Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 3 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Standard Set 3 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.

The sum of three non-zero prime numbers is 100. One of them exceed the other by 36. Find the largest number. [1]

(a) 73

(b) 91

(c) 67

(d) 57

Answer:

(c) 67

If the sum of three prime numbers is even, then one of the numbers must be 2. [1]

Let the second number be x.

Then, as per the given condition,

x + (x + 36) + 2 = 100

⇒ 2x + 38 = 100 ⇒ x = 31

So, the numbers are 2, 31,67.

Hence, largest number is 67.

Question 2.

In the given figure, AD is the bisector of ∠A. If BD = 4 cm, DC = 3 cm and AB = 6 cm, then the value of AC is [1]

(a) 4 cm

(b) 4.5 cm

(c) 6 cm

(d) 5 cm

Answer:

(b) 4.5 cm

In ∆ABC, AD is the bisector of ∠A.

∴ \(\frac{B D}{D C}\) = \(\frac{A B}{A C}\) [by angle bisector theorem]

⇒ \(\frac{4}{3}\) = \(\frac{6}{A C}\) ⇒ 4 AC = 18

∴ AC = \(\frac{9}{2}\) = 4.5 cm

Question 3.

The sum of the first 7 terms of AP sequence, 27, 30, 33, is [1]

(a) 152

(b) 252

(c) 100

(d) 150

Answer:

(b) 252

In the given AP sequence, 27, 30, 33, …….

first term, a = 27 and

common difference, d = 30 – 27 = 3

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

∴ S_{7} = \(\frac{7}{2}\) [2 × 27 + (7 – 1) × 3]

= \(\frac{7}{2}\) [54 + 18] = \(\frac{7}{2}\) × 72 = 7 × 36 = 252

Question 4.

If the pair of linear equations 3x + y = 3 and 6x + ky = 8 does not have a solution, then the value of k is [1]

(a) 2

(b) -3

(c) 0

(d) 1

Answer:

(a) 2

Given, 3x + y = 3 and 6x + ky = 8

On comparing the above equation with standard equation ax + by + c = 0, we get

a_{1} = 3, b_{1} = 1 and c_{1} = – 3

and a_{2} = 6 b_{2} =k and c_{2} = -8

Since, given equations have no solution,

\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) ⇒ \(\frac{3}{6}=\frac{1}{k}\) ⇒ k = 2

Question 5.

If HCF (306,657) = 9, what will be the LCM (306, 657)? [1]

(a) 12338

(b) 22338

(c) 23388

(d) 22388

Answer:

(b) 22338

We know that,

LCM × HCF = Product of two numbers

∴ LCM (306, 657) × HCF (306, 657) = 306 × 657

⇒ LCM × 9 = 306 × 657

⇒ LCM =22338

Question 6.

In ∆PQR, if PS is the internal bisector of ∠P meeting QR at S and PQ =13 cm, QS = (3 + x) cm, SR = (x – 3) cm and PR = 7 cm, then the value of x is [1]

(a) 9 cm

(b) 10 cm

(c) 13 cm

(d) 12 cm

Answer:

(b) 10 cm

Since, PS is the internal bisector of ∠P and it meets QR at S.

\(\frac{P Q}{P R}=\frac{Q S}{R S}\) [by angle bisector theorem]

⇒ \(\frac{13}{7}=\frac{3+x}{x-3}\)

⇒ 13(x – 3) = 7(3 + x)

⇒ 13x – 39 = 21 + 7x

⇒ 6x = 60

⇒ x = 10 cm

Question 7.

If a sector of a circle has area 100 cm³, then the perimeter of this sector is [1]

(a) r + \(\frac{200}{r}\)

(b) 2r + \(\frac{200}{r}\)

(c) r + \(\frac{100}{r}\)

(d) 3r + \(\frac{200}{r}\)

Answer:

(b) 2r + \(\frac{200}{r}\)

We know that, the area of a sector of angle θ in a

circle of radius r is \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

\(\frac{\theta}{360^{\circ}} \times \pi r^2\) = 100 ⇒ \(\frac{\theta}{360^{\circ}}=\frac{100}{\pi r^2}\) …….(i)

Now, perimeter of a sector of angle θ in a circle of radius r is

2r + \(\frac{\theta}{360^{\circ}}\) × 2πr = 2r + \(\frac{100}{\pi r^2}\) × 2πr [using Eq.(i)]

= 2r + \(\frac{200}{r}\)

Question 8.

cos^{4} x – sin^{2} x is equal to [1]

(a) 2 sin^{2}x – 1

(b) 1 – 2 cos^{2}x

(c) sin^{2}x – cos^{2}x

(d) 2 cos^{2}x – 1

Answer:

(d) 2 cos^{2}x – 1

Given, cos^{4}x – sin^{4}x

⇒ (cos^{2}x + sin^{2}x) (cos^{2}x – sin^{2}x)

[∵ a^{2} – b^{2} = (a + b) (a – b)]

⇒ 1 . (cos^{2}x – (1 – cos^{2}x))

[∵ sin^{2} A + cos^{2} A = 1 and sin^{2}A = 1 – cos^{2}A]

⇒ cos^{2}x – 1 + cos^{2}x = 2 cos^{2}x – 1

Question 9.

Given that, sin α = \(\frac{1}{2}\) and cos β = \(\frac{1}{2}\), then the value of α + β is [1]

(a) 60°

(b) 30°

(c) 0°

(d) 90°

Answer:

(d) 90°

Given, sin α = \(\frac{1}{2}\) and cos β = \(\frac{1}{2}\)

⇒ sin α = sin 30° and cos β = cos 60°

[∵ sin 30° = cos 60° = \(\frac{1}{2}\)]

⇒ α = 30° and β = 60°

∴ α + β = 30° + 60° = 90°

Question 10.

The non zero value of k for which the quadratic equation 3x² – kx + k = 0 has equal roots, is [1]

(a) 10

(b) 11

(c) 12

(d) 14

Answer:

(c) 12

Given, quadratic equation is 3x² – kx + k = 0.

On comparing with ax² + bx + c = 0, we get

a = 3,b = -k and c = k

Condition for equal roots,

b² – 4ac = 0

⇒ (- k)² – 4 × 3 × k = 0

⇒ k(k – 12) = 0 ⇒ k = 0,12

Hence, non-zero value of k is 12.

Question 11.

For what value of k, -4 is a zero of the polynomial x² – x -(2k + 2)? [1]

(a) 7

(b) 8

(c) 9

(d) 10

Answer:

(c) 9

Let p(x) = x² – x – (2k + 2)

Since, – 4 is a zero of p(x).

p(-4) = 0

⇒ (- 4)² – (-4) – (2k + 2) = 0

⇒ 16 + 4 – 2k – 2 = 0

⇒ 2k = 18 ⇒ k =9

Question 12.

The probability of passing a certain test is \(\frac{x}{24}\). If the probability of not passing is \(\frac{7}{8}\) then x is equal to [1]

(a) 2

(b) 3

(c) 4

(d) 6

Answer:

(a) 2

Let E be the event of passing the test, P(E) = \(\frac{x}{24}\)

Also, P (not passing the test), \(P(\bar{E})\) = \(\frac{7}{8}\)

Now, P(E) + \(P(\bar{E})\) = 1

\(\frac{x}{24}\) + \(\frac{7}{8}\) = 1 ⇒ \(\frac{x}{24}\) = 1 – \(\frac{7}{8}\)

⇒ \(\frac{x}{24}\) = \(\frac{8-7}{8}\) ⇒ \(\frac{x}{24}\) = \(\frac{1}{8}\)

⇒ x = \(\frac{24}{8}\) ⇒ x = 3

Question 13.

If point P lies inside the circle, then the number of tangent(s) drawn from point P, is [1]

(a) 0

(b) 1

(c) 2

(d) None of these

Answer:

(a) 0

If point P lies inside the circle, then no tangent can be drawn.

Question 14.

The area of shaded portion is [1]

(a) 9.625 cm²

(b) 6.925 cm²

(c) 9 cm²

(d) 6 cm²

Answer:

(a) 9.625 cm²

Required area

= Area of sector of angle 30° with radius 7 cm – Area of sector of angle 30° with radius 3.5 cm

= \(\left(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 7^2\right)\) – \(\left(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 3.5^2\right)\)

= \(\frac{77}{6}-\frac{77}{24}\) = \(\frac{77}{24}\) × 3 = \(\frac{77}{8}\) = 9.625 cm²

Question 15.

If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is [1]

(a) 4πr²

(b) 6πr²

(c) 3πr²

(d) 8πr²

Answer:

(a) 4πr²

Because curved surface area of a hemisphere is 2πr² and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere.

Hence, the curved surface area of new solid

= 2πr² + 2πr² = 4πr²

Question 16.

The mean and median of a distribution are 14 and 15, respectively. The value of mode is [1]

(a) 16

(b) 17

(c) 13

(d) 18

Answer:

(b) 17

Given, mean = 14 and median = 15

By using empirical relationship,

Mode = 3 Median -2 Mean

= 3 × 15 – 2 × 14 = 45 – 28 = 17

Question 17.

In an equilateral ΔABC, G is the centroid. Each side of the triangle is 6 cm. Then, the length of AG is [1]

(a) 2√2 cm

(b) 3√2 cm

(c) 2√3 cm

(d) 3√3 cm

Answer:

(c) 2√3 cm

In an equilateral triangle,

AG : GD = 2 : 1

⇒ AG = 2x and GD = x

Now, AD = \(\frac{\sqrt{3}}{2}\) a = \(\frac{\sqrt{3}}{2}\) × 6 = 3√3 cm

Again AG + GD = 3√3

2x + x = 3√3

3x = 3√3

x = √3 cm

AG = 2x = 2√3 cm

Question 18.

If the point C(k, 4) divides the join of points A(2,6) and 5(5,1) in the ratio 1 : 3, then the value of k is ? [1]

(a) 11

(b) \(\frac{29}{4}\)

(c) \(\frac{11}{4}\)

(d) \(\frac{9}{4}\)

Answer:

(c) \(\frac{11}{4}\)

Given, line joining points A (2, 6) and B (5,1) is

divided by point C in 1 : 3.

AC : BC = 1 : 3

Thus, the x-coordinate of the point C is given by

k = \(\frac{3 \times 2+1 \times 5}{1+3}\) = \(\frac{6+5}{4}\) = \(\frac{11}{4}\)

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): Points A(6,4), B(-4, – 6) and C(4,6) are such that AB = \(\sqrt{200}\), BC = \(\sqrt{208}\) and AC = \(\sqrt{8}\). Since, AB + BC > AC, points A, B and C form a triangle.

Reason (R): If BC^{2} = AB^{2} + AC^{2}, then ∆ABC is a right angled triangle, right angled at A. [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).

∴ AB + BC > AC and AB +AC > BC

[∵ sum of the two sides is greater than third side]

∴ ABC is a triangle.

Also, BC^{2} = AB^{2} + AC^{2}

[∵ \(\sqrt{208}^2\) = \(\sqrt{200}^2\) + \(\sqrt{8}^2\)]

∴ ABC is a right angled triangle.

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

Question 20.

Assertion (A): If LCM =182, product of integers is 26 × 91, then HCF = 13.

Reason (R): LCM × Product of integers = HCF [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:

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

We know, LCM × HCF = Product of two numbers

⇒ 182 × HCF = 26 × 91

⇒ HCF = \(\frac{26 × 91}{182}\) = 13

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

Section B

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

Question 21.

Check whether the lines x + y = 1 and 2x + y = x + 2 are either parallel or perpendicular. [2]

Solution:

We have, x + y = 1

and 2x + y = x +2 =>x + y = 2

On comparing both equations with ax + by + c = 0,

we get,

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

Here \(\frac{a_1}{a_2}=\frac{1}{1}\), \(\frac{b_1}{b_2}=\frac{1}{1}\) and \(\frac{c_1}{c_2}=\frac{1}{2}\)

∴ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Question 22.

If two tangents inclined at an angle of 60° are drawn to a circle of radius 5 cm, then find the length of each tangent.

Or

In the given figure, if the angle between two radii of a circle is 130°, then find the angle between the tangents at the ends of the radii. [2]

Solution:

In the given figure, OA = 5 cm and ∠APB = 60°

In ∆APO, ∠PAO = 90°

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

Since, ∠APB = 60°

⇒ ∠APO= \(\frac{60°}{2}\) = 30°

[∵ ∆OAP ≅ ∆OBP ⇒ ∠OPA = ∠OPB]

∴ tan 30° = \(\frac{A O}{A P}\) ⇒ \(\frac{1}{\sqrt{3}}=\frac{5}{A P}\)

⇒ AP = 5√3 cm

Hence, the length of each tangent is 5√3 cm.

Or

From figure, ∠AOB = 130° … (i)

Clearly, ∠PAO = ∠PBO = 90° … (ii)

[∵ radius is perpendicular to the tangent at the point of contact]

Now, applying angle sum property of quadrilateral in OAPB, we have

∠PAO + ∠AOB + ∠OBP + ∠APB = 360°

⇒ 90° +130° + 90 °+ ∠APB = 360° [from Eqs. (i) and (ii)]

∴ ∠APB = 360° – 310° = 50°

Hence, the angle between the tangent at the ends of the radii is 50°.

Question 23.

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 min. [2]

Solution:

Given, length of minute hand = radius of the circle

⇒ r = 14 cm

∵ Angle swept by the minute hand in 60 min = 360°

∴ Angle swept by the minute hand in 5 min

= \(\frac{360°}{60°}\) × 5 = 30°

Now, area of the sector with, r = 14 cm and θ = 30°

= \(\frac{\theta}{360^{\circ}}\) × πr² = \(\frac{360°}{60°}\) × \(\frac{22}{7}\) × 14 × 14

= \(\frac{11 \times 14}{3}\) = \(\frac{154}{3}\) cm²

Thus, the required area swept by the minute hand by

5 min = \(\frac{154}{3}\) cm².

Question 24.

If tan(A + B) = √3 and tan (A – B) = \(\frac{1}{\sqrt{3}}\). 0° < A + B ≤ 90°, A >B, then find A and B.

Or

If cosec A = √2, then find the value of \(\frac{2 \sin ^2 A+3 \cot ^2 A}{4 \tan ^2 A-\cos ^2 A}\). [2]

Solution:

We have, tan (A + B) = √3

⇒ tan (A + B) = tan 60° [∵ tan 60° = √3]

∴ A + B = 60° ……….(i)

and tan (A – B) = \(\frac{1}{\sqrt{3}}\)

⇒ tan (A – B) = tan 30° [∵ tan 30° = \(\frac{1}{\sqrt{3}}\)]

∴ A – B = 30° ……….(ii)

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

A + B + A – B = 60°+ 30°

⇒ 2A = 90°

⇒ A = 45°

On putting, A = 45° in Eq. (i), we get

45° + B = 60°

∴ B = 60° – 45° = 15°

Hence, A = 45° and B = 15°.

Or

We have, cosec A = √2

⇒ cosec A = cosec 45° ⇒ A = 45°

Question 25.

In the given figure, line BD || CE. If AB = 1.5 cm, BC = 6 cm and AD =2 cm. Find DE. [2]

Solution:

In ∆ADB and ∆AEC,

∠ADB = ∠AEC

[corresponding angles as BD || CE]

∠ABD = ∠ACE

[corresponding angles as BD || CE]

∠A = ∠A [common]

∴ ∆ADB ~ ∆AEC

[by AAA similarity criterion]

⇒ \(\frac{A D}{A E}=\frac{A B}{A C}\)

∴ \(\frac{2}{2+DE}=\frac{1.5}{1.5+6}\) ⇒ \(\frac{2}{2+DE}=\frac{1.5}{7.5}\)

⇒ \(\frac{2}{2+DE}=\frac{1}{5}\) ⇒ 2 + DE = 10

⇒ DE = 8cm

Section C

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

Question 26.

Prove that 5√2 is irrational. [3]

Solution:

Let us assume that 5√2 is a rational number.

Then, there exist coprime positive integers a and b such that

5√2 = \(\frac{a}{b}\)

⇒ √2 = \(\frac{a}{5b}\)

∵ 5 a and feare integers, so \(\frac{a}{5b}\) is a rational number.

⇒ √2 is a rational number.

But this contradicts the fact that √2 is irrational. So, our assumption is not correct.

Hence, 5√2 is an irrational number.

Question 27.

Prove that

\(\frac{1}{2}\) = 1 + sec θ cosec θ.

Or

If sec θ = x + \(\frac{1}{4x}\), then prove that

sec θ + tan θ = 2 x or \(\frac{1}{2x}\). [3]

Solution:

Or

We have sec θ = x + \(\frac{1}{4x}\) ……(i)

∵ tan² θ = sec² θ – 1

Question 28.

Show graphically that the following system of equations is inconsistent i.e. it has no solution 3x – Ay -1 = 0 and 2x – \(\frac{8}{3}\) y + 5=0. [3]

Solution:

We have, 3x – 4y -1 = 0

and 2x – \(\frac{8}{3}\) y + 5 = 0

Table for equation 3x – 4y -1 = 0 ⇒ y = \(\frac{3x-1}{4}\)

Now, we plot all these points on a graph paper and join them.

Table for equation 2x – \(\frac{8}{3}\) y + 5 = 0

⇒ 6x – 8y + 15= 0

⇒ y = \(\frac{6x+15}{8}\)

x | 0 | -2.5 | 1.5 |

y | 1.875 | 0 | 3 |

Points | D(0, 1.875) | E(-2.5, 0) | F(1.5,3) |

Now, we plot all these points on a graph paper and join them.

So, the two lines have no common point. Hence, the given system of equations is inconsistent.

Question 29.

In a AP, if pth term is q and the qth term is p, then show that the nth term is (p + q – n). [3]

Solution:

Let a be the first term and d be the common difference of the given AP Then,

T_{p} = a + (p – 1)d and T_{q} = a + (q – 1)d

Now, T_{p} = q and T_{q} = p [given]

∴ a + (p – 1)d = q …(i)

and a + (q – 1)d = p ……(ii)

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

(q – p)d = (p – q) =* d = -1

On putting d = -1 in Eq. (i), we get a = (p + q – 1)

Thus, a = (p + q – 1) and d = -1

∴ nth term = a + (n – 1)d = (p + q – 1) + (n – 1) × (-1)

= p + q – n

Hence, nth term = (p + q – n).

Question 30.

If a and p are the zeroes of the quadratic polynomial f(x) = 3x² – 5x – 2, then find the value of α^{3} + β^{3}. [3]

Solution:

Given, a and p are the zeroes of quadratic polynomial.

f(x) = 3x² – 5x – 2

On comparing with f(x) = ax² + bx + c, we get

a = 3, b = – 5 and c = – 2

We know that

α + β = \(\frac{b}{a}\) = \(\frac{-(-5)}{3}\) = \(\frac{5}{3}\)

and = \(\frac{c}{a}\) = \(-\frac{2}{3}\)

∴ α^{3} + β^{3} = (α + β) (α^{2} – α . β + β^{2})

= (α + β) [(α + β)^{2} – 3α . β]

= \(\frac{5}{3}\)[(\(\frac{b}{a}\))^{2} – 3 × (\(-\frac{2}{3}\))]

= \(\frac{125}{27}\) + 2 × \(\frac{5}{3}\)

= \(\frac{125+90}{27}\) = \(\frac{215}{27}\)

Question 31.

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?

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, then prove that BC – BD. [3]

Solution:

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

Here, P is an external point.

∴ PS = PT = 4cm

So, ∠PTS = ∠PST

[∵ angles opposite to equal sides are equal]

In APTS, we have

∠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 = \(\frac{120°}{2}\) = 60°

∴ ∆PTS is an equilateral triangle.

Hence, TS = 4 cm

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

Or

Given AB is a diameter of the circle with centre O and Dt 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 in a semi-circle]

∴ ∠OCA + ∠OCB = 90°

⇒ ∠OCB = 60° and ∠BCD = 30°

In ∆ACD, ∠ACD + ∠CAD + ∠ADC = 180°

⇒ 120° + 30° + ∠ADC = 180°

⇒ ∠ADC = 30°

∴ In ∆BCD,

∠BCD = ∠BDC = 30°

∴ BC = BD Hence proved.

Section D

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

Question 32.

Find the nature of roots of the following quadratic equations. In case real roots exist, find them

(i) 4x² + 12x + 9 = 0

(ii) 3x² + 5x – 7 = 0 [5]

Solution:

(i) Given, quadratic equation is

4x² + 12x + 9 = 0

On comparing with ax² + bx + c = 0, we get

a = 4, b = 12 and c = 9

Now, D = b² – 4ac

= (12)² – 4 (4) (9)

= 144 – 144 = 0

Since, D = 0, so given quadratic equation has two equal and real roots which are given by

x = \(\frac{-b \pm \sqrt{D}}{2 a}\) = \(\frac{-12 \pm 0}{2(4)}\)

⇒ x = \(\frac{-12+0}{8}\)

or x = \(\frac{-12-0}{8}\)

⇒ x = \(-\frac{3}{2}\) or x = \(-\frac{3}{2}\)

Hence, the roots are \(-\frac{3}{2}\) and \(-\frac{3}{2}\).

(ii) Given, quadratic equation is

3x² + 5x – 7 = 0

On comparing with ax² + bx + c = 0, we get

a = 3, b = 5 and c = – 7

Now, D = b² – 4ac = (5)² – 4(3) (-7)

= 25 + 84=109

Since, D > 0, so given quadratic equation has two distinct real roots which are given by

x = \(\frac{-b \pm \sqrt{D}}{2 a}\) = \(\frac{-5 \pm \sqrt{109}}{2(3)}\)

⇒ x = \(\frac{-5 + \sqrt{109}}{6}\) [taking positive sign]

or x = \(\frac{-5 – \sqrt{109}}{6}\) [taking negative sign]

Hence, the roots are \(\frac{-5 + \sqrt{109}}{6}\) and \(\frac{-5 – \sqrt{109}}{6}\).

Question 33.

The angle of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m. [5]

Solution:

Let the tower be represented by AB in the figure.

Let AB = h m.

∴ In right angled ∆ABC,

we have

\(\frac{A B}{A C}\) = tan θ

⇒ \(\frac{h}{9}\) = tan θ ……..(i)

In right angled ∆ABD,

we have

\(\frac{A B}{A D}\) = tan(90°- θ) = cot θ

⇒ \(\frac{h}{4}\) = cot θ ……..(i)

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

\(\frac{h}{9}\) × \(\frac{h}{4}\) = tan θ × cot θ = 1 [∵ tan x × cot x = 1]

⇒ \(\frac{h^2}{36}\) = 1

⇒ h² = 36

⇒ h = ± 6m

∴ h = 6m [∵ height is positive only]

Thus, the height of the tower is 6 m.

Question 34.

The weights (in kg) of 50 wrestlers are recorded in the following table.

Weight (in kg) | Number of wrestlers |

100-110 | 4 |

110-120 | 14 |

120-130 | 21 |

130-140 | 8 |

140-150 | 3 |

Find the mean weight of the wrestlers.

Or

If mode of the following series is 54, then find the value of f.

Class interval | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 |

Frequency | 3 | 5 | f | 16 | 12 | 7 |

Find the modal class in which the given mode lies and find the value of f. [5]

Solution:

We first find the class mark x, of each class and then proceed as follows

Here, Assumed mean (A) = 125,

Class width (h) = 10

and total observation (N) = 50

By assumed mean method,

Mean (x) = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)

= 125 + \(\frac{(-80)}{50}\) = 125 – 1.6

= 1234 kg

Or

Here, given mode is 54, which lies between 45-60.

Therefore, the modal class is 45-60.

l = 45, f_{1} = 16, f_{0} = f, f_{2} = 12 and h = 15

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

54 = 45 + \(\left(\frac{16-f}{2(16)-f-12}\right)\) × 15

⇒ 9 = \(\frac{16-f}{20-f}\) × 15

⇒ 9(20 – f) = 15(16 – f)

⇒ 180 – 9f = 240 – 15f

⇒ 6f = 240 – 180 = 60

⇒ f = 10

Hence, required value of f is 10.

Question 35.

A right angled triangle whose sides other than hypotenuse are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and surface area of the double cone, so formed. [take, π =3.14]

Or

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 4 cm and the diameter of the base is 8 cm. If a right circular cylinder circumscribes the solid. Find how much more space it will cover? [5]

Solution:

Let BAC be a right angled triangle such that

AB = 15 cm and AC = 20 cm.

Using Pythagoras theorem, we have

(BC)^{2} = (AB)^{2} + (AC)^{2}

⇒ (BC)^{2} = 15^{2} + 20^{2}

⇒ (BC)^{2} = 225 + 400 = 625

∴ SC = 25 cm

[taking positive square root]

Let OB = x and OA = y.

Again, using Pythagoras theorem in AOAB and AOAC, we have

AB^{2} = OB^{2} + OA^{2}

⇒ 15^{2} = x^{2} + y^{2}

⇒ x^{2} + y^{2} = 225 …(i)

and AC^{2} = OA^{2} + OC^{2}

⇒ 20^{2} = y^{2} + (25 – x)

⇒ (25- x)^{2} + y^{2} = 400 …(ii)

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

⇒ {(25 – x)^{2} + y^{2}} – {x^{2} + y^{2}} = 400 – 225

⇒ (25 – x)^{2} – x^{2} = 175

⇒ (25 – x – x)(25 – x + x) = 175

[∵ a^{2} – b^{2} = (a – b) (a + b)]

⇒ (25 – 2x) × 25 = 175

⇒ 25 – 2x = 7

⇒ 2x = 18

On putting x = 9 in Eq. (i), we get

81 + y^{2} = 225

⇒ y^{2} = 144

∴ y = 12

[taking positive square root]

Thus, we have OA = 12 cm and OB = 9 cm

∴ Volume of the double cone

= Volume of cone CAA’ + Volume of cone BAA’

= \(\frac{1}{3}\) π (OA)^{2} × OC + \(\frac{1}{3}\) π (OA)^{2} × OB

= \(\frac{1}{3}\) π × 12^{2} × 16 + \(\frac{1}{3}\) π × 12^{2} × 9

= \(\frac{1}{3}\) π × 144(16 + 9)

= \(\frac{1}{3}\) × 3.14 × 144 × 25

= 3768 cm³

∴Total surface area of the double cone

= Curved surface area of cone CAA’ + Curved surface area of cone BAA’

= π × OA × AC + π × OA × AB

= π × 12 × 20 + π × 12 × 15= 420 π

= 420 × 3.14= 1318.8 cm²

Or

Let 6PC be the hemisphere and ABC be the cone mounted on the base of the hemisphere.

Let EFGH be the right circular cylinder circumscribing the given toy.

We have, height of cone, OA = 4 cm

Diameter of the base of the cone, d =8 cm

∴ Radius of the base of cone, r = \(\frac{d}{2}\) = \(\frac{8}{2}\) = 4 cm

Here, AP = AO + OP = 4 + 4 = 8 cm

∴ Required space

= Volume of cylinder – (Volume of cone + Volume of hemisphere)

Hence, the right circular cylinder covers 64π cm³ more space than the solid toy.

Section E

(Case study based questions are compulsory )

Question 36.

Spinning Wheel Game

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7 and 8 (see figure) and these are equally likely outcomes.

Based on the above information, answer the following questions.

(i) Find the probability that the arrow point lies on the number 8. [1]

(ii) What is the probability that arrow point lies on an odd number? [2]

Or

What is the probability that it will point at a number greater than 2? [2]

(iii) Find out the probability that the pointer lies on 9. [1]

Solution:

∵ Total number of points on the circle = 8

(i) Let E_{1}= Event of getting arrow at number 8.

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

Probability that arrow points at number 8,

P(E_{1}) = \(\frac{\text { Outcomes favourable to E_1 }}{\text { Total number of outcomes }}\)

= \(\frac{1}{8}\)

(ii) Let E_{2} = Event of getting arrow at an odd number

Here, odd numbers are 1, 3, 5 and 7.

∴ Number of outcomes favourable to E_{2} = 4

Probability that arrow points at an odd number,

P(E_{2}) = \(\frac{\text { Outcomes favourable to E_2 }}{\text { Total number of outcomes }}\)

= \(\frac{4}{8}\) = \(\frac{1}{2}\)

Let E_{3} = Event of getting arrow at a number greater than 2, i.e. at 3, 4, 5,6,7 and 8

∴ Number of outcomes favourable to E_{3} = 6

Probability that arrow points at a number greater than 2,

P(E_{3}) = \(\frac{\text { Outcomes favourable to E_3 }}{\text { Total number of outcomes }}\)

= \(\frac{6}{8}\) = \(\frac{3}{4}\)

(iii) Let E_{4} = Event of getting arrow at number 9.

∴ Number of outcomes favourable to E_{4} = 0

Probability that arrow points at number 9.

P(E_{4}) = \(\frac{\text { Outcomes favourable to E_3 }}{\text { Total number of outcomes }}\)

= \(\frac{0}{8}\) = 0

Question 37.

No Smoking Campaign

All of them know that smoking is injurious for health. So, college students decide to make a campaign.

To raise social awareness about hazards of smoking, a school decided to start “No SMOKING” campaign.

10 students are asked to prepare campaign banners in the shape of triangle (as show in the figure)

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

(i) If cost of per cm² of banner is ₹ 2, then find the overall cost incurred on such campaign. [2]

Or

If we want to draw a circumscribed circle of given, then find the coordinates of the centre of circle. [2]

(ii) If we draw the image of figure about the line BC, then find the total area. [1]

(iii) Find the centroid of the given triangle. [1]

Solution:

(i) Here, from the figure,

coordinates of A = (1, 1),

coordinates of B = (6, 1)

and coordinates of C = (1, 5)

∴ Area of banner = Area of ∆ABC

Now, area of one banner = \(\frac{1}{2}\) × AB × AC

= \(\frac{1}{2}\) × 5 × 4

= 10 sq units

Then, area of 10 banners

= 10 × Area of one banner

∴ Cost of 10 banners at the rate of ₹ 2 per cm²

= 2 × Area of 10 banners

= 2 × 10 × 10

= ₹ 200

Or

The centre of circumscribed circle of given triangle is the mid-point of hypotenuse.

Centre of circle = Mid-point of BC

= \(\left(\frac{1+6}{2}, \frac{5+1}{2}\right)\)

= \(\left(\frac{7}{2}, \frac{6}{2}\right)\)

= (3.5, 3)

(ii) Total area of the required figure

= 2 × Area of ∆ABC

= 2 × 10 = 20 sq units

(iii) Centroid of the triangle (x, y)

= \(\left(\frac{1+1+6}{3}, \frac{5+1+1}{3}\right)\)

= \(\left(\frac{8}{3}, \frac{7}{2}\right)\)

= (2.6, 2.3)

Question 38.

Beehive

A beehive is an enclosed cell structure in which some honeybee species of the subgenus apis live and raise their young. Each cell is in the shape of a hexagon.

In a regular hexagon, there are six edges of equal lengths. Take O as centre and join all the vertices from the centre.

Similarity of Triangle

Two triangles are said to be similar, if their all corresponding angles are equal and all corresponding sides are proportional. Based on the above information, answer the following questions.

(i) Find the number of equilateral triangles in the given figure. [1]

(ii) If area of two triangles are equal, then they are always congruent or not. [1]

(iii) How many triangles are similar in the given figure? [2]

Or

Find the area of the hexagon, if each edge is of length a. [2]

Solution:

(i) Total number of equilateral triangles in the given figure is 6.

(ii) If area of two triangles are equal, then they are always congruent.

(iii) As we know that there are six equilateral triangle all having equal sides.

Hence, we get six similar triangles.

Or

Area of hexagon

= 6 × Area of one equilateral triangle having side a

= 6 × \(\frac{\sqrt{3}}{4}\) × (a)^{2}

= \(\frac{3 \sqrt{3}}{2}\) (a)^{2} sq units.