ICSE Class 10 Mathematics 2022 Semester 2 Question Paper with Solutions

Option (b) is correct.

Explanation: All possible out comes of a dice \(=\{1,2,3,4,5,6\}\)

Favourable outcomes (divisible by 3 )

Probability of getting a number divisible by 3

Option (c) is correct.

Explanation: Given, Volume of conical tent \(=462 \mathrm{~m}^{3}\) and Area of the base \(=154 \mathrm{~m}^{2}\) We know that

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Option (c) is correct.

Explanation:

We can see that cumulative frequency just greater than 7 is 9, which is lie in the class interval 20-30. So, the median class is 20-30.

Option (d) is correct.

Explanation: If two lines, having slopes \(m_{1}\) and \(m_{2}\), perpendicular to one another, then \(m_{1}\) \(\times m_{2}=-1\)

Option (b) is correct.

Explanation: Height of the light house \(=80 \mathrm{~m}\) Distance of the point from the foot of the light house \(=80 \mathrm{~m}\) Let the angle of elevation be \(\theta\), then \(\tan \theta\)

So, \(\theta=45^{\circ}\)

Option (a) is correct.

Option (b) is correct.

Explanation: Given point (-3, 5) lies in II \({ }^{\text {nd }}\) quadrant. So, after reflection sign of Y-coordinate will invert and sign of X-coordinate remain same. Reflection of this point will be in \(I I \mathrm{II}^{\text {rd }}\)-quadrant, i.e., (-3, -5).

Option (c) is correct.

Explanation: Given, \(\angle \mathrm{BAD}=(2 x+5)^{\circ}\) and

So,

(Sum of opposite angles of a cyclic quadrilateral are supplementary)

Option (c) is correct.

Explanation: Coordinates of centroid of a triangle is given by

So, coordinates of the centroid of triangle ABC is given by \(\left(\frac{1+4+x}{3}, \frac{4+1+4}{3}\right)\) or \(\left(\frac{5+x}{3}, 3\right)\)

Also, coordinates of centroid of \(\Delta \mathrm{ABC}(4,3)\) So, \(4=\frac{5+x}{3} \Rightarrow 12=5+x \Rightarrow x=7\)

Option (d) is correct.

Explanation: We have, radius of a roller \(=14 \mathrm{~cm}\) height of the roller \(=100 \mathrm{~cm}\) Now, C.S.A. of roller \((\) Cylinder \()=2 \pi r h\) \(=2 \times \frac{22}{7} \times 14 \times 100=8800 \mathrm{~cm}^{2}\) (Attempt any three questions from this Section.)

L.H.S.

Hence Proved.

Given A \((2 a+2,3), \mathrm{B}(7,4)\) and C \((2 a+5,2)\) are collinear. When three points are colliner, then

or \(x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0\) We have

We know, Mean \((\bar{x})=\frac{\sum f x}{\sum f}\)

PQOR is a quadrilateral.

(a)

(supplementary angles)

(b)

( ∵ The angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle.)

Given, No. of white balls= 5 No. of red balls \(=2\) No. of black balls \(=3\) Total no. of balls \(=5+2+3\)

Probability of getting a red ball

Given, For cone, Radius \((\mathrm{R})=5 \mathrm{~cm}\), Height \((\mathrm{H})=9 \mathrm{~cm}\) For cylinder, Radius \((r)=0.5 \mathrm{~cm}\), Height \((h)=1.5 \mathrm{~cm}\) No. of cylinders \(=\frac{\text { Volume of Cone }}{\text { Volume of cylinder }}\)

Given, Heights of the lamp posts \(\mathrm{AB}=\mathrm{CD}=100 \mathrm{~m}\) and

Now, in \(\triangle \mathrm{APR}\)

Correction-In this question paper \(\angle \mathrm{APB}=\) 40° is given, which should be 30° to solve it. Now, In \(\triangle \mathrm{PCD}\)

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Mode = 35 Marks

To find mode, (i) First identify the rectangle with highest frequency (Modal class); here 30-40. (ii) Join the top corners of the modal rectangle with immediate next corners of the adjacent rectangles. (iii) Let the point where the joining lines cut each other (here A). Draw a perpendicular from A to \(X\)-axis. The point ‘P’, where the perpendicular meet the X-axis will give the mode.

Given points are

and

Let \(\mathrm{P}(x, y)\) divides the line segment AB in the ratio 1:3. Then by using section formula,

Given word is ‘SECONDARY’

Probability of not selecting a vowel = Probability of selecting a consonant

(a) See graph

(b) Reflected points are : B’ (-2, 2), C'(-5, 2) and D’ (-4, 0)

(c) On joining the points ABCDD’C’B’A, the shape obtain is not proper geometrical shape, but collection of a quadrilateral and triangle.

Given: For Cylinder,

We know,

For cone,

We know,

Remaining Volume = Volume of cylinder – 2 × Volume of a cone

From the given figure,

We have \(\mathrm{EA}=3 \mathrm{~cm}\),

When two chords intersect externally, then (from figure)

or

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(a) Slope of line CD is given by

(b) Slope of line AB, Which is perpendicular to CD given by

Now, equation of a line (AB), when slope and a coordinates is given

where

So,

L.H.S.

We know, Mean \((\bar{x})=\frac{\sum f x}{\sum f}\)

(given)

L.H.S.

(a) Given, \(\angle \mathrm{ABC}=62^{\circ}\)

(∵ Angles in the same segment are equal)

(b) In \(\triangle \mathrm{ABC}\),

(∵ Angles in a semicircle is a right angle) Now,

(Angle sum property)

Given,

and

On solving above two equations we get the intersection points as (4, 0). Now, equation of line parallel to \(2 \mathrm{x}+\mathrm{y}-7=\) 0 is given by

which is passing through \((4,0)\) So,

So,the required equation of the line is 2x + y \(=8\) or,

Other than the given class intervals, we assume a class interval 0-10 prior the first class with zero frequency. Now, plot the points (10, 0), (20, 3), (30, 11), (40, 25), (50, 34), (60, 38), \((70,40)\) on the graph paper. In order to obtain ogive, we draw a smooth curve passing through these points. In order to find the median, we first locate the point \(\frac{\mathrm{N}}{2}=\frac{40}{2}=20\) on Y-axis. Let the point P from this draw a line parallel to X-axis cutting the curve at Q. From Q draw a line parallel to \(Y\)-axis meeting the X-axis at point M. The coordinate of M is 37.5. Hence, the median is 37.5. Now, plot the points (10, 0), (20, 3), (30, 11), (40, 25), (50, 34), (60, 38), \((70,40)\) on the graph paper. In order to obtain ogive, we draw a smooth curve passing through these, points.