Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2022 Semester 2 is the best way to boost your preparation for the board exams.
ICSE Class 10 Maths Question Paper 2022 Solved Semester 2
Time : 1 1/2 hour
Maximum Marks : 40
- Answers to this Paper must be written on the paper provided separately.
- You will not be allowed to write during the first 10 minutes.
- 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 three questions from Section B.
- The marks intended for questions are given in brackets [ ].
- Mathematical tables are provided.
Section – A
(Attempt all questions)
Question 1.
Choose the correct answers to the questions from the given options. (Do not copy the question. Write the correct answer only.) [10]
(i) The probability of getting a number divisible by 3 in throwing a dice is:
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Answer:
(b) \(\frac{1}{3}\)
[Here, S = {1, 2, 3, 4, 5} and E = {3, 6}]
Required probability = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(ii) The volume of a conical tent is 462 m3 and the area of (he base is 154 m2. The height of the cone is:
(a) 15 m
(b) 12 m
(c) 9 m
(d) 24 m
Answer:
(c) 9 m
[Here, \(\frac{1}{3}\) πr2h = 462
\(\frac{1}{3}\) × 154 × h = 462 ⇒ h = \(\frac{462}{154}\) × 3 = 9 m]
(iii) The median class for the given distribution is :
| Class Interval | Frequency |
| 0 – 10 | 2 |
| 10 – 20 | 4 |
| 20 – 30 | 3 |
| 30 – 40 | 5 |
(a) 0-10
(b)10-20
(c) 20 – 30
(d) 30-40
Answer:
(c) 20 – 30
[Here, N = 2 + 4 + 3 + 5 = 14 and \(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{14}{2}\) = 7
Class interval corresponding to \(\frac{\mathrm{N}}{\mathrm{2}}\) is the median class = 20 – 30]
(iv) If two lines are perpendicular to one another then the relation between their slopes m1 and m2 is :
(a) m1 = m2
(b) m1 = \(\frac{1}{m_2}\)
(c) m1 = -m2
(d) m1 × m2 = -1
Answer:
(d) m1 × m2 = -1
(v) A lighthouse is 80 m high. The angle of elevation of its top from a point 80 m away from its foot along the same horizontal line is :
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer:
(b) 45°
[Here, tan θ = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{80}{80}\)
= 1
= tan 45°
(vi) The modal class of a given distribution always corresponds to the :
(a) interval with highest frequency
(b) interval with lowest frequency
(c) the first interval
(d) the last interval
Answer:
(a) interval with highest frequency
(vii) The coordinates of the point P (- 3, 5) on reflecting on the x-axis are :
(a) (3, 5)
(b) (- 3, -5)
(c) (3, – 5)
(d) (-3, 5)
Answer:
(b) (- 3, -5)
(viii) ABCD is a cyclic quadrilateral. If ∠BAD = (2x + 5)° and ∠BCD = (x + 10)° then x is equal to:
(a) 65°
(b) 450
(c) 55°
(d) 50
Answer:
(c) 55°
[Here, ∠A + ∠C = 180°
2x + 5 + x + 10 = 180
⇒ 3x = 180 – 15 = 165
⇒ x = 55
(ix) A(1, 4), B(4, 1) and C(x, 4) are the vertices of ∆ABC. 1f the centroid of the triangle is G(4, 3) then x is equal to
(a) 2
(b) 1
(c) 7
(d) 4
Answer:
(c) 7
[Here, \(\left(\frac{1+4+x}{3}, \frac{4+1+4}{3}\right)\) = (4, 3)
⇒ \(\left(\frac{5+x}{3}, \frac{9}{3}\right)\) = (4, 3)
⇒ 5 + x = 12
⇒ x = 7]
(x) The radius of a roller 100 cm long is 14 cm. The curved surface area of the roller is :
(Take π = \(\frac{22}{7}\))
(a) 13200 cm2
(b) 15400 cm2
(c) 4400 cm2
(d) 8800 cm2
Answer:
(d) 8800 cm2
[Given that, r = 14 cm, h = 100 cm
Curved surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 14 × 100 = 8800 cm2]
Section – B
(Attempt any three questions from this section.)
Question 2.
(i) Prove that: \(\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}\) = 2 sec2 θ [2]
Answer:
L.H.S.
Hence Proved.
(ii) Find ‘a’, if A (2a + 2, 3), B (7, 4) and C (2a + 5, 2) are collinear. [2]
Answer:
Here, ar(∆ABC) = 0
\(\frac{1}{2}\) [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
(2a + 2) (4 – 2) + 7(2 – 3) + (2a + 5)(3 – 4) = 0
4a + 4 – 7 – 2a – 5 = 0
2a = 8
⇒ a = 4
(iii) Calculate the mean of the following frequency distribution. [3]
| Class Interval | Frequency |
| 5 – 15 | 2 |
| 15 – 25 | 6 |
| 25 – 35 | 4 |
| 35 – 45 | 8 |
| 45 – 55 | 4 |
Answer:
| Class Interval | Class Marks (xi) | Frequency (fi) | fixi |
| 5 – 15 | 10 | 2 | 20 |
| 15 – 25 | 20 | 6 | 120 |
| 25 – 35 | 30 | 4 | 120 |
| 35 – 45 | 40 | 8 | 320 |
| 45 – 55 | 50 | 4 | 200 |
| Total | 24 | 780 |
Mean = \(\frac{\sum x_i f_i}{\sum f_i}\) = \(\frac{780}{24}\) = 32.5
(iv) In the given figure O is the centre of the circle. PQ and PR are tangents and ZQPR = 70°. Calculate : [3]
(a) ∠QOR
(b) ∠QSR
Answer:
(a) In quad. PROQ, ∠PQO = 90°, ∠PRO = 90°
∠QOR + ∠P = 180°
∠QOR + 70 = 180°
∠QOR = 180° – 70° = 110°
(b) ∠QTR = \(\frac{1}{2}\)∠QOR = \(\frac{1}{2}\) × 110°
= 55°
Since QSRT is a cyclic quadrilateral
∠QSR + ∠QTR = 180°
∠QSR = 180° – ∠QTR
= 180° – 55°
= 125°
Question 3.
(i) A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is a red ball ? [2]
Answer:
We have, 5 white, 2 red and 3 black balls
Total number of balls = 5 + 2 + 3 = 10
Number of balls of red colour = 2
∴ Required probability = \(\frac{2}{10}=\frac{1}{5}\)
(ii) A solid cone of radius 5 cm and height 9 cm is melted and made into small cylinders of radius of 0.5 cm and height 1.5 cm. Find the number of cylinders so formed. |2]
Answer:
Let n be the number of cylinders
∴ n × Volume of cylinder = Volume of cone
n × πR2H = \(\frac{1}{3}\) πr2h
n × 0.5 × 0.5 × 1.5 = \(\frac{1}{3}\) × 5 × 5 × 9
n = \(\frac{5 \times 5 \times 3}{5 \times 5 \times 15}\) × 1000
= 200
(iii) Two lamp posts AB and CD each of height 100 m are on either side of the road. P is a point on the road between the two lamp posts. The angles of elevation of the top of the lamp posts from the point P are 60° and 40°. Find the distances PB and PD. [3]
Answer:
In rt. ∠ed ∆DPC
\(\frac{\mathrm{CD}}{\mathrm{PD}}\) = tan 60° ⇒ \(\frac{\mathrm{100}}{\mathrm{PD}}\) = \(\sqrt{3}\)
⇒ PD = \(\frac{100}{\sqrt{3}}\) m
In rt. ∠ed ∆BPA
\(\frac{\mathrm{AB}}{\mathrm{PB}}\) = tan 40° ⇒ \(\frac{\mathrm{100}}{\mathrm{PB}}\) = 0.8391
⇒ PD = \(\frac{100}{0.8391}\) = 119.18 m
(iv) Marks obtained by 100 students in an examination are given below : [3]
| Marks | No. of Students |
| 0 – 10 | 5 |
| 10 – 20 | 15 |
| 20 – 30 | 20 |
| 30 – 40 | 28 |
| 40 – 50 | 20 |
| 50 – 60 | 12 |
Draw a histogram for the given data using a graph paper and find the mode.
Take 2 cm = 10 marks along one axis and 2 cm = 10 students atong the other axis.
Answer:
Mode = 35
Question 4.
(i) Find a point P which divides internally the line segment joining the points A(-3, 9) and B(1, -3) in the ratio 1 : 3. [2]
Answer:
P(x, y) = P \(\left(\frac{1-9}{1+3}, \frac{-3+27}{1+3}\right)\)
= P\(\left(\frac{-8}{4}, \frac{24}{4}\right)\) = P(-2, 6)
(ii) A letter of the word ‘SECONDARY’ is selected at random. What Is the probability that the letter selected Is not a vowel ? [2]
Answer:
Here, S = {S, E, C, O, N, D, A, R, Y} = 9
Event (E) = {S, C, N, D, R, Y} = 6
P(E) = \(\frac{6}{9}\) = \(\frac{2}{3}\)
(iii) Use a graph paper for this question. Take 2 cm = I unit along both the axes. [3]
(a) Plot the points A(0, 4), B(2, 2), C(5, 2) and D(4, 0), E(0, 0) is the origin.
(b) Reflect B, C, D on the y-axis and name them as B’, C’ and D’ respectively.
(c) Join the points ABCDD’C’B’ and A in order and give a geometrical name to the closed figure.
Answer:
(a)
(b) B'(-2, 2)
C'(-5, 2)
D’ (-4, 0)
(c) Boat
(iv) A solid wooden cylinder is of radius 6 cm and height 16 cm. Two cones each of radius 2 cm and height 6 cm are drilled out of the cylinder. Find the volume of the remaining solid. Take π = \(\frac{22}{7}\)
Solution:
Volume of the remaining solid
= \(\frac{22}{7}\) × 6 × 6 × 16 – \(\frac{2}{3}\) × \(\frac{22}{7}\) × 2 × 2 × 6
= \(\frac{12672}{7}\) – \(\frac{352}{7}\) = \(\frac{12320}{7}\) = 1760 cm3
Question 5.
(i) Two chords AB and CD of a circle intersect externally at E. If EC = 2 cm, EA = 3 cm and AB = 5 cm, find the length of CD. [2]
Solution:
Here, two chords AB and CD intersect externally at E.
∴ BE × AE = DE × CE
(5 + 3) × 3 = (CD + 2) × 2
12 = CD + 2
⇒ CD = 12 – 2
= 10 cm
(ii) Line AB is perpendicular to CD. Coordinates of B,C and D are respectively (4, 0), (0, -1) and (4, 3).
Find : [2]
(a) Slope of CD
(b) Equation of AB
Solution:
(a) Slope of CD = \(\frac{3+1}{4-0}=\frac{4}{4}\) = 1
(b) Slope of AB =\(\frac{-1}{\text { Slope of } C D}\) = -1
(∵ AB ⊥ CD)
Equation of AB
y – 0 = 1(x – 4)
= -x + 4
x + y – 4 = 0
(iii) Prove that : \(\frac{(1+\sin \theta)^2+(1-\sin \theta)^2}{2 \cos ^2 \theta}\) = sec2θ + tan2θ [3]
Solution:
L.H.S.
= sec2θ + tan2θ
= R.H.S
(iv) The mean of the following distribution is 50. Find the unknown frequency. [3]
| Class Interval | Frequency |
| 0 – 20 | 6 |
| 20 – 40 | f |
| 40 – 60 | 8 |
| 60 – 80 | 12 |
| 80 – 100 | 8 |
Solution:
| Class Interval | Class Marks (xi) | Frequency (fi) | fixi |
| 0 – 20 | 10 | 6 | 60 |
| 20 – 40 | 30 | f | 30 f |
| 40 – 60 | 50 | 8 | 400 |
| 60 – 80 | 70 | 12 | 840 |
| 80 – 100 | 90 | 8 | 720 |
| Total | 34+ f | 30f + 2020 |
Mean = 50 (given)
\(\frac{\sum f_i x_i}{\sum f_i}\) = 50
\(\frac{30 f+2020}{34+f}\) = 50
30f + 2020 = 1700 + 50f ⇒ 20f = 320
⇒ f = 16
Question 6.
(i) Prove that: 1 + \(\frac{\tan ^2 \theta}{1+\sec \theta}\) = sec2θ [2]
Solution:
L.H.S. = 1 + \(\frac{\tan ^2 \theta}{1+\sec \theta}\) = 1 + \(\frac{\sec ^2 \theta-1}{\sec \theta+1}\)
= 1 + \(\frac{(\sec \theta-1)(\sec \theta+1)}{(\sec \theta+1)}\)
= 1 + sec θ – 1 = sec θ = R.H.S.
(ii) In the given figure A, B, C and D are points on the circle with centre O. Given ∠ABC = 62°.
Find: [2]
(a) ∠ADC
(b) ∠CAB
Solution:
(a) ∠ADC = ∠ABC (angles in the same segment AC)
= 62°
(b) Since AOB is a diameter
∴ ∠BCA = 90° (angle in the semi circle)
In ∆ABC
∠ABC + ∠BCA + ∠CAB = 180°
62° + 900 + LCAB = 180°
∠CAB = 180° – 90° – 62° = 28°
(iii) Find the equation of a line parallel to the line 2x + y – 7 = 0 and passing through the intersection of the lines x + y – 4 = 0 and 2x – y = 8. [3]
Solution:
x + y – 4 = 0 …………… (i); 2x – y = 8 …………….. (ii)
Adding (i) and (ii), we have
3x = 12 ⇒x = 4
From (i), we obtain y = 0
∴ Point of intersection of (i) and (ii) is (4, 0)
Slope of line 2x + y – 7 = 0 = – \(\frac{2}{1}\) = -2
Hence, the equation of required line, is :
y – 0 = -2 (x – 4) or 2x + y – 8 = 0.
(iv) Marks obtained by 40 students in an examination are given below:
| Marks | No. of Students |
| 10 – 20 | 3 |
| 20 – 30 | 8 |
| 30 – 40 | 14 |
| 40 – 50 | 9 |
| 50 – 60 | 4 |
| 60 – 70 | 2 |
Using graph paper draw an ogive and estimate the median marks.
Take 2 cm = 10 marks along one axis and 2 cm = 5 students along the other axis. [3]
Solution:
| Marks | No. of Students | c.f | Points |
| 10 – 20 | 3 | 3 | (20, 3) |
| 20 – 30 | 8 | 11 | (30, 11) |
| 30 – 40 | 14 | 25 | (40, 25) |
| 40 – 50 | 9 | 34 | (50, 34) |
| 50 – 60 | 4 | 38 | (60, 38) |
| 60 – 70 | 2 | 40 | (70, 40) |
Plot the points shown in the table on the graph and joining these points by a free hand curve, starting from the lower limit of first class interval and ending at the upper limit of last class, we have the required ogive.
Median = \(\left(\frac{\mathrm{N}}{2}\right)^{\text {th }}\) term = \(\frac{40}{2}\) = 20th term
Draw NP || x-axis and PM ⊥ x-axis
The value of point M on x-axis is the median.
Hence, the median marks = 37 marks.
