Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2019 is the best way to boost your preparation for the board exams.
ICSE Class 10 Maths Question Paper 2019 Solved
Section – A (40 MARKS)
(Attempt all questions from this Section)
Question 1.
(a) Solve the following inequation and write down the solution set: [3]
11x – 4 < 15x + 4 ≤ 13x + 14, x ∈ W
Represent the solution on a real number line.
Solution:
11x – 4 ≤ 15x + 4 ;
15x + 4 ≤ 13x + 4, x ∈ W
-4 – 4 < 15x – 11x;
15x – 13x ≤ 4 – 4
– 8 < 4x ;
2x <0
– 2 < x
x ≤ 0
Thus, x = 0, x ∈ W
(b) A man invests ₹ 4500 in shares of a company which is paying 7.5% dividend. [3]
If ₹ 100 shares are available at a discount of 10%. Find :
(i) Number of shares he purchases.
(ii) His annual income.
Solution:
Total investment = ₹ 4500
Face value of a share = ₹ 100
Discount = 10%
Market value of a share = ₹ (100 – 10)
= ₹ 90
Now, Number of shares purchased
= \(\frac{4500}{90}\)
= 60
Annual income = ₹ 7.5%
of 50 × 100= ₹ \(\frac{7.5}{100}\) × 5000
= ₹ 375
(c) In a class of 40 students, marks obtained by the students in a class test (out of 10) are given below: [4]
Calculate the following for the given distribution :
| Marks | Number of Students |
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
| 4 | 3 |
| 5 | 6 |
| 6 | 10 |
| 7 | 5 |
| 8 | 4 |
| 9 | 3 |
| 10 | 3 |
(i) Median
(ii) Mode
Solution:
| Marks | Number of Students | Cumulative |
| 1 | 1 | 1 |
| 2 | 2 | 3 |
| 3 | 3 | 6 |
| 4 | 3 | 9 |
| 5 | 6 | 15 |
| 6 | 10 | 25 |
| 7 | 5 | 30 |
| 8 | 4 | 34 |
| 9 | 3 | 37 |
| 10 | 3 | 40 |
| Total | 40 |
Here, N = 40 and \(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{40}{2}\) = 20
Marks corresponding to cumulative frequency 20 is 6.
Thus, the required median is 6.
Clearly, 6 occurs 10 times which is maximum.
Hence, mode is 6.
Question 2.
(a) Using the factor theorem, show that (x – 2) is a factor of x3 + x2 – 4x – 4. [3]
Hence, factorise the polynomial completely.
Solution:
Given polynomial is p(x) = x3 + x2 – 4x – 4
x – 2 is its factor, if p(2) = 0
∴ p(2) = (2)3 + (2)2 – 4(2) – 4
= 8 + 4 – 8 – 4
= 0
Thus, x – 2 is a factor of p(x).
Now, x3 + x2 – 4x – 4 = x2(x + 1) – 4(x + 1)
=(x + 1)(x2 – 4)
=(x + 1)(x + 2)(x – 2)
Hence, the required factors are (x + 1), (x + 2) and (x – 2).
(b) Prove that: (cosec θ – sin θ) (sec θ – cos θ)(tan θ + cot θ) = 1 [3]
Solution:
L.H.S. = (cosec θ – sin θ)(sec θ – cos θ)(tan θ + cot θ)
= 1 = R.H.S.
(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the : [4]
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Solution:
Here, a4 = 8
⇒ a + 3d = 8 ……………… (i)
a6 = 14
⇒ a + 5d = 14 …………….. (ii)
Subtracting (i) from (ii), we have
2d = 6
⇒ d = 3
From (i), we have
a + 3(3) = 8
⇒ 0 = 8 – 9 = -1
Now, Sn = \(\frac{\mathrm{N}}{\mathrm{2}}\) [2a + (n – 1)d]
S20 = \(\frac{20}{2}\) [2(-1) + (20 – 1)(3)]
= 10 [-2 + 57]
= 10 [55]
= 550
Hence, first term is -1, common difference is 3 and sum of the first 20 terms is 550.
Question 3.
(a) Simplify : sin A \(\left[\begin{array}{ll}
\sin A & -\cos A \\
\cos A & \sin A
\end{array}\right]\) + cos A \(\left[\begin{array}{ll}
\cos A & \sin A \\
-\sin A & \cos A
\end{array}\right]\) [3]
Solution:
(b) M and N are two points on the X axis and Y axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. [3]
Find: (i) the coordinates of M and N
(ii) slope of the line MN.
Solution:
Let the coordinates of M and N be (x, 0) and (0, y)
P(3, 2) = P \(\left(\frac{2 \times 0+3 x}{2+3}, \frac{2 y+3 \times 0}{2+3}\right)\)
= P \(\left(\frac{3 x}{5}, \frac{2 y}{5}\right)\)
∴ 3 = \(\frac{3 x}{5}\) ⇒ x = 5 and 2 = \(\frac{2 y}{5}\) ⇒ y = 5
Thus, the coordinates of M and N are M(5, 0) and N(0, 5).
Slope of the line MN = \(\frac{y_2-y_1}{x_2-x_1}=\frac{y-0}{0-x}\)
= – \(\frac{y}{x}\) = – \(\frac{5}{5}\) – 1
Hence, the slope of the line MN is -1.
(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the: [4]
(i) radius of the cylinder .
(ii) curved surface area of the cylinder Take π = 3.1
Solution:
Radius of metallic sphere (R) = 6 cm
Height of cylinder (h) = 32 cm
∴ Volume of cylinder= Volume of metallic sphere
πr2h = \(\frac{4}{3}\) πR3
⇒ r2(32) = \(\frac{4}{3}\) (6)3
⇒ r2 = \(\frac{4}{3} \times \frac{6 \times 6 \times 6}{32}\) = 9
∴ r = 3 cm
Curved surface area of the cylinder = 2πrh
= 2 × 3.1 × 3 × 32 = 595.2 cm2
Question 4.
(a) The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. Find K. [3]
Solution:
Here, \(\frac{\mathrm{K}+3}{\mathrm{~K}+2}=\frac{3 \mathrm{~K}-7}{2 \mathrm{~K}-3}\)
⇒ (K + 3) (2K – 3) = (K + 2) (3K – 7)
⇒ 2K2 – 3K + 6K – 9 = 3K2 – 7K + 6K – 14
⇒ K2 – 4K – 5 = 0
⇒ (K – 5) (K + 1) = 0
⇒ K = 5 or K = -1
(b) Solve for x the quadratic equation x2 – 4x – 8 = 0 [3]
Give your answer correct to three significant figures.
Solution:
Given quadratic equation is x2 – 4x – 8 = 0
By using quadratic formula, we have
∴ x = \(\frac{-(-4) \pm \sqrt{(-4)^2-4(1)(-8)}}{2(1)}\)
= \(\frac{4 \pm \sqrt{16+32}}{2}\)
= \(\frac{4 \pm \sqrt{48}}{2}=\frac{4 \pm 4 \sqrt{3}}{2}\) = 2 ± 2\(\sqrt{3}\)
= 2(1 ± \(\sqrt{3}\)) = 2(1 ± 1.73205)
= 2(2.73205) or 2(-0.73205)
= 5.46410 or -1.4641
= 5.46 or -1.46
(c) Use ruler and compass only for answering this question. [4]
Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre. Construct two tangents to the circle from the external point P.
Measure and write down the length of any one tangent.
Solution:
Steps of Construction :
- Draw a circle of radius 4 cm and centre O.
- Draw a radius and produce it to P, such that OP = 7 cm.
- Bisect OP at M.
- With M as centre and MP as radius, draw a circle to intersect the given circle at Q and R.
- Join PQ and PR.
PQ and PR are the required tangents and length of the tangents is 5.74 cm.
Section – B (40 MARKS)
(Attempt any four questions)
Question 5.
(a) There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. [3]
Find the probability that the number on the disc is :
(i) an odd number
(ii) divisible by 2 and 3 both
(iii) a number less than 16.
Solution:
Sample space = 25 discs numbered from 1 to 25.
(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25
Probability (an odd number) = \(\frac{13}{24}\)
(ii) Numbers divisible by 2 and 3 both are 6, 12, 18, 24
Probability (divisible by 2 and 3 both) = \(\frac{4}{25}\)
(iii) Numbers less than 16 are 1 to 15
Probability (a no. less than 16) = \(\frac{15}{25}\) or \(\frac{3}{5}\)
(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives ₹ 441 as interest at the time of maturity.
Find the amount Rekha deposited each month. [3]
Solution:
Here, n = 20, R = 9% p.a., Interest = ₹ 441
Let the monthly deposit be of ₹ x.
∴ x\(\left(\frac{20 \times 21}{2 \times 12}\right)\) × \(\frac{9}{100}\) = 441
x = \(\frac{441 \times 100 \times 24}{20 \times 21 \times 9}\)
= 280
Hence, the monthly deposit is ₹ 280.
(c) Use a graph sheet for this question. [4]
Take 1 cm = 1 unit along both x and y axis.
(i) Plot the following points :
A(0, 5), B(3, 0), C(1, 0) and D(1, -5)
(ii) Reflect the points B, C and D on the y-axis and name them as B’, C’, D’ respectively.
(iii) Write down the coordinates of B’, C’ and D’.
(iv) Join the points A, B, C, D, D’, C’, B’, A in order and give a name to the closed figure ABCDD’C’B’.
Solution:
(i) and (ii)
(iii) B’ (-3, 0), C'(-1, 0) and D'(-1, -5)
(iv) Arrow Head.
Question 6.
(a) In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm. [3]
(i) Prove that ∆PQR ~ ∆PST.
(ii) Find Area of ∆PQR : Area of quadrilateral SRQT.
Solution:
In rt. ∠ed ∆PQR and ∆PST
∠P = ∠P [common]
∠Q = ∠S [each = 90°]
∴ ∆PQR ~ ∆PST [by AA similarity rule]
(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find: [3]
(i) ‘n’ the number of terms of the G.P.
(ii) Sum of the n terms.
Solution:
Given that, a = 3 and an = 96, r = 2
∴ arn – 1 = 96
⇒ 3(2)n – 1 = 96
⇒ (2)n – 1 = 32 = 25
⇒ n – 1= 5
⇒ n = 6
Now, Sn = \(\frac{a\left(r^n-1\right)}{r-1}\), r > 1
= \(\frac{3\left(2^n-1\right)}{2-1}\)
= 3(2n – 1)
(c) A hemispherical and conical hole is scooped out of a solid wooden cylinder. [4]
Find the volume of the remaining solid where the measurements are as follows :
The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm.
Give your answer correct to the nearest whole number. Take π = \(\frac{22}{7}\)
Solution:
Given that:
Radius of each of hemisphere, cone and cylinder (r) = 3 cm
Height of cylinder = 7 cm
Height of cone = 3 cm
Volume of remaining solid
= Vol. of cylinder – Vol. of cone – Vol. of hemisphere
= πr2h \(\frac{1}{3}\) πr2h – \(\frac{2}{3}\) πr3
= πr2 [h – \(\frac{\mathrm{h}}{\mathrm{3}}\) – \(\frac{\mathrm{2r}}{\mathrm{3}}\)]
= \(\frac{22}{7}\) × 3 × 3 [7 – \(\frac{3}{2}\) – \(\frac{2}{3}\) × 3]
= \(\frac{198}{7}\) × 4
= 113.14 cm3
Question 7.
(a) In the given figure, AC is a tangent to the circle with centre O.
If ∠ADB = 55°, find x and y. Give reasons for your answers. [3]
Solution:
We know that angle between the radius and the tangent at the point of contact is right angle.
∴ ∠A = 90°
Also, in ∆OBE, OB = OE = radius (r)
∴ ∠B = ∠OEB ………….. (i)
In ∆ABD,
∠A + ∠B + ∠ADB = 180°
90° + ∠B + 55° = 180°
∠B = 180° – 90° – 55°
= 35°
Thus, ∠B = ∠OEB = 35°
∠OEB = ∠DEC = 35° [vertically opp. ∠s]
∠EDC + ∠ADE = 180°
∠EDC + 55° = 180°
⇒ ∠EDC = 180° – 55°
= 125°
In ∆EDC,
∠DEC + ∠EDC + x = 180°
35° +125° + x = 180°
⇒ x = 180° – 35° – 125°
= 20°
In ∆AOC, ∠A + y° + x° = 180°
⇒ 90° + y° + 20° = 180°
⇒ y° = 180° – 90° – 20°
= 70°
Hence, x – 20° and y – 70°.
(b) The model of a building is constructed with the scale factor 1: 30. [3]
(i) If the height of the model is 80 cm, find the actual height of the building in metres.
(ii) If the actual volume of a tank at the top of the building is 27m3, find the volume of the tank on the top of the model.
Solution:
Here, scale factor (k) = \(\frac{1}{30}\)
(i) Height of the model = k (Actual height of the building)
⇒ 80 cm = \(\frac{1}{30}\) (Actual height of the building)
⇒ Actual height of the building = 30 × 80 = 2400 cm
(ii) Volume of the tank at the top of the model
= k3 (Actual volume of the tank)
⇒ Volume of the tank at the top of the model
= \(\left(\frac{1}{30}\right)^3\) × 27 m3 = 0.001 m3
(c) Given \(\left[\begin{array}{ll}
4 & 2 \\
-1 & 1
\end{array}\right]\) M = 6I, where M is a matrix and I is unit matrix of order 2 × 2.
(i) State the order of matrix M.
(ii) Find the matrix M. [4]
Solution:
= 4 – 6
= -2
Hence, M = \(\left[\begin{array}{rr}
1 & -2 \\
1 & 4
\end{array}\right]\)
Question 8.
(a) The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference. [3]
Solution:
Let the first three terms of an A.P. be a – d, a and a + d.
According to the statement, we have
a – d + a + a + d = 42
3 a = 42
a = 14
Now, (a – d) (a + d) = 52
a2 – d2 = 52
142 – d2 = 52
⇒ d2 = 196 – 52 = 144
⇒ d = ± 12
Hence, the first term is 14 and common difference is +12.
(b) The vertices of a ∆ABC are A (3, 8), B(-1, 2) and C(6, -6). Find : [3]
(i) Slope of BC.
(ii) Equation of a line perpendicular to BC and passing through A.
Solution:
Vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, – 6)
Slope of BC = \(\frac{-6-2}{6+1}=\frac{-8}{7}\)
Slope of the line perpendicular to BC = \(\frac{7}{8}\)
Now, equation of the line perpendicular to BC and passing through A is
y – 8 = \(\frac{7}{8}\) (x – 3)
8y – 64 = 7x – 21
⇒ 7x – 8y + 43 = 0
(c) Using ruler and a compass only construct a semi-circle with diameter BC = 7 cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure ∠ADC and write it down. [4]
Solution:
Steps of Construction :
- Draw a line segment BC = 7 cm.
- Draw its perpendicular bisector l and let it intersect BC in M.
- With M as centre and radius equal to BM or CM, draw a semi-circle and let the semi-circle intersect the perpendicular bisector of line segment BC in A. Join BA.
- Draw the angle bisector of ∠ABC and let it intersect the semi-circle in D.
- Join AD and CD.
Hence, ∠ADC = 135°
Question 9.
(a) The data on the number of patients attending a hospital in a month are given below. [3]
Find the average (mean) number of patients attending the hospital in a month by using the shortcut method. Take the assumed mean as 45. Give your answer correct to 2 decimal places.
| Number of Patients | Number of Days |
| 10 – 20 | 5 |
| 20 – 30 | 2 |
| 30 – 40 | 7 |
| 40 – 50 | 9 |
| 50 – 60 | 2 |
| 60 – 70 | 5 |
Solution:
Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\sum f_i d_i}{\sum f_i}\)
Mean (\(\overline{\mathrm{X}}\)) = 45 + \(\frac{-140}{30}\)
= 45 – 4.67
= 40.33
(b) Using properties of proportion solve for x, given [3]
\(\frac{\sqrt{5 x}+\sqrt{2 x-6}}{\sqrt{5 x}-\sqrt{2 x-6}}\) = 4
Solution:
\(\frac{\sqrt{5 x}+\sqrt{2 x-6}}{\sqrt{5 x}-\sqrt{2 x-6}}=\frac{4}{1}\)
Applying componendo and dividendo, we have
\(\frac{\sqrt{5 x}+\sqrt{2 x-6}+\sqrt{5 x}-\sqrt{2 x-6}}{\sqrt{5 x}+\sqrt{2 x-6}-\sqrt{5 x}+\sqrt{2 x-6}}=\frac{4+1}{4-1}\)
\(\frac{2 \sqrt{5 x}}{2 \sqrt{2 x-6}}=\frac{5}{3}\)
\(\frac{\sqrt{5 x}}{\sqrt{2 x-6}}=\frac{5}{3}\)
Squaring both sides, we obtain
\(\frac{5 x}{2 x-6}=\frac{25}{9}\)
45x = 50x – 150
50x – 45x = 150
5x = 150
x = 30
(c) Sachin invests ₹ 8500 in 10%, ₹ 100 shares at ₹ 170. He sells the shares when the price of each share rises by ₹ 30. He invests the proceeds in 12% ₹ 100 shares at ₹ 125. Find: [4]
(i) the sale proceeds.
(ii) the number of ₹ 125 shares he buys.
(iii) the change in his annual income.
Solution:
Total investment = ₹ 8500
Market value of each share = ₹ 170
Number of shares purchased = \(\frac{8500}{170}\) = 50
Dividend received = ₹ \(\frac{10}{100}\) × 50 × 100 = ₹ 500
Now, market value of each share
= ₹ (170 + 30) = ₹ 200
Amount received on selling = ₹ (50 × 200)
= ₹ 10000
Market value of new shares = ₹ 125 each
Number of shares purchased = \(\frac{10000}{125}\) = 80
Dividend received = ₹ \(\frac{12}{100}\) × 80 × 100
= ₹ 960
Change in income = ₹ (960 – 500)
= ₹ 460
Question 10.
(a) Use graph paper for this question. [6]
The marks obtained by 120 students in an English test are given below:
| Marks | No. of Students |
| 0-10 | 5 |
| 10-20 | 9 |
| 20-30 | 16 |
| 30-40 | 22 |
| 40-50 | 26 |
| 50-60 | 18 |
| 60-70 | 11 |
| 70-80 | 6 |
| 80-90 | 4 |
| 90-100 | 3 |
Draw the ogive and hence, estimate :
(i) the median marks.
(ii) the number of students who did not pass the test if the pass percentage was 50.
(iii) the upper quartile marks.
Solution:
| Marks | No, of Students | c.f. Less Than Type | |
| 0-10 | 5 | Less than 10 | 5 |
| 10-20 | 9 | Less than 20 | 14 |
| 20-30 | 16 | Less than 30 | 30 |
| 30-40 | 22 | Less than 40 | 52 |
| 40-50 | 26 | Less than 50 | 78 |
| 50-60 | 18 | Less than 60 | 96 |
| 60-70 | 11 | Less than 70 | 107 |
| 70 – 80 | 6 | Less than 80 | 113 |
| 80-90 | 4 | Less than 90 | 117 |
| 90 – 100 | 3 | Less than 100 | 120 |
| Total | 120 | ||
Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120).
On the graph paper by taking upper limits on x-axis and number of students ony-axis. Join them free hand to get smooth curve.
Here, N = 120
\(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{120}{2}\) = 60
Median marks = 42 marks
Number of students who did not pass = 78 students
Upper quartile marks = 57 marks
(b) A man observes the angle of elevation of the top of the tower to be 45°. He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figures. [4]
Solution:
Let AB be the tower of height h m. P and Q are the two observing points, such that
∠APB = 45°, ∠AQB = 60°, PQ = 20 m
In rt. ∠ed ∆QBA,
\(\frac{\mathrm{AB}}{\mathrm{QB}}\) = tan 60°
= \(\frac{20(3+\sqrt{3})}{2}\)
= 10(3 + 1.73) = 47.3
Hence, the height of the tower is 47.3 m.
Question 11.
(a) Using the Remainder Theorem find the remainders obtained when x3 + (kx + 8)x + k is divided by x + 1 and x – 2. [3]
Hence, find k if the sum of the two remainders is 1.
Solution:
Given polynomial is
p(x) = x3 +(kx + 8)x + k
g(x) = x + 1
∴ R1 = P(-1)
= (-1)3 + {k(-1)} + 8} (-1) + k
= -1 + k – 8 + k
= 2k – 9
h(x) = x – 2
∴ R2 = P(2)
= (2)3 + (2k + 8)2 + k
= 8 + 4k + 16 + k
= 5k + 24
Now, R1 + R2 = 1
⇒ 2k – 9 + 5k + 24 = 1
⇒ 7k = 1 + 9 – 24
⇒ 7k = – 14
⇒ k = – 2
(b) The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers. [3]
Solution:
Let the two consecutive natural numbers which are multiples of 3 be 3x and 3(x + 1).
Now, 3x(3x + 3) = 810
⇒ x2 + x = 90
⇒ x2 + x – 90 = 0
⇒ (x + 10) (x – 9) = 0
⇒ x = 9 or
x = – 10
Rejecting negative value of x, because numbers are natural. We have x = 9.
Hence, the required numbers are 27 and 30.
(c) In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC||AE. If ∠BAC = 50°, find giving reasons: [4]
(i) ∠ACB
(ii) ∠EDC
(iii) ∠BEC
Hence, prove that BE is also a diameter.
Solution:
Since AC is a diameter and angle in a semi-circle is right
∴ ∠B = 90°
and ∠ACB = 40°
Also, BC || AE
∴ ∠EAC = ∠ACB
= 40° [alt. int. ∠s]
In cyclic quadrilateral ACDE
∠EAC + ∠EDC = 180°
40° + ∠EDC = 180°
⇒ ∠EDC = 180° – 40° = 140°
∠BEC = ∠BAC
= 50°
[∠s in the same segment]
Also, ∠EAC = ∠EBC
= 40°
[∠s in the same segment]
∴ ∠ABE = ∠ABC – ∠EBC
= 90° – 40°
= 50°
Again, ∠ABE = ∠ACE = 50°
[∠s in the same segment]
Now, ∠ACE + ∠ACB
= 50° + 40°
= 90°
⇒ ∠BCE = 90°
Hence, BE is a diameter, because angle is a semi-circle is right angle.