ICSE 2020 Maths Question Paper Solved for Class 10

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2020 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2020 Solved

Time Allowed (Two hours and a half)

General Instructions:

Answers to this Paper must be written on he paper provided separately.
You will Not be allowed to write during the first 15 munutes.
This time is to be spent in reading the question paper.
The time given at the head of this paper is the time allowed for writing the answers.
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 of parts of questions are given in brackets [ ].

Section – A (40 MARKS)
(Attempt all questions from this Section)

Question 1.
(a) Solve the following Quadratic Equation : x² – 7x + 3 = 0 [3]
Give your answer correct to two decimal places.
Solution:
Given quadratic equation is
x² – 7x + 3 = 0
By using quadratic formula, we have

(b) Given A = \(\left[\begin{array}{ll}
x & 3 \\
y & 3
\end{array}\right]\)
If A² = 31,where I is the identity matrix of order 2, find x and y. [3]
Solution:
A² = 3I
⇒ \(\left[\begin{array}{ll}
x & 3 \\
y & 3
\end{array}\right]\left[\begin{array}{ll}
x & 3 \\
y & 3
\end{array}\right]=3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
x^2+3 y & 3 x+9 \\
x y+3 y & 3 y+9
\end{array}\right]=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
Equating the corresponding terms, we have
3x + 9 = 0 and 3y + 9 = 3
3(x + 3) = 0 and 3y = 3 – 9
⇒ x + 3 = 0 and 3y = -6
⇒ x = -3 and y = -2

(c) Using ruler and compass construct a triangle ABC where AB = 3 cm, BC= 4cm and ∠ABC = 90°. Hence construct a circle circumscribing the triangle ABC. Measure and write down the radius of the circle. [4]
Solution:
Steps of Construction :

Draw a line segment BC of length 4 cm.
Through B, construct an angle of 90° and cut off BA = 3 cm.
Join AC.
Draw the perpendicular bisector of AC and let it intersect AC in O.
With O, as centre and radius equal to OC or OA, draw a circle circumscribing the ΔABC. Radius of the circle = 2.5 cm.

Question 2.
(a) Use factor theorem to factorise 6x³ + 17x² + 4x -12 completely. [3]
Solution:
Let f(x) = 6x³ + 17x² + 4x – 12
Foe x = -2, f(-2) = 6(-2)³ + 17(-2)² + 4(-2) – 12
= -48 + 68 – 8 – 12 = 0
⇒ x + 2 is a factor of f(x)
Now, by using long division, we obtain

∴ f(x) = 6x³ + 17x² + 4x – 12
= (x + 2) (6x² + 5x – 6)
= (x + 2) (6x² + 9x – 4x – 6)
= (x + 2) [3x (2x +3) – 2 (2x + 3)]
= (x + 2) (3x – 2) (2x + 3)

(b) Solve the following inequation and represent the solution set on the number line : [3]
\(\frac{3x}{2}\) + 2 < x + 4 ≤ \(\frac{x}{2}\) + 5, x ∈ R [3]
Solution:
Given that: \(\frac{3x}{2}\) + 2 < x + 4 ≤ \(\frac{x}{2}\) + 5, x ∈ R
⇒ \(\frac{3x}{2}\) < x + 4 and x + 4 ≤ \(\frac{x}{2}\) + 5
⇒ 3x + 10 < 5x + 20 and 2x + 8 ≤ x + 10
⇒ 10 – 20 < 5x – 3x and 2x – x ≤ 10 – 8
⇒ – 10 < 2x and x ≤ 2 .
⇒ -5 < x and x ≤ 2
∴ -5 < x ≤ 2
Solution set on number line is :

Weekly Wages (in ₹)	No. of People
3000 – 4000	4
4000 – 5000	9
5000 – 6000	18
6000 – 7000	6
7000 – 8000	7
8000 – 9000	2
9000 – 10000	4

Estimate the mode from the graph.
Solution:

We draw the histogram by taking 1 cm = 1000 units along x – axis and 1 cm = 2 units along y – axis. Now, inside the highest rectangle, we join DA and BC. Let DA and BC cut each other at M. From M, we draw the perpendicular on x – axis meeting it at N. We find that ON is equal to 5500 units approximately.
Hence, the required mode = ₹ 5500.

Question 3.
(a) In the given figure O is the centre of the circle and AB is a diameter. [3]
If AC = BD and ∠AOC = 72°. Find:
(i) ∠ABC
(ii) ∠BAD
(iii) ∠ABD

Solution:
Since angle subtended at the centre is double the angle subtended at the remaining part of the circle.
∴ ∠ABC = – \(\frac{1}{2}\) ∠AOC
= \(\frac{1}{2}\) × 72° = 36°
Also, AC =BD [given]
∴ ∠BAD = ∠ABC
[∠s subtended by equal chords]
= 36°
Again, AOB is a diameter of the circle
∴ ∠ADB = 90°
[∠ subtended in a semi-circle]
∠BAD + ∠ADB + ∠ABD = 180°
36° + 90° + ∠ABD = 180°
∠ABD = 180° – 36° – 90° = 54°

(b) Prove that : \(\frac{\sin A}{1+\cot A}-\frac{\cos A}{1+\tan A}\) = sin A = cos A [3]
Solution:
L.H.S.

= \(\frac{(\sin A-\cos A)(\sin A+\cos A)}{\sin A+\cos A}\)
= sin A – cos A
= R.H.S.

(c) In what ratio is the line joining P (5, 3) and Q(-5, 3) divided by the y-axis ? Also find the coordinates of the point of intersection. [4]
Solution :
Any point on y-axis is R(0, y).
R(0, y) = \(\left(\frac{-5 k+5}{k+1}, \frac{3 k+3}{k+1}\right)\)

⇒ 0 = \(\frac{-5 k+5}{k+1}\)
⇒ 5k = 5
⇒ k = 1.
Also, y = \(\frac{3(1)+3}{1+1}=\frac{6}{2}\) = 3
Hence, the required ratio is 1 : 1 and coordinates of the point of intersection are (0, 3).

Question 4.
(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]
Solution:
Let r be the radius of each marble.
∴ 64 × Volume of each marble = Volume of spherical ball
64 × \(\frac{4}{3}\) π r³ = \(\frac{4}{3}\) π (6)³
r³ = \(\frac{6^3}{64}=\left(\frac{6}{4}\right)^3\)
r = \(\frac{6}{4}\)
r = 1.5 cm
Hence, the radius of each 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]
Solution:
Sample space = {A,U,T, H, O, R, I, E, S, Z}
(i) P(a vowel) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
(ii) P (one of the first 9 letters of English alphabet)
= \(\frac{4}{10}\) = \(\frac{2}{5}\)
(iii) P (one of the last 9 letters of English alphabet)
= \(\frac{5}{10}\) = \(\frac{1}{2}\)

(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]
Solution:
Cost of Medicines = ₹ 950
GST = ₹ \(\frac{5}{100}\) × 950 = ₹ 47.5
Cost of pair of shoes = ₹ 3000
GST = ₹ \(\frac{18}{100}\) × 3000 = ₹ 540
Cost of Laptop bag = ₹ 1000
Discount = ₹ \(\frac{30}{100}\) × 1000 = ₹ 300
∴ Sale price of Laptop bag = ₹ (1000 – 300)
= ₹ 700
GST = ₹ \(\frac{18}{100}\) × 700
= ₹ 126
Total amount of GST paid = ₹ (47.5 + 540 + 126)
= ₹ 713.5
Total bill amount including GST paid
= ₹ (950 + 3000 + 700 + 713.5 )
= ₹ 5363.5

Section – B (40 Marks)

Question 5.
(a) A company with 500 shares of nominal value ₹ 120 declares an annual dividend.of 15%. Calculate: [3]
(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 (c) is 10%, find the market value of each share.
Solution:
No. of shares = 500
Nominal value of a share = ₹ 120
Annual dividend per share = 15%
∴Total amount of dividend paid
= \(\frac{15}{100}\) × 120 × 500
= ₹ 9000
Annual income of Mr. Sharma
= ₹ \(\frac{15}{100}\) × 120 × 80
= ₹ 1440
Mr. Sharma received only 10% on his investment
∴10% of investment per share
= 15% of ₹ 120
\(\frac{10}{100}\) × x = \(\frac{15}{100}\) × 120
x = ₹ 180
∴ Mr. Sharma paid for each share = ₹ 180.

(b) The mean of the following data is 16. Calculate the value of f. [3]

Marks	5	10	15	20	25
No. of Students	3	7	f	9	6

Solution:
Here, \(\frac{5 \times 3+10 \times 7+15 \times f+20 \times 9+25 \times 6}{3+7+f+9+6}\) = 16
\(\frac{15+70+15 f+180+150}{25+f}\) = 16
415 + 15f = (25 + f) 16
415 + 15f = 400 + 16f
f = 15

(c) The 4th, 6th 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]
Solution:
Let a and r be the first term and common ratio of given GP.
∴ ar³ = 10 ……………… (i)
ar⁵ = 40 ……………… (ii)
ar^n-1 = 640 ……………………. (iii)
Dividing (ii) by (i), we obtain
\(\frac{a r^5}{a r^3}=\frac{40}{10}\)
⇒ r² = 4
⇒ r = 2 [∵ common ratio is + ve]
From (i), we have
a(2)³ = 10
a(8) = 10
a = \(\frac{10}{8}\) = \(\frac{5}{4}\)
From (iii), we have \(\frac{5}{4}\) (2)^n-1 = 640
2^n-1 = 640 × \(\frac{4}{5}\) = 512
2^n-1 = 2⁹
n – 1 = 9
n = 9 + 1 = 10

Question 6.
(a) If A = \(\left[\begin{array}{ll}
3 & 0 \\
5 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-4 & 2 \\
1 & 0
\end{array}\right]\), find A² – 2AB + B² [3]
Solution:

(b) In the given figure, AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P.
(i) Prove that ∆PAB ~ ∆PCD.
(ii) Find the length of CD,
(iii) Find area of ∆PAB : area of ∆PCD: [3]

Solution:
In ∆PAB and ∆PCD
∠APB = ∠CPD [∠s is the same segment]
∠APB = ∠CPD [vertically opp. ∠s]
∴ ∆PAB ~ ∆PCD [by AA similarity rule]
⇒ \(\frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{CP}}{\mathrm{AP}}\)
⇒ \(\frac{\mathrm{CD}}{9}=\frac{5}{7.5}\)
⇒ CD = \(\frac{9 \times 5}{7.5}\) = 6 cm
\(\frac{{ar}(\triangle \mathrm{PAB})}{{ar}(\triangle \mathrm{PCD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2}=\left(\frac{9}{6}\right)^2=\frac{9}{4}\)
ar (∆PAB) : ar (∆PCD) = 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]
Solution:
Let AB and CD be the tower and cliff, such that

∠CAE = 45°, ∠CBD = 60°,
AB = 20 m and CD = h m
CE = (h – 20) m
In rt. ∠ed ∆BDC,
\(\frac{\mathrm{CD}}{\mathrm{BD}}\) = tan 60°
CD = \(\sqrt{3}\) BD
h = \(\sqrt{3}\) BD
or BD = \(\frac{h}{\sqrt{3}}\) ……………. (i)
In rt. ∠ed ∆AEC,
\(\frac{\mathrm{CE}}{\mathrm{AE}}\) = tan 45°
CE = AE (1)
h – 20 = \(\frac{h}{\sqrt{3}}\) [∵ AE = BD = \(\frac{h}{\sqrt{3}}\)]
\(\sqrt{3}\) h – 20\(\sqrt{3}\) = h
(\(\sqrt{3}\) – 1)h = 20\(\sqrt{3}\)
h = \(\frac{20 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)
= \(\frac{20(3+\sqrt{3})}{2}\)
h = 10 (3 + \(\sqrt{3}\)) m
BD = \(\frac{h}{\sqrt{3}}=\frac{10(3+\sqrt{3})}{\sqrt{3}}\)
= 10(\(\sqrt{3}\) + 1) m
Hence, the height of the cliff is 10(3 + \(\sqrt{3}\)) m and distance between the cliff and the tower is 10 (\(\sqrt{3}\) +1) m.

Question 7.
(a) Find the value of ‘p’ if the lines, 5x – 3y + 2 = 0 and 6x – py + 7 = 0 are perpendicular to each other. Hence find the equation of a line passing through (-2,-1) and parallel to 6x – py + 7 = 0. [3]
Solution:
Here, given lines are 5x – 3y + 2 = 0 and bx – py + 7 = 0
or y = \(\frac{5}{3}\) x + \(\frac{2}{3}\) and y = \(\frac{6}{p}\) x + \(\frac{7}{p}\)
Since these two lines are perpendicular.
∴ \(\left(\frac{5}{3}\right)\left(\frac{6}{p}\right)\) = -1
⇒ 3p = -30 ⇒ p = – 10
Now, equation of the line passing through (-2, -1) and parallel to 6x – py + 7 = 0 or 6x + 10y + 7 = 0 is
y + 1 = – \(\frac{6}{10}\) (x + 2)
5y + 5 = -3x – 6
3x + 5y + 11 = 0

(b) Using properties of proportion find x: y, given:
\(\sqrt{3}\)
Solution:
Given that: \(\frac{x^2+2 x}{2 x+4}=\frac{y^2+3 y}{3 y+9}\)
Applying componendo and dividendo, we have
\(\frac{x^2+2 x+2 x+4}{x^2+2 x-2 x-4}=\frac{y^2+3 y+3 y+9}{y^2+3 y-3 y-9}\)

Hence, x : y = 2 : 3

(c) In the given figure, TP and TQ are two tangents to the circle with centre O, touching at A and C respectively. If ∠BCQ = 55° and ∠BAP = 60°, find: [4]
(i) ∠OBA and ∠OBC
(ii) ∠AOC
(iii) ∠ATC

Solution:
∠OCQ = 90°
∠OCB + ∠BCQ = 90°
∠OCB + 55° = 90°
∠OCB = 90° – 55° = 35°
In ∆OBC, OC = OB
= radii of same circle
∴ ∠OBC = ∠OCB = 35°
Again, ∠OAP = 90°
∠OAB + ∠BAP = 90°
∠OAB + 60° = 90°
∠OAB = 90°- 60° = 30°
In ∆OBA, OB = OA
= radii of same circle
∴ ∠OBA = ∠OAB = 30°
Now, ∠ABC = ∠OBA + ∠OBC
= 30°+ 35° = 65°
∠AOC = 2 ∠ABC
[∠ subtended at the centre is double the ∠ subtended at the remaining part of the circle]
= 2(65°)
= 130°
In quad. ∆OCT
∠OAT + ∠AOC + ∠OCT + ∠ATC = 360°
90° + 130° + 90° + ∠ATC = 360°
∠ATC = 360° – 130° – 180°
= 50°
Hence, ∠OBA = 30°, ∠OBC = 35°,
∠AOC = 130° and ∠ATC = 50°.

Question 8.
(a) What must be added to the polynomial 2x³ – 3x² – 8x, so that it leaves a remainder 10 when divided by 2x + 1? [3]
Solution:
Let f(x) = 2x³ – 3x² – 8x
Put 2x + 1 = 0 ⇒ x = – \(\frac{1}{2}\)
∴ Remainder = f\(\left(-\frac{1}{2}\right)\)
= 2\(\left(-\frac{1}{2}\right)^3\) – 3\(\left(-\frac{1}{2}\right)^2\) – 8 \(\left(-\frac{1}{2}\right)\)
= – 2 × \(\frac{1}{8}\) – 3 × \(\frac{1}{4}\) + 8 × \(\frac{1}{2}\)
= –\(\frac{1}{4}\) – \(\frac{3}{4}\) + 4
= \(\frac{-1-3}{4}\) + 4
= -1 + 4 = 3
But, remainder = 10
Hence, the required number added is 10 – 3 i.e., 7.

(b) Mr. Sonu 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]
Solution:
Here, P = ₹ 750, T = 2 years
n = 2 × 12 = 24
Maturity amount = ₹ 19125
Let the rate of interest per annum be r %
So, A = n × P + P × \(\frac{n(n+1) \times r}{2 \times 12 \times 100}\)
19125 = 24 × 750 + 750 × \(\frac{24(24+1)}{24}\) × \(\frac{r}{100}\)
= 18000 + \(\frac{750 \times 25}{100}\) × r
19125 – 18000 = \(\frac{375}{2}\) × r
\(\frac{1125 \times 2}{375}\) = r
6 = r
Hence, the rate of interest is 6% per annum.

(c) Use graph paper for this question. [4]
Take : 1 cm = 1 unit on both x and y axes.
(i) Plot the following points on your graph sheets : A(-4, 0), B(-3, 2), C(0, 4), D(4, 1) andE(7, 3)
(ii) Reflect the points B, C, D and E on the x-axis and name them as B’, C’, D’ and E’ respectively.
(iii) Join the points A, B, C, D, E’, D’, B’ and A’ in order.
(iv) Name the closed figure formed.
Solution:
Fish

Question 9.
(a) 40 students enter for a game of shot-put competition. The distance thrown (in metres) is recorded below: [6]

Distance in m	12 – 13	13 – 14	14 – 15	15 – 16	16 – 17	17 – 18	18 – 19
Number of Students	3	9	12	9	4	2	1

Use a graph paper to draw an ogive for the above distribution.
Use a scale of 2 cm = 1 m on one axis and 2 cm = 5 students on the other axis.
Hence using your graph find :
(i) the median
(ii) Upper Quartile
(iii) Number of students who cover a distance which is above 16 \(\frac{1}{2}\)m .
Solution:

Distance (in m)	No. of Students	cf
12-13	3	Less than 13	3
13-14	9	Less than 14	12
14-15	12	Less than 15	24
15-16	9	Less than 16	33
16-17	4	Less than 17	37
17-18	2	Less than 18	39
18-19	1	Less than 19	40

Points (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39) and (19, 40) are plotted, joined free-hand to get a smooth curve.
It is a less than type ogive. Here, N = 40 and \(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{40}{2}\) = 20
Take OP = 20 units, draw PQ || to x-axis meeting the ogive at Q, draw QR||OP.
Median = 14.65 m
Upper Quartile = 15.5 m
No. of students who cover a distance above 16 \(\frac{1}{2}\) m
= 40 – 36 = 4

(b) If x = \(\), prove that x² – 4ax + 1 = 0 [4]
Solution:

Question 10.
(a) If the 6th term of an AP is equal to four times its first term and the sum of first six terms is 75, find the first term and the common difference. [3]
Solution:
Let a and d be the first term and common difference.
a₆ = 4a
⇒ a + 5d = 4a
⇒ 3 a = 5 d …………….. (i)
S₆ = 75
⇒ \(\frac{6}{2}\)(2a + (6 -1)d) = 75
⇒ 3 (2a + 5d) = 75
2a + 3a = \(\frac{75}{3}\) [using (i)]
5a = 25
a = 5
From (i) 5 d = 3(5) ⇒ d = 3
Hence, first term is 5 and common difference is 3.

(b) The difference of two natural numbers is 7 and their product is 450. Find the numbers. [3]
Solution:
Let the two natural numbers be x and y such that x > y.
x – y = 1 ⇒ x = y + 1 …………. (i)
and xy = 450 ………….. (ii)
From (i) and (ii), we have
(y + 7)y = 450
y² + 7y – 450 = 0
y² + 25y – 18y – 450 = 0
y(y + 25) – 18(y + 25) = 0
(y – 18) (y + 25) = 0
⇒ y = 18 or y = -25
Since x andy are natural numbers.
∴ y = 18 and x = y + 7 = 18 + 7 = 25
Hence, the required numbers are 25 and 18.

(c) Use ruler and compass for this question. Construct a circle of radius 4.5 cm. Draw a chord AB = 6 cm. [4]
(i) Find the locus of points equidistant from A and B. Mark the point where it meets the circle as D.
(ii) Join AD and find the locus of points which are equidistant from AD and AB. Mark the point where it meets the circle as C.
(iii) Join BC and CD. Measure and write down the length of side CD of the quadrilateral ABCD.
Solution:

Steps of Construction :

Draw a circle of radius 4.5 cm and centre O.
Draw a chord AB of length 6 cm.
Draw the perpendicular bisector of the chord AB and let it intersect the circle in D.
Join AD and BD.
Draw the perpendicular bisector of the chord BD and let it intersect the circle in C.
Join BC and CD.
On measurement of side CD of the quadrilateral ABCD is ; CD = 5 cm.

Question 11.
(a) A model of a high rise building is made to a scale of 1:50. [3]
(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 20 m², find the floor area of that in the model.
Solution:
Here, the scale factor = k = \(\frac{1}{50}\)
(i) The height of the model = 0.8 m
Also, height of the model = k (actual height of the building)
0.8 = \(\frac{1}{50}\) × actual height of the building
∴ Actual height of the building
= 0.8 × 50 = 40 m

(ii) Area of the floor in the model
= k² (actual floor area)
= \(\left(\frac{1}{50}\right)^2\) × 20 m²
= \(\frac{20}{2500}\) = 0.008 m²
∴ The floor area of the flat in the model is 0.008 m²

(b) From a solid wooden cylinder of height 28 cm and diameter 6 cm, two concial cavities are hollowed out. The diameters of the cones are also of 6cm and height 10.5 cm. [3]

Taking π = \(\frac{22}{7}\), find the volume of the remaining solid.
Solution:
Radius of the solid (R) = \(\frac{6}{2}\) = 3 cm
Height of the solid (H) = 28 cm
Radius of each conical cavities (r) = \(\frac{6}{2}\) = 3 cm
Height of each conical cavities (h) = 10.5 cm
Volume of the remaining solid = Volume of wooden cylinder – 2 × Volume of conical cavity
= πR²H – \(\frac{2}{3}\) π r²h
= \(\frac{22}{7}\) × 3 × 3 × 28 – \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 10.5
= 22 × 9 × 4 – 2 × 22 × 3 × 1.5
= 792 – 198 = 594 cm³