cosec}^{2} \mathrm{~A}\)
B\(\sin ^{2} \mathrm{~A}\)
C\(\sec ^{2} \mathrm{~A}\)
D\(\cos ^{2} \mathrm{~A}\)
Answer & Solution

Option (d) is correct.

Explanation: \((1+\sin \mathrm{A})(1-\sin \mathrm{A})\)

\[ \begin{aligned} & =(1)^{2}-(\sin \mathrm{A})^{2} \\ & =1-\sin ^{2} \mathrm{~A} \\ & =\sin ^{2} A+\cos ^{2} A-\sin ^{2} A \\ & =\cos ^{2} A \end{aligned} \]
ix

The coordinates of the vertices of \(\boldsymbol{\triangle} \mathbf{A B C}\) are respectively (-4, -2), \((6,2)\) and \((4,6)\). The centroid G of \(\triangle \mathrm{ABC}\) is: [1]

A\((2,2)\)
B\((2,3)\)
C\((3,3)\)
D(0, -1)
Answer & Solution

Option (a) is correct.

Explanation: \(G\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)

\[ \begin{aligned} & =G\left(\frac{-4+6+4}{3}, \frac{-2+2+6}{3}\right) \\ & =G(2,2) \end{aligned} \]
x

The \(\mathbf{n}^{\text {th }}\) term an Arithmetic Progression (A.P.) is \(\mathbf{2} \boldsymbol{n}+\mathbf{5}\). The \(\mathbf{1 0}^{\text {th }}\) term is: [1]

A7
B15
C25
D45
Answer & Solution

Option (c) is correct.

Explanation:

\[ \begin{aligned} t_{n} & =2 n+5 \\ t_{10} & =2 \times 10+5=25 \end{aligned} \]
xi

The mean proportional between 4 and 9 is: [1]

A4
B6
C9
D36
Answer & Solution

Option (b) is correct.

Explanation: Mean proportional \(=\sqrt{4 \times 9}=6\)

xii

Which of the following cannot be determined graphically for a grouped frequency distribution? [1]

AMedian
BMode
CQuartiles
DMean
Answer & Solution

Option (d) is correct.

Explanation: Mean is a numerical measure that requires the individual data values, which cannot be determined graphically for a grouped frequency distribution.

xiii

Volume of a cylinder of height 3 cm is \(\mathbf{4 8} \pi \mathbf{~ c m}^{\mathbf{3}}\). Radius of the cylinder is: [1]

A48 cm
B16 cm
C4 cm
D24 cm
Answer & Solution

Option (c) is correct.

Explanation: Vol. of cylinder \(=48 \pi \mathrm{~cm}^{3}\)

\[ \begin{aligned} \pi r^{2} h & =48 \pi \\ r^{2}(3) & =48 \\ r & =4 \mathrm{~cm} \end{aligned} \]
xiv

Naveen deposits ₹800 every month in a recurring deposit account for 6 months. If he receives ₹4884 at the time of maturity, then the interest he earns is: [1]

A₹84
B₹42
C₹24
D₹284
Answer & Solution

Option (a) is correct.

Explanation: Amount = Principle + Interest

\[ 4884=800 \times 6+\text { Interest } \]

∴

\[ \begin{aligned} \text { Interest } & =4884-4800 \\ & =\text{₹} 84 \end{aligned} \]
xv

The solution set for the inequation \(\mathbf{2} \boldsymbol{x}+\mathbf{4} \leq \mathbf{1 4}\), \(x \varepsilon \mathbf{W}\) is: [1]

A\{1, 2, 3, 4, 5\}
B\{0, 1, 2, 3, 4, 5\}
C\{1, 2, 3, 4\}
D\{0, 1, 2, 3, 4\}
Answer & Solution

Option (b) is correct.

Explanation:

\[ \begin{aligned} 2 x+4 & \leq 14 \\ 2 x & \leq 14-4 \\ x & \leq 5 \end{aligned} \]

∴ Solution set \(\{0,1,2,3,4,5\}\)

2
Question 2
ICSE Class 10 Mathematics 2023
i

Find the value of ‘ \(\boldsymbol{a}\) ‘ if \(\boldsymbol{x}-\boldsymbol{a}\) is a factor of the polynomial \(3 x^{3}+x^{2}-a x-81\). [4]

Answer & Solution
\[ p(x)=3 x^{3}+x^{2}-a x-81 \]

\((x-a)\) is a factor of \(p(x)\)

\[ \begin{aligned} p(a) & =0 \\ 3(a)^{3}+(a)^{2}-a(a)-81 & =0 \\ 3 a^{3}+a^{2}-a^{2}-81 & =0 \\ a & =3 \end{aligned} \]
ii

Salman deposits ₹1000 every month in a recurring deposit account for 2 years. If he receives ₹26000 on maturity, find:

(a) the total interest Salman earns.

(b) the rate of interest. [4]

Answer & Solution

\(\mathrm{P}=\text{₹} 1000, n=2\), years \(=24\) months, \(r=\) rate of interest

\[ \begin{aligned} I & =P \times \frac{n(n+1)}{2} \times \frac{r}{100} \times \frac{1}{12} \\ I & =1000 \times \frac{24(24+1)}{2} \times \frac{r}{1200} \\ I & =250 r \\ A & =P+I \\ 26000 & =1000 \times 24+I \\ 2000 & =I \qquad \text{…(i)} \end{aligned} \]

…(ii)

(a) Salman earns interest as ₹2000

From (i) and (ii)

\[ \begin{aligned} 250 r & =2000 \\ r & =8 \end{aligned} \]

∴ (b) rate of interest is 8%

iii

In the given figure O, is the centre of the circle. CE is a tangent to the circle at \(\mathbf{A}\). If \(\angle \mathbf{A B D}=\mathbf{2 6}^{\boldsymbol{\circ}}\), then find:

ICSE Class 10 Mathematics 2023 Question 2(iii) figure

(a) ∠BDA

(b) ∠BAD

(c) ∠CAD

(d) ∠ODB [4]

Answer & Solution

(a)

ICSE Class 10 Mathematics 2023 Question 2(iii) solution figure
\[ \angle B D A=90^{\circ} \text { (Angle in a semicircle) } \]

(b) In \(\triangle B A D\)

\[ \angle B A D+\angle A B D+\angle A D B=180^{\circ} \]

(Sum of all angles of \(\Delta\) )

\[ \begin{aligned} & & \angle B A D+26^{\circ}+90^{\circ} & =180^{\circ} \\ & \therefore & \angle B A D & =64^{\circ} \end{aligned} \]

(c)

\[ \begin{aligned} & \angle C A D=\angle A B D \\ & \angle C A D=26^{\circ} \end{aligned} \]

(alternate segment angle) (d)

\[ \begin{aligned} O D & =O B & & \text { (radius of circle) } \\ \therefore & \angle O D B & =\angle O B D & \end{aligned} \]

(Angles opposite to equal sides of a triangle are equal)

\[ \therefore \quad \angle O D B=26^{\circ} \]
3
Question 3
ICSE Class 10 Mathematics 2023
i

Solve the following quadratic equation:

\[ x^{2}+4 x-8=0 \]

Give your answer correct to one decimal place. (Use mathematical tables if necessary.) [4]

Answer & Solution
\(x^{2}+4 x-8=0\)

Compare the equation \(a x^{2}+b x+c=0\) ∴

\[ \begin{aligned} a & =1, b=4 \text { and } c=-8 \\ D & =b^{2}-4 a c=(4)^{2}-4(1)(-8) \\ & =16+32=48 \end{aligned} \]

∴ Roots are real and distinct

\[ \begin{aligned} x & =\frac{-b \pm \sqrt{D}}{2 a} \\ & =\frac{-4 \pm \sqrt{48}}{2 \times 1}=\frac{-4 \pm 4 \sqrt{3}}{2} \end{aligned} \]
\[ \begin{aligned} & x=-2 \pm 2 \sqrt{3} \\ & x=-2 \pm 2 \times 1.732 \\ & x=-2 \pm 3.464 \\ & x=-2+3.464 \end{aligned} \]

and

\[ \begin{aligned} & x=-2-3.464 \\ & x=1.464 \text { and } x=-5.464 \\ & x=1.5 \text { and } x=-5.5 \end{aligned} \]
ii

Prove the following identity: [4]

\[ \left(\sin ^{2} \theta-1\right)\left(\tan ^{2} \theta+1\right)+1=0 \]
Answer & Solution
\[ \begin{aligned} & \text { L.H.S. }=\left(\sin ^{2} \theta-1\right)\left(\tan ^{2} \theta+1\right)+1 \\ &=\left(\sin ^{2} \theta-\sin ^{2} \theta-\cos ^{2} \theta\right)\left(\sec ^{2} \theta\right)+1 \\ & \begin{array}{r} 1+\tan ^{2} \theta=\sec ^{2} \theta \\ =\left(-\cos ^{2} \theta\right)\left(\frac{1}{\cos ^{2} \theta}\right)+1 \\ =-1+1 \\ =0=\text { R.H.S. } \end{array} \\ & \quad \text { Hence Proved. } \end{aligned} \]
iii

Use graph sheet to answer this question. Take 2 cm \(=1\) unit along both the axes.

(a) Plot A, B, C where A(0, 4), B(1, 1) and C(4, 0)

(b) Reflect A and B on the \(\boldsymbol{x}\)-axis and name them as E and D respectively.

(c) Reflect B through the origin and name if F. Write down the coordinates of F.

(d) Reflect B and C on the \(\boldsymbol{y}\)-axis and name them as H and G respectively.

(e) Join points A, B, C, D, E, F, G, H and A in order and name the closed figure formed. [5]

Answer & Solution
ICSE Class 10 Mathematics 2023 Question 3(iii) solution figure

(c) F(-1, -1)

(e) Star

✍️
Section B

ICSE Mathematics 2023 — Section B

40 marks • Attempt any four questions from this Section.
40
Marks to Attempt
4
Question 4
ICSE Class 10 Mathematics 2023
i

If \(\boldsymbol{A}=\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right], \boldsymbol{B}=\left[\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right], \boldsymbol{C}=\left[\begin{array}{ll}4 & 1 \\ 1 & 5\end{array}\right]\) and \(\boldsymbol{I}=\) \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\). Find \(\boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C})-\mathbf{1 4} \boldsymbol{I}\) [3]

Answer & Solution
\[ \begin{aligned} A & =\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right], B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right], C=\left[\begin{array}{ll} 4 & 1 \\ 1 & 5 \end{array}\right] \\ & A(B+C)-14 I \\ & I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left\{\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{ll} 4 & 1 \\ 1 & 5 \end{array}\right]\right\}-14\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 5 & 3 \\ 3 & 9 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right] \\ & =\left[\begin{array}{cc} 5+9 & 3+27 \\ 10+12 & 6+36 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right] \\ & =\left[\begin{array}{ll} 14 & 30 \\ 22 & 42 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right] \\ & =\left[\begin{array}{cc} 0 & 30 \\ 22 & 28 \end{array}\right] \end{aligned} \]
ii

ABC is a triangle whose vertices are A(1, -1), B(0, 4) and C(-6, 4). D is the midpoint of BC. Find the:

(a) coordinates of D.

(b) equation of the median AD. [3]

Answer & Solution
ICSE Class 10 Mathematics 2023 Question 4(ii) solution figure

(a)

\[ \begin{aligned} \mathrm{D} & =\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \\ & =\left(\frac{0-6}{2}, \frac{4+4}{2}\right) \end{aligned} \]

therefore coordinates of D is (-3, 4)

\[ \text { Slope of median } A D=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{4+1}{-3-1}=\frac{-5}{4} \]

Equation of the AD

\[ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ y+1 & =\frac{-5}{4}(x-1) \end{aligned} \]

⇒

\[ 5 x+4 y-1=0 \]
iii

In the given figure, O is the centre of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P.

ICSE Class 10 Mathematics 2023 Question 4(iii) figure
\[ \begin{aligned} \mathrm{AB} & =9 \mathrm{~cm}, \mathrm{BP}=16 \mathrm{~cm}, \angle \mathrm{PTB}=50^{\circ} \\ \angle \mathrm{OBA} & =45^{\circ} \end{aligned} \]

Find:

(a) length of PT

(b) ∠BAT

(c) ∠BOT

(d) ∠ABT [4]

Answer & Solution

(a)

ICSE Class 10 Mathematics 2023 Question 4(iii) solution figure
\[ \begin{aligned} P T^{2} & =P A \times P B \\ & =(16+9) \times 16 \\ P T & =\sqrt{25 \times 16} \end{aligned} \]

∴

\[ P T=5 \times 4=20 \mathrm{~cm} \]

(b)

\[ \angle \mathrm{BAT}=\angle \mathrm{PTB}=50^{\circ} \]

(alternate segment angle) (c)

\[ \angle \mathrm{BOT}=2 \angle \mathrm{BAT} \]

(Degree measure theorem)

\[ \angle \mathrm{BOT}=2 \times 50=100^{\circ} \]

(d) In \(\triangle \mathrm{OBT}\)

\[ \angle \mathrm{OBT}+\angle \mathrm{OTB}+\angle \mathrm{BOT}=180^{\circ} \]

(Sum of all angles of \(\Delta\) )

\[ \begin{aligned} \angle \mathrm{OBT}+\angle \mathrm{OBT}+100^{\circ} & =180^{\circ} \\ \angle \mathrm{OBT} & =\angle \mathrm{OTB} \end{aligned} \]

(angles opposite to equal sides of a triangle are equal)

\[ \begin{aligned} 2 \angle \mathrm{OBT} & =180^{\circ}-100^{\circ}=80^{\circ} \\ \angle \mathrm{OBT} & =\frac{80^{\circ}}{2}=40^{\circ} \\ \angle \mathrm{ABT} & =\angle \mathrm{ABO}+\angle \mathrm{OBT} \\ & =45^{\circ}+40^{\circ} \end{aligned} \]

∴

\[ \angle \mathrm{ABT}=85^{\circ} \]
5
Question 5
ICSE Class 10 Mathematics 2023
i

Mrs. Arora bought the following articles from a departmental store:

S.No.ItemPriceRate of GSTDiscount
1.Hair oil₹120018%₹100
2.Cashew nuts₹60012%–

Find the:

(a) Total GST paid.

(b) Total bill amount including GST. [3]

Answer & Solution

Mrs Arora paid for

1. Hair oil \(\text{₹}(1200-100)=\text{₹} 1100\)

\[ \text { G.S.T. }=\frac{18}{100} \times 1100=\text{₹} 198 \]

2. Cashew nuts

\[ \text { G.S.T. }=600 \times \frac{12}{100}=\text{₹} 72 \]

(a) Total G.S.T. paid \(=\text{₹} 198+72=\text{₹} 270\)

(b) Total Bill Amount including G.S.T.

\[ \begin{aligned} & =\text{₹} 1100+600+270 \\ & =\text{₹} 1970 \end{aligned} \]
ii

Solve the following inequation. Write down the solution set and represent it on the real number line. [3]

\[ -5(x-9) \geq 17-9 x>x+2, x \in \mathbf{R} \]
Answer & Solution
\[ \begin{aligned} -5(x-9) & \geq 17-9 x>x+2 \\ -5(x-9) & \geq 17-9 x \text { and } 17-9 x>x+2 \\ -5 x+9 x & \geq 17-45 \\ 4 x & \geq-28 \\ x & \geq-7 \end{aligned} \]
ICSE Class 10 Mathematics 2023 Question 5(ii) solution figure
\[ \begin{aligned} 17-9 x & >x+2 \\ -9 x-x & >2-17 \\ -10 x & >-15 \\ x & <\frac{-15}{-10} \\ x & <1.5 \end{aligned} \]
ICSE Class 10 Mathematics 2023 Question 5(ii) solution figure

Solution set

ICSE Class 10 Mathematics 2023 Question 5(ii) solution figure
\[ -7 \leq x<1.5 \]
iii

In the given figure, AC || DE || BF. If \(\mathbf{A C}=\mathbf{2 4 ~ c m}, \mathbf{E G}=\mathbf{8 ~ c m}, \mathbf{G B}=\mathbf{1 6 ~ c m}, \mathbf{B F}=\mathbf{3 0 ~ c m}\).

ICSE Class 10 Mathematics 2023 Question 5(iii) figure

(a) Prove \(\Delta \mathrm{GED}-\triangle \mathrm{GBF}\)

(b) Find DE

(c) DB : AB [4]

Answer & Solution

(a)

ICSE Class 10 Mathematics 2023 Question 5(iii) solution figure

In \(\triangle G E D\) and \(\triangle G B F\)

\[ \angle E G D=\angle B G F \]

(Vertically opposite angles)

\[ \angle D E G=\angle F B G \]

(Alternative angles \(D E \| B F\) )

\(\therefore \quad \triangle G E D \sim \triangle G B F\)

(AA Similarity Criterion) Hence Proved. (b)

\[ \triangle G E D \sim \triangle G B F \]

(Proved)

\[ \frac{G E}{G B}=\frac{E D}{B F}=\frac{G D}{G F} \quad \text { (by BPT) } \]
\[ \begin{aligned} \frac{E D}{B F} & =\frac{G E}{G B} \\ \frac{E D}{30} & =\frac{8}{16} \\ E D & =\frac{8}{16} \times 30=15 \mathrm{~cm} \\ D E & =15 \mathrm{~cm} \end{aligned} \]

(c) In \(\triangle A B C, D E \| A C\)

\[ \angle B D E=\angle B A C \]

and

\[ \angle B E D=\angle B C A \]

by AA Similarity criterion, triangle \(B D E \sim\) triangle \(B A C\).

\[ \begin{aligned} & \frac{D B}{A B}=\frac{B E}{B C}=\frac{D E}{A C} \\ & \frac{D B}{A B}=\frac{D E}{A C} \\ & \frac{D B}{A B}=\frac{15}{24}=\frac{5}{8} \end{aligned} \]

∴

\[ D B: A B=5: 8 \]
6
Question 6
ICSE Class 10 Mathematics 2023
i

The following distribution gives the daily wages of 60 workers of a factory.

Daily income ₹Number of worker (f)
200-3006
300-40010
400-50014
500-60016
600-70010
700-8004

Use graph paper to answer this question. Take \(2 \mathrm{~cm}=\text{₹} 100\) along one axis and \(2 \mathrm{~cm}=2\) workers along the other axis. Draw a histogram and hence find the mode of the given distribution. [3]

Answer & Solution
ICSE Class 10 Mathematics 2023 Question 6(i) solution figure
ii

The \(\mathbf{5}^{\text {th }}\) term and the \(\mathbf{9}^{\text {th }}\) term of an Arithmetic Progression are 4 and -12 respectively. Find:

(a) the first term

(b) common difference

(c) sum of 16 terms of the AP. [3]

Answer & Solution

Given,

\[ t_{5}=4 \text { and } t_{9}=-12 \]

Let the first term be \(a\) and common difference be \(d\) respectively

\[ \begin{aligned} a_{n} & =a+(n-1) d \\ t_{5} & =4 \\ a+(5-1) d & =4 \\ a+4 d & =4 \qquad \text{…(i)} \end{aligned} \]

Similarly, \(a+8 d=-12\) …(ii) from (i) and (ii)

\[ d=-4 \text { and } a=20 \]

(a) first term is 20

(b) Common difference \(=-4\)

(c)

\[ \begin{aligned} \mathrm{S}_{n} & =\frac{n}{2}[2 a+(n+1) d] \\ \mathrm{S}_{16} & =\frac{16}{2}[2 \times 20+(16-1)(-4)] \\ & =8[40-60)] \\ & =-160 \end{aligned} \]

∴ Sum of 16 terms is -160

iii

A and B are two points on the \(\boldsymbol{x}\)-axis and \(\boldsymbol{y}\)-axis respectively.

ICSE Class 10 Mathematics 2023 Question 6(iii) figure

(a) Write down the coordinates of A and B.

(b) P is a point on AB such that \(\mathbf{A P}: \mathbf{P B}=\mathbf{3}: \mathbf{1}\). Using section formula find the coordinates of point P.

(c) Find the equation of a line passing through P and perpendicular to AB. [4]

Answer & Solution

(a)

ICSE Class 10 Mathematics 2023 Question 6(iii) solution figure

Coordinates of A(4, 0) Coordinates of B(0, 4) (b)

\[ \begin{aligned} & P=\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right) \\ & =\left(\frac{3 \times 0+1 \times 4}{3+1}, \frac{3 \times 4+1 \times 0}{3+1}\right) \\ & P=(1,3) \end{aligned} \]

(c) Slope of line \(A B=\frac{4-0}{0-4}=-1\)

∴ Slope of perpendicular line \(A B=1\)

\[ \begin{aligned} y-y_{1} & =m\left(x-x_{1}\right) \\ y-3 & =1(x-1) \\ y-3 & =x-1 \\ x-y+2 & =0 \end{aligned} \]
7
Question 7
ICSE Class 10 Mathematics 2023
i

A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is:

(a) multiple of 5

(b) a perfect square

(c) a prime number? [3]

Answer & Solution
\[ \mathrm{S}=\{1,2,3,4, \ldots \ldots \ldots . . .25\} \]

∴

\[ n(\mathrm{~S})=25 \]

(a) multiple of 5

\[ \begin{aligned} & \qquad \mathrm{E} & =\{5,10,15,20,25\} \\ n(\mathrm{E}) & =5 \\ & \text { Probability of multiple of } 5=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{5}{25} \\ & =\frac{1}{5} \end{aligned} \]

(b) Perfect square \(=(1,4,9,16,25)\)

\[ \text { Required probability }=\frac{5}{25}=\frac{1}{5} \]

(c) a prime number \{2, 3, 5, 7, 11, 13, 17, 19, 23\}

\[ \text { Required probability }=\frac{9}{25} \]
ii

A man covers a distance of 100 km, travelling with a uniform speed of \(\boldsymbol{x}\) km/hr, had the speed been 5 km/hr more it would have taken 1 hour less. Find \(\boldsymbol{x}\) the original speed. [3]

Answer & Solution

Let the original speed be \(x\) km/h

\[ \begin{aligned} \text { Time } & =\frac{\text { Distance }}{\text { Speed }} \\ t_{1} & =\frac{100}{x} \mathrm{~h} \\ t_{2} & =\frac{100}{x+5} \mathrm{~h} \\ t_{1}-t_{2} & =1 \\ \frac{100}{x}-\frac{100}{x+5} & =1 \\ \frac{100 x+500-100 x}{x(x+5)} & =1 \\ 500 & =x^{2}+5 x \\ x^{2}+5 x-500 & =0 \\ (x+25)(x-20) & =0 \end{aligned} \]

If \(x+25=0 \Rightarrow x=-25\) it is not possible because speed can not be negative if \(x-20=0\)

\[ x=20 \]

∴ original speed be 20 km/h

iii

A solid in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height. If the radius of the cylinder is 14 cm, find the rise in the water level. [4]

ICSE Class 10 Mathematics 2023 Question 7(iii) figure
Answer & Solution

Volume of solid = Vol. of hemisphere + Vol. of Cone.

\[ \begin{aligned} & =\frac{2}{3} \pi r^{3}+\frac{1}{3} \pi r^{2} h=\frac{\pi r^{2}}{3}(2 r+h) \\ & =\frac{\pi \times 7 \times 7}{3}(2 \times 7+4)=\frac{49 \pi}{3}(18) \\ & =294 \pi \mathrm{~cm}^{3} \end{aligned} \]

Let the level of water rise in cylinder be \(x \mathrm{~cm}\) ∴ Rise in Vol. of water \(=294 \pi\)

\[ \begin{aligned} \pi R^{2} x & =294 \pi \\ (14)^{2} x & =294 \\ x & =\frac{294}{14 \times 14}=\frac{21}{14} \\ x & =\frac{3}{2} \mathrm{~cm}=1.5 \mathrm{~cm} \end{aligned} \]

∴ level of water rise in cylinder \(=1.5 \mathrm{~cm}\)

8
Question 8
ICSE Class 10 Mathematics 2023
i

The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method. [3]

MarksNumber of Students ( \(\boldsymbol{f}\) )
0-103
10-208
20-3014
30-409
40-504
50-602
Answer & Solution
MarksNumber of Students (f)\(x\)\(d=\) \(x-\mathrm{A}\)\(\boldsymbol{f} \boldsymbol{d}\)
0-1035-20-60
10-20815-10-80
20-301425 = A00
30-409351090
40-504452080
50-602553060
4090
\[ \begin{aligned} \bar{x} & =A+\frac{\Sigma f d}{\Sigma f} \\ & =25+\frac{90}{40} \\ & =25+2.25 \\ \text { Mean } & =27.25 \end{aligned} \]
ii

What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion? [3]

Answer & Solution

Let \(x\) be added to get the four numbers in proportion

\[ \begin{aligned} & \therefore(4+x):(6+x)::(8+x):(11+x) \\ &(4+x)(11+x)=(6+x)(8+x) \\ & x^{2}+15 x+44=x^{2}+14 x+48 \\ & 15 x-14 x=48-44 \\ & x=4 \end{aligned} \]

∴ 4 should be added to get the four numbers in proportion

iii

Using ruler and compass construct a triangle ABC in which \(\mathbf{A B}=\mathbf{6 ~ c m}, \angle \mathbf{B A C}=\mathbf{1 2 0}^{\circ}\) and \(\mathbf{A C}=\mathbf{5}\) cm. Construct a circle passing through A, B and C. Measure and write down the radius of the circle. [4]

Answer & Solution
ICSE Class 10 Mathematics 2023 Question 8(iii) solution figure

Step of construction:

(a) Draw \(A B=6 \mathrm{~cm}\)

(b) Make \(\angle B A D=120^{\circ}\)

(c) Taking centre A and radius of \(A C 5\) cm draw an arc which intersect at the point \(C\)

(d) join \(B C\). triangle \(A B C\) is required triangle

(e) Draw perpendicular bisectors of side \(A B\) and \(A C\) respectively which intersect at the point \(O\)

(f) Taking \(O\) as centre and \(O A\) as radius draw a circle which passes through the points \(A, B\) and C respectively.

Now, measured the radius \(O A=5.5 \mathrm{~cm}\)

9
Question 9
ICSE Class 10 Mathematics 2023
i

Using Componendo and Dividendo solve for \(\boldsymbol{x}\). [3]

\[ \frac{\sqrt{2 x+2}+\sqrt{2 x-1}}{\sqrt{2 x+2}-\sqrt{2 x-1}}=3 \]
Answer & Solution
\[ \frac{\sqrt{2 x+2}+\sqrt{2 x-1}}{\sqrt{2 x+2}-\sqrt{2 x-1}}=3 \]

Applying componendo and dividendo

\[ \frac{\sqrt{2 x+2}+\sqrt{2 x-1}+\sqrt{2 x+2}-\sqrt{2 x-1}}{\sqrt{2 x+2}+\sqrt{2 x-1}-\sqrt{2 x+2}+\sqrt{2 x-1}}=\frac{3+1}{3-1} \]
\[ \begin{aligned} & \frac{2 \sqrt{2 x+2}}{2 \sqrt{2 x-1}}=\frac{4}{2}=2 \\ & \frac{\sqrt{2 x+2}}{\sqrt{2 x-1}}=2 \end{aligned} \]

squaring both sides

\[ \begin{aligned} \frac{2 x+2}{2 x-1} & =4 \\ 2 x+2 & =8 x-4 \\ 2 x-8 x & =-4-2 \\ -6 x & =-6 \\ x & =1 \end{aligned} \]
ii

Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60 … is 300? Hence find the sum of the terms of the Arithmetic Progression (A.P.) [3]

Answer & Solution

15, 30, 45, 60, …………….. 300

\[ \begin{aligned} a & =15, d=30-15=15 \\ a_{n} & =300 \\ a+(n-1) d & =300 \\ 15+(n-1)(15) & =300 \\ 1+(n-1) & =20 \\ n & =20 \end{aligned} \]

\(20^{\text {th }}\) term is 300

\[ \begin{aligned} & S_{n}=\frac{n}{2}\left[a+a_{n}\right]=\frac{20}{2}[15+300] \\ & S_{20}=3150 \end{aligned} \]
iii

From the top of a tower 100 m high a man observes the angles of depression of two ships A and B, on opposite sides of the lower as 45° and 38° respectively. If the foot of the tower and the ships are in the same horizontal line find the distance between the two ships A and B to the nearest metre. (Use Mathematical Tables for this question.) [4]

ICSE Class 10 Mathematics 2023 Question 9(iii) figure
Answer & Solution
ICSE Class 10 Mathematics 2023 Question 9(iii) solution figure

In \(\triangle A C D\)

\[ \begin{aligned} \tan A & =\frac{C D}{A C} \\ \tan 45^{\circ} & =\frac{100}{A C} \\ 1 & =\frac{100}{A C} \\ \therefore \quad \mathrm{AC} & =100 \mathrm{~m} \end{aligned} \]

In \(\triangle B C D\)

\[ \begin{aligned} \tan B & =\frac{C D}{B C} \\ \tan 38^{\circ} & =\frac{100}{B C} \\ 0.7813 & =\frac{100}{B C} \\ \therefore \quad B C & =\frac{100}{0.7813} \\ & =127.99 \mathrm{~m} \end{aligned} \]

Distance between two ships \(=A C+B C\)

\[ \begin{aligned} = & 100+127.99 \\ 227.99 \mathrm{~m} & =228 \mathrm{~m} \end{aligned} \]
10
Question 10
ICSE Class 10 Mathematics 2023
i

Factorize completely using factor theorem: [4]

\[ 2 x^{3}-x^{2}-13 x-6 \]
Answer & Solution
\(2 x^{3}-x^{2}-13 x-6\)

Put

\[ \begin{aligned} x & =-2 \\ & =2(-2)^{3}-(-2)^{2}-13(-2)-6 \\ & =-16-4+26-6 \\ & =0 \end{aligned} \]

\(\therefore(x+2)\) is a factor of polynomial

\[ \begin{aligned} & x+2) 2 x^{3}-x^{2}-13 x-6\left(2 x^{2}-5 x-3\right. \\ & 2 x^{3}+4 x^{2} \\ & \frac{(-)(-)}{-5 x^{2}-13 x-6} \\ & -5 x^{2}-10 x \\ & \begin{array}{c} (+) \quad(+) \\ \hline-3 x-6 \\ -3 x-6 \\ (+)(+) \\ \hline \times \end{array} \\ & 2 x^{3}-x^{2}-13 x-6=(x+2)\left(2 x^{2}-5 x-3\right) \\ & =(x+2)\left\{2 x^{2}-6 x+x-3\right\} \\ & =(x+2)\{2 x(x-3)+1(x-3)\} \\ & =(x+2)(x-3)(2 x+1) \\ & \therefore 2 x^{3}-x^{2}-13 x-6=(x-3)(2 x+1)(x+2) \end{aligned} \]
ii

Use graph paper to answer this question. During a medical checkup of 60 students in a school, weights were recorded as follows:

Weight (in kg)Number of Students
28-302
30-324
32-3410
34-3613
36-3815
38-409
40-425
42-442

Taking \(\mathbf{2} \mathbf{~ c m}=\mathbf{2 ~ k g}\) along one axis and \(\mathbf{2 ~ c m}=\mathbf{1 0}\) students along the other axis draw an ogive. Use your graph to find the:

(a) median

(b) upper Quartile

(c) number of students whose weight is above 37 kg [6]

Answer & Solution
Weight (in kg)Number of Students (f)Cumulative frequency (cf)
28-3022
30-3246
32-341016
34-361329
36-381544
38-40953
40-42558
42-44260
Weight in kgs. (less than)Number of Students
280
302
326
3416
3629
3844
4053
4258
4460

(a) median \(=36.2 \mathrm{~kg}\)

(b) upper quartile \(=38.2 \mathrm{~kg}\)

(c) Number of students where weight is above \(37 \mathrm{~kg}=60-37=23\) students

ICSE Class 10 Mathematics 2023 Question 10(ii) solution figure

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