Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 4 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 4 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions:

1. This question paper has 5 Sections A-E.

2. Section A has 20 MCQs carrying 1 mark each.

3. Section B has 5 questions carrying 2 marks each.

4. Section C has 6 questions carrying 3 marks each.

5. Section D has 4 questions carrying 5 marks each.

6. Section E has 3 Case Based integrated units of assessment (4 marks each).

7. All questions are compulsory. However, an internal choice in 2 questions of 2 marks, 2 questions of 3 marks and 2 questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.

8. Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section A

(Section-A consists of 20 Questions of 1 mark each)

Question 1.

Which term of the AP 2, -1,-4 , -7, …… is -40 ? [1]

(a) 8th

(b) 11th

(c) 15th

(d) 23rd

Answer:

(c) 15th

Here, a = 2 and d = (-1 – 2) = – 3

Let a_{n} = – 40

Then, a + (n – 1)d = a_{n}

⇒ 2 + (n – 1) × (-3) = – 40

⇒ n = 15th

Question 2.

If a semi-circle is rolled along its diameter, then the figure formed is [1]

(a) spherical

(b) conical

(c) cylindrical

(d) None of these

Answer:

(b) conical

When we roll a semi-circle along its diameter, then conical is formed.

Question 3.

The area of the shaded portion is [1]

(a) 940.5 cm²

(b) 930.5 cm²

(c) 400.5 cm²

(d) 510.5 cm²

Answer:

(a) 940.5 cm²

Radius of smaller circle = 31 cm

Radius of larger circle = 31 + 4.5= 35.5 cm

∴ Area of path = [π(35.5)2 – π(31)2]

= π [(35.5)^{2} – (31)^{2}]

= π (35.5 + 31) (35.5 – 31)

= π × 66.5 × 4.5

= \(\frac{22}{7}\) × 66.5 × 4.5= 940.5 cm²

Question 4.

The distance between two parallel tangents to a circle of radius 7 cm, is [1]

(a) 7 cm

(b) 0 cm

(c) 14 cm

(d) None of these

Answer:

(c) 14 cm

The distance between two parallel tangents to a circle of radius 7 cm is 2 × 7 = 14cm.

Question 5.

If θ = 15°, then \(\frac{\cos 3 \theta-2 \cos 4 \theta}{\sin 3 \theta+2 \cos 4 \theta}\) is equal to [1]

(a) \(\frac{1}{\sqrt{2}}\)

(b) \(\frac{1+\sqrt{2}}{1-\sqrt{2}}\)

(c) \(\frac{2+\sqrt{2}}{2-\sqrt{2}}\)

(d) \(\frac{1-\sqrt{2}}{1+\sqrt{2}}\)

Answer:

(d) \(\frac{1-\sqrt{2}}{1+\sqrt{2}}\)

Given, θ = 15°

Question 6.

HCF of two number is 23 and their LCM is 1449. If one of the number is 161, then the other number is [1]

(a) 207

(b) 307

(c) 1449

(d) None of these

Answer

(a) 207

Let the required number be x.

As, product of two numbers

= (HCF × LCM) of two numbers

∴ x × 161 = 23 × 1449

⇒ x = \(\frac{23 × 1449}{161}\) = 207

Question 7.

In a right angled ∆ABC, right angled at C, if tan A = 1, then the value of 2 sin A cos A, is [1]

(a) 0

(b) 1

(c) 2

(d) \(\frac{1}{2}\)

Answer:

(b) 1

We have, tan A = 1 ⇒ ∠A = 45°

∴ 2 sin A cos A = 2 sin45°cos45°

= 2 × \(\frac{1}{\sqrt{2}}\) × \(\frac{1}{\sqrt{2}}\) = \(\frac{1}{2}\) = 1

Question 8.

If ∆ADE ~ ∆ACB, ∠DEC = 105° and ∠ECB = 65°, then the value of x is [1]

(a) 60°

(b) 90°

(c) 30°

(d) 40°

Answer:

(d) 40°

In ∆ADE and ∆ACB,

∵ ∆ADE ~ ∆ACB [given]

∴ ∠ACB = ∠ADE = 65°

Also, ∠AED = ∠ABC = 180° – 105° = 75°

In ∆ADE, ∠ADE + ∠AED + ∠DAE = 180°

⇒ 65° + 75° + ∠DAE =180°

⇒ ∠DAE = 180° – 140° = 40°

Question 9.

If 2x² + bx + 8 =0 to have non-real roots, then the interval for b is [1]

(a) -8 < b < 8

(b) -6 < b < 6 (c) -8 > b > 8

(d) None of these

Answer:

(a) -8 < b < 8

Given, 2×2 + bx+8 = 0 …(i)

We have, for non-real roots D < 0

⇒ b² – 4ac < 0

⇒ b² – 64 < 0

⇒ b² < 64

⇒ |b| < 8

⇒ b < 8 and b > -8

For -8 < b < 8, Eq. (i) has non-real roots.

Question 10.

Two numbers are in the ratio of 15 : 11. If their HCF is 13, then numbers will be [1]

(a) 195 and 143

(b) 190 and 140

(c) 185 and 163

(d) 185 and 143

Answer:

(a) 195 and 143

Let the required numbers be 15x and 11x.

Then, their HCF is x, so x = 13

[∵ given HCF = 13]

∴ The numbers are 15 × 13 and 11 × 13 i.e. 195 and 143.

Question 11.

If the common difference of an AP is 5, then the value of a_{18} – a_{13} is [1]

(a) 5

(b) 20

(c) 25

(d) 30

Answer:

(c) 25

Given, common difference of an AP , d = 5

Now, a_{18} – a_{13} = (a + 17d) – (a + 12d)

= a + 17d – a – 12d

= 5d = 5(5) = 25

Question 12.

If the point P(k, 0) divides the line segment joining the points A(2, – 2) and B(-7, 4) in the ratio 1:2, thenjhc value of k is [1]

(a) 1

(b) 2

(c) – 2

(d) -1

Answer:

(d) -1

When a point P(x, y) divides (internally) the line segment joining the points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) in the ratio m : n, then by section-formula

Question 13.

Suppose mean of 10 observations is 20, if we add 5 in each observation, then the new mean is [1]

(a) 25

(b) 10

(c) 20

(d) 5

Answer:

(a) 25

If we add 5 in each observation, then new mean will be 20 + 5 i.e. 25.

Question 14.

Gayatri was making a mathematical model, in which she placed 4 cubes each of edge 20 cm one above the another. Then, the surface area of the resulting cuboid, is

(a) 7000 cm²

(b) 7200 cm²

(c) 6000 cm²

(d) 7250 cm²

Answer:

(b) 7200 cm²

Given, edge of a cube is 20 cm.

The dimensions of the cuboid when 4 cubes each of edge 20 cm are placed one above another.

Length of the cuboid (l) = 20 cm

Breadth of the cuboid (b) = 20 cm

and Height of the cuboid (h) = 20 + 20 + 20 + 20

= 80 cm

∴ Surface area of the resulting cuboid formed = 2 (lb + bh + hl)

= 2(20 × 20 + 20 × 80 + 80 × 20)

= 2(400 + 1600 + 1600)

= (2 × 3600) = 7200 cm²

Question 15.

A solid ball is exactly fitted inside a cubical box of side a. The volume of the ball is [1]

(a) \(\frac{1}{6}\) πa^{3}

(b) \(\frac{4}{3}\) πa^{3}

(c) \(\frac{1}{3}\) πa^{3}

(d) None of these

Answer:

(a) \(\frac{1}{6}\) πa^{3}

Because solid ball is exactly fitted inside the cubical box of side a. So, a is the diameter for the solid ball.

∴ Radius of the ball = \(\frac{a}{2}\)

So, volume of the ball = \(\frac{4}{3}\) π(\(\frac{a}{2}\))^{3} = \(\frac{1}{6}\) πa^{3}

Question 16.

Which of the following is a measure of central tendency?

(a) Frequency

(b) Cumulative frequency

(c) Mean

(d) Class-limit

Answer:

(c) Mean

Mean is a measure of central tendency.

Question 17.

Write the value of k for which the system of equations x + ky = 0 and 2x – y = 0 has unique solution. [1]

(a) k = \(\frac{1}{2}\)

(b) k = \(– \frac{1}{2}\)

(c) k ≠ \(\frac{-1}{2}\)

(d) k ≠ \(\frac{1}{2}\)

Answer:

(c) k ≠ \(\frac{-1}{2}\)

Given system of equations,

x + ky = 0 and 2x – y = 0

On comparing these equations with

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{1} = 0, we get

a_{1} = 1, b_{1} =k, c_{1} = 0

and a_{2} =2, b_{2}= -1, c_{2} = 0

Condition for unique solution,

\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{1}{2} \neq \frac{k}{-1}\)

⇒ k ≠ \(\frac{-1}{2}\)

Question 18.

A parallelogram has vertices P(1, 4), Q(7, 11), R(a, 4) and S(1, -3). Then, the value of a is [1]

(a) 6

(b) 7

(c) 5

(d) 4

Answer:

(b) 7

Let P(1, 4), O(7, 11), R(a, 4) and S(1, -3) be the vertices of a parallelogram PQRS, respectively.

Join PR and OS.

Let PR and QS intersect at a point T.

We know that the diagonals of a parallelogram bisect each other.

So, T is the mid-point of PR as well as that of QS.

∴ Mid-point of PR = Mid-point of QS

⇒ (\(\frac{1 + a}{2}\), \(\frac{4+4}{2}\)) = (\(\frac{7+1}{2}\), \(\frac{11-3}{2}\))

⇒ (\(\frac{1+a}{2}\), 4) = (4, 4)

On comparing x-coordinate from both sides, we get

\(\frac{1+a}{2}\) = 4

⇒ 1 + a = 8 ⇒ a = 7

Hence, the value of a is 7.

Directions In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Question 19.

Assertion (A): All regular polygons of the same number of sides such as equilateral triangle, squares etc. are similar.

Reason (R): Two polygons are said to be similar, if their corresponding angles are equal and lengths of corresponding sides are proportional.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)

(c) Assertion (A) is true but Reason (R) is false

(d) Assertion (A) is false but Reason (R) is true

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

Two polygons are similar, if their corresponding angles are equal and sides are proportional.

∵ In equilateral triangle and square, each angle are equal and sides are also proportional therefore, regular polygons are similar.

Hence, both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 20.

Assertion (A): Mid-point of a line segment divides line in the ratio 1:1.

Reason (R): If area of triangle is zero that means points are collinear.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)

(c) Assertion (A) is true but Reason (R) is false

(d) Assertion (A) is false but Reason (R) is true

Answer:

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Both the Assertion (A) and Reason (R) are true but the Reason (R) is not the correct explanation of the Assertion (A).

Section B

(Section B consists of 5 questions of 2 marks each)

Question 21.

Prove that

(cosec A – sin A) (sec A – cos A) = \(\frac{1}{tan A + cot A}\)

Or

If cos θ + sin θ = √2 cos θ, then prove that cos θ – sin θ = √2 sin θ. [2]

Solution:

LHS = (cosec A – sin A) (sec A – cos A)

Or

Let cos θ – sin θ = x …(i)

Given, cos θ + sin θ = √2 cos θ …….(ii)

On squaring Eqs. (i) and (ii) and adding, we get

cos²θ + sin²θ – 2 cos0 ⋅ sin0 + cos²θ + sin²θ + 2 cosθ ⋅ sinθ = x² + 2 cos²θ

⇒ 1 + 1 = x² + 2 cos²θ

⇒ 2 = x² + 2(1 – sin²θ)

⇒ 2 = x² + 2 – 2 sin²θ

⇒ x² = 2 sin²θ

⇒ x = √2 sin θ

Hence, cos θ – sin θ = √2 sin θ

Alternate :

Given, cos θ + sin θ = √2 cos θ

sin θ = √2 cos θ – cos θ

⇒ sin θ = cos0(√2 – 1)

⇒ (√2 + 1) sin θ = cos θ (√2 – 1) (√2 + 1)

⇒ √2 sin θ + sinθ = cos θ(2 – 1)

⇒ √2 sin θ = cos θ – sin θ

Hence, cos θ – sin θ = √2 sin θ

Question 22.

In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80°, 40° at the centre are shown. Find the area (in cm2) of the shaded region. [2]

Solution:

Given, radius of circle, r = 7 cm

Now, area of shaded region = Area of three sectors

Question 23.

In given ∆ABC, DE || AC. If DC || AP, where point P lies on BC produced, then prove that \(\frac{B E}{E C}=\frac{B C}{C P}\).

Or

In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, show that the two triangles are similar but not congruent. [2]

Sol. Given, in ∆ABC,

DE || AC [given]

So, \(\frac{B E}{E C}=\frac{B D}{D A}\) ……(i)

[by basic proportionality theorem]

DC || AP [given]

So, \(\frac{B C}{C P}=\frac{B D}{D A}\) ……..(ii)

[by basic proportionality theorem]

From Eqs. (i) and (ii), we get

\(\frac{B E}{E C}=\frac{B C}{C P}\)

Hence Proved.

Or

In ∆ABC and ∆DEF, we have ∠B = ∠E, ∠F = ∠C

⇒ ∆ABC ~ ∆DEF

[by AA similarity criterion]

Since, AB and DE are corresponding sides.

But AB = 3DE [given]

We know that two triangles are congruent,

if they have the same shape and size.

But here, AB = 3DE i.e. two triangles are not of same size.

∴ ∆ABC is not congruent to ∆DEF.

Hence, the two triangles are similar but not congruent.

Question 24.

In the given figure, PA is a tangent from an external point P to a circle with centre O. If ∠POB = 125°, then find ∠APO. [2]

Solution:

∠OAP = 90°, AO ⊥ PA

Now, ∠AOP + ∠BOP = 180°

⇒ ∠AOP + 125° =180°

⇒ ∠AOP = 180° – 125° = 55°

Also, in ∆APO,

∠APO + ∠AOP + ∠PAO = 180°

⇒ ∠APO + 55° + 90° = 180°

⇒ ∠APO + 145° = 180°

∴ ∠APO = 180° – 145° = 35°

Question 25.

The students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2 rows less. If 4 students are less in each row, there would be 4 rows more, then find the number of students in the class. [2]

Solution:

Let the number of rows = x

and the number of students in each row = y

Then, the total number of students = xy

When there are 4 more students in each row,

number of students in each row = y + 4

and number of rows = x – 2

Now, total number of students = (x – 2) (y + 4)

Given, (x – 2)(y + 4) = xy

⇒ 4x – 2y = 8

⇒ 2x – y = 4 … (i)

When 4 students are removed from each row, number of students in each row = (y – 4)

and number of rows = (x + 4)

Total number of students = (x + 4) (y – 4)

Given, {x + 4) (y – 4) = xy

⇒ 4y – Ax = 16

⇒ 4(y – x) = 16

⇒ y – x = 4 …(ii)

Adding Eqs. (i) and (ii), we get x = 8

On putting x = 8 in Eq. (ii), we get

y – 8 = 4 ⇒ y = 12

x = 8 and y = 12

Total number of students in the class = (12 × 8) = 96

Section C

(Section C consists of 6 questions of 3 marks each)

Question 26.

If the 3rd and 9th terms of an AP are 4 and – 8 respectively, then which term of this AP is zero?

Or

Find the common difference of an AP, whose first term is 1/2 and the 8th term is 17/6. Also, find the ratio of 4th term and 50th term. [3]

Solution:

Let a be the first term and d be the common difference of an AP.

∵ The nth term of an AP is a_{n} = a + (n – 1)d

∴ a_{3} = a + 2d = 4 [∵ a_{3} = 4, given]…(i)

and a_{9} = a + 8d = -8 [∵ a_{9} = -8, given]…(ii)

On subtracting Eq. (i) from Eq. (ii), we get

6d = -12

⇒ d = \(\frac{-12}{6}\) = -2

On putting the value of cf in Eq. (i), we get

a + 2 × (-2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Let the nth term of this AP be zero,

i.e. a_{n} = 0

⇒ a + (n – 1)d = 0

⇒ 8 + (n – 1) (-2) = 0

⇒ (n – 1)(-2) = -8

⇒ n – 1 = \(\frac{-8}{-2}\) = 4

∴ n = 4 + 1 = 5

Hence, 5th term of this AP is zero.

Or

Let a be the first term and d be the common difference of an AP.

Then, nth term, T_{n} = a + (n – 1) d

Given, 8th term,

T_{8} = \(\frac{17}{6}\) ⇒ a + 7d = \(\frac{17}{6}\)

⇒ \(\frac{1}{2}\) + 7d = \(\frac{17}{6}\)

⇒ 7d = \(\frac{17}{6}\) – \(\frac{1}{2}\)

⇒ 7d = \(\frac{14}{6}\)

⇒ d = \(\frac{1}{3}\)

Now, 4th term, T_{4} = a + 3d

= \(\frac{1}{2}\) +3(\(\frac{1}{3}\)) = \(\frac{1}{2}\) + 1 = \(\frac{3}{2}\)

and 50th term, 7_{50} = a + 49d

= \(\frac{1}{2}\) + 49 × \(\frac{1}{3}\) = \(\frac{101}{3}\)

Required ratio = \(\frac{3/2}{101/6}\) = \(\frac{3}{2}\) × \(\frac{6}{101}\)

= 9 : 101

Question 27.

Find the value(s) of k for which the following equations has equal roots (k – 12)x^{2} + 2(k – 12)x + 2 = 0 [3]

Solution:

Given quadratic equation is

(k – 12)x^{2} + 2(k – 12)x + 2 = 0

On comparing with ax^{2} + bx + c = 0, we get

a = k – 12, b = 2(k – 12) and c = 2

∴ D = b^{2} – 4ac = 4(k – 12)^{2} – 4(k – 12) × 2

= 4(k – 12)[k – 12 – 2]

= 4(k – 12)(k – 14)

The given equation will have equal roots, if

D = 0

⇒ 4(k – 12)(k – 14) = 0

⇒ (k – 12) = 0 and (k – 14) = 0

⇒ k = 12 and k = 14

Hence, the value of k is 12 and 14.

Question 28.

If a and 3 are zeroes of the polynomial x^{2} – 2x – 15, then form a quadratic polynomial whose zeroes are 2α and 2ß.

Or

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers. [3]

Solution:

Let p(x) = x^{2} – 2x -15

On comparing with ax^{2} + bx + c, we get

a = 1, b = -2 and c = -15

Given, α and ß are the zeroes of p(x).

Sum of zeroes,

(α + β) = \(-\frac{b}{a}\)

⇒ α + β = \(-\frac{(-2)}{1}\)

⇒ α + β = 2 …………(i)

and product of zeroes, (αß) = \(\frac{c}{a}\)

⇒ αß = \(\frac{-15}{1}\)

⇒ αß = -15 …(ii)

We have to form a polynomial whose zeroes are 2α and 2β.

∴ Sum of zeroes = 2α + 2β = 2(a + β)

= 2 × 2 = 4 [usingEq. (i)]

and product of zeroes = 2α ⋅ 2 β

= 4 αβ

= 4 × (-15)

= – 60 [using Eq. (ii)]

∴ Required polynomial

= x² – (Sum of zeroes) x + (Product of zeroes)

= x² – 4x + (-60) = x² – 4x – 60

Or

Let the required numbers be x and y, where x > y.

Given, difference of squares of two numbers = 180

We have, x² – y² = 180 …(i)

and also it is given that the square of smaller number

= 8 × larger number

We have, y² = 8x …(ii)

From Eqs. (i) and (ii), we get

x² – 180 = 8x

⇒ x² – 8x – 180 = 0

⇒ x² – 18x + 10x – 180 = 0 [by factorisation]

⇒ x(x – 18) + 10(x – 18) = 0

⇒ (x – 18) (x + 10) = 0

⇒ x – 18 = 0 or x + 10 = 0

⇒ x = 18 or x = -10

Now, if x = 18, then square of smaller number

= 8 × 18 = 144 [from Eq. (ii)]

⇒ Smaller number = ± 12

⇒ Smaller number = 12 or -12

and if x = -10, then square of smaller number

= [8 × (-10)] = – 80, which is not possible as square of a number cannot be negative.

Hence, the required numbers are 18 and 12 or 18 and -12.

Question 29.

In the given figure, PQ is a chord of a circle and PT is tangent at P such that ∠QPT = 60°, then find the measure of ∠PRQ.

Solution:

Take a point Q’ on circle and join PQ’ and QQ’.

Now, ∠OPQ = 90° – 60° = 30°

[∵ ∠OPT = 90°, as radius OP is perpendicular to the tangent]

∠OQP = 30° [angles opposite to equal sides are equal]

In ∆OPQ, using angle sum property of a triangle,

∠POQ + ∠OPQ + OQP = 180°

⇒ ∠POQ + 30°+ 30° =180°

⇒ ∠POQ = 180°- 60° = 120°

⇒ ∠PQ’Q = 60°

[angle subtended by an arc at centre is twice the angle subtended at remaining part of circle]

⇒ ∠PRQ = 180° – ∠PQ’Q = 180°- 60° = 120°

[∵ opposite angles are supplementary in a cyclic quadrilateral PQ’QR]

Question 30.

If(- 3, 2), (1, -2) and (5, 6) are the mid-point of the sides of a triangle, then find the coordinates of the vertices of the triangle. [3]

Solution:

Let A = (x_{1}, y_{1}), B = (x_{2}, y_{2}) and C = (x_{3}, y_{3}) be the coordinates of vertices of the triangle.

Let D (- 3, 2), E (1, -2) and F (5, 6) be the mid-points of the sides BC, CA and AB respectively.

Since, D(- 3,2) is the mid-point of BC

∴ \(\frac{x_2}{x_3}=-3\) and \(\frac{y_2}{y_1}=2\)

⇒ x_{2} + x_{3} = -6 ……..(i)

and y_{2} + y_{3} = 4 ……..(ii)

As, E(1, – 2) is the mid-point of AC

∴ \(\frac{x_1}{x_3}=1\) and \(\frac{y_1}{y_3}=-2\)

⇒ x_{1} + x_{2} = 2 …………(iii)

and y_{1} + y_{3} = -4 …………(iv)

Also, F(5, 6) is the mid-point of AB

∴ \(\frac{x_1}{x_2}=5\) and \(\frac{y_1}{y_2}=6\)

⇒ x_{1} + x_{2} = 10 ……….(v)

and y_{1} + y_{2} = 12 ……….(vi)

On adding Eqs. (i), (iii) and (v), we get

2(x_{1} + x_{2} + x_{3}) = 6

⇒ x_{1} + x_{2} + x_{3} = 3 ……….(vii)

On subtracting Eqs. (i), (iii) and (v) from Eq. (vii) in turn, we get

x_{1} = 9, x_{2} = 1 and x_{3} = -1

On adding Eqs. (ii), (iv) and (vi), we get

2(y_{1} + y_{2} + y_{3}) = -12

⇒ y_{1} + y_{2} + y_{3} = -6 …(viii)

Subtracting Eqs. (ii), (iv) and (vi) from Eq. (viii) in turn, we get

y_{1} = 2, y_{2} = 10 and y_{3} = – 6

Hence, the vertices of ∆ABC are A(9, 2), B(1, 10) and C(- 7, – 6).

Question 31.

If tan A = n tanB and sin A = m sin B, then prove that cos² A = \(\frac{m^2-1}{n^2-1}\)

Solution:

We have to find cos²A in terms of m and n . This means that ∠B is to be eliminated from the given relations.

Now, tan A = n tan B ⇒ tan B = \(\frac{1}{n}\) tan A

Section D

(Section D consists of 4 questions of 5 marks each.)

Question 32.

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°, then find the height of the tower. [5]

Solution:

Let the height of the tower PR be h m, the angle of elevation at point 0 is 30° i.e. ∠PQR = 30° and S be the position of observer after moving 20 m towards the tower.

Question 33.

Draw the graphs of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and X-axis. Find the area of the shaded region.

Or

Places P_{1} and P_{2} are 250 km apart from each other on a national highway. A car starts from P_{1} and another from P_{2} at the same time. If they go in the same direction, then they meet in 5 h and if they go in opposite directions they meet in \(\frac{25}{13}\) h, then find their speeds. [5]

Solution:

We have, 2x+ y = 6 and 2x – y + 2 = 0

Table for equation y = 6 – 2x is

x | 0 | 3 |

y | 6 | 0 |

On plotting the points A(0, 6) and B(3, 0) on a graph paper and join them, we obtain the graph of line represented by the equation 2x + y = 6 as shown in the figure.

On plotting the pointsC(0,2)and D(-1, 0)on the same graph paper and join them, we obtain the graph of line represented by the equation 2x – y + 2 = 0 as shown in the figure.

The two lines intersect at point P( 1, 4).

Thus, x = 1 and y = 4 is the solution of the given system of equations. The area enclosed by the lines and X-axis is shaded part in the figure. Draw PM perpendicular from P on X-axis.

Clearly, we have

PM = y-coordinate of point P(1, 4)

⇒ PM = 4 and DB = 4

Area of the shaded region = Area of A PSD

= \(\frac{1}{2}\) × Base × Height

= \(\frac{1}{2}\)(DB × PM)

= \(\frac{1}{2}\) × 4 × 4

= 8 sq units

Or

Let A and B be the two cars. A starts from P_{1} with constant speed of x km/h and B starts from P_{2} with constant speed of y km/h.

Case I When the two cars move in same directions as shown in figure, the cars meet at the position Q.

Here, P_{1}Q = 5x km, i.e. the distance travelled by car A in 5 h with x km/h speed.

P_{2}Q = 5y km, i.e. the distance travelled by car B in 5 h with y km/h speed.

We have, P_{1}Q – P_{2}Q = 250

5x – 5y = 250

⇒ x – y = 50 …(i)

Case II When two cars move in opposite directions as shown in figure, the cars meet at the position R.

Here, P_{1}R = \(\frac{25}{13}\) x km and P_{2}R = \(\frac{25}{13}\) y km

So, P_{1}R + P_{2}R = 250

\(\frac{25}{13}\) x + \(\frac{25}{13}\) y = 250

⇒ x + y = 130 ………(ii)

On adding Eq. (i) and Eq. (ii), we get 2x = 180

⇒ 2x = 180

⇒ x = 90

On subtracting Eqs. (i) from (ii), we get

⇒ 2y = 80

⇒ y = 40

∴ Their speeds are 90 km/h and 40 km/h.

Question 34.

ABCD is a trapezium with AB || DC. E and F are two points on non-parallel sides AD and BC respectively, such that EF is parallel to AB. Show that \(\frac{A E}{E D}=\frac{B F}{F C}\) [5]

Solution:

Question 35.

Find the unknown entries a, b, c, d, e and f in the following distribution of heights of students in a class

Height (in cm) | Frequency | Cumulative frequency |

150-155 | 12 | a |

155-160 | b | 25 |

160-165 | 10 | c |

165-170 | d | 43 |

170-175 | e | 48 |

175-180 | 2 | f |

Total | 50 |

Or

Find the median for the following frequency distribution

Class | 0-8 | 8-16 | 16-24 | 24-32 | 32-40 | 40-48 |

Frequency | 8 | 10 | 16 | 24 | 15 | 7 |

Solution:

The cumulative frequency table for the given continuous distribution is given below

Height (in cm) | Frequency (f) | Cumulative frequency (given) |
Cumulative frequency
(cf) |

150-155 | 12 | a | 12 |

155-160 | b | 25 | 12 + 5 |

160-165 | 10 | c | 22 + 5 |

165-170 | d | 43 | 22 +b + d |

170-175 | e | 48 | 22 + 5 + d + e |

175-180 | 2 | f | 24 + b + d + e |

Total | 50 |

On comparing last two tables, we get

a = 12

12 + b = 25

⇒ b = 25 – 12 = 13

22 + b = c

⇒ c = 22 + 13 = 35

22 + b + d =43

⇒ 22 + 13 + d = 43

⇒ d = 43 – 35 = 8

22 + b + d + e = 48

⇒ 22 + 13 + 8 + e = 48

⇒ e = 48 – 43 = 5

and 24 + b + d + e = f

⇒ 24 + 13 + 8 + 5 = f

f = 50

Or

We prepare the cumulative frequency table, as given below.

Section E

(Case study based questions are compulsory)

Question 36.

Traffic Light Signal

In our daily life we all see traffic lights. A traffic controller set the timmings of traffic lights in such a way that all light are not green at the same time or specially not in the rush hour. It may create problem in an hour because lights are for few minutes only. So, he take the timmings of nearby places in same area and calculate 1 cm of all traffic stops and he easily manage the traffic by increasing the duration or set at different times. There are two traffic lights on a particular highway which shows green light on time of 90 sec and 144 sec, respectively.

On the basis of above information, answer the following questions.

(i) Find the HCF of the time interval between two green lights. [2]

Or

Find the LCM of the time interval between two green lights.[2]

(ii) Factor tree is used for determining the ……. [1]

(iii) Identify whether the term HCF (a, b) × LCM (a, b) – ab is true or not. [1]

Solution:

(i) Using the factor tree for the prime factorisation of 90 and 144, we have

90 = 2 × 3² × 5and 144 = 2^{4} × 3^{2}

To find the HCF, we list the common prime factors and their least exponents in 90 and 144 as under

Table

∴ HCF = 2^{1} × 3^{2} = 2 × 9 = 18

Or

To find the LCM, we list all prime factors of 90 and 144 and their greatest exponents as follows

Table

∴ LCM =24 × 32 × 51 =16 × 9 × 5 = 720

(ii) Factor tree is used for determining the prime factor.

(iii) The term is true,

HCF (a, b) × LCM (a, b) = ab

If a = 90, b = 144, then

LCM (a, b) = 720 and HCF (a, b) = 18

Now, LCM × HCF = 720 × 18 = 12960

and a × b = 90 × 144 = 12960

Thus, HCF (a, b) × LCM (a, b) = ab = 12960.

Question 37.

Cylindrical Wooden Article

A wooden article was made by scooping out a hemisphere from one end of a cylinder and cone from the other end as shown in the figure. If the height of the cylinder is 40 cm, radius of cylinder is 7 cm and height of the cone is 24 cm.

On the basis of above information, answer the following questions.

(i) Find the slant height of the cone and volume of hemisphere. [2]

Or

Find the total volume of the article. [2]

(ii) Find the curve surface area of cone. [1]

(iii) Find the surface area of the article. [1]

Solution:

Given, height of the cylinder (H) = 40 cm

Radius of the cylinder (r) = 7 cm

Radius of the hemisphere (r) = 7 cm

Radius of the cone (r) = 7 cm

Height of the cone (h) = 24 cm

(i) Slant height of the cone (l) = \(\sqrt{h^2 + r^2}\)

= \(\sqrt{(24)^2 + (7)^2}\)

= \(\sqrt{576 + 49}\)

= \(\sqrt{625}\)

= 25 cm

and volume of hemisphere

= \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3}\) × \(\frac{22}{7}\) × (7)^{3}

= \(\frac{44}{3}\) × 49

= \(\frac{2156}{3}\) = 718.67 cm³

Or

Volume of the article

= Volume of the cylinder – Volume of the cone – Volume of the hemisphere

(ii) Curve surface area of cone = πrl

= \(\frac{22}{7}\) × 7 × 25 = 22 × 25 = 550 cm²

(iii) Total surface area of the article

= Curved surface area of the cylinder

+ Curved surface area of the cone

+ Surface area of the hemisphere

= 2πrH + πrl + 2πr² = πr [2H + l + 2r]

= \(\frac{22}{7}\) × 7 × [2 × 40 + 25 + 2 × 7]

= 22 × [80 + 25 + 14] = 22 × 119 = 2618 cm²

Question 38.

On the basis of above information answer the following questions.

(i) Find the probability of getting a king of red colour. [1]

(ii) Find the probability of getting a face card. [1]

(iii) Find the probability of getting a jack of hearts. [2]

Or

Find the probability of getting a spade. [2]

Solution:

Total number of cards in one deck of cards is 52.

∴ Total number of outcomes =52

(i) Let E_{1} = Event of getting a king of red colour

∴ Number of outcomes favourable to £-, = 2

[∴ there are four kings in a deck of playing cards out of which two are red and two are black]

Hence, the probability of getting a king of red colour.

P(E_{1}) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(ii) Let E_{2} = Event of getting a face card

∴ Number of outcomes favourable to E_{2} = 12

[∴ in a deck of cards, there are 12 face cards, namely 4 kings, 4 jacks, 4 queens]

Hence, the probability of getting a face card,

P(E_{2}) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(iii) Let E_{3}= Event of getting a jack of heart

∴ Number of outcomes favourable to E_{3} = 1

[∵ there are four jack cards in a deck, namely 1 of heart, 1 of club, 1 of spade and 1 of diamond]

Hence, the probability of getting a jack of heart,

P(E_{3}) = \(\frac{1}{52}\)

Or

Let E_{4} = Event of getting a spade

∴ Number of outcomes fourable to E_{4} = 13

[∵ in a deck of cards, there are 13 spades, 13 clubs, 13 hearts and 13 diamonds]

Hence, the probability of getting a spade,

P(E_{4}) = \(\frac{13}{52}\) = \(\frac{1}{4}\)