Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 11 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 11 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.

If the mean of first n natural numbers is \(\frac{5 n}{9}\), then the value of n is (1)

(a) 5

(b) 7

(c) 9

(d) 0

Answer:

(c) 9

According to the given condition, mean of first n natural numbers = \(\frac{5 n}{9}\)

Question 2.

The centroid of APQR whose vertices are P(-8, 0), Q(5, 5) and R(-3, -2) is (1)

(a) (-2, 1)

(b) (1, -2)

(c) (2, 1)

(d) (1, 2)

Answer:

(a) (-2, 1)

Here, (x_{1}, y_{1}) = (-8, 0), (x_{2}, y_{2}) = (5, 5) and (x_{3}, y_{3}) = (-3, -2)

∴ Centroid of ΔPQR = (x, y)

= \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

= \(\left(\frac{-8+5-3}{3}, \frac{0+5-2}{3}\right)\)

= (-2, 1)

Question 3.

The value of 2tan^{2}θ + cos^{2}θ – 2, if θ is an acute angle and sin θ = cos θ, is (1)

(a) 0

(b) 1

(c) –\(\frac{1}{2}\)

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

Answer:

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

Given, sinθ = cosθ

⇒ \(\frac{\sin \theta}{\cos \theta}\) = 1 ⇒ tanθ = 1

⇒ tanθ = tan 45° ⇒ θ = 45°

∴ 2tan^{2}θ + cos^{2}θ – 2 = 2tan^{2} 45° + cos^{2}45° – 2

= 2(1)^{2} + \(\left(\frac{1}{\sqrt{2}}\right)^2\) – 2

= 2 × 1 + \(\frac{1}{2}\) – 2 = \(\frac{5}{2}\) – 2 = \(\frac{1}{2}\)

Question 4.

30th term of the AP 10, 7, 4, ……, is (1)

(a) 97

(b) 77

(c) -77

(d) -87

Answer:

(c) -77

Here, a = 10, n = 30

∴ T_{n} = a + (n – 1)d and d = 7 – 10 = -3

Then, T_{30} = 10 + (30 – 1) × (- 3)

⇒ T_{30} = 10 + 29 × (-3)

⇒ T_{30} = 10 – 87 = – 77

Question 5.

In a rectangle ABCD, AB = 40 cm, ∠BAC = 30°, then the side BC is (1)

(a) \(\frac{40 \sqrt{3}}{3}\) cm

(b) \(\frac{20 \sqrt{3}}{3}\) cm

(c) \(\frac{20}{\sqrt{3}}\) cm

(d) None of these

Answer:

(a) \(\frac{40 \sqrt{3}}{3}\) cm

In ΔABC, we have AB =40 cm and ∠BAC = 30°

∴ tan 30° = \(\frac{B C}{A B}\) ⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{B C}{40}\)

⇒ BC = \(\frac{40}{\sqrt{3}}\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{40 \sqrt{3}}{3}\) cm

Question 6.

If α and β are the zeroes of the polynomial f(x) = x^{2} -5x + k such that α – β = 1, then the value of 4k is (1)

(a) 12

(b) 24

(c) 10

(d) 20

Answer:

(b) 24

Given, α and β are the zeroes of the given polynomial f(x) = x^{2} – 5x + k

α + β = –\(\left(-\frac{5}{1}\right)\) = 5 and αβ = \(\frac{k}{1}\) = k

Since, α – β = 1 ⇒ (α – β)^{2} = 1

⇒ (α + β)^{2} – 4αβ = 1 ⇒ (5)^{2} – 4 × k = 1

⇒ 25 – 4k = 1 ⇒ 4k = 24

Question 7.

If the HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, then the other number is (1)

(a) 99

(b) 33

(c) 112

(d) 94

Answer:

(a) 99

Let first number = 77

We know that

LCM × HCF = First number × Second number

⇒ Second number = \(\frac{11 \times 69}{77}\) = 99

Question 8.

The HCF of 96 and 404 by prime factorisation method is (1)

(a) 0

(b) 4

(c) 2

(d) 3

Answer:

(b) 4

We have, 96 = 2^{5} × 3

and 404 = 2^{2} × 101

∴ HCF = Common factor in both numbers = 2^{2} = 4

Question 9.

The HCF of two numbers p and q is 5 and their LCM is 200, then the product of pq is equal to (1)

(a) 100

(b) 1000

(c) 200

(d) 500

Answer:

(b) 1000

Given, HCF (p, q) = 5 and

LCM (p, g) = 200

∵ LCM × HCF = Product of two numbers

⇒ 5 × 200 = pq

⇒ pq = 1000

Question 10.

ΔABC and ΔDEF are similar such that 2AB = DE and BC = 8 cm, then the value of EF is (1)

(a) 15 cm

(b) 16 cm

(c) 18 cm

(d) None of these

Answer:

(b) 16 cm

Since, ΔABC ~ ΔDEF

⇒ \(\frac{A B}{D E}\) = \(\frac{B C}{E F}\) = \(\frac{A C}{D F}\) ⇒ \(\frac{2 A B}{2 D E}\) = \(\frac{B C}{E F}\)

⇒ \(\frac{D E}{2 D E}\) = \(\frac{8}{E F}\) [∵ 2AB = DE]

∴ EF = 16 cm

Question 11.

If the circumference of a circle exceeds it diameter by 30 cm, then the radius of the circle is (1)

(a) 7 cm

(b) 8 cm

(c) 4 cm

(d) 6 cm

Answer:

(a) 7 cm

Let C be the circumference and d be the diameter.

Then, C = d + 30

⇒ 2π ᐧ \(\frac{d}{2}\) = (d + 30) ⇒ 2πd = 2d + 60

⇒ πd = d + 30 ⇒ d(π – 1) = 30

⇒ d\(\left(\frac{22-7}{7}\right)\) = 30 ⇒ d\(\left(\frac{15}{7}\right)\) = 30 ⇒ d = 7 × 2

∴ r = 7 cm

Question 12.

If the first term of an AP is 2 and common difference is 4, then the sum of its 40 terms is (1)

(a) 3000

(b) 2800

(c) 3200

(d) None of these

Answer:

(c) 3200

Given, a = 2 and d = 4

Then, S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d]

⇒ S_{40} = \(\frac{40}{2}\)[2 × 2 + (40 – 1) × 4]

= 20[4 + 39 × 4]

= 20 × 160 = 3200

Question 13.

If P(-2, 2) is the mid-point of the line segment joining A(-5, b) and B(b, 3), then the value of b is (1)

(a) 0

(b) -1

(c) 1

(d) 2

Answer:

(c) 1

Mid-point of A(-5, b) and B(b, 3) is \(\left(\frac{-5+b}{2}, \frac{b+3}{2}\right)\) = (-2, 2)

⇒ \(\frac{-5+b}{2}\) = -2

⇒ -5 + b = -4 ⇒ b = 5 – 4 = 1

and \(\frac{b+3}{2}\) ⇒ b = 4 – 3 = 1

∴ b = 1

Question 14.

In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. (1)

Then, the value of AT is

(a) 2\(\sqrt{3}\) cm

(b) 4 cm

(c) 2 cm

(d) None of these

Answer:

(a) 2\(\sqrt{3}\) cm

From the given figure,

∠OAT = 90°

[∵ tangent and radius are perpendicular to each other at the point of contact]

In right angled ΔOAT,

\(\frac{A T}{O T}\) = cos 30°

⇒ \(\frac{A T}{4}\) = \(\frac{\sqrt{3}}{2}\) ⇒ AT = 2\(\sqrt{3}\)cm

Question 15.

In a lottery ticket, there are 10 prizes and 25 blanks, the probability of not getting a prize is (1)

(a) \(\frac{7}{5}\)

(b) \(\frac{5}{7}\)

(c) 1

(d) 0

Answer:

(b) \(\frac{5}{7}\)

Total number of lottery ticket = 10 + 25 = 35

∴ P(not getting prize) = \(\frac{25}{35}\) = \(\frac{5}{7}\)

Question 16.

A wire is in the shape of a circle of radius 21 cm. It is bent in the form of a square. The side of the square is ( take π = \(\frac{22}{7}\)) (1)

(a) 22 cm

(b) 33 cm

(c) 44 cm

(d) 66 cm

Answer:

(b) 33 cm

As, length of wire is same, so both figures have same perimeter,

∴ Circumference of circle = Perimeter of the square.

Let r be the radius of the circle and a is the side of the square.

So, 2πr = 4a

⇒ 4a = 2 × \(\frac{22}{7}\) × 21 [∵ r = 21 cm]

⇒ 4a = 132

⇒ a = 33 cm

Question 17.

\(\frac{1+\cot ^2 A}{1+\tan ^2 A}\)is equal to (1)

(a) tan^{2} A

(b) sec^{2} A

(c) cot^{2}A

(d) 1 – sin^{2}A

Answer:

(c) cot^{2}A

we have, \(\frac{1+\cot ^2 A}{1+\tan ^2 A}\)

= \(\frac{1+\cot ^2 A}{1+\frac{1}{\cot ^2 A}}\) [∵ tanθ = \(\frac{1}{\cot \theta}\)]

= \(\frac{\cot ^2 A\left(1+\cot ^2 A\right)}{\cot ^2 A+1}\) = cot^{2} A

Question 18.

The area of a quadrant of circle whose circumference is 12 cm, is (1)

(a) 102.34 cm^{2}

(b) 95.15 cm^{2}

(c) 34 cm^{2}

(d) 45.81 cm^{2}

Answer:

(d) 45.81 cm^{2}

Given, circumference of a quadrant = 12 cm

⇒ \(\frac{2 \pi r}{4}\) = 12 ⇒ πr = 24 ⇒ r = \(\frac{24}{\pi}\)

Now, area of a quadrant = \(\frac{\pi r^2}{4}\)

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) The distance of a points P (x, y) from the origin is \(\sqrt{x^2-y^2}\).

Reason (R) If P(-1, 1) is the mid-point of the line segment joining A(-3, b) and B(1, b + 4), then value of b is -1. (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:

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

The distance of point P(x, y)from origin

= \(\sqrt{(x-0)^2+(y-0)^2}\)

= \(\sqrt{x^2+y^2}\)

Mid-point of A(-3, b) and B(1, b + 4) is

\(\left(\frac{-3+1}{2}, \frac{b+b+4}{2}\right)\) = (-1, b + 2)

Hence, Assertion is false but Reason is true.

Question 20.

Assertion (A) 3y^{2} + 17y – 30 = 0 have distinct roots.

Reason (R) The quadratic equation ax^{2} + bx + c = 0 have distinct roots (real roots), if D > 0. (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:

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

Assertion True

3y^{2} + 17y – 30 = 0

D = b^{2} – 4ac = (17)^{2} – 4 × 3 (-30)

= 289+ 360 = 649 > 0

So, roots are real and distinct.

Reason True and is the correct explanation of assertion.

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

Section B

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

Question 21.

One root of the quadratic equation 2x^{2} – 8x – k – 0 is 5/2. Find the other root and the value of k. (2)

Answer:

Given, 5/2 is one of the roots of the equation

2x^{2} – 8x – k = 0

On putting x = \(\frac{5}{2}\) in Eq.(i), we get

2(\(\frac{5}{2}\))^{2} – 8(\(\frac{5}{2}\)) – k = 0

⇒ \(\frac{50}{4}\) – \(\frac{40}{2}\) – k = 0 ⇒ \(\frac{50-80-4 k}{4}\) = 0

⇒ -30 – 4k = 0

⇒ 4k = -30

⇒ k = \(\frac{-30}{4}\) = \(\frac{-15}{4}\)

Now, putting the value of k in Eq. (i), we get

2x^{2} – 8x + \(\frac{15}{2}\) = 0

⇒ 4x^{2} – 16x + 15 = 0

⇒ 4x^{2} – 6x – 10x + 15 = 0 [by factorisation]

⇒ 2x (2x – 3) – 5 (2x – 3) = 0

⇒ (2x – 5) (2x – 3) = 0

∴ x = \(\frac{5}{2}\) or \(\frac{3}{2}\)

Hence, the other root is \(\frac{3}{2}\) and the value of k is –\(\frac{15}{2}\).

Or

Using quadratic formula, solve for x

9x^{2} – 3(a + b)x + ab = 0.

Answer:

Given, equation is 9x^{2} – 3(a + b)x + ab = 0.

On comparing it with standard form

Ax^{2} + Bx + C = 0, we get

A = 9, B = -3(a + b) and C = ab

Discriminant, D = B^{2} – 4AC

= 9(a + b)^{2} – 4(9)(ab)

= 9{(a + b)^{2} – 4ab}

= 9(a – b)^{2} ≥ 0

[∵ (a + b)^{2} – 4ab = (a – b)^{2}]

Therefore, the two real roots of the equation are given by

Hence, the two roots are \(\frac{a}{3}\) and \(\frac{b}{3}\).

Question 22.

A conical tent is to accommodate 77 persons. Each person must have 4m^{3} of air to breathe. Given, the radius of the tent as 7 m. Find the height of the tent. (2)

Answer:

Given, in the conical tent, each person requires 16 m^{3} of air to breathe.

∴ Volume of the conical tent = 77 × 4 = 308 m^{3}

⇒ \(\frac{1}{3}\)πr^{2}h = 308

⇒ \(\frac{1}{3}\) × \(\frac{22}{7}\) × (7)^{2} × h = 308

[given, radius of the tent = 7 m]

⇒ \(\frac{1}{3}\) × 22 × 7 × h = 308

⇒ h = \(\frac{308 \times 3}{22 \times 7}\) = 24m

Thus, height of the conical tent = 6 m.

Question 23.

If ΔABC – ΔDEF such that DE = 6 cm, EF = 4 cm, DF = 5 cm and BC = 8 cm, then find perimeter of ΔABC. (2)

Answer:

Given, ΔABC ~ ΔDEF

and DE = 6 cm, EF = 4 cm, DF = 5 cm

and BC = 8 cm

Now, perimeter of ΔABC = AB + BC + CA

= 12 + 8 + 10 = 30 cm

Question 24.

Find the value of θ in the following sin 2θ = sin 60°.cos 30° – cos 60° sin 30°. (2)

Answer:

We have, sin2θ = sin 60°.cos 30° – cos 60° sin 30°

⇒ sin2θ = \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2}\) × \(\frac{1}{2}\)

[∵ sin 60° = \(\frac{\sqrt{3}}{2}\), cos 30° = \(\frac{\sqrt{3}}{2}\)]

⇒ sin2θ = \(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{2}{4}\)

⇒ sin2θ = \(\frac{1}{2}\)

⇒ sin2θ = sin30° [∵ sin 30° = \(\frac{1}{2}\)]

⇒ 2θ = 30°

⇒ θ = 10°

Question 25.

In the given figure, PT and PS are tangents to a circle from a point P, such that PT = 4 cm and ∠TPS = 60°. Find the length of chord TS. How many lines of same length TS can be drawn in the circle? (2)

Answer:

We know that, tangents drawn from external point to the circle are equal in length.

Here, P is an external point.

∴ PS = PT = 4 cm

So, ∠PTS = ∠PST

[∵ angle opposite to equaP sides are equal]

∠PTS + ∠PST + ∠TPS = 180°

[by angle sum property of triangle]

⇒ ∠PTS + ∠PTS + 60° = 180°

[∵ ∠PST = ∠PTS and ∠TPS = 60°].

⇒ 2∠PTS = 180° – 60°

⇒ 2∠PTS = 120°

⇒ ∠PTS = \(\frac{120^{\circ}}{2}\) = 60°

∴ ΔPTS is an equilateral triangle.

Hence, TS = 4 cm

Here, infinite lines of same length TS can be drawn in a circle.

Or

AB is a diameter and AC is a chord of a circle such that ∠BAC = 30°. If the tangent at C intersects AB produced at D, prove that BC = BD.

Answer:

Given AB is a diameter of the circle with centre O and DC is the tangent of circle and ∠BAC = 30°.

To prove BC = BD

Construction Join O to C.

Proof Since, OC ⊥ CD

[∵ the tangent at any point of a circle is perpendicular to the radius through the point of contact]

∴ ∠OCB + ∠BCD = 90°

Now, OC = OA [radii]

⇒ ∠OCA = ∠OAC [angles opposite to equal sides are equal]

∴ ∠OCA = 30°

Now, ∠ACB = 90° [angle on a semi-circle]

∴ ∠OCA + ∠OCB = 90°

⇒ ∠OCB = 60° and ∠BCD = 30°

In ΔACD, ∠ACD + ∠CAD + ∠ADC = 180°

⇒ 120° + 30° + ∠ADC = 180°

⇒ ∠ADC = 30°

∴ In ΔBCD, ∠BCD = ∠BDC = 30°

[angles opposite to equal sides are equal]

∴ BC = BD Hence proved.

Section C

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

Question 26.

Prove that sin θ(1 + tan θ) + cos θ(1 + cotθ) = sec θ + cosec θ. (3)

Answer:

LHS = sinθ(1 + tanθ) + cos θ(1 + cotθ)

Or

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

Answer:

Question 27.

From a bus stand in Delhi, if we buy 2 tickets to Pitampura and 3 tickets to Dilshad Garden, the total cost is ₹ 46 but if we buy 3 tickets to Pitampura and 5 tickets to Dilshad Garden, the total cost is ₹ 74. Then, find the fares from the bus stand to Pitampura and to Dilshad Garden. (3)

Answer:

Let fare from bus stand to Pitampura be ₹ x and to Dilshad Garden be ₹ y.

According to the question, we have

Cost of 2 tickets to Pitampura + 3 tickets to Dilshad Garden = ₹ 46

∴ 2x + 3y = 46 ……..(i)

and cost of 3 tickets to Pitampura + 5 tickets to Dilshad Garden = ₹ 74

and 3x + 5y = 74 …….(ii)

Multiplying Eq. (i) by 3 and Eq. (ii) by 2, we get

6x + 9y = 138 ………(iii)

and 6x + 10y = 148 …………..(iv)

Subtracting Eq. (iii) from Eq. (iv), we get

y = 10

From Eq. (i),

2x + 3(10) = 46 ⇒ 2x + 30 = 46

⇒ 2x = 16

⇒ x = 8

Hence, the fare from the bus stand in Delhi to Pitampura is ₹ 8 and the fare to Dilshad Garden is ₹ 10.

Question 28.

Find the HCF and LCM of 84, 90 and 120 by prime factorisation method. (3)

Answer:

The prime factorisation of 84, 90 and 120 gives

84 = 2^{2} × 3 × 7

90 = 2 × 3^{2} × 5

and 120 = 2^{3} × 3 × 5

To find the HCF, we list the common prime factors and their smallest exponents in 84, 90 and 192.

Here, 2^{1} and 3^{1} are the smallest exponents of the common factors 2 and 3, respectively.

So, HCF (84, 90, 120) = 2^{1} × 3^{1} = 2 × 3 = 6

To find the LCM, we list all prime factors of 84, 90 and 120 and their greatest exponents.

Here, 2^{3}, 3^{2}, 5^{1} and 7^{1} are the greatest exponents of the prime factors 2, 3, 5 and 7, respectively involved in three numbers.

So, LCM (84, 90, 120) = 2^{3} × 3^{2} × 5^{1} × 7^{1}

= 8 × 9 × 5 × 7 = 2520

Question 29.

Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax^{2} + bx + c, a ≠ 0, c ≠ 0. (3)

Answer:

Let α and β be the zeroes of the polynomial

f(x) = ax^{2} + bx + c

Then, α + β = –\(\frac{b}{a}\) = αβ = \(\frac{c}{a}\)

Let S and P denote respectively, the sum and product of the zeroes of a polynomial whose zeroes are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)

∴ Required polynomial g(x) is given by

g(x) = k(x^{2} – Sx + P) = k(x^{2} + \(\frac{b}{c}\)x + \(\frac{a}{c}\)x)

where k is any non-zero constant.

Or

If α, β are the zeroes of the polynomial p(x) = 2x^{2} + 5x + k satisfying the relation α^{2} + β^{2} + αβ = \(\frac{21}{4}\), then find the value of k for this to be possible.

Answer:

Given, α and β are zeroes of polynomial p(x) = 2x^{2} + 5x + k.

Question 30.

If the pth, qth and rth terms of an AP are a, b and c respectively, then show that a(q – r) + b(r – p) + c(p – q) = 0. (3)

Answer:

Let m be the first term and d the common difference of the given AR Then,

pth term = m + (p – 1)d = a ……..(i)

qth term = m + (q – 1)d = b ……..(ii)

and rth term = m + (r – 1)d = c …….(iii)

On multiplying (q – r), (r – p) and (p – q) in Eqs. (i), (ii) and (iii), respectively and then adding, we get

a(q – r) + b(r – p) + c(p – q)

= (q – r){m + (p – 1)d} + (r – p){m + (q – 1 )d} + (p – q){m + (r – 1)d}

= m(q – r + r – p + p – q) + d[(p – 1 )(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]

= m × 0 + d[pq – pr – q + r + qr – qp – r + p + rp – qr – p + q]

= 0 + d (0) = 0

⇒ a(q – r) + b(r – p) + c(p – q ) = 0

Hence proved.

Question 31.

If the point R (x, y) is equidistant from the points P(a + b, a – b)and Q(b – a, a + b), then find distance of P from origin, mid-point of PQ and also prove that xa = yb. (3)

Answer:

P(a + b, a – b) be the given points and O(0, 0) be the origin

∴ OP = \(\sqrt{(a+b-0)^2+(a-b-0)^2}\) [by distance formula]

= \(\sqrt{(a+b)^2+(a-b)^2}\)

= \(\sqrt{a^2+b^2+2 a b+a^2+b^2-2 a b}\) = \(\sqrt{2\left(a^2+b^2\right)}\)

Mid-point of PQ

= \(\left(\frac{a+b+b-a}{2}, \frac{a-b+a+b}{2}\right)\)

= (b, a) [by mid-point formula]

Given, R (x, y) is equidistant from the points P (a + b, a – b) and Q (b – a, a + b), i.e. we have

PR = QR

⇒ \(\sqrt{[x-(a+b)]^2+[y-(a-b)]^2}\) = \(\sqrt{[x-(b-a)]^2+[y-(a+b)]^2}\) [by distance formula]

On squaring both sides, we get

(x – a – b)^{2} + (y – a + b)^{2} = (x – b + a)^{2} + (y – a – b)^{2}

⇒ x^{2} + a^{2} + b^{2} – 2xa + 2ab – 2xb + y^{2} + a^{2} + b^{2} – 2ya – 2ab + 2yb

= x^{2} + b^{2} + a^{2} – 2xb – 2ab + 2xa + y^{2} + a^{2} + b^{2} – 2ya + 2ab – 2yb

[∵ (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx]

∴ -2xa + 2yb = 2xa – 2yb ⇒ 4yb = 4xa

⇒ xa = yb Hence proved.

Section D

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

Question 32.

A girl of height 80 cm is walking away from the base of a lamp-post at a speed of 2.1 m/sec. If the lamp is 4 m above the ground, then find the length of her shadow after 4 sec. (5)

Answer:

Let AB be the lamp-post, CD be the girl and D be the position of girl after 4 sec.

Again, let DE = x m be the length of shadow of the girl.

Given, CD = 80 cm = 0.8 m, AS = 4m and speed of the girl = 2.1 m/sec

∴ Distance of the girl from lamp-post after 4 sec,

BD = 2.1 × 4 = 8.4 m

[∵ distance = speed × time]

In ΔABE and ΔCDE,

∠B = ∠D [each 90°]

∠E = ∠E [common angle]

ΔABE ~ ΔCDE [by AA similarity criterion]

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

[since, corresponding sides of similar triangles are proportional]

On substituting all the values in Eq, (i), we get

\(\frac{8.4+x}{x}\) = \(\frac{4}{0.8}\) [∵ BE = BD + DE = 8.4 + x]

⇒ \(\frac{8.4+x}{x}\) = \(\frac{40}{8}\)

⇒ \(\frac{8.4+x}{x}\) = 5

⇒ 8.4 + x = 5x

⇒ 4x = 8.4 ⇒ x = 2.1 m

Hence, the length of her shadow after 4 sec is 2.1 m.

Question 33.

Find the unknown entries m, n, o, p, q and r in the following distribution of heights of students in a class. The total number of students is 50. (5)

Height (in cm) | 150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 – 175 | 175 – 180 |

Frequency | 12 | n | 10 | P | q | 2 |

Cumulative frequency | m | 25 | 0 | 43 | 48 | r |

Answer:

Given, distribution table is

We know that, cumulative frequency of an interval is equal to the sum of frequency of that interval and of previous intervals.

Here, m = 12

∴ m + n = 25

⇒ 12 + n = 25

⇒ n = 25 – 12

⇒ n = 13

and 25 + 10 = o

⇒ o = 35

Also, 0 + p = 43

⇒ 35 + p = 43

⇒ p = 43 – 35 = 8

and 43 + q = 48

⇒ q = 48 – 43 = 5

and 48 + 2 = r ⇒ r = 50

Hence, the required values are m = 12, n = 13, 0 = 35, p = 8, q = 5 and r = 50.

Or

The following table gives the wages of workers in a factory’.

Wages (in ₹) | Number of workers |

45 – 50 | 5 |

50 – 55 | 8 |

55 – 60 | 30 |

60 – 65 | 25 |

65 – 70 | 14 |

70 – 75 | 12 |

75 – 80 | 6 |

Calculate the mean by the step deviation method.

Answer:

Let assumed mean, A = 62.5

To make a table for the product of frequency and step-deviation

We know that

Mean \((\bar{x})\) = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × h = 62.5 + \(\frac{(-5) \times 5}{100}\)

= 62.5 – 0.25 = 62.25

Hence, the mean wages are ₹ 62.25.

Question 34.

The height of a cone is 40 cm. A small cone is cut at the top by a plane parallel to the base. If the volume of the small cone be \(\frac{1}{27}\) of the volume of the given cone, at what height above the base is the sections made? (5)

Answer:

Let R be the radius of the given cone, r the radius of the small cone, h be the height of the frustum and be the height of the small cone. In figure, ∆ONC and ∆OMA are similar by AA similarity

⇒ \(\left(\frac{r}{R}\right)^3\) = \(\frac{1}{27}\) = \(\left(\frac{1}{3}\right)^3\)

⇒ \(\frac{r}{R}\) = \(\frac{1}{3}\) ….. (ii)

From Eqs. (i) and (ii) h_{1} = \(\frac{1}{3}\) × 40 = \(\frac{40}{3}\) cm

= 13.33 cm

Therefore, h = 40 – h_{1} = 40 – \(\frac{40}{3}\) = 26.67 cm

Hence, the section is made at a height of 26.67 cm above the base of the cone.

Question 35.

Find the roots of the following equation \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -47. (5)

Answer:

Given, equation is \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7

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

a = 1, b = -3 and c = 2

By using quadratic formula, we get

Hence, the roots of the given equation are 2 and 1.

Or

Solve for x, 2\(\left(\frac{2 x+3}{x-3}\right)\) – 25\(\left(\frac{x-3}{2 x+3}\right)\) = 5

Answer:

Let \(\frac{2 x+3}{x-3}\) = y ……… (i)

Then, \(\frac{x-3}{2 x+3}\) = \(\frac{1}{y}\)

Therefore, the given equation reduces to 2y – 25\(\frac{1}{y}\) = 5 ⇒ 2y^{2} – 25 = 5y

⇒ 2y^{2} – 5y – 25 = 0 ⇒ 2y^{2} – 10y + 5y – 25 = 0

[by factorisation method]

⇒ 2y(y – 5) + 5(y – 5) = 0 ⇒ (y – 5)(2y + 5) = 0

⇒ y = 5 or y = \(\frac{-5}{2}\)

Now, putting y = 5 in Eq. (i), we get

\(\frac{2 x+3}{x-3}\) = \(\frac{5}{1}\)

⇒ 5x – 15 = 2x + 3

⇒ 3x = 18 ⇒ x = 6

Again, putting y = –\(\frac{5}{2}\) in Eq. (i), we get

\(\frac{2 x+3}{x-3}\) = –\(\frac{5}{2}\)

⇒ 5x + 15 = 4x + 6

∴ 9x = 9

⇒ x = 1

Hence, the values of x are 1 and 6.

Section E

(Case Study Based Questions)

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

Question 36.

Fun Game

In summer vacation Sachin and Rahul invite some friends in a park and they want to play some fun game. So, they consider block in the shapes of a cube with one letter/number written on each face as shown below

While through the cube, they want to know the change of getting some particular number or alphabet.

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

(i) Find the probability of getting an alphabet. (2)

Answer:

(i) When a cube is thrown once, all possible outcomes are

Total number of possible outcomes = 6

Let E_{1} be the event of getting an alphabet.

Then, the favourable outcomes are A and B. Number of favourable outcomes = 2

∴ P (getting an alphabet) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Or

Find the probability of getting a prime number. (2)

(ii) Find the probability of getting a consonant. (1)

(iii) When we have no reason to believe that one is more likely to occur than the other, then what is it said? (1)

Answer:

Let E_{2} be the event of getting a prime number. Then, the favourable outcomes are 2, 3, 5, 7

Number of favourable outcomes = 4

∴ P (getting a prime number) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

(ii) Let E_{3} be the event of getting a consonant.

Then, the favourable outcome is B.

Number of favourable outcomes = 1

∴ P (getting a consonant) = \(\frac{1}{6}\)

(iii) When we have no reason to believe that one is more likely to occur, then it is said to be equally likely outcomes.

Question 37.

Number of Tangents on a Circle

The number of tangents drawn from a point on a circle depends upon the position of the point with respect to the circle.

Suppose O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle at E and AB is the tangent to the circle at E.

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

(i) Find the length of the tangent at point T. (2)

Answer:

(i) Given, radius of circle OP = 5cm and OT = 13cm.

We know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

In right angled ΔOPT,

OT^{2} = OP^{2} + PT^{2}

[by Pythagoras theorem]

⇒ 13^{2} = 5^{2} + PT^{2}

⇒ PT^{2} = 169 – 25 = 144

∴ PT = 12 cm

[taking positive square root as length cannot be negative]

Or

Find the length of the tangent AB, if the length of the tangent at point T is 12 cm. (2)

(ii) How many tangents can be drawn from point O ? (1)

(iii) How many tangents can be drawn from point T? (1)

Answer:

Also, we know that length of tangents drawn from a point to a circle are equal.

Therefore, AP = AE = x (say)

Now, AT = PT – AP = (12 – x) cm

Since, AB is the tangent to the circle at point E and OE is radius.

Therefore, OE ⊥ AB ⇒ ∠OEA = 90°

⇒ ∠AET = 90°

[∵ OT is a straight line, so ∠OEA + ∠AET = 180°]

Now, in right angled ΔAET,

AT^{2} = AE^{2} + ET^{2}

[by Pythagoras theorem]

⇒ (12 – x)^{2} = x^{2} + (13 – 5)^{2}

[∵ ET =OT – OE = 13 – 5]

⇒ 144 – 24x + x^{2} = x^{2} + 64

[∵ (a – b)^{2} = a^{2} + b^{2} – 2ab]

⇒ -24x = -80 ⇒ x = \(\frac{10}{3}\) cm

⇒ AE = \(\frac{10}{3}\) cm

Similarly, BE = \(\frac{10}{3}\) cm

Hence, AB = AE + BE = \(\frac{10}{3}\) + \(\frac{10}{3}\) = \(\frac{20}{3}\) cm

(ii) No tangent can be drawn from point O.

(iii) 2 tangents can be drawn from point T.

Question 38.

Sattelite Towers in Himalayas

The sattellite image of Himalaya Mountain is shown below. In this image there are many signal towers are standing. The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower with height 50 m from the foot of the hill is 30°.

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

(i) Find the horizontal distance between hill and tower. (2)

Answer:

(i) Given, AB = 50 m is the height of the tower.

Let CD = h m be the height of the hill.

Then, ∠ACB = 30° and ∠CAD = 60°

In right angled ΔBAC,

cot 30° = \(\frac{A C}{A B}\) [∵ cotθ = \(\frac{\text { base }}{\text { perpendicular }}\)]

Or

Find the height of the hill, if the distance between bottom of hill and tower is 50\(\sqrt{3}\) m. (2)

(ii) Find the distance from foot of tower to the top of the hill. (1)

(iii) Find the distance from foot of the hill to the top of the tower. (1)

Answer:

(ii) ∴ Required distance,

(ii) ∴ Required distance,