Class 10: ICSE SEMESTER 2 EXAMINATION SQP – MATHEMATICS – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Maximum Marks: 40
Time allowed: One and a half hours
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 intended marks for questions or parts of questions are given in brackets [ ].

SECTION A
(Attempt all questions from this Section.)

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 point (3,0) is invariant under reflection in:

(a) The origin (b) x-axis (c) y-axis (d) both x and y axes

Answer:

$\fbox{ B }$

We know that, every point in a line is invariant under the reflection in the same line.

As the points (3,0) lie on the x-axis, (3,0) is invariant under reflection in x-axis.

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(ii) In the given figure, AB is a diameter of the circle with center ‘O’. If $\angle COB = 55^{\circ}$ then the value of x is:

(a) $27.5^{\circ}$ (b) $55^{\circ}$ (c) $110^{\circ}$ (d) $125^{\circ}$

Answer:

$\fbox{ A }$

$\angle AOC = 180^{\circ} - 55^{\circ} = 125^{\circ}$

$AO = OC$ (radius) $\therefore \angle OAC = \angle OCB$

$\Rightarrow 2\angle OCB + 125 = 180$

$\displaystyle \Rightarrow \angle OCB = \frac{55}{2} = 27.5^{\circ}$

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(iii) If a rectangular sheet having dimensions 22 cm x 11 cm is rolled along its shorter side to form a cylinder. Then the curved surface area of the cylinder so formed is:

(a) $968 \text{ cm}^2$ (b) $424 \text{ cm}^2$ (c) $121 \text{ cm}^2$ (d) $242 \text{ cm}^2$

Answer:

$\fbox{ D }$

$\text{Circumference of the cylinder } = 2 \pi r = 22 \text{ cm}$

$\text{Height of the cylinder } = 11 \text{ cm}$

$\text{Therefore, Curved Surface Area } = 2\pi r h = 22 \times 11 = 242 \text{ cm}^2$

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(iv) If the vertices of a triangle are (1,3), (2, – 4) and (-3, 1). Then the co-ordinate of its centroid is:

(a) (0, 0) (b) (0, 1) (c) (1, 0) (d) (1, 1)

Answer:

$\fbox{ A }$

Centroid $(x, y)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is $\displaystyle \text{calculated by the formula } \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}$

$\displaystyle x = \frac{x_1+x_2+x_3}{3} = \frac{1+2-3}{3} = 0$

$\displaystyle y = \frac{y_1+y_2+y_3}{3} = \frac{3-4+1}{3} = 0$

Hence the centroid $\displaystyle = (0, 0)$

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(v) $\tan \theta \times \sqrt{1 - \sin^2 \theta}$ is equal to

(a) $\cos \theta$ (b) $\sin \theta$ (c) $\tan \theta$ (d) $\cot \theta$

Answer:

$\fbox{ B }$

$\tan \theta \times \sqrt{1 - \sin^2 \theta} = \tan \theta \times \cos \theta = \sin \theta$

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(vi) The median class for the given distribution is:

$\displaystyle \begin{array} {|l | c| c| c| c |} \hline \text{Class Interval } & 1-5 & 6-10 & 11-15 & 16-20 \\ \hline \text{Cumulative Frequency: } &2 & 6 & 11 & 18 \\ \hline \end{array}$

(a) 1 – 5 (b) 6 – 10 (c) 11 – 15 (d) 11 – 20

Answer:

$\fbox{ C }$

$\begin{array} {|c|c|c|c| } \hline \text{Class Interval} & \text{Class Boundaries} & \text{Frequency} & \text{Cumulative Frequency} \\ \hline 1-5 & 0.5-5.5 & 2 & 2 \\ \hline 6-10 & 5.5 - 10.5 & 6 & 8 \\ \hline 11-15 & 10.5-15.5 & 11 & 19 \\ \hline 16-20 & 15.5 - 20.5 & 8 & 27 \\ \hline & & \sum f_i = N = 27 & \\ \hline \end{array}$

$\displaystyle \frac{N}{2} = \frac{27}{2} = 13.5$

Therefore the median class is $\displaystyle 11-15.$

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(vii) If the lines $7y = ax + 4 \text{ and } 2y = 3 - x$ , are parallel to each other, then the value of ‘a’ is:

$\displaystyle \text{(a) } -1 \hspace{2.0cm} \text{(b) } \frac{-7}{2} \hspace{2.0cm} \text{(c) } \frac{-2}{7} \hspace{2.0cm} \text{(d) } 14$

Answer:

$\fbox{ B }$

$\displaystyle 7y = ax + 4 \hspace{0.5cm} \Rightarrow y = \frac{a}{7} x + \frac{4}{7} \hspace{0.5cm} \Rightarrow m_1 = \frac{a}{7}$

$\displaystyle 2y = 3 - x \hspace{0.5cm} \Rightarrow y = \frac{-1}{2} x + \frac{3}{2} \hspace{0.5cm} \Rightarrow m_2 = \frac{-1}{2}$

Since the two line are parallel,

$\displaystyle m_1 = m_2 \hspace{0.5cm} \Rightarrow \frac{a}{7} = \frac{-1}{2} \hspace{0.5cm} \Rightarrow a = \frac{-7}{2}$

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(viii) Volume of a cylinder is $330 \text{ cm}^3$ The volume of the cone having same radius and height as that of the given cylinder is:

(a) $330 \text{ cm}^3$ (b) $165 \text{ cm}^3$ (c) $110 \text{ cm}^3$ (d) $220 \text{ cm}^3$

Answer:

$\fbox{ C }$

Let $r$ be the radius and $h$ be the height of the cylinder and the cone.

$\displaystyle \text{Therefore } \pi r^2 h = 330$

$\displaystyle \text{Volume of cone } = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 330 = 110 \text{ cm}^3$

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(ix) In the given graph, the modal class is the class with frequency:

(a) 72 (b) 21 (c) 48 (d) 36

Answer:

$\fbox{ A }$

Modal class = 20 – 30 (This is the class with the highest frequency).

Therefore the frequency is 72

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(x) If the probability of a player winning a game is 0.56. The probability of his losing this game is:

(a) 0.56 (b) 1 (c) 0.44 (d) 0

Answer:

$\fbox{ C }$

Probability of losing $= 1 - 0.56 = 0.44$

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SECTION B
(Attempt any three questions from this Section.)

Question 2

(i) Find the ratio in which the x-axis divides internally the line joining points A (6, -4) and B ( -3, 8). [2]

Answer:

Let the line segment $A (6, -4)$ and $B ( -3, 8)$ is divided at point $P(x,0)$ by x-axis in ratio $m:n.$

$\displaystyle \therefore x = \frac{mx_2+nx_1}{m+n} \text{ and } y = \frac{my_2+ny_1}{m+n}$

$\displaystyle \text{ Here } (x, y) = ( x, 0), \ \ \ \ (x_1, y_1) = (6, -4) \text{ and } (x_2, y_2) = ( -3, 8)$

$\displaystyle \therefore 0 = \frac{m(8)+n(-4)}{m+n} \Rightarrow 8m - 4 n = 0 \Rightarrow \frac{m}{n} = \frac{1}{2}$

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(ii) Three rotten apples are accidently mixed with twelve good ones. One apple is picked at random. What is the probability that it is a good one? [2]

Answer:

Total no. of apples = 15 No. of good apples = 12 No. of bad apples = 3

One apple is picked at random.

$\displaystyle \text{Probability that it is a good one } = \frac{9}{15} = \frac{3}{5}$

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(iii) In the given figure , AC is a tangent to circle at point B. $\triangle EFD$ is an equilateral triangle and $\angle CBD = 40^{\circ}$ . Find: [3]

(a) $\angle BFD$ (b) $\angle FBD$ (c) $\angle ABF$

Answer:

$\triangle EFD$ is an equilateral triangle

Therefore $\angle DFE = \angle FED = \angle EDF = 60^{\circ}$

$\angle CBD = 40^{\circ}$ (given)

$\angle FED + \angle FBD = 180 \Rightarrow \angle FBD = 120^{\circ}$

$\angle ABF = 180 - 120 - 40 = 20^{\circ}$

$\angle BFD = 180 - 60 - 90 = 30^{\circ}$

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(iv) A drone camera is used to shoot an object P from two different positions R and S along the same vertical line QRS. The angle of depression of the object P from these two positions are $35^{\circ}$ and $60^{\circ}$ respectively as shown in the diagram. If the distance of the object P from point Q is 50 meters, then find the distance between R and S correct to the nearest meter. [3]

Answer:

$\displaystyle \text{In } \triangle PQR:$

$\displaystyle \frac{RQ}{50} = \tan 35^{\circ} \Rightarrow RQ = 50 \tan 35^{\circ} = 50 \times 0.7002 = 35.01$

$\displaystyle \text{In } \triangle PQS:$

$\displaystyle \frac{SR + RQ}{50} = \tan 60^{\circ} \Rightarrow SR + 35.01 = 50 \sqrt{3} \\ \\ \Rightarrow SR = 50 \times 1.732 - 35.01 = 51.59$

Hence SR nearest meter is 52 m.

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Question 3

(i) In the given figure, PT is a tangent to the circle at T, chord BA is produced to meet the tangent at P. Perpendicular BC bisects the chord TA at C. If PA = 9 cm and TB = 7 cm, find the lengths of: [2]

(a) AB (b) PT

Answer:

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(ii) How many solid right circular cylinders of radius 2 cm and height 3 cm can be made by melting a solid right circular cylinder of diameter 12 cm and height 15 cm? [2]

Answer:

$\displaystyle \text{Number of cylinders that can be produced }= \frac{\pi (6)^2 \times 15}{\pi (2)^2 \times 3 } = \frac{36 \times 5}{4} = 45$

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(iii) Prove that: [3]

$\displaystyle \frac{\cos^2 A}{\cos A - \sin A }+ \frac{\sin A}{1- \cot A} = \sin A + \cos A$

Answer:

$\displaystyle \text{LHS }= \frac{\cos^2 A}{\cos A - \sin A }+ \frac{\sin A}{1- \cot A}$

$\displaystyle = \frac{\cos^2 A}{\cos A - \sin A }+ \frac{\sin^2 A}{\sin A - \cos A}$

$\displaystyle = \frac{\cos^2 A}{\cos A - \sin A }- \frac{\sin^2 A}{\cos A - \sin A}$

$\displaystyle = \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A }$

$\displaystyle = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A }$

$\displaystyle = (\cos A + \sin A) = \text{ RHS. Hence proved. }$

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(iv) Use graph paper for this question, take 2 cm = 10 marks along one axis and 2 cm = 10 students along the other axis.

The following table shows the distribution of marks in a 50 marks test in Mathematics:

$\displaystyle \begin{array} {|l | c| c| c| c | c | } \hline \text{Marks } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{No. of Students } & 6 & 10 & 13 & 7 & 4 \\ \hline \end{array}$

Draw the ogive for the above distribution and hence estimate the median marks. [3]

Answer:

$\\$

Question 4

(i) Find the equation of the perpendicular dropped from the point P (-1,2) onto the line joining A (1,4) and B (2,3). [2]

Answer:

$\displaystyle \text{Let } A = (1, 4), B = (2, 3), \text{ and } P = (-1, 2).$

$\displaystyle \text{ Slope of } AB = \frac{3-4}{2-1} = -1$

$\displaystyle \text{Slope of equation perpendicular to } AB = \frac{-1}{\text{Slope of } AB} = 1$

The equation of the perpendicular drawn through P onto AB is given by:

$\displaystyle y - y_1 = m (x - x_1)$

$\displaystyle \Rightarrow y - 2 = 1 ( x -(-1))$

$\displaystyle \Rightarrow y - 2 = x + 1$

$\displaystyle \Rightarrow y = x +3$

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(ii) Find the mean for the following distribution: [2]

$\displaystyle \begin{array} {|l | c| c| c| c |} \hline \text{Class Interval } & 20-40 & 40-60 & 60-80 & 80-100 \\ \hline \text{Frequency: } &4 & 7 & 6 & 3 \\ \hline \end{array}$

Answer:

$\displaystyle \begin{array} {|c|c|c|c|} \hline \text{Class Interval } & x_i & f_i & f_ix_i \\ \hline 20-40 & 30 & 4 & 120 \\ \hline 40-60 & 50 & 7 & 350 \\ \hline 60-80 & 70 & 6 & 420 \\ \hline 80-100 & 90 & 3 & 270 \\ \hline & &\sum f_i = 20 & \sum f_ix_i = 1160\\ \hline \end{array}$

$\displaystyle \text{Mean = } \overline{X} = \frac{\sum f_ix_i}{\sum f_i} = \frac{1160}{20} = 58$

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(iii) A solid piece of wooden cone is of radius OP = 7 cm and height OQ = 12 cm. A cylinder whose radius and height equal to half of that of the cone is drilled out from this piece of wooden cone. Find the volume of the remaining piece of wood. $( \text{ use } \frac{\pi}{2} )$ [3]

Answer:

$\displaystyle \text{Volume of large cone } = \frac{1}{3} \pi (7)^2 (12) = 196 \pi$

$\displaystyle \text{Volume of cylinder } = \pi (3.5)^2 (6) = 73.5 \pi$

$\displaystyle \text{Volume left } = 196 \pi - 73.5 \pi = 122.5 \pi = 122.5 \times \frac{22}{7} = 385 \text{ cm}^3$

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(iv) Use a graph sheet for this question, take 2cm = 1 unit along both x and y axis: [3]

(a) Plot the points A (3,2) and B (5,0). Reflect point A on the y-axis to A’. Write co-ordinates of A’.

(b) Reflect point B on the line AA’ to B’. Write the co-ordinates of B’.

Answer:

The figure is an arrow head.

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Question 5

(i) In the given figure, the sides of the quadrilateral PQRS touches the circle at A,B,C and D. If RC = 4 cm, RQ = 7 cm and PD = 5 cm. Find the length of PQ: [2]

Answer:

We know that the length of tangents from an external point to a circle are equal.

$PD = PA = 5 \text{ cm }$

$RC = RB = 4 \text{ cm }$

$\therefore QB = 7 - 4 = 3 \text{ cm }$

$QA = QB = 3 \text{ cm }$

$\therefore PQ = PA + AQ = 5 + 3 = 8 \text{ cm }$

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(ii) Prove that: [2]

$\displaystyle \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} = 1 - \sin \theta \cos \theta$

Answer:

$\displaystyle \text{We know: } a^3 + b^3 = (a+b)(a^2 + b^2 - ab)$

$\displaystyle \text{LHS } = \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta}$

$\displaystyle = \frac{(\sin \theta + \cos \theta)(\sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta) }{\sin \theta + \cos \theta}$

$\displaystyle = \sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta$

$\displaystyle = 1 - \sin \theta \cos \theta = \text{ RHS. Hence proved. }$

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(iii) In the given diagram, $OA = OB, \angle OAB = \theta$ and the line AB passes through point P (-3, 4). [3]

Find:

(a) Slope and inclination $(\theta)$ of the line AB (b) Equation of the line AB

Answer:

(a) $\displaystyle OA = OB$

$\displaystyle \text{slope } = \tan \theta = \frac{OB}{OA} = 1$

$\displaystyle \Rightarrow \theta = 45^{\circ}$

(b) Equation of $\displaystyle AB$

$\displaystyle y - 4 = 1 ( x -(-3))$

$\displaystyle \Rightarrow y = x+ 7$

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(iv) Use graph paper for this question. Estimate the mode of the given distribution by plotting a histogram. [Take 2 cm = 10 marks along one axis and 2 cm = 5 students along the other axis] [3]

$\displaystyle \begin{array} {|l | c| c| c| c | c | } \hline \text{Daily wages (in Rs.) } & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{No. of Workers } & 6 & 12 & 20 & 15 & 9 \\ \hline \end{array}$

Answer:

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Question 6

(i) A box contains tokens numbered 5 to 16. A token is drawn at random. Find the probability that the token drawn bears a number divisible by: [2]
(a) 5 (b) Neither by 2 nor by 3

Answer:

The bag contains the following tokens: $\displaystyle \{ 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 \}$

$\displaystyle \text{(a) } P( \text{Token is divisible by 5}) = \frac{3}{12} = \frac{1}{4}$

$\displaystyle \text{(b) } P( \text{divisible neither by 2 nor by 3 } ) = 1 - \frac{8}{12} = \frac{4}{12} = \frac{1}{3}$

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(ii) Point M (2, b) is the mid-point of the line segment joining points P (a, 7) and Q (6, 5). Find the values of ‘a’ and ‘b’. [2]

Answer:

Point $\displaystyle M (2, b)$ is the mid-point of the line segment joining points $\displaystyle P (a, 7)$ and $\displaystyle Q (6, 5).$

$\displaystyle \therefore \frac{a+6}{2} = 2 \Rightarrow a = -2$

$\displaystyle \text{and } \frac{7+5}{2} = b \Rightarrow b = 6$

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(iii) An aeroplane is flying horizontally along a straight line at a height of 3000 m from the ground at a speed of 160 m/s. Find the time it would take for the angle of elevation of the plane as seen from a particular point on the ground to change from 60⁰ to 45⁰. Give your answer correct to the nearest second. [3]

Answer:

In $\displaystyle \triangle ABC$

$\displaystyle \frac{3000}{AB} = \tan 60^{\circ} \Rightarrow AB = \frac{3000}{\sqrt{3}} = 1000\sqrt{3} m$

In $\displaystyle \triangle ADE$

$\displaystyle \frac{3000}{1000\sqrt{3}+ BD} = \tan 45^{\circ} \Rightarrow BD = 3000 - 1000\sqrt{3} = 1267.9$

$\displaystyle \text{Time taken by the aeroplane } = \frac{1267.9}{160} = 7.92 = 8 s$

Hence the time taken by the aeroplane to nearest second is 8 seconds

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(iv) Given that the mean of the following frequency distribution is 30, find the missing frequency ‘f’ [3]

$\displaystyle \begin{array} {|l | c| c| c| c | c | c|} \hline \text{Class Interval } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text{Frequency: } &4 & 6 & 10 & f & 6 & 4 \\ \hline \end{array}$

Answer:

$\displaystyle \text{Given } \overline{X} = 30$

$\displaystyle \begin{array} {|c|c|c|c|} \hline \text{Class Interval } & x_i & f_i & f_ix_i \\ \hline 0-10 & 5 & 4 & 20 \\ \hline 10-20 & 15 & 6 & 90 \\ \hline 20-30 & 25 & 10 & 250 \\ \hline 30-40 & 35 & f & 35f \\ \hline 40-50 & 45 & 6& 270 \\ \hline 50-60 & 55 & 4 & 200 \\ \hline & &\sum f_i = 30 + f & \sum f_ix_i = 850+35f\\ \hline \end{array}$