ICSE Class 10 Mathematics 2020 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 2020 Question Paper with Solutions, including algebra, geometry, trigonometry, statistics, graphs, constructions and commercial mathematics. Use this solved paper to revise important concepts, practise board-style questions and strengthen your exam preparation.

Year
2020
Maximum Marks
80
Time Allowed
Two hours and a half
📄
Paper Instructions
ICSE Class 10 Mathematics 2020

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 question are given in brackets [ ].

Mathematical tables are provided.

✓
Section A

ICSE Mathematics 2020 — Section A

40 marks • Attempt all questions from this Section.
40
Marks
1
Question 1
ICSE Class 10 Mathematics 2020
a

Solve the following Quadratic Equation :

\[ x^{2}-7 x+3=0 \]

Give your answer correct to two decimal places. [3]

Answer & Solution
\[ x^{2}-7 x+3=0 \]

Compare the equation by \(a x^{2}+b x+c=0\) then, \(a=1, b=-7\) and \(c=3\)

\[ \begin{aligned} x & =\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ & =\frac{7 \pm \sqrt{49-4 \times 1 \times 3}}{2 \times 1} \\ & =\frac{7 \pm \sqrt{37}}{2} \\ & =\frac{7 \pm \sqrt{37}}{2}=\frac{7 \pm 6.08}{2} \end{aligned} \]

∴

\[ \begin{aligned} x & =\frac{7+6.08}{2}=\frac{13.08}{2} \\ & =6.54 \end{aligned} \]

and

\[ \begin{aligned} x & =\frac{7-6.08}{2}=\frac{0.92}{2} \\ & =0.46 \end{aligned} \]

Hence, \(x=6.54\) (Approx.) and \(x=0.46\) (Approx).

b

\(\quad\) Given \(\mathbf{A}=\left[\begin{array}{cc}\boldsymbol{x} & \mathbf{3} \\ \boldsymbol{y} & \mathbf{3}\end{array}\right]\) If \(\boldsymbol{A}^{\mathbf{2}}=\mathbf{3} \boldsymbol{I}\), where \(\boldsymbol{I}\) is the identity matrix of order 2, find \(x\) and \(y\). [3]

Answer & Solution
\[ \begin{aligned} \mathrm{A} & =\left[\begin{array}{ll} x & 3 \\ y & 3 \end{array}\right] \\ A^{2} & =\mathrm{A} \times \mathrm{A} \end{aligned} \]
\[ \begin{aligned} & =\left[\begin{array}{ll} x & 3 \\ y & 3 \end{array}\right]\left[\begin{array}{ll} x & 3 \\ y & 3 \end{array}\right] \\ & =\left[\begin{array}{ll} x^{2}+3 y & 3 x+9 \\ x y+3 y & 3 y+9 \end{array}\right] \\ A^{2} & =3 I \text { (Given) } \\ {\left[\begin{array}{ll} x^{2}+3 y & 3 x+9 \\ x y+3 y & 3 y+9 \end{array}\right]=} & 3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \end{aligned} \]

Comparing the elements of matrices

\[ \begin{aligned} & 3 x+9=0 \Rightarrow x=-3 \\ & 3 y+9=3 \Rightarrow y=-2 \end{aligned} \]
c

Using ruler and compass construct a triangle \(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}\) where \(\boldsymbol{A} \boldsymbol{B}=\mathbf{3 ~ c m}, \boldsymbol{B} \boldsymbol{C}=\mathbf{4 ~ c m}\) and \(\angle \boldsymbol{A} \boldsymbol{B} \boldsymbol{C}\) \(=90^{\circ}\). Hence construct a circle circumscribing triangle \(A B C\). Measure and write down the radius of the circle. [4]

Answer & Solution
ICSE Class 10 Mathematics 2020 Question 1(c) solution figure

Steps of construction

(i) Draw \(B C=4 \mathrm{~cm}\)

(ii) Make \(\angle P B C=90^{\circ}\)

(iii) Taking centre \(B\) and radius 3 cm draw an arc which intersect BP at the point A.

(iv) Join \(A C, \triangle A B C\) is required triangle. \(\angle B=90^{\circ}\) \(\therefore A C\) is the diameter of circle.

(v) Draw perpendicular bisector of \(A C\).

(vi) Taking centre \(O\) and radius equal to \(B O\) draw a circle which passes through vertex \(A, B\) and \(C\). Radius of circumscribe circle \(=2.5 \mathrm{~cm}\).

2
Question 2
ICSE Class 10 Mathematics 2020
a

Use factor theorem to factorise \(\mathbf{6} \boldsymbol{x}^{\mathbf{3}}+\mathbf{1 7} \boldsymbol{x}^{\mathbf{2}}+\mathbf{4} \boldsymbol{x}-\) 12 completely. [3]

Answer & Solution
\[ \begin{aligned} p(x) & =6 x^{3}+17 x^{2}+4 x-12 \\ p(-2) & =6(-2)^{3}+17(-2)^{2}+4(-2)-12 \\ & =-48+68-8-12 \\ & =68-68=0 \end{aligned} \]

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

\[ \begin{aligned} & x + 2 \longdiv { 6 x ^ { 3 } + 1 7 x ^ { 2 } + 4 x – 1 2 } ( 6 x ^ { 2 } + 5 x – 6 \\ & \begin{array}{c} 6 x^{3}+12 x^{2} \\ \frac{(-)}{(-)} \\ 5 x^{2}+4 x-12 \\ 5 x^{2}+10 x \\ \frac{(-) \quad(-)}{-6 x-12} \\ -6 x-12 \\ \frac{(+)(+)}{0} \end{array} \\ & \therefore 6 x^{3}+17 x^{2}+4 x-12 \\ & =(x+2)\left(6 x^{2}+5 x-6\right) \\ & =(x+2)\left\{6 x^{2}+9 x-4 x-6\right\} \\ & =(x+2)\{3 x(2 x+3)-2(2 x+3)\} \\ & =(x+2)(2 x+3)(3 x-2) \end{aligned} \]
b

Solve the following inequation and represent the solution set on the number line. [3]

\[ \frac{3 x}{5}+2<x+4 \leq \frac{x}{2}+5, x \in R \]
Answer & Solution
\[ \frac{3 x}{5}+2<x+4 \leq \frac{x}{2}+5, \quad x \in R \]
\[ \begin{aligned} & \frac{3 x}{5}+2<x+14 \\ & \frac{3 x}{5}-x<14-2 \\ & \frac{3 x-5 x}{5}<12 \\ & \text { Multiply by } 5 \\ & -2 x<12 \times 5 \\ & x>\frac{60}{-2} \\ & x>-30 \\ & \therefore x>-30<x \leq 2 \end{aligned} \]
ICSE Class 10 Mathematics 2020 Question 2(b) solution figure
c

Draw a Histogram for the given data, using a graph paper :

Weekly Wages (in ₹)No. of People
3000-40004
4000-50009
5000-600018
6000-70006
7000-80007
8000-90002
9000-100004

Estimate the mode from the graph. [4]

Answer & Solution
ICSE Class 10 Mathematics 2020 Question 2(c) solution figure

= Mode -5400 (approx)

3
Question 3
ICSE Class 10 Mathematics 2020
a

In the figure given below, O is the centre of the circle and \(A B\) is a diameter. If \(\boldsymbol{A} \boldsymbol{C}=\boldsymbol{B} \boldsymbol{D}\) and \(\angle \boldsymbol{A} \boldsymbol{O} \boldsymbol{C}=\mathbf{7 2} \boldsymbol{°}^{\boldsymbol{\circ}}\). Find:

(i) \(\angle \mathrm{ABC}\)

(ii) ∠BAD

(iii) ∠ABD [3]

ICSE Class 10 Mathematics 2020 Question 3(a) figure
Answer & Solution

\(A B\) is the diameter of circle (i)

\[ \begin{array}{ll} \therefore \quad \angle A D B=90^{\circ} & \text { (semicircle angle) } \\ \angle A B C=\frac{1}{2} \angle A O C & \text { (same arc angles at } \end{array} \]

(same arc angles at circumference and centres)

\[ \therefore \quad \angle A B C=\frac{1}{2} \times 72^{\circ}=36^{\circ} \]

(ii)

\[ \begin{array}{rlrl} A C & =B D & \\ \therefore & & \angle A B C & =\angle B A D \text { (equal arcs make equal } \\ & & & \text { angle) } \\ \therefore & \angle B A D & =36^{\circ} & \end{array} \]

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

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

(sum of all angles of \(\Delta\) )

\[ \begin{aligned} \angle A B D+36^{\circ} & +90^{\circ}=180^{\circ} \\ \angle A B D & =180^{\circ}-126^{\circ} \\ \angle A B D & =54^{\circ} \end{aligned} \]
b

Prove that : [3]

\[ \frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\tan A}=\sin A-\cos A \]
Answer & Solution
\[ \begin{aligned} \text { L.H.S. }= & \frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\tan A} \\ = & \frac{\sin A}{1+\frac{\cos A}{\sin A}}-\frac{\cos A}{1+\frac{\sin A}{\cos A}} \\ = & \frac{\sin ^{2} A}{\sin A+\cos A}-\frac{\cos ^{2} A}{\cos A+\sin A} \\ = & \frac{\sin ^{2} A-\cos ^{2} A}{\sin A+\cos A} \\ = & \frac{\left(a^{2}-b^{2}=(a+b)(a-b)\right]}{(\sin A+\cos A)(\sin A-\cos A)} \\ = & \sin \mathrm{A}-\cos \mathrm{A} \\ = & \text { Hence proved. } \end{aligned} \]
c

In what ratio is the line joining \(\boldsymbol{P}(\mathbf{5}, \mathbf{3})\) and \(Q(-5,3)\) divided by the \(y\)-axis ? Also find the coordinates of the point of intersection. [4]

Answer & Solution

Let the line joining the \((5,3)\) and (-5, 3) divided by the \(y\)-axis in the ratio \(k: 1\). (-5, 3)

ICSE Class 10 Mathematics 2020 Question 3(c) solution figure

Let the coordinate of the point of intersection \(R\) be \((0, y)\).

\[ \begin{aligned} R_{x} & =\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \\ 0 & =\frac{1 \times 5+k(-5)}{1+k} \\ 5-5 k=0 & \Rightarrow k=1 \\ R_{y} & =\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \\ y & =\frac{1 \times 3+k \times 3}{1+k} \\ y & =\frac{3+3}{1+1}=3 \quad[\mathrm{Put}, k=1] \\ \text { Ratio } & =k: 1=1: 1 \end{aligned} \]

Point of intersection of the line (0, 3).

4
Question 4
ICSE Class 10 Mathematics 2020
a

A solid spherical ball of radius 6 cm is melted and recast into 64 identical spherical marbles. Find the radius of each marble. [3]

Answer & Solution

Volume of solid spherical ball \(=\frac{4}{3} \pi r^{3}\)

\[ \begin{aligned} & =\frac{4}{3} \pi(6)^{3} \\ & =\frac{4}{3} \pi \times 216 \\ & =\frac{216 \times 4}{3} \pi \mathrm{~cm}^{3} \end{aligned} \]

Let the radius of spherical marble be \(R\) cm. Then 64 × volume of spherical marbles = volume of spherical ball

\[ \begin{aligned} 64 \times \frac{4}{3} \pi R^{3} & =\frac{4 \times 216}{3} \pi \mathrm{~cm}^{3} \\ R^{3} & =\frac{216}{64} \\ R & =\sqrt[3]{\frac{216}{64}}=\frac{6}{4}=\frac{3}{2} \mathrm{~cm} \\ & =1.5 \mathrm{~cm} \end{aligned} \]

∴ Radius of spherical marble is 1.5 cm.

b

Each of the letters of the word ‘AUTHORIZES’ is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is :

(i) a vowel ?

(ii) one of the first 9 letters of the English alphabet which appears in the given word ?

(iii) one of the last 9 letters of the English alphabet which appears in the given word ? [3]

Answer & Solution

‘AUTHORIZES’ Total number of disc \(n(\mathrm{~S})=10\)

(i) Number of discs a letter of vowel = 5

(A, U, O, I, E) \(\therefore n(\mathrm{E})=5\) Probability of a disc is vowel

\[ =\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{5}{10}=\frac{1}{2} \]

(ii) One of the first 9 letter in english alphabet \(=4(A, E, H, \mathrm{I})\)

∴

\[ n(E)=4 \]

Probability of a disc is drawn is written one of first 9 letters of english alphabet

\[ \begin{aligned} & =\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{10} \\ & =\frac{2}{5} \end{aligned} \]

(iii) One of the last \(9(R, S, T, U, Z)\) letters of english alphabet

\[ n(E)=5 \]

Probability of a disc is drawn is written one of the last 9 letters of english alphabet

\[ =\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{5}{10} \]
\[ =\frac{1}{2} \]
c

Mr. Bedi visits the market and buys the following articles : Medicines costing ₹ 950, GST @ 5% A Pair of shoes costing ₹ 3000, GST @ 18% A Laptop bag costing ₹ 1000 with a discount of 30% GST @ 18%

(i) Calculate the total amount of GST paid.

(ii) The total bill amount including GST paid by Mr. Bedi. [4]

Answer & Solution

Medicine costing \(=\text{₹} 950\)

\[ \text { GST @ } 5 \%=\frac{5}{100} \times 950=\text{₹} 47.50 \]

A pair of shoes costing \(=\text{₹} 3000\)

\[ \text { GST @ } 18 \%=\frac{18}{100} \times 3000=\text{₹} 540 \]

Laptop bag costing ₹ 1000 with a discount of 30%

\[ \begin{aligned} \therefore \quad \text { Net } \text { cost } & =1000-\frac{30}{100} \times 1000 \\ & =\text{₹} 700 \end{aligned} \]

GST @ \(18 \%=\frac{18}{100} \times 700=\text{₹} 126\)

(i) Total amount of GST = ₹ 47.50 + 540 + 126

\[ \text { = \text{₹} } 713.50 \]

(ii) Total bill amount including GST

\[ \begin{aligned} & =\text{₹} 950+3000+700+713.50 \\ & =\text{₹} 4650+713.50 \\ & =\text{₹} 5363.50 \end{aligned} \]
✍️
Section B

ICSE Mathematics 2020 — Section B

40 marks • Attempt any four questions from this Section.
40
Marks to Attempt
5
Question 5
ICSE Class 10 Mathematics 2020
a

A company with 500 shares of nominal value ₹ 120 declares an annual dividend of \(15 \%\). Calculate :

(i) the total amount of dividend paid by the company.

(ii) annual income of Mr. Sharma who holds 80 shares of the company.

If the return percent of Mr. Sharma from his shares is 10%. Find the market value of each share. [3]

Answer & Solution

Number of shares \(=500\)

\[ \begin{aligned} \text { Nominal value } & =\text{₹} 120 \\ \text { Annual dividend } & =15 \% \end{aligned} \]

(i) Total amount of dividend paid by the company = Rate of dividend × Nominal value of share × Number of share

\[ \begin{aligned} & =\frac{15}{100} \times 120 \times 500 \\ & =15 \times 600=\text{₹} 9000 \end{aligned} \]

∴ Total amount paid by the company = ₹ 9000

(ii) Mr sharma’s Annual Income = Rate of dividend × Nominal value of share × Number of share

\[ \begin{aligned} & =\frac{15}{100} \times 120 \times 80 \\ & =15 \times 96=\text{₹} 1440 \end{aligned} \]

∴ Mr sharma’s Annual Income \(=\text{₹} 1440\)

\[ \begin{aligned} \text { Return of Investment } & =\frac{\text { Income }}{\text { Investment }} \times 100 \\ 10 & =\frac{1440}{\text { Investment }} \times 100 \end{aligned} \]

\(\therefore \quad\) Investment \(=\frac{144000}{10}=\text{₹} 14400\)

\[ \begin{aligned} \text { Market value of share } & =\frac{\text { Investment }}{\text { No. of share }} \\ & =\frac{14400}{80} \end{aligned} \]

∴ Market value of share \(=\text{₹} 180\)

b

The mean of the following data is 16. Calculate the value of \(\boldsymbol{f}\). [3]

Marks510152025
No. of Students37\(f\)96
Answer & Solution
MarksNo. of Students
\((x)\)(f)\(f \times x\)
5315
10770
15\(f\)\(15 f\)
209180
256150
\(25+f\)\(415+15 f\)
\[ \begin{aligned} \text { Mean } & =16 \\ \frac{\Sigma f x}{\Sigma f} & =16 \\ \frac{415+15 f}{25+f} & =16 \\ 415+15 f & =400+16 f \\ 16 f-15 f & =415-400 \end{aligned} \]

∴

\[ f=15 \]
c

The \(\mathbf{4}^{\text {th }}, \mathbf{6}^{\text {th }}\) and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series. [4]

Answer & Solution

Let the first term of G.P. be \(a\), common ratio be \(r\) and number of terms be \(n\).

\[ \begin{aligned} a_{4} & =10, a_{6}=40 \text { and } a_{n}=640 \text { (given) } \\ a_{4} & =10 \\ a r^{3} & =10 \qquad \text{…(i)} \end{aligned} \]

Again,

\[ \begin{aligned} a_{6} & =40 \\ a r^{5} & =40 \end{aligned} \]

From eqs. (i) \& (ii)

\[ \begin{aligned} \frac{a r^{5}}{a r^{3}} & =\frac{40}{10} \\ \Rightarrow r^{2} & =4 \\ \therefore \quad r & = \pm 2 \\ \therefore \quad r & =2 \end{aligned} \]

( \(r>0\) ) From eq. (i)

\[ \begin{aligned} a r^{3} & =10 \\ a(2)^{3} & =10 \\ a & =\frac{10}{8}=\frac{5}{4} \end{aligned} \]

Again,

\[ \begin{aligned} a_{n} & =640 \\ a r^{n-1} & =640 \\ \frac{5}{4}(2)^{n-1} & =640 \end{aligned} \]
\[ \begin{aligned} & 2^{n-1}=\frac{640 \times 4}{5} \\ & 2^{n-1}=128 \times 4=2^{9} \end{aligned} \]

Comparing the power

\[ n-1=9 \Rightarrow n=10 \]

∴ First term of G.P. \(=\frac{5}{4}\), common ratio \(=2\) and number of terms \(=10\)

6
Question 6
ICSE Class 10 Mathematics 2020
a

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

Answer & Solution

\(A=\left[\begin{array}{ll}3 & 0 \\ 5 & 1\end{array}\right]\) and \(B=\left[\begin{array}{cc}-4 & 2 \\ 1 & 0\end{array}\right]\)

\[ \begin{aligned} A^{2} & =A \times A=\left[\begin{array}{ll} 3 & 0 \\ 5 & 1 \end{array}\right] \times\left[\begin{array}{ll} 3 & 0 \\ 5 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} 9+0 & 0+0 \\ 15+5 & 0+1 \end{array}\right] \\ \therefore \quad A^{2} & =\left[\begin{array}{cc} 9 & 0 \\ 20 & 1 \end{array}\right] \\ A B & =\left[\begin{array}{ll} 3 & 0 \\ 5 & 1 \end{array}\right]\left[\begin{array}{cc} -4 & 2 \\ 1 & 0 \end{array}\right] \end{aligned} \]
\[ \begin{aligned} &=\left[\begin{array}{cc} -12+0 & 6+0 \\ -20+1 & 10+0 \end{array}\right] \\ & A B=\left[\begin{array}{cc} -12 & 6 \\ -19 & 10 \end{array}\right] \\ & \quad B^{2}=B \times B=\left[\begin{array}{cc} -4 & 2 \\ 1 & 0 \end{array}\right]\left[\begin{array}{cc} -4 & 2 \\ 1 & 0 \end{array}\right] \\ & \quad=\left[\begin{array}{cc} 16+2 & -8+0 \\ -4+0 & 2+0 \end{array}\right] \\ & \therefore B^{2}=\left[\begin{array}{cc} 18 & -8 \\ -4 & 2 \end{array}\right] \\ & \therefore A^{2}-2 A B+B^{2} \\ & {\left[\begin{array}{cc} 9 & 0 \\ 20 & 1 \end{array}\right]-2\left[\begin{array}{cc} -12 & 6 \\ -19 & 10 \end{array}\right]+\left[\begin{array}{cc} 18 & -8 \\ -4 & 2 \end{array}\right] } \\ & {\left[\begin{array}{cc} 9 & 0 \\ 20 & 1 \end{array}\right]+\left[\begin{array}{cc} 24 & -12 \\ 38 & -20 \end{array}\right]+\left[\begin{array}{cc} 18 & -8 \\ -4 & 2 \end{array}\right] } \\ & {\left[\begin{array}{cc} 9+24+18 & 0-12-8 \\ 20+38-4 & 1-20+2 \end{array}\right]=\left[\begin{array}{cc} 51 & -20 \\ 54 & -17 \end{array}\right] } \end{aligned} \]
b

In the given figure \(\boldsymbol{A} \boldsymbol{B}=\mathbf{9 ~ c m}, \boldsymbol{P} \boldsymbol{A}=\mathbf{7 . 5 ~ c m}\) and \(P C=5 \mathrm{~cm}\). Chords \(\boldsymbol{A} \boldsymbol{D}\) and \(\boldsymbol{B} \boldsymbol{C}\) intersect at \(\boldsymbol{P}\).

ICSE Class 10 Mathematics 2020 Question 6(b) figure

(i) Prove that \(\triangle P A B \sim \triangle P C D\).

(ii) Find the length of \(C D\).

(iii) Find area of \(\triangle P A B\) : area of \(\triangle P C D\). [3]

Answer & Solution

(i) \(\triangle P A B\) and \(\triangle P C D\)

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

(Vertically opposite angle)

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

(same arc \(\stackrel{\triangle}{\text { AC }}\) angles)

\[ \therefore \quad \triangle P A B \sim \triangle P C D \]

(AA similarly test)

ICSE Class 10 Mathematics 2020 Question 6(b) solution figure

(ii)

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

(Property of similar triangle)

\[ \begin{aligned} \frac{P A}{P C} & =\frac{A B}{C D} \\ \frac{7.5}{5} & =\frac{9}{C D} \end{aligned} \]

∴

\[ C D=\frac{9 \times 5}{7.5}=\frac{9 \times 5 \times 10}{75} \]

∴

\[ C D=6 \mathrm{~cm} \]

(iii)

\[ \begin{aligned} \frac{a r \triangle P A B}{a r \triangle P C D} & =\frac{A P^{2}}{C P^{2}}=\left(\frac{A P}{C P}\right)^{2} \\ & =\left(\frac{7.5}{5}\right)^{2} \\ & =\frac{9}{4} \end{aligned} \]
\(\therefore \operatorname{ar} \triangle P A B: \operatorname{ar} \triangle P C D=9: 4\)
c

From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. If the height of the tower is 20 m. Find :

(i) the height of the cliff.

(ii) the distance between the cliff and the tower. [4]

Answer & Solution
ICSE Class 10 Mathematics 2020 Question 6(c) solution figure

\(O P\) is the cliff and \(A B\) is the tower. In \(\triangle P B C\),

\[ \begin{aligned} \tan B & =\frac{C P}{B C} \\ \tan 45^{\circ} & =\frac{C P}{B C} \\ 1 & =\frac{C P}{B C} \end{aligned} \]

∴

\[ C P=B C \qquad \text{…(i)} \]

In \(\triangle O A P\),

\[ \begin{aligned} \tan A= & \frac{O P}{O A} \\ & \left(\because \angle Q P A=\angle P A O=60^{\circ}\right) \\ \tan 60^{\circ}= & \frac{O C+C P}{O A} \\ \sqrt{3}= & \frac{A B+C P}{O A} \\ \sqrt{3}= & \frac{20+C P}{B C} \quad(O A=B C) \\ \sqrt{3}= & \frac{20+C P}{C P} \quad(B C=C P) \\ \sqrt{3} C P= & 20+C P \\ (\sqrt{3}-1) C P= & 20 \end{aligned} \]
\[ \begin{aligned} C P & =\frac{20}{\sqrt{3}-1} \mathrm{~m} \\ C P & =\frac{20}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ C P & =\frac{20(\sqrt{3}+1)}{3-1} \\ & =10(\sqrt{3}+1) \mathrm{m} \\ C P & =10(1.732+1) \mathrm{m} \\ C P & =27.32 \mathrm{~m} \end{aligned} \]

(i) height of the cliff

\[ O P=O C+C P \]
\[ =20+27.32 \]

∴ Height of the cliff \(=47.32 \mathrm{~m}\)

(ii) Distance between the cliff and the tower

\[ \begin{aligned} & O A=B C=C P \\ & O A=27.32 \mathrm{~m} \end{aligned} \]
7
Question 7
ICSE Class 10 Mathematics 2020
a

Find the value of ‘ \(\boldsymbol{p}\) ‘ if the lines, \(\mathbf{5} \boldsymbol{x}-\mathbf{3} \boldsymbol{y}+\mathbf{2}=\mathbf{0}\) and \(6 x-p y+7=0\) are perpendicular to each other. Hence, find the equation of a line passing through (-2, -1) and parallel to \(\mathbf{6} \boldsymbol{x}-\boldsymbol{p} \boldsymbol{y}+\mathbf{7}=\) 0. [3]

Answer & Solution

Slope of line \(5 x-3 y+2=0\) is

\[ \begin{aligned} & =-\frac{\text { coefficent of } x}{\text { coefficent of } y} \\ & =\frac{-5}{-3}=\frac{5}{3} \end{aligned} \]

Slope of line \(6 x-p y+7=0\) is \(=\frac{-6}{-p}=\frac{6}{p}\) Lines are perpendicular each other ∴ Product of slopes = -1

\[ \begin{aligned} \frac{5}{3} \times \frac{6}{p} & =-1 \\ 3 p & =-30 \Rightarrow p=-10 \end{aligned} \]

∴

\[ p=-10 \]

Given line \(6 x-p y+7=0\) is \(6 x+10 y+7=0\) Equation of the line, parallel to \(6 x+10 y+7=0\) is \(6 x+10 y+k=0\) Since, line passes through (-2,-1),

\[ \begin{aligned} & \therefore 6(-2)+10(-1)+k=0 \\ & \quad-12-10+k=0 \Rightarrow k=22 \end{aligned} \]

Equation of the line parallel to \(6 x+10 y+7\) is \(6 x+10 y+22=0\) or \(3 x+5 y+11=0\)

b

Using properties of proportion find \(x: y\), given : [3]

\[ \frac{x^{2}+2 x}{2 x+4}=\frac{y^{2}+3 y}{3 y+9} \]
Answer & Solution
\[ \frac{x^{2}+2 x}{2 x+4}=\frac{y^{2}+3 y}{3 y+9} \]

(using compodendo dividendo \(\frac{a}{b}=\frac{c}{d}\)

\[ \left.\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}\right) \]
\[ \begin{aligned} \frac{x^{2}+2 x+2 x+4}{x^{2}+2 x-2 x-4} & =\frac{y^{2}+3 y+3 y+9}{y^{2}+3 y-3 y-9} \\ \frac{x^{2}+4 x+4}{x^{2}-4} & =\frac{y^{2}+6 y+9}{y^{2}-9} \\ \frac{(x+2)^{2}}{(x+2)(x-2)} & =\frac{(y+3)^{2}}{(y+3)(y-3)} \\ \frac{x+2}{x-2} & =\frac{y+3}{y-3} \end{aligned} \]

(using compodendo – dividendo)

\[ \begin{aligned} \frac{x+2+x-2}{x+2-x+2} & =\frac{y+3+y-3}{y+3-y+3} \\ \frac{2 x}{4} & =\frac{2 y}{6} \\ \frac{x}{2} & =\frac{y}{3} \\ \frac{x}{y} & =\frac{2}{3} \end{aligned} \]

∴

\[ x: y=2: 3 \]
c

In the given figure \(\boldsymbol{T} \boldsymbol{P}\) and \(\boldsymbol{T} \boldsymbol{Q}\) are two tangents to the circle with centre \(\boldsymbol{O}\), touching at \(\boldsymbol{A}\) and \(\boldsymbol{C}\) respectively. If \(\angle B C Q=55^{\circ}\) and \(\angle B A P=60^{\circ}\), find :

(i) \(\angle O B A\) and \(\angle O B C\)

(ii) \(\angle A O C\)

(iii) \(\angle A T C\) [4]

ICSE Class 10 Mathematics 2020 Question 7(c) figure
Answer & Solution

Given, \(\angle B C Q=55^{\circ}\) and \(\angle B A P=60^{\circ}\)

ICSE Class 10 Mathematics 2020 Question 7(c) solution figure

(i) ∵

\[ \angle P A O=90^{\circ} \]

(As PAT is the tangent) ∴

\[ \angle O A B=90^{\circ}-60^{\circ}=30^{\circ} \]

In \(\triangle O A B, O A=O B\) (radii of circle)

\[ \angle O B A=\angle O A B=30^{\circ} \]

Similarly

\[ \therefore \angle Q C O=90^{\circ} \]

(As QCT is the tangent) ∴

\[ \angle O C B=90^{\circ}-55^{\circ}=35^{\circ} \]

In \(\triangle O C B, O C=O B\) (radii of circle)

\[ \angle O B C=\angle O C B=35^{\circ} \]

(ii) Now,

\[ \begin{aligned} \angle A B C & =\angle O B A+\angle O B C \\ & =30^{\circ}+35^{\circ}=65^{\circ} \end{aligned} \]

Again, \(\angle A O C=2 \angle A B C=2 \times 65^{\circ}=130^{\circ}\) ( ∵ Angle subtended at the center of the circle by an arc is twice the angle subtended at the circle) (iii)

\[ \begin{aligned} \angle A O C+\angle A T C & =180^{\circ} \\ 130^{\circ}+\angle A T C & =180^{\circ} \\ \therefore \quad \angle A T C & =180^{\circ}-130^{\circ}=50^{\circ} \\ \therefore \quad \angle A T C & =50^{\circ} \end{aligned} \]
8
Question 8
ICSE Class 10 Mathematics 2020
a

What must be added to the polynomial \(\mathbf{2} \boldsymbol{x}^{\mathbf{3}}-\) \(\mathbf{3} \boldsymbol{x}^{\mathbf{2}}-\mathbf{8} \boldsymbol{x}\), so that it leaves a remainder 10 when divided by \(\mathbf{2} \boldsymbol{x}+\mathbf{1}\) ? [3]

Answer & Solution
\[ \begin{aligned} p(x) & =2 x^{3}-3 x^{2}-8 x \\ q(x) & =2 x+1 \\ \text { Remainder } & =10 \end{aligned} \]

Let \(k\) be added to get remainder 10 when divided by \((2 x+1)\).

\[ \begin{aligned} & p(x)=2 x^{3}-3 x^{2}-8 x+k \\ & \text { Remainder }=10 \\ & p\left(-\frac{1}{2}\right)=10 \\ & 2\left(-\frac{1}{2}\right)^{3}-3\left(-\frac{1}{2}\right)^{2}-8\left(-\frac{1}{2}\right)+k=10 \\ &-2 \times \frac{1}{8}-3 \times \frac{1}{4}+4+k=10 \\ &-1+4+k=10 \\ & k=7 \end{aligned} \]

Hence, 7 be added to get remainder 10 when divided by \(2 x+1\) to given polynomial.

b

Mr. Sona has a recurring deposit account and deposits ₹ 750 per month for 2 years. If he gets ₹ 19125 at the time of maturity, find the rate of interest. [3]

Answer & Solution

Maturity amount (M.A.) = ₹ 19125 Monthly deposit (P)= ₹ 750 Time \((n)=2\) years \(=24\) months

\[ \begin{aligned} & \text { M.A. }=P \times n+P \times \frac{r}{100} \times \frac{n(n+1)}{2} \times \frac{1}{12} \\ & 750 \times 24+\frac{750 \times r}{2400}(24)(24+1)=19125 \\ & 18000+\frac{750}{4} r=19125 \\ & \frac{750 r}{4}=19125-18000 \\ & r=\frac{1125 \times 4}{750}=\frac{4500}{750} \\ & \begin{array}{l} r=6 \\ \therefore \quad \text { Rate of interest }=6 \% \end{array} \end{aligned} \]
c

Use graph paper for this question. Take \(\mathbf{1 ~ c m}=\mathbf{1}\) unit on both \(\boldsymbol{x}\) and \(\boldsymbol{y}\) axes.

(i) Plot the following points on your graph sheets. \(A(-4,0), B(-3,2), C(0,4), D(4,1)\) and E \((7,3)\)

(ii) Reflect the points \(B, C, D\) and \(E\) on the \(x\)-axis and name them as \(\boldsymbol{B}^{\prime}, \boldsymbol{C}^{\prime}, \boldsymbol{D}^{\prime}\) and \(\boldsymbol{E}^{\prime}\) respectively.

(iii) Join the points \(\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}, \boldsymbol{D}, \boldsymbol{E}, \boldsymbol{E}^{\prime}, \boldsymbol{D}^{\prime}, \boldsymbol{C}^{\prime}, \boldsymbol{B}^{\prime}\) and \(\boldsymbol{A}\) in order.

(iv) Name the closed figure formed. [4]

Answer & Solution

(i)

ICSE Class 10 Mathematics 2020 Question 8(c) solution figure

(ii) B'(-3, -2), C'(0, -4), D'(4, -1) and E'(7, -3) respectively are the reflected points through the x-axis.

(iii) On joining the points A, B, C, D, E, E’, D’, C’, B’ and A in order, we get a closed figure.

(iv) Closed figure formed is nine sided polygon or nonagon, polygon fish or kite.

9
Question 9
ICSE Class 10 Mathematics 2020
a

40 Students enter for a game of shot-put competition. The distance thrown (in metres) is recorded below :

Distance in \(\boldsymbol{m}\)12-1313-1414-1515-1616-1717-1818-19
Number of Students39129421

Use a graph paper to draw an ogive for the above distribution. Use a scale of \(2 \mathrm{~cm}=1 \mathrm{~m}\) on one axis and 2 cm \(=5\) students on the other axis. Hence, using your graph find :

(iii) number of students who cover a distance which is above \(\mathbf{1 6} \frac{\mathbf{1}}{\mathbf{2}} \mathrm{m}\).

(i) the median.

(ii) upper qartile. [6]

Answer & Solution
ICSE Class 10 Mathematics 2020 Question 9(a) solution figure
Distance in ( \(\boldsymbol{m}\) )Number of studentsLess thanCumulative frequency
12-133133
13-1491412
14-15121524
15-1691633
16-1741737
17-1821839
18-1911940

(i) Median = 14.6± 0.2

(ii) Upper quartile \(=15.75 \pm 0.2\)

(iii) Number of students above \(16 \frac{1}{2} \mathrm{~m}=40-35=5\)

b
\[ \begin{aligned} & \text { If } x=\frac{\sqrt{2 a+1}+\sqrt{2 a-1}}{\sqrt{2 a+1}-\sqrt{2 a-1}} \text {, prove that } x^{2}-4 a x+1 \\ & =0 . \end{aligned} \]

[4]

Answer & Solution
\[ \begin{aligned} & x=\frac{\sqrt{2 a+1}+\sqrt{2 a-1}}{\sqrt{2 a+1}-\sqrt{2 a-1}} \\ & \frac{x+1}{x-1}=\frac{\sqrt{2 a+1}+\sqrt{2 a-1}+\sqrt{2 a+1}-\sqrt{2 a-1}}{\sqrt{2 a+1}+\sqrt{2 a-1}-\sqrt{2 a+1}+\sqrt{2 a-1}} \end{aligned} \]

(Apply compodendo-dividendo \(\frac{A}{B}=\frac{C}{D}\)

\[ \begin{aligned} & \left.\Rightarrow \frac{A+B}{A-B}=\frac{C+D}{C-D}\right) \\ \frac{x+1}{x-1}=\frac{2 \sqrt{2 a+1}}{2 \sqrt{2 a-1}}=\frac{\sqrt{2 a+1}}{\sqrt{2 a-1}} & \end{aligned} \]

Squaring both sides

\[ \begin{aligned} & \frac{(x+1)^{2}}{(x-1)^{2}}=\left(\frac{\sqrt{2 a+1}}{\sqrt{2 a-1}}\right)^{2} \\ & \frac{x^{2}+2 x+1}{x^{2}-2 x+1}=\frac{2 a+1}{2 a-1} \end{aligned} \]

Apply compodendo – dividendo

\[ \begin{aligned} \frac{x^{2}+2 x+1+x^{2}-2 x+1}{x^{2}+2 x+1-x^{2}+2 x-1} & =\frac{2 a+1+2 a-1}{2 a+1-2 a+1} \\ \frac{2\left(x^{2}+1\right)}{4 x} & =\frac{4 a}{2} \\ \frac{x^{2}+1}{2 x} & =\frac{2 a}{1} \\ x^{2}+1 & =4 a x \\ x^{2}-4 a x+1 & =0 \quad \text { Hence proved. } \end{aligned} \]
10
Question 10
ICSE Class 10 Mathematics 2020
a

If the \(\mathbf{6}^{\text {th }}\) term of an A.P. is equal to four times its first term and the sum of fist six terms is 75, find the first term and the common difference. [3]

Answer & Solution

Let the first term of an A.P. be \(a\) and common difference be \(d\) respectively

\[ \begin{aligned} a_{6} & =4 \times a \text { (given) } \\ a_{6} & =4 a \qquad \text{…(i)}\\ a+5 d & =4 a \quad\left\{a_{n}=a+(n-1) d\right\} \\ 3 a & =5 d \\ a & =\frac{5}{3} d \qquad \text{…(ii)}\\ S_{6} & =75 \text { (given) } \\ \frac{n}{2}\left[a+a_{6}\right] & =75 \\ \frac{6}{2}\left(a+a_{6}\right) & =75 \\ a+a_{6} & =\frac{75}{3} \\ a+4 a & =25 \\ a & =\frac{25}{5}=5 \\ a & =5 \end{aligned} \]

From equation (i)

\[ \begin{aligned} & & a & =\frac{5}{3} d \\ & & 5 & =\frac{5}{3} d \\ \therefore & & d & =\frac{5 \times 3}{5} \times 3 \\ \therefore & & d & =3 \end{aligned} \]

First term of an A.P. \(=5\) and common difference = 3

b

The difference of two natural numbers is 7 and their product is 450. Find the numbers. [3]

Answer & Solution

Let the numbers be \(x\) and \(y\). According the given condition

\[ \begin{aligned} x-y=7 & \Rightarrow y=x-7[x>y] \qquad \text{…(i)}\\ x y & =450 \qquad \text{…(i)} \end{aligned} \]

From (i) \& (ii)

\[ \begin{aligned} x(x-7) & =450 \\ x^{2}-7 x-450 & =0 \\ x^{2}-25 x+18 x-450 & =0 \end{aligned} \]
\[ \begin{aligned} x(x-25)+18(x-25) & =0 \\ (x-25)(x+18) & =0 \end{aligned} \]

If \(x+18=0 \Rightarrow x=-18\) it is not a natural number.

\[ \text { If } x-25=0 \Rightarrow x=25, y=25-7=18 \]

Thus, numbers 25 and 18.

c

Use ruler and compass for this question. Construct a circle of radius 4.5 cm . Draw a chord \(\boldsymbol{A} \boldsymbol{B}=\mathbf{6 ~ c m}\).

(i) Find the locus of points equidistant from \(\boldsymbol{A}\) and \(\boldsymbol{B}\).

Mark the point where it meets the circle as \(\boldsymbol{D}\).

(ii) Join \(\boldsymbol{A} \boldsymbol{D}\) and find the locus of points which are equidistant from \(A D\) and \(A B\). Mark the point where it meets the circle as \(\boldsymbol{C}\).

(iii) Join \(\boldsymbol{B} \boldsymbol{C}\) and \(\boldsymbol{C} \boldsymbol{D}\), Mesuare and write down the length of side \(\boldsymbol{C} \boldsymbol{D}\) of the quadrilateral \(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C} \boldsymbol{D}\). [4]

Answer & Solution
ICSE Class 10 Mathematics 2020 Question 10(c) solution figure

\(B C=C D=5.1 \mathrm{~cm}\) (Approx.)

11
Question 11
ICSE Class 10 Mathematics 2020
a

A model of a high rise building is made to a scale of 1 : 50.

(i) If the height of the model is 0.8 m, find the height of the actual building.

(ii) If the floor area of a flat in the building is \(\mathbf{2 0 ~ \mathbf { m } ^ { 2 }}\), find the floor area of that in the model. [3]

Answer & Solution

Scale \(=1: 50\) Height of the model 0.8 m ∴ Height of the building \(=0.8 \times 50 \mathrm{~m}=40 \mathrm{~m}\) Floor area of building \(=20 \mathrm{~m}^{2}\) Floor area of the model \(=\frac{20}{50 \times 50} \mathrm{~m}^{2}\)

\[ \begin{aligned} & =\frac{1}{125} \mathrm{~m}^{2} \\ & =0.008 \mathrm{~m}^{2} \end{aligned} \]

∴ Floor area of the model

\[ \begin{aligned} & =0.008 \times 10000 \mathrm{~cm}^{2} \\ & =80 \mathrm{~cm}^{2} \end{aligned} \]
b

From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Taking \(\pi=\frac{22}{7}\) find the volume of the remaining solid. [3]

ICSE Class 10 Mathematics 2020 Question 11(b) figure
Answer & Solution

Dimension of cylinder Height \((h)=28 \mathrm{~cm}\) Diameter \((2 r)=6 \mathrm{~cm}\)

ICSE Class 10 Mathematics 2020 Question 11(b) solution figure

Dimension of conical cavities Height \((\mathrm{H})=10.5 \mathrm{~cm}\) diameter \((2 r)=6 \mathrm{~cm}\) Volume of remaining solid = Volume of cylinder – 2 × Volume of cavity

\[ \begin{aligned} & =\pi r^{2} h-2 \times \frac{1}{3} \pi r^{2} H \\ & =\pi r^{2}\left(h-\frac{2}{3} \mathrm{H}\right) \\ & =\frac{22}{7} \times 3 \times 3\left(28-\frac{2}{3} \times 10.5\right) \\ & =\frac{22}{7} \times 3 \times 3(28-7) \\ & =\frac{22}{7} \times 3 \times 3 \times 21 \\ & =22 \times 3 \times 3 \times 3=594 \mathrm{~cm}^{3} \end{aligned} \]

Hence, volume of remaining solid \(=594 \mathrm{~cm}^{3}\)

c

Prove the identity [4]

\[ \left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\tan ^{2} \theta \]
Answer & Solution
\[ \begin{aligned} & \left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\tan ^{2} \theta \\ \text { L.H.S. } & =\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2} \\ & =\left(\frac{1-\tan \theta}{1-\frac{1}{\tan \theta}}\right)^{2} \end{aligned} \]
\[ =\left(\frac{1-\tan \theta}{\frac{\tan \theta-1}{\tan \theta}}\right)^{2} \]
\[ \begin{aligned} & =\left\{\frac{\tan \theta(1-\tan \theta)}{-(1-\tan \theta)}\right\}^{2} \\ & =\left\{\frac{\tan \theta}{-1}\right\}^{2}=\tan ^{2} \theta \end{aligned} \]

= R.H.S. Hence proved. \(\square\)

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