Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 8 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 8 with Solutions

Time: 3 hrs

Max. Marks: 80

Instructions

- This question paper has 5 Sections A-E.
- Section A has 20 MCQs carrying 1 mark each.
- Section B has 5 questions carrying 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 Case Based integrated units of assessment (4 marks each).
- 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.
- 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.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 sec, B in 308 sec and C in 198 sec, all starting at the same point. After what time will they meet again at the starting point? [1]

(a) 46 min 12 sec

(b) 12 min 46 sec

(c) 44 min 12 sec

(d) 46 min 44 sec

Answer:

(a) 46 min 12 sec

Required time = LCM(252, 308, 198) = 2772 sec

Now, 1 min = 60 sec ⇒ 1 sec = \(\frac{1}{60}\) min

∴ 2772 sec = \(\frac{2772}{60}\) min = 46 min 12 sec

Question 2.

In the given figure, PQ || NO. If \(\frac{M P}{P N}\) = \(\frac{4}{13}\) and QO = 15.6 cm, then the value of MQ is [1]

(a) 5.8 cm

(b) 6.2 cm

(c) 4.8 cm

(d) 8.4 cm

Answer:

(c) 4.8 cm

In ∆MNO, we have

PQ || NO

Question 3.

The sum of all natural numbers from 1 to 100, is [1]

(a) 5050

(b) 5000

(c) 5100

(d) 5200

Answer:

(a) 5050

We know that natural numbers forms an AP.

Let s = 1 + 2 + 3 + ……. + 100

Here, a = 1, n = 100 and l = 100

∵ S_{n} = \(\frac{n}{2}\)(a + l)

∵ S_{100} = \(\frac{100}{2}\) (1 + 100) = 50 × 101 = 5050

Question 4.

If one of the zeroes of a quadratic polynomial (k – 1)x^{2} + kx + 1 is -3, then the value of k is [1]

(a) \(\frac{3}{4}\)

(b) 1

(c) \(\frac{4}{3}\)

(d) 2

Answer:

(c) \(\frac{4}{3}\)

Let p(x) = (k – 1)x^{2} + kx + 1

Since, -3 is a zero of polynomial.

∴ p(-3) = 0

∴ (k – 1)(-3)^{2} + k (-3) + 1 = 0

⇒ 9(k – 1) – 3k + 1 = 0

⇒ 9k – 9 – 3k + 1 = 0

⇒ 6k – 8 = 0

⇒ 6k = 8

∴ k = \(\frac{8}{6}\) ⇒ k = \(\frac{4}{3}\)

Question 5.

What is the nature of roots of the quadratic equation 5y^{2} – 4y + 3 = 0? [1]

(a) Four real roots

(b) Two real roots

(c) No real root

(d) One real root

Answer:

(c) No real root

Given, quadratic equation is 5y^{2} – 4y + 3 = 0.

On comparing with ay^{2} + by + c = 0, we get

a = 5, b = -4 and c = 3

Now, D = b^{2} – 4ac

= (-4)^{2} – 4(5)(3)

= 16 – 60 = -44 < 0

Since, D < 0, so the given quadratic equation has no real root.

Question 6.

PAQ is tangent to a circle with centre O at point A. If ∠OBA = 40°, ∠BOA = 100°, then ∠BAP is equal to [1]

(a) 45°

(b) 60°

(c) 50°

(d) 55°

Answer:

(c) 50°

In ∆OAB,

OB = OA [radius of circle]

⇒ ∠OAB = ∠OBA = 40°

Now, ∠OAP = 90°

[since, radius ⊥ tanget at point of contact]

∠BAP = ∠OAP – ∠OAB

= 90° – 40° = 50°

Question 7.

If sec A = \(\frac{17}{8}\), then the value of \(\frac{3-4 \sin ^2 A}{4 \cos ^2 A-3}\) is [1]

(a) \(\frac{33}{611}\)

(b) \(\frac{-33}{611}\)

(c) \(\frac{611}{33}\)

(d) \(\frac{-611}{33}\)

Answer:

(a) \(\frac{33}{611}\)

Given, sec A = \(\frac{17}{8}\)

Question 8.

The point at which the pair of equations 4^{x + y} = 256 and 256^{x – y} = 4 will intersect, is [1]

(a) (\(\frac{17}{8}\), \(\frac{15}{8}\))

(b) (\(\frac{15}{8}\), \(\frac{17}{8}\))

(c) (\(\frac{8}{17}\), \(\frac{8}{15}\))

(d) (\(\frac{8}{15}\), \(\frac{8}{17}\))

Answer:

(\(\frac{17}{8}\), \(\frac{15}{8}\))

Given, 4^{x + y} = 256

⇒ 4^{x + y} = (4)^{4}

On comparing the powers, we get

x + y = 4 …… (i)

Also, (256)^{x – y} = 4 ⇒ (4^{4})^{x – y} = (4)^{1}

On comparing the powers, we get

4(x – y) = 1

⇒ x – y = \(\frac{1}{4}\) ……..(ii)

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

2x = 4 + \(\frac{1}{4}\) = \(\frac{17}{4}\)

∴ x = \(\frac{17}{8}\)

On putting x = \(\frac{17}{8}\) in Eq. (i), we get

\(\frac{17}{8}\) + y = 4

∴ y = 4 – \(\frac{17}{8}\) = \(\frac{15}{8}\)

Question 9.

The area swept by the minute hand of a clock of length 15 cm in 10 min is [1]

(a) \(\frac{75}{2} \pi\) cm^{2}

(b) \(\frac{2}{75} \pi\) cm^{2}

(c) \(\frac{75}{4} \pi\) cm^{2}

(d) \(\frac{75}{6} \pi\) cm^{2}

Answer:

(a) \(\frac{75}{2} \pi\) cm^{2}

Angle made by minute hand in 10 min

= \(\frac{360^{\circ} \times 10}{60}\) = 60°

∴ Area swept by the minute hand in 10 min

= \(\frac{\pi r^2}{360^{\circ}}\) × 60°

[∵ length of minute hand = r = 15 cm]

Question 10.

The pair of linear equation 2x + ky – 3 = 0, 6x + \(\frac{2}{3}\)y + 7 = 0 has a unique solution, if [1]

(a) k = \(\frac{2}{3} \pi\)

(b) k ≠ \(\frac{2}{3} \pi\)

(c) k ≠ 5

(d) k ≠ \(\frac{75}{6} \pi\)

Answer:

(d) k ≠ \(\frac{75}{6} \pi\)

The pair of linear equations 2x + ky – 3 = 0 and 6x + \(\frac{2}{3}\)y + 7 = 0 has a unique solution, if

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

⇒ \(\frac{2}{6}\) ≠ \(\frac{k}{2 / 3}\)

⇒ \(\frac{1}{3}\) ≠ \(\frac{3 k}{2}\) ⇒ k ≠ \(\frac{2}{9}\)

Question 11.

The point of intersection of the coordinate axes is [1]

(a) X-axis

(b) F-axis

(c) origin

(d) (1, 2)

Answer:

(c) origin

Since, X-axis and Y-axis intersect at (0, 0).

So, point of intersection of the coordinate axes is origin.

Question 12.

If 3 tan θ = 5, then the value of \(\frac{3 \sin \theta-5 \cos \theta}{3 \sin \theta+5 \cos \theta}\) is [1]

(a) 1

(b) 2

(c) -1

(d) 0

Answer:

(d) 0

Given, 3 tan θ = 5 ⇒ tan θ = \(\frac{5}{3}\)

Question 13.

A wire is in the form of a circle of radius 28 cm. It is re-bent into a square form. The length of the side of the square is [1]

(a) 44 cm

(b) 50 cm

(c) 46 cm

(d) 48 cm

Answer:

(a) 44 cm

Let the side of the square be x cm.

Length of the wire = Circumference of the circle

= 2πr = 2 × \(\frac{22}{7}\) × 28

= 176 cm [∵ r = 28 cm]

∵ Perimeter of the square = Length of the wire

⇒ 4x = 176

∴ x = \(\frac{176}{4}\) = 44

Hence, length of the side of the square is 44 cm.

Question 14.

If the difference of mode and median of a data is 24, then the difference of median and mean is [1]

(a) 12

(b) 24

(c) 8

(d) 36

Answer:

(a) 12

We have, Mode – Median = 24

We know that, Mode = 3 Median – 2 Mean

∴ Mode – Median = 2 Median – 2 Mean

⇒ 24 = 2(Median – Mean)

⇒ Median – Mean = 12

Question 15.

In tossing 3 coins, the probability of getting at least 2 head is [1]

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

(b) \(\frac{2}{3}\)

(c) \(\frac{1}{3}\)

(d) \(\frac{1}{4}\)

Answer:

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

All possible outcomes are

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Favourable outcomes are { HHT, HTH, THH, HHH}

∴ P(E) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 16.

If, sin A + sin^{2} A = 1, then (cos^{2}A + cos^{4}A) is equal to [1]

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

(b) 1

(c) 2

(d) 3

Answer:

(b) 1

Given, sin^{2}A + sin^{2}A = 1 ………. (i)

⇒ sin A = 1 – sin^{2}A

⇒ sin A = cos^{2}A …….. (ii)

Now, cos^{2}A + cos^{4}A = sin A + sin^{2}A

[∵ using cos^{2}A = sinA from Eq. (ii)]

= 1 [using Eq.(i)]

Question 17.

Three metallic solid cubes whose edges are 4 cm, 5 cm and 6 cm are melted and formed into a single cube. The edge of the cube formed is (in cm) [1]

(a) 3(15)^{1/3}

(b) 3(15)^{1/3}

(c) 3(15)^{1/4}

(d) 3(15)^{1/5}

Answer:

(a) 3(15)^{1/3}

Let the length of the edge of the new cube formed be x cm.

Then, volume of the new cube

= Sum of the volume of three metallic cubes

⇒ x^{3} = (4)^{3} + (5)^{3} + (6)^{3}

= 64 + 125 + 216 = 405

⇒ x^{3} = 5 × 3^{3} × 3 ⇒ x = 3(15)^{1/3}

Hence, edge of the cube is 3(15)^{1/3} cm.

Question 18.

The value of k such that the quadratic polynomial x^{2} – (k + 6)x + 2(2k + 1) has sum of the zero as half of their product, is [1]

(a) 2

(b) 3

(c) -5

(d) 5

Answer:

(d) 5

On comparing given quadratic polynomial with ax^{2} + bx + c whose roots are α and β,

α + β = \(\frac{-b}{a}\) = \(\frac{-\{-(k+6)\}}{1}\) = k + 6

αβ = \(\frac{c}{a}\) = \(\frac{2(2 k+1)}{1}\) = 2(2k + 1)

But, \(\frac{\alpha \beta}{2}\) = α + β

⇒ \(\frac{2(2 k+1)}{2}\) = k + 6 ⇒ k = 5

Direction 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) If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.

Reason (R) Line segment joining the mid-point of any two sides of a triangle is parallel to the third side. [1]

(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).

Assertion is basic proportionality theorem and Reason is mid-point theorem. But mid-point theorem is not the correct explanation of BPT.

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

Question 20.

Assertion (A) Three points A, B and C are such that AB + BC > AC, then they are collinear. [1]

Reason (R) Three points are collinear if they lie on a straight line.

(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:

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

Three points A, B and C are collinear, if and only if AB + BC = AC but here AB + BC > AC

∴ A, B and C are not collinear.

Hence, Assertion (A) is false but Reason (R) is true.

Section B

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

Question 21.

In ∆ABC, D and E are points on the sides AB and AC respectively, such that DE || BC. If AD = -4x -3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, then find the value of x. [2]

Answer:

Given, in ∆ABC, DE || BC

⇒ 20x^{2} – 12x + 9 – 15x = 24x^{2} – 21x – 8x + 7

⇒ 4x^{2} – 2x – 2 = 0

⇒ 2x^{2} – x – 1 = 0

[dividing on both sides by 2]

⇒ 2x^{2} – 2x + x – 1 = 0

[by splitting the middle term]

⇒ 2x (x – 1) + 1 (x – 1) = 0

⇒ (2x + 1) (x – 1) = 0

⇒ x = –\(\frac{1}{2}\) or x = 1

If x = –\(\frac{1}{2}\) then AD = 4 × \(\left(-\frac{1}{2}\right)\) – 3 = -5 < 0

[not possible; since, length cannot be negative]

Hence, x = 1 is the required value.

Question 22.

From an external point A, two tangents AP and AQ are drawn to a circle with centre O. At one point B on the circle tangent is drawn, which intersects AP and AQ at C and D, respectively. If AP = 20 cm, find the perimeter of the ∆ACD. [2]

Answer:

Two tangents AP and AQ are drawn to a circle with centre 0 from an external point A

Perimeter of ∆ACD = AC + CD + AD

= AC + CB + BD + AD

= AC + CP + DQ + AD

[∵ CB = CP and BD = DQ]

= AP + AQ

= 2 AP

[AP = AQ, tangents from external point to a circle are equal]

= 2(20)

= 40 cm

Question 23.

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the area of the sector. [2]

Answer:

Let ACB be the given arc subtending an angle of 60° at the centre, As shown in figure.

Then, r = 21 cm and θ = 60°

Area of the sector OACBO

= \(\frac{\theta}{360^{\circ}}\) × πr^{2}

= \(\frac{60^{\circ}}{360^{\circ}}\) × \(\) × 21 × 21

= 231 cm^{2}

Question 24.

The difference between two numbers is 26 and one number is three times the other number. Find the numbers. [2]

Answer:

Let the two numbers be x and y (x > y).

According to the question,

x – y = 26 …….(i)

and x = 3y …….(ii)

On substituting the value of x from Eq. (ii) in Eq. (i), we get

3y – y = 26

⇒ 2y = 26

⇒ y = 13

On substituting y = 13 in Eq. (ii), we get

x = 3 × 13

⇒ x = 39

Hence, the two numbers are 39 and 13.

Or

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5, then find the numbers. [2]

Answer:

Let the two numbers be x and y.

Then, by first condition, ratio of these two numbers = 5 : 6

x : y = 5: 6

⇒ \(\frac{x}{y}\) = \(\frac{5}{6}\) ⇒ y = \(\frac{6 x}{5}\) …… (i)

and by second condition, when 8 is subtracted from each of the numbers, then ratio becomes 4 : 5.

\(\frac{x-8}{y-8}\) = \(\frac{4}{5}\)

⇒ 5x – 4y = 8

Now, put the value of y in Eq. (ii), we get

5x – 4\(\left(\frac{6 x}{5}\right)\) = 8

⇒ 25x – 24x = 40 ⇒ x = 40

Put the value of x in Eq. (i), we get

y = \(\frac{6}{5}\) × 40 = 6 × 8 = 48

Hence, the required numbers are 40 and 48.

Question 25.

Prove that \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ. [2]

Answer:

Or

Prove that cot A + tan A = sec A cosec A. [2]

Answer:

Section C

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

Question 26.

The angles of a triangle are in AP. If the greatest angle equals to the sum of the other two, then find the angles. Also, find these angles are multiple of which angle. [3]

Answer:

Let the angles of a triangle are (a – d), a and (a + d).

Since, sum of the three angles of a triangle = 180°

∴ a – d + a + a + d = 180°

⇒ 3a = 180°

⇒ a = 60°

So, the angles are 60° – d, 60° and 60° + d.

According to the question,

Greatest angle = Sum of two smaller angles

∴ 60° + d = 60° – d + 60°

⇒ 2d = 60°

∴ d = 30°

Hence, the required angles are 60° – 30°, 60° and 60° + 30° i.e. 30°, 60° and 90°.

Here, we see that above angles are multiple of 30°.

Question 27.

If α and β are the zeroes of the quadratic polynomial f(x) = x^{2} + x -2, then find the polynomial whose zeroes are 2α + 1 and 2β + 1. [3]

Answer:

Given, quadratic polynomial

f(x) = x^{2} + x – 2

α + β = -1 and α ∙ β = -2

Sum of zeroes i.e. (2α + 1, 2β + 1)

= 2α + 1 + 2β + 1 = 2(α + β + 1)

= 2(-1 + 1)

= 0

Product of zeroes i.e. (2α + 1, 2β + 1)

= (2α + 1)(2β + 1)

= 4αβ + 2(α + β) + 1

= 4(-2) + 2(-1) + 1

= -8 – 2 + 1

= -9

The quadratic polynomial whose zeroes are (2α + 1), (2β + 1) is

= x^{2} – (sum of zeroes) x + product of zeroes

= x^{2} – 0 × x + (-9)

= x^{2} – 9

Question 28.

Prove that \(\sqrt{n}\) is not a rational number, if n is not a perfect square. [3]

Answer:

Let \(\sqrt{n}\) be a rational number.

Then, assume \(\sqrt{n}\) = \(\frac{p}{q}\)

[where p, q are coprime and q ≠ 0]

⇒ n = \(\frac{p^2}{q^2}\) [squaring on both sides]

⇒ p^{2} = nq^{2} …….. (i)

⇒ n divides p^{2}

⇒ n divides p [by using theorem]…………. (ii)

Let p = nm ⇒ p^{2} = n^{2}m^{2} [squaring on both sides]

On putting the value of p^{2} in Eq. (i), we get

n^{2}m^{2} = nq^{2} =q^{2} ⇒ nm^{2}

⇒ n divides q^{2}

⇒ n divides q …….. (iii)

From Eq. (ii), n divides p

and from Eq. (iii) n divides q.

It means n is a common factor of both p and q.

This contradicts the assumption that p and q are co-prime.

So, our supposition is wrong.

Hence, \(\sqrt{n}\) cannot be a rational number.

Or

The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Find the length of the longest rod that can measure the three dimensions of the room exactly.

Answer:

Given, length of the room = 8 m 25 cm = 825 cm

[∵ 1m = 100 cm]

Breadth of the room = 6m 75 cm = 675 cm

and height of the room = 4m 50 cm = 450 cm

Clearly, the length of the longest rod (in cm) is the

HCF of 825, 675 and 450.

Thus, 825 = 3 × 5^{2} × 11, 675 = 3^{3} × 5^{2}

and 450 = 2 × 3^{2} × 5^{2}

Now, HCF (825, 675, 450) = Product of the smallest power of each common prime factor = 3 × 5^{2} = 75

Hence, the required length of the longest rod is 75 cm.

Question 29.

In the given figure, from an external point P, a tangent PT and a line segment PAB drawn to a circle with centre O. ON is perpendicular on the chord AB. [3]

Prove that

(a) PA – PB = PN^{2} – AN^{2}

(b) PN^{2} – AN^{2} = OP^{2} – OT^{2}

(c) PA ∙ PB = PT^{2}

Answer:

Given, PT is a tangent drawn from an external point P and a line segment P48 is drawn to a circle with centre O. ON is perpendicular on the chord AB.

To prove

(i) PA ∙ PB = PN^{2} – AN^{2}

(ii) PN^{2} – AN^{2} = OP^{2} – OT^{2}

(iii) PA ∙ PB = PT^{2}

Proof

(i) PA ∙ PB = (PN – AN)(PN + BN)

⇒ PA ∙ PB = (PN – AN)(PN + AN)

[∵ AN = BN as ON is perpendicular to chord AB and so on bisect it]

= PN^{2} – AN^{2}

(ii) PN^{2} – AN^{2} = (OP^{2} – ON^{2}) – AN^{2}

[∵ In ∆ONP, OP^{2} = ON^{2} + PN^{2}]

⇒ PN^{2} – AN^{2} = OP^{2} – ON^{2} – AN^{2}

= OP^{2} – (ON^{2} + AN^{2}) = OP^{2} – OA^{2}

[∵ In ∆ONA, OA^{2} = ON^{2} + AN^{2}]

= OP^{2} – OT^{2}

[∵ OA = OT = radii of circle]

(iii) From parts (i) and (ii), we get

PA ∙ PB = OP^{2} – OT^{2}

⇒ PA ∙ PB = PT^{2} [∵ In ∆OTP, OP^{2} = OT^{2} + PT^{2}]

⇒ PT^{2} = OP^{2} – OT^{2}]

Hence proved.

Question 30.

Three chairs and two tables cost ₹ 1850. Five chairs and three tables cost ₹ 2850. Find the cost of seven chairs and three tables. [3]

Answer:

Let the cost of one chair and one table are ₹ x and ₹ y, respectively. Then,

According to the question.

3x + 2y = 1850 ………. (i)

and 5x + 3y = 2850 ……. (ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 2 and then subtracting Eq. (ii) from Eq. (i), we get

3(3x + 2y) – 2 (5x + 3y) = 1850 × 3 – 2850 × 2

⇒ 9x – 10x = 5550 – 5700

⇒ -x = -150

⇒ x = 150

On putting x =150 in Eq. (i). we get

3 × 150 + 2y = 1850

⇒ 450 + 2y = 1850

⇒ 2y = 1850 – 450 = 1400

⇒ y = \(\frac{1400}{2}\) = 700

∴ Cost of one chair is ₹ 150 and cost of one table is ₹ 700.

Hence, the cost of seven chairs and three tables

= 7x + 3y = 7 × 150 + 3 × 700

= 1050 + 2100

= ₹ 3150

Or

The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2. Find the number. How many such numbers are there? [3]

Answer:

Let the ten’s and the units digit be x and y respectively.

Then, the number = 10x + y

After reversing the order of the digits, the number is 10y + x.

According to the question,

(10x + y) + (10y + x) = 66

⇒ 11x + 11y = 66

⇒ x + y = 6 ……… (i)

Also, it is given that the digits differ by 2, therefore either

x – y = 2 ……..(ii)

or y – x = 2 ……… (iii)

If x – y = 2 then adding Eqs. (i) and (ii), we get

2x = 8 ⇒ x = 4

From Eq. (i).

4 + y = 6 ⇒ y = 2

Thus, the number = 10 × 4 + 2 = 42

If y – x = 2 then adding Eq. (i) and (iii), we get

2y = 8 ⇒ y = 4

From Eq. (i),

x + 4 = 6 ⇒ x = 2

Thus, the number = 10 × 2 + 4 = 24

Hence, there are two such numbers 42 and 24.

Question 31.

An army pilot is flying an aeroplane at an altitude of 1800 m observes some suspicious activity of two ships which are sailing towards it in the same direction and immediately report it to the navy chief. The angles of depression of the ships as observed from the aeroplane are 60° and 30°, respectively. Find the distance between two ships. [3]

Answer:

Let the aeroplane is at B and let the two ships are at point C and D such that their angles of depression from B are 60° and 30°, respectively.

We have, AB = 1800 m

Hence, the distance between the two ships is 1200\(\sqrt{3}\)m.

Section D

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

Question 32.

A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC [3]

Answer:

Since, the sides of quadrilateral ABCD. i.e. AB, BC, CD and DA touch the circle at P, Q, R and S respectively, and the length of two tangents to a circle from an external point are equal.

∴ AP = AS and BP = BQ

DR = DS and CR = CQ

Adding them, we get

(AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)

⇒ AB + CD = BC + DA

Or

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see figure). Find the length of TP. [3]

Answer:

JoinOT. Let it intersect PQ at the point R. Then, ∆TPQ is isosceles and TO is the angle bisector of ∠PTQ. So, OT perpendicular PQ and, therefore OT bisects PQ which gives PR = RQ = 4 cm

Also, QR = \(\sqrt{O P^2-P R^2}\) = \(\sqrt{5^2-4^2}\) cm = 3 cm

Now, ∠TPR + ∠RPO = 90° = ∠TPR + ∠PTR

So, ∠RPO = ∠PTR

Therefore, right angled ∆TRP is similar to the right angled ∆PRO by AA similarity.

⇒ \(\frac{T P}{P O}\) = \(\frac{R P}{R O}\) i.e., \(\frac{T P}{5}\) = \(\frac{4}{3}\) or TP = \(\frac{20}{3}\) cm

Let TP = x and RT = y.

Then, in right angle ∆PRT, by Pythagoras theorem.

x^{2} = y^{2} + 16 (taking right angled ∆PRT) …….. (i)

(taking right angled ∆OPT) ………. (ii)

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

25 = 6y – 7 or y = \(\frac{32}{6}\) = \(\frac{16}{3}\)

Therefore, x^{2} = \(\left(\frac{16}{3}\right)^2\) + 16

= \(\frac{16}{9}\)(16 + 9) = \(\frac{16 \times 25}{9}\) [From Eq.(i)]

or x = \(\frac{20}{3}\)

Question 33.

Find the median distribution in the following [3]

Monthly Consumption of electricity (in units) | Number of consumers |

65 – 84 | 4 |

85 – 104 | 5 |

105 – 124 | 13 |

125 – 144 | 20 |

145 – 164 | 14 |

165 – 184 | 7 |

185 – 204 | 4 |

Answer:

The given frequency distribution is discontinuous. So, converting it into a continuous distribution using adjustment factor of \(\frac{85-84}{2}\) or 0.5, we get the following cumulative frequency table.

Here, N = 67

Now, \(\frac{N}{2}\) = \(\frac{67}{2}\) = 33.5

Since, the cumulative frequency just greater than 33.5 is 42 and corresponding class interval is 124.5—144.5

∴ Median class is 124.5 – 144.5.

Here, l = 124.5, f = 20, cf = 22, \(\frac{N}{2}\) = 33.5

and h = 20

Median = l + \(\left(\frac{N / 2-c f}{f}\right)\) × h

= 124.5 + \(\left(\frac{33.5-22}{20}\right)\) × 20

= 12.45 + 11.5 = 136

Or

Find the mode of the following distribution [3]

Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |

Number of students | 4 | 6 | 7 | 12 | 5 | 6 |

Answer:

Given, distribution table is

The highest frequency in the given distribution is 12, whose corresponding class is 30 – 40.

Thus, 30 – 40 is the required modal class.

Here, l = 30, f_{1} = 12, f_{0} = 7, f_{2} = 5 and h = 10

∴ Mode = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h

= 30 + \(\frac{12-7}{2 \times 12-7-5}\) × 10

= 30 + \(\frac{50}{24-12}\)

= 30 + \(\frac{50}{12}\) = 30 + 4.14

= 34.17

Hence, mode of the given distribution is 34.17.

Question 34.

A quadratic equation can be defined as an equation. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax^{2} + bx + c =0, where a, b and c are real numbers and a ≠ 0. Every quadratic equation has two roots depending on the nature of its discriminant, D = b^{2} – 4ac. [3]

Based on the above information, answer the following questions

(i) Which of the quadratic equations have no real roots, if the quadratic equations are

-4x^{2} + 7x – 4 = 0 and 2x^{2} + 3x – 2 = 0

Answer:

Given, quadratic equation -4x^{2} + 7x – 4 = 0

On comparing with ax^{2} + bx + c = 0, we get a = -4, b = 1 and c = -4

Discriminant,

D = b^{2} – 4ac

= (7)^{2} – 4(-4)(-4)

= 49 – 64 = -15 < 0

Thus, we know that when D < 0 the quadratic equation has no real roots.

Hence, the above quadratic equation has no real roots.

Now, for 2x^{2} + 3x – 2 = 0,

Here, a = 2, b = 3 and c = -2

D = (3)^{2} – 4(2)(-2)

= 9 + 16 = 25 > 0

Thus, D > 0

Hence, the quadratic equation have two distinct real roots.

(ii) Which of the quadratic equations have irrational roots, if the quadratic equations are

4x^{2} – 7x + 3 = 0 and 6x^{2} – 3x – 5 = 0

Answer:

We know that, when discriminant is not a perfect square, then ax^{2} + bx + c = 0 has irrational roots.

For 4x^{2} – 7x + 3 = 0

a = 4, b = -7 and c = 3

D = (-7)^{2} – 4(4)(3)

= 49 – 48 = 1 [it’s a perfect square]

Therefore, the quadratic equation have no irrational roots.

Now, for 6x^{2} – 3x – 5 = 0,

a = 6, b = -3 and c = -5

D = (-3)^{2} – 4(6)(-5)

= 9 + 120 = 129

[it’s not a perfect square]

Therefore, the quadratic equation have irrational roots.

Question 35.

The volume of two spheres are in the ratio 64 : 27. Find their radii, if the sum of their radii is 21 cm. [3]

Answer:

Let the volume of two spheres are V_{1} and V_{2}, respectively and corresponding radii are r_{1} and r_{2}.

On putting r_{2} = 9 cm in Eq. (i), we get

r_{1} = \(\frac{4}{3}\) × 9 = 12 cm

Hence, the radii of the two spheres are 12 cm and 9 cm.

Section E

(Case Study Based Questions)

(Section E consists of 3 questions. All are compulsory.)

Question 36.

Fishing

Reshma is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from the initial and 2.4 m from a point directly under the tip of the rod.

Based on the above information, answer the following questions.

(i) Find the area of triangle formed in the question. (1)

(ii) Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? (2)

Answer:

(i) Area of right angled triangle = \(\frac{1}{2}\) × BC × AB

= \(\frac{1}{2}\) × 2.4 × 1.8

= 2.16 m^{2}

(ii) Let the tip of her fishing rod be A. So, its distance from surface of water B is AB = 1.8 m

Again, let C be the point at 2.4 m away from B.

Then length of the string that she has out

Or

If string of the rod makes angle 0 with the horizontal, then find sin^{2}θ + cos^{2}θ. (2)

Answer:

(iii) If we decrease the width of the triangle, then what is the effect on the length of hypotenuse. (1)

Answer:

If we decrease the width of the triangle, then the length of hypotenuse decreases.

Question 37.

Sports Day Activity in School

In sports day activities of Delhi Public School, the lines have been drawn with chalk powder in rectangular shaped field OBCD. Each line is 1/2 m apart from each other. 60 flower pots have been placed at a distance of 1/2 m from

each other along OD. Yamini runs \(\frac{1}{4}\)th of the distance OD on the 3rd line and plants a red flower. Kamla runs \(\frac{1}{5}\)th of the distance OD on the 7th line and plants a yellow flower.

Based on the above information, answer the following questions

(i) Find the distance between red and yellow flowers. (1)

(ii) Find the area of rectangular field. (2)

Answer:

(i) Distance between red flower (1.5, 7.5) and yellow flower (3.5, 6) = \(\sqrt{(3.5-1.5)^2+(6-7.5)^2}\)

= \(\sqrt{4+2.25}\) = \(\sqrt{6.25}\) = 2.5 m

(ii) The width of the rectangular field = \(\frac{1}{2}\) × 60

= 30 m

∴ Area of rectangular field = Length × Breadth

= 5.5 × 30

= 165 m^{2}

Or

Find the length of the diagonal of the rectangular field. (2)

(iii) What is the length of the rectangle field? (1)

Answer:

Coordinates of point O are (0, 0)

and coordinates of point C are (\(\frac{1}{2}\) × 11, \(\frac{1}{2}\) × 60)

i.e. C(5.5, 30)

∴ Length of diagonal of rectangle

= \(\sqrt{(0-5.5)^2+(0-30)^2}\)

= \(\sqrt{(5.5)^2+(30)^2}\)

= \(\sqrt{30.25+900}\)

= \(\sqrt{930.25}\)

= 30.5 m

(iii) Length of the rectangle field = \(\frac{1}{2}\) × 11 = 5.5 m

Question 38.

Probability of Space

A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and very small size of die, is dropped on the rectangular paper without seeing towards it.

Based on the above information, answer the following questions

(i) Write the interval in which the probability of an event lies. (1)

(ii) What is the probability of an impossible event? (1)

(iii) In tossing of die, what is the probability of getting a number 4 and the probability of getting a number greater than 4. (2)

Answer:

(i) Probability of an event lies in the interval [0, 1].

(ii) The probability of an impossible event is 0.

(iii) When a die is tossed once, then sample space,

S = {1, 2, 3, 4, 5, 6}

So, the total number of possible outcomes = 6

Let E_{1} = Event of getting a number 4.

Then, the number of outcomes favourable to E_{1} = 1

∴ P(getting a number 4) = P(E_{1})

\(=\frac{\text { Number of outcomes favourable to } E_1}{\text { Total number of possible outcomes }}\) = \(\)

= \(\frac{1}{6}\)

Let E_{2} = Event of getting a number greater than 4, i.e. 5 and 6.

Then, the number of outcomes favourable to E_{2} = 2

∴ P(getting a number greater than 4)

= P(E_{2}) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Or

What is the probability that tossing die falls inside the circle? (2)

Answer:

∵ Area of rectangle = 40 × 30 = 1200 cm^{2}

Area of circle = πr^{2} = \(\frac{22}{7}\) × 10 × 10 = \(\frac{2200}{7}\) cm^{2}

P (die will fall inside the circle) = \(\frac{2200}{7 \times 1200}\) = \(\frac{11}{42}\)