# CBSE Class 10 Maths Question Paper 2019 (Series: JMS/4) with Solutions

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## CBSE Class 10 Maths Question Paper 2019 (Series: JMS/4) with Solutions

Time allowed: 3 hours

Maximum marks: 80

General Instructions:

Read the following instructions carefully and follow them:

- All questions are compulsory.
- This question paper consists of 30 questions divided into four sections – A, B, C and D.
- Section A contains 6 questions ofl mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in 2 questions ofl mark, 2 questions of 2 marks each, 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculator is not permitted.

SET I Code No. 30/4/1

Section – A

Questions number 1 to 6 carry 1 mark each.

Question 1.

Find the value of k for which the quadratic equation kx (x – 2) + 6 = 0 has two equal roots.

Answer:

We have kx(x – 2) + 6 = 0

∴ kx^{2} – 2kx + 6 = 0, k ≠ 0

For two equal roots …[Given

b^{2} – 4ac = 0, a = k, b = -2k, c = 6

⇒ (-2k)^{2} – 4(k) (6) = 0

⇒ 4k^{2} – 24k = 0 ⇒ 4k(k – 6) = 0

⇒ 4k = 0 or k – 6 = 0

⇒ k = 0 or k = 6

But k ≠ 0

∴ k = 6

Question 2.

Find the number of terms in the A.P.: 18, 15\(\frac{1}{2}\), 13 ………. , – 47.

Answer:

We have, a = 18,

d = 15\(\frac{1}{2}\) – 18 = \(\frac{31}{2}\) – 18 = \(\frac{31-36}{2}\) = \(\frac{-5}{2}\)

a_{n} = -47 …[Given

a_{n} = [a + (n -1) d] = – 47

18 + (n – 1) (\(\frac{-5}{2}\)) = -47

(n – 1) (\(\frac{-5}{2}\)) = – 47 – 18

(n – 1) (\(\frac{-5}{2}\)) = -65

(n -1) = – 65 × (\(\frac{-2}{5}\))

(n – 1) = 13 × 2

n = 26 + 1 = 27

∴ Number of terms = 27

Question 3.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 4.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 5.

Find the distance between the points (a, b) and (-a,-b).

Answer:

Let P(a, b) and Q (- a, – b)

Question 6.

Find a rational number between √2 and √7.

Or

Question 6.

Write the number of zeroes in the end of a number whose prime factorization is 2^{2} × 5^{3} × 3^{2} × 17.

Answer:

1^{st} Method:

√2 = 1.414 and = √7 = 2.646

∴ √2 < 1.5 < √7

⇒ √2< \(\frac{15}{10}\) < √7 or √2 < \(\frac{3}{2}\) < √7

∴ \(\frac{1}{2}\) is a rational no. between √2 and √7.

II^{nd} Method:

√2 < \(\sqrt{6.25}\) < √7

√2 < 2.5 < √7

√2 < \(\frac{25}{10}\) < √7 ⇒ √2 < \(\frac{5}{2}\) < √7

∴ \(\frac{2}{2}\) is a rational no. between √2 and √7.

Or

2^{2} × 5^{3} × 3^{2} × 17

= 2^{2} × 5^{2} × 5 × 3^{2} × 17

= (2 × 5)^{2} × 5 × 3^{2} × 17

= (10)^{2} × 5 × 3^{2} × 17

The number of zeroes in the end of the given number = 2

Section – B

Questions number 7 to 12 carry 2 marks each.

Question 7.

How many multiples of 4 lie between 10 and 205?

Or

Question 7.

Determine the A.P. whose third term is 16 and 7^{th} term exceeds the 5 term by 12.

Answer:

Multiples of 4 between 10 and 205 are 12, 16, 20,… 204.

Here, a = 12, d = 16 – 12 = 4, a_{n} = 204

Now, a_{n} = a + (n – 1)d = 204

⇒ 12 + (n – 1) (4) = 204

⇒ (n – 1) (4) = 204 – 12 = 192

⇒ (n – 1) = \(\frac{192}{4}\) = 48

∴ n = 48 + 1 = 49

Or

a_{3} = 16

a + 2d = 16

a + 2(6) – 16 ……[From (i)

a + 12 = 16

a = 4

a_{7} = a_{5} + 12

a + 6d = a + 4d + 12

a + 6d – a – 4d = 12

2d = 12

d = 6 …(i)

Thus, AP is

Question 8.

The point R divides the line segment AB, where A (-4, 0) and B (0, 6) such that AR = \(\frac{3}{4}\) AB.

Find the coordinates of R.

Answer:

Given: AR = \(\frac{3}{4}\) AB ⇒ \(\frac{\mathrm{AR}}{\mathrm{AB}}\) = \(\frac{3}{4}\)

Let AR = 3k, AB = 4k

∴ RB = AB – AR = 4k – 3k = 1k

\(\frac{\mathrm{AR}}{\mathrm{AB}}\) = \(\frac{3 k}{1 k}\) = \(\frac{3}{1}\)

∴ AR : RB = 3 : 1

Using section formula,

Coordinates of R = (\(\frac{m x_2+n x_1}{m+n}\), \(\frac{m y_2+n y_1}{m+n}\))

R = (\(\frac{3(0)+1(-4)}{3+1}, \frac{3(6)+1(0)}{3+1}\)) = (\(\frac{-4}{4}\), \(\frac{18}{4}\))

∴ Required coordinates of point R,

R = (-1, \(\frac{9}{2}\))

Question 9.

Not in Current Syllabus.

Answer:

Not in Current Syllabus.

Question 10.

Three different coins are tossed simultaneously. Find the probability of getting exactly one head.

Answer:

Total number of outcomes = 2^{n} = 2^{3} = 8

S = {HHH, TTT, HTH, THT, HHT, THH, TTH, HTT)

Possible outcomes of getting exactly one head = THT, TTH, HTT„ i.e., 3.

∴ P(exactly one head) = \(\frac{3}{8}\)

Question 11.

A card is drawn at random from a pack of 52 playing cards. Find the probability of drawing a card which is neither a spade nor a king.

Answer:

Total number of cards = 52

∴ P(neither a spade nor a king)

= 1 – P(spade cards) – P(kings excluding king of spades)

1 – \(\frac{13}{52}\) – \(\frac{3}{52}\)

= \(\frac{52-13-3}{52}\) = \(\frac{36}{52}\) = \(\frac{9}{13}\)

Question 12.

Find the solution of the pair of equations : \(\frac{3}{x}\) + \(\frac{8}{y}\) = -1; \(\frac{1}{x}\) – \(\frac{2}{y}\) = 2, x, y ≠ 0.

Or

Question 12.

Find the value(s) of k for which the pair of equations kx + 2y = 3, 3x + 6y = 10 has a unique solution.

Answer:

Putting q = \(\frac{-1}{2}\) in (i), we get p = 1

Now, \(\frac{1}{x}\) = p

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

∴ x = 1

\(\frac{1}{y}\) = q

\(\frac{1}{y}\) = \(\frac{-1}{2}\)

∴ y = -2

Or

We have,

kx + 2y = 3

3x + 6y = 10

Here,

a_{1} = k, b_{1} = 2, c_{1} = 3,

a_{2} = 3, b_{2} = 6, c_{2} = 10

For a unique solution, \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\)

⇒ \(\frac{k}{3}\) ≠ \(\frac{2}{6}\) ⇒

\(\frac{k}{3}\) ≠ \(\frac{1}{3}\)

⇒ k ≠ 1

The given system of equations will have unique solution for all real values of k except k = 1.

Section – C

Questions number 13 to 22 carry 3 marks each.

Question 13.

Prove that (3 + 2√5) is an irrational number, given that √5 is an irrational number.

Answer:

Let us assume, to the contrary, that 3 + 2√5 is rational.

So that we can find integers a and b(b ≠ 0),

such that 3 + 2 = \(\frac{a}{b}\),

where a and b are coprime

Rearranging this equation, we get

√5 = \(\frac{a-3 b}{2 b}\)

√5 = \(\frac{a}{2 b}-\frac{3 b}{2 b}\) √5 = \(\frac{a}{2 b}-\frac{3}{2}\)

Since a and b are integers, we get that √5 = \(\frac{a}{2 b}-\frac{3}{2}\) is rational

and so V5 is rational.

But this contradicts the fact that √5 is irrational.

So we conclude that 3 + 2√5 is irrational.

Question 14.

A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.

Answer:

Let the usual speed of the train = x km/hr

∴ decreased speed of the train = (x – 8) km/hr.

According to the question,

\(\frac{480}{x-8}-\frac{480}{x}\) = 3 … [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)

⇒ 480(\(\frac{1}{x-8}-\frac{1}{x}\)) = 3

⇒ 160(\(\frac{x-(x-8)}{(x-8)(x)}\)) = 1

⇒ x (x – 8) = 1280

⇒ x^{2} – 8x – 1280 = 0

⇒ x^{2} – 40x + 32x – 1280 = 0

⇒ x (x – 40) + 32 (x – 40) = 0

⇒ (x – 40) (x + 32) = 0

⇒ x – 40 = 0 or x + 32 = 0

⇒ x = 40 or x = – 32

(as speed of train cannot be negative)

∴ The usual speed of the train = 40 km/hr.

Question 15.

If α and β are the zeroes of the quadratic polynomial f(x) = x^{2} – 4x + 3, find the value of (α^{4}β^{2} + α^{2}β^{4}).

Answer:

We have, f(x) = x^{2} – 4x + 3

a = 1, b = – 4, c = 3

Sum of the zeroes, (a + β) = \(\frac{b}{a}\) = (\(\frac{-4}{1}\)) = 4

Product of the zeroes, (αβ = \(\frac{c}{a}\) = \(\frac{3}{1}\)

α^{4}β^{2} + α^{2}β^{4}

⇒ = α^{2}β^{2} (α^{2} + β^{2})

⇒ = (αβ)^{2}[(α + β)^{2} – 2αβ)]

= (3)^{2} (4^{2} – 2(3)) = 9 (16 – 6)

= 9 (10) = 90

Question 16.

Prove that: (sin θ + 1 + cos θ) (sin θ – 1 + cos θ). sec θ cosec θ = 2.

Or

Question 16.

Prove that: [later]\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}[/later] = 2 cosec θ

Answer:

L.H.S. = (sin θ + 1 + cos θ) (sin θ – 1 + cos θ) sec θ cosec θ

= [(sin θ + cos θ) + 1] [(sin θ + cos θ) – 1] sec θ cosec θ

= [(sin θ + cos θ)^{2} – (1)^{2}] sec θ cosec θ …[∵ (a + b) (a – b) = a^{2} – b^{2}

= [(sin^{2} θ + cos^{2} θ + 2 sin θ cos θ – 1]. sec θ cosec θ

= (1 + 2 sin θ cos θ – 1). sec θ cosec θ …..[∵ sin^{2} θ + cos^{2} θ = 1]

= (2 sin θ cos θ) . \(\frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}\)

= 2 = R.H.S [Hence proved]

Or

= 2 cosec θ = R.H.S [Hence proved]

Question 17.

In what ratio does the point P (-4, y) divide the line segment joining the points A (-6, 10) and B(3, -8)? Hence find the value of y.

Answer:

Let AP : PB = k : 1

Using section formula,

Coordinates of P = Coordinates of P

(\(\frac{3 k-6}{k+1}, \frac{-8 k+10}{k+1}\)) = (-4, y)

\(\frac{3 k-6}{k+1}\) = -4

3k – 6 = – 4k – 4

3k + 4k = – 4 + 6

7k = 2

k = \(\frac{2}{7}\) ……….. (i)

\(\frac{-8 k+10}{k+1}\) = y

\(\frac{-8\left(\frac{2}{7}\right)+10}{\frac{2}{7}+1}\) = y … [From (i)

\(\frac{-16+70}{2+7}\) = y

∴ Required ratio = 2 : 7 ∴ y = \(\frac{54}{9}\) = 6

Question 18.

ABC is a right triangle in which ∠B = 90°. If AB = 8 cm and BC = 6 cm, find the diameter of the circle inscribed in the triangle.

Answer:

Construction: Join AO, OB, CO.

Proof: ABC is a rtΔ

∴ AC^{2} = AB^{2} + BC^{2} …[Pythagoras’ Theorem]

∴ AC^{2} = 64 + 36

∴ AC = 10

Area of ΔABC

= \(\frac{1}{2}\) × BC × AB

= \(\frac{1}{2}\) × 6 × 8 = 24 sq. cm

Now, Area of ΔABC

= ar(ΔAOB) + ar(ΔBOC) + ar(ΔAOC)

24 = \(\frac{1}{2}\) × r × AB + \(\frac{1}{2}\) × r × BC + \(\frac{1}{2}\) × r × AC

24 = \(\frac{1}{2}\) r [AB + BC + AC]

24 = \(\frac{1}{2}\) r [8 + 6 + 10]

24 = \(\frac{1}{2}\) r × 24 = 12 ∴ r = \(\frac{24}{12}\)

Now, diameter = 2r = 2(2) = 4 cm

Question 19.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 20.

Not in Current Syllabus.

Answer:

Here θ = 60°

∴ Area of the shaded region

= ar(major sector of large circle) – ar (major sector of small circle)

= \(\frac{360-60}{360}\) × \(\frac{22}{7}\) × (42)^{2} – \(\frac{360-60}{360}\) × \(\frac{22}{7}\) × (21)^{2} …[∵ area of major sector = \(\frac{360^{\circ}-\theta}{360^{\circ}}\)πr^{2}

= \(\frac{300}{360}\) × \(\frac{22}{7}\) × 42 × 42 \(\frac{300}{360}\) × \(\frac{22}{7}\) × 21 × 21

= \(\frac{300}{360}\) × \(\frac{22}{7}\) × 21 × 21 (2 × 2 – 1)

= \(\frac{5}{6}\) × 22 × 3 × 21(4 – 1)

= 55 × 21 × 3 = 3,465 cm^{2}

Question 21.

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/hr, in how much time will the tank be filled?

Answer:

(Volume of cyl. pipe) × Required time = Volume of cylindrical tank

= \(\frac{5}{3}\) × 60 min.

= 100 min. or 1 hr. and 40 min.

Question 22.

Calculate the mode of the following distribution :

Answer:

Class | Frequency |

10 – 15 | 4 |

15 – 20 | 7 f_{0} |

20 – 25 | 20 f_{1} |

25 – 30 | 8 f_{2} |

30 – 35 | 1 |

Maximum frequency = 20,

∴ Modal class is 20 – 25.

∴ Mode = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h …[l = 20, f_{1} = 20, f_{0} = 7, f_{2} = 8, h = 5

= 20 + \(\frac{20-7}{2(20)-7-8}\) × 5

= 20 + \(\frac{13}{25}\) × 5 = 20 + \(\frac{13}{5}\)

= 20 + 2.6 = 22.6

Section – D

Questions number 23 to 30 carry 4 marks each.

Question 23.

Solve for x : \(\frac{1}{2 a+b+2 x}\) = \(\frac{1}{2 a}\) + \(\frac{1}{b}\) + \(\frac{1}{2 x}\); x ≠ 0, x \(\frac{-2 a-b}{2}\), a, b ≠ 0

Or

Question 23.

The sum of the areas of two squares is 640 m^{2}. If the difference of their perimeter is 64 m, find the sides of the square.

Answer:

⇒ 2ax + bx + 2x^{2} = -ab

⇒ 2x^{2} + 2ax + bx + ab = 0

⇒ 2x(x + a) + b(x + a) = 0

⇒ (x + a) (2x + b) = 0

⇒ x + a = 0 or 2x + b = 0

⇒ x = -a or x = \(\frac{-b}{2}\)

Or

Let the side of large and small square be x m and y m respectively.

According to Question,

(x)^{2} + (y)^{2} = 640 ……. (i) …[∵ Area of square = (side)^{2}

4x – 4y = 64 ……. (ii) …[∵ Perimeter of square =4 × side

⇒ 4(x – y) = 64 64

⇒ x – y = \(\frac{64}{4}\) = 16

⇒ x = 16 + y …(iii)

Substituting x = 16 + y in equation (i),

(16 + y)^{2} + y^{2} = 640

256 + y^{2} + 32y + y^{2} = 640

2y^{2} + 32y – 384 = 0

y^{2} + 16y – 192 = 0 …[divide by 2)

y^{2} + 24y – 8y – 192 = 0

y (y + 24) – 8 (y + 24) = 0

(y – 8) (y + 24) = 0

y – 8 = 0 or y + 24 = 0

y = 8 or y = – 24

(rejected as side of square cannot be negative)

Putting y = 8 in (iii), we get x = 16 + 8 = 24

Therefore, side of the small square y = 8 m and side of the large square x = 24 m.

Question 24.

If the sum of the first p terms of an. A.P. is the same as the sum of its first q terms (where p ≠ q), then show that the sum of first (p + q) terms is zero.

Answer:

Let the first term be a and the common difference be d.

S_{p} = S_{q} …..[Given

\(\frac{p}{2}\)[2a + (p – 1)d] = \(\frac{q}{2}\)[2a + (q – 1)d]

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

⇒ 2ap + p (p – 1) d = 2aq + q (q – 1) d

⇒ 2ap – 2aq + p (p – 1) d – q(q – 1) d = 0

⇒ 2a (p – q) + d [p(p – 1) – q (q – 1)] = 0

⇒ 2a(p – q) + d (p^{2} – p – q^{2} + q) = 0

⇒ 2a (p – q) + d (p^{2} – q^{2} – p + q) = 0

⇒ 2a (p – q) + d [(p + q) (p – q) – (p – q)] = 0

⇒ 2a (p – q) + d (p – q)[(p + q – 1)] = 0

⇒ (p – q) [2a + (p + q-d] = 0

⇒ p – q = 0 or 2a + (p + q – 1)d = 0 …… (i)

p = q (not possible)

Now, S_{p+q} = (\(\frac{p+q}{2}\)) [2a + (p + q – 1)d]

= (\(\frac{p+q}{2}\)) (0) …[From (i)

= 0 [Hence Proved]

Question 25.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 26.

A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.

Or

Question 26.

There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find»the width of the rivfer and height of the other pole.

Answer:

Let AB be the cliff.

In rt. ΔABC,

tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

⇒ \(\frac{\sqrt{3}}{1}=\frac{150}{B C}\)

⇒ √3 BC = 150

BC = \(\frac{150}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{150 \sqrt{3}}{3}\) = 50√3

In rt. ΔABD, tan 45° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)

⇒ 1 = \(\frac{150}{B D}\)

⇒ BD = 150 m

Now, Distance covered by Boat,

CD = BD – BC

= 150 – 50√3 =50 (3 – √3)

= 50(3 – 1.732) (∵ √3 – 1.732)

= 50 × 1.268

= 63.4 m

∴ Speed of Boat = \(\frac{\text { Distance }}{\text { Time }}\) \(\frac{63.4}{2}\)

= 31.7 metres/minute

Or

Let CE be one pole and AB be the other pole. AD = BC – distance between them.

Width of the river BC = 20√3 m …[From (i)

= 20(1.732)m = 34.64m …[∵√3 = 1.732

Height of other pole, AB = CD = CE – DE

= (60 – 20)m = 40 m ……[From (ii)

Question 27.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 28.

Prove that: sin^{8} θ – cos^{8} θ = (1 – 2 cos^{2} θ) (1 – 2 sin^{2} θ cos^{2} θ).

Answer:

L.H.S. sin^{8} θ – cos^{8} θ

= (sin^{4} θ)^{2} – (cos^{4} θ)^{2}

= (sin^{4} θ – cos^{4} θ) (sin^{4} θ + cos^{4} θ) …[∵ a^{2} – b^{2} = (a – b) (a + b)]

= [(sin^{2} θ)^{2} – (cos^{2} θ)^{2}] (sin^{2} θ . sin^{2} θ + cos^{2} θ . cos^{2} θ)

= (sin^{2} θ – cos^{2} θ) (sin^{2} θ + cos^{2} θ) (sin^{2} θ .

(1 – cos^{2} θ) + cos^{2} θ (1 – sin^{2} θ)) …. [∵ sin^{2} θ = 1 – cos^{2} θ, cos^{2} θ = 1 – sin^{2} θ, sin^{2} θ + cos^{2} θ = 1

= (1 – cos^{2} θ – cos^{2} θ) . 1 . (sin^{2} θ – sin^{2} θ cos^{2} θ + cos2 θ – sin^{2} θ cos^{2} θ)

– (1 – 2 cos^{2} θ) (sin^{2} θ + cos^{2} θ – 2 sin^{2} θ cos^{2} θ)

– (1 – 2 cos^{2} θ) (1 – 2 sin^{2} θ cos^{2} θ) = R.H.S. [Hence Proved]

Question 29.

Not in Current Syllabus.

Answer:

Not in Current Syllabus

Question 30.

Calculate the mean of the following frequency distribution :

Answer:

Class | Frequency (f_{i}) |
Class Mark (x_{i}) |
f_{i}x_{i} |

10 – 30 | 5 | 20 | 100 |

30 – 50 | 8 | 40 | 320 |

50 – 70 | 12 | 60 | 720 |

70 – 90 | 20 | 80 | 1600 |

90 – 110 | 3 | 100 | 300 |

110 – 130 | 2 | 120 | 240 |

Σf_{i} = 50 |
Σf_{i}x_{i} = 3280 |

∴ Mean = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac{3280}{50}\) = 65.6

SET II Code No. 30/4/2

Note: Except for the following questions, all the remaining questions have been asked in Set – I.

Question 1.

For what values of k does the quadratic equation 4x^{2} – 12x – k = 0 have no real roots?

Answer:

We have, 4x^{2} – 12x – k = 0

Here, a = 4, b = – 12, c = -k

For No Real Roots,

D < 0

∴ b^{2} – 4ac < 0

⇒ (-12)^{2} – 4 (4) (-k) < 0

⇒ 144 + 16k < 0

⇒ 16k < – 144

⇒ k < \(\frac{-144}{2}\) 0

∴ k < – 9

Question 7.

A bag contains 15 balls, out of which some are white and the others are black. If the probability of drawing a black ball at random from the bag is \(\frac{2}{3}\), then find how many white balls are there in the bag. 3

Answer:

Let number of white balls = x

Number of black balls = (15 – x)

According to the Question,

P (black ball) = \(\frac{2}{3}\) …[Given

\(\frac{15-x}{15}\) = \(\frac{2}{3}\)

⇒ 30 = 45 – 3x

⇒ 3x = 45 – 30

⇒ 3x = 15

x = \(\frac{15}{3}\) = 5

∴ Number of white balls = 5

Question 20.

Prove that 2 + 3√ 3 is an irrational number when it is given that √ 3 is an irrational number.

Answer:

Let us assume, to the contrary, that 2 + 3√3 is rational.

So that, we can find integers a and b(b ≠ 0).

Such that 2 + 3√3 = \(\frac{a}{b}\), where a and b are coprime.

Rearranging the equation, we get

3√3 = \(\frac{a}{b}\) – 2

⇒ 3√3 = \(\frac{a-2 b}{b}\)

√3 = \(\frac{a-2 b}{3 b}\)

⇒ √3 = \(\frac{a}{3 b}\) – \(\frac{2}{3}\)

Since a and b are integers, we get \(\frac{a}{3 b}\) – \(\frac{2}{3}\) is rational and so √3 is rational.

But this contradicts the fact that √3 is irrational. (Given)

This contradiction has arisen because of our incorrect assumption that 2 + 3√3 is rational.

So, we conclude that 2 + 3√3 is an irrational number.

Question 21.

Sum of the areas of two squares is 157 m^{2}. If the sum of their perimeters is 68 m, find the sides of the two squares.

Answer:

Let x and y be the side of large and small squares respectively.

According to the Question,

x^{2} + y^{2} = 157 …(i) …[∵ Area of square = side^{2}

4x + 4y = 68 …………(ii) ……[∵ Perimeter of square = 4 × side

⇒ 4 (x + y) = 68

⇒ x + y = \(\frac{68}{4}\) = 17 ⇒ x = 17 – y …..(iii)

Substituting x = 17 – y in (i), we get

(17 – y)^{2} + y^{2} = 157

⇒ 289 + y^{2} – 34y + y^{2} = 157

⇒ 2y^{2} – 34y + 132 = 0

⇒ y^{2} – 17y + 66 = 0 …[divided by 2)

⇒ y^{2} – 11y – 6y + 66 = 0

y(y – 11 -6(y – 11) = o

(y – 6) (y – 11) = 0

y – 6 = 0 or y – 11 = 0

y = 6 or y = 11 (Rejected x > y)

Putting y = 6 in (iii), we get x = 17 – 6 = 11

∴ Side of the large square = 11 m and side of the small square = 6 m.

Question 22.

Find the quadratic polynomial, sum and product of whose zeroes are – 1 and – 20 respectively. Also find the zeroes of the polynomial so obtained.

Answer:

As we know, Quadratic polynomial p(x)

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

= [(x^{2} – (- 1)x + (-20)]

= (x^{2} + 1x – 20)

Zeroes of quadratic polynomial,

(x^{2} + 5x – 4x – 20) = 0

[x(x + 5) – 4 (x + 5)] = 0

(x – 4) (x + 5) = 0

x – 4 = 0 or x + 5 = 0

x = 4 or x = – 5

Therefore, zeroes are 4 and – 5.

Question 23.

A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time, it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane.

Or

Question 23.

Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m^{2}.

Answer:

Let the usual speed of the plane = x km/hr.

∴ The increased speed of the plane

= (x + 250) km/ hr

As we know, Time = \(\frac{\text { Distance }}{\text { Speed }}\)

According to Question,

⇒ x^{2} + 250x = 2 (375000)

⇒ x^{2} + 250x = 750000

⇒ x^{2} + 250x – 750000 = 0

⇒ x^{2} + 1000x – 750x – 750000 = 0

⇒ x(x + 1000) – 750 (x + 1000) = 0

⇒ (x – 750) (x + 1000) = 0

⇒ x – 750 = 0 or x + 1000 = 0

⇒ x = 750 or x = -1000

(rejected as speed cannot be negative)

Therefore, the usual speed of the plane = 750 km/hr.

Or

Let length and breadth of a rectangular park be x m and y m respectively.

According to the Question,

Perimeter of Rectangular Park = 60 m

⇒ 2(x + y) = 60 …[Perimeter of Rectangle = 2(l + b)

⇒ x + y = \(\frac{60}{2}\) = 30 ∴ x + y = 30

y = 30 – x …(i)

Now, Area of rectangular park = 200 m^{2} …[Given

xy = 200 …[Area of Rectangle = (side)^{2}

⇒ x (30 – x) = 200 …[From (i)

⇒ 30x – x^{2} = 200

⇒ x^{2} – 30x + 200 = 0

⇒ x^{2} – 20x – 10x + 200 = 0

⇒ x (x – 20) – 10 (x – 20) = 0

⇒ (x – 10) (x – 20) =0

⇒ x – 10 = 0 or x – 20 = 0

⇒ x = 10 or x = 20

From (i), when x = 10, y = 30 – 10 = 20

When x = 20, y = 30 – 20 = 10

Question 24.

Find the value of x, when in the A.P. given below 2 + 6 + 10 + … + x = 1800.

Answer:

First term, a = 2

Common difference, d = 6 – 2 = 4

n^{th} term, a_{n} = x

S_{n} = 1800 …[Given

⇒ \(\frac{n}{2}\) [2a + (n – 1) d] = 1800

⇒ \(\frac{n}{2}\) [2(2) + (n – 1) 4] = 1800

⇒ n(4 + 4n – 4) = 3600

⇒ 4n^{2} = 3600

⇒ n^{2} = \(\frac{3600}{4}\) = 900

n = \(\sqrt{900}\) = ± 30

∴ n = 30 …[∵ number of terms cannot be negative

Now, n^{th} term,

a_{n} = a + (n – 1) d

= 2 + (30 – 1) 4 = 2 + 29 (4)

= 2 + 116 = 118

Question 25.

If sec θ + tan θ = m, show that [later][/latex] = sin θ .

Answer:

sec θ + tan θ = m ….[Given

1 + sin θ = m^{2} – m^{2} sin θ

sin θ + m^{2} sin θ = m^{2} – 1

(1 + m^{2}) sin θ = m^{2} – 1

∴ sin θ = \(\frac{m^2-1}{m^2+1}\) [Hence proved]

SET III Code No. 30/4/3

Note: Except for the following questions, all the remaining questions have been asked in Set – I and Set – II.

Question 1.

Which term of the A.P. – 4, -1, 2,… is 101?

Answer:

First term, a = – 4

Common difference, d = – 1 – (- 4)

= – 1 + 4 = 3

a_{n} = 101

⇒ a + (n – 1 )d = 101

⇒ – 4 + (n – 1) 3 = 101

⇒ (n – 1) 3 = 101 + 4

⇒ (n – 1) = \(\frac{105}{3}\) = 35

∴ n = 35 + 1 = 36

Question 12.

A die is thrown once. Find the probability of getting

(a) a prime number

(b) an odd number.

Answer:

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

(i) Prime numbers are 2, 3, 5 i.e., 3

P (a prime number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

(ii) Odd numbers are = 1, 3, 5 i.e., 3

P (an odd number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 17.

Show that \(\frac{2+3 \sqrt{2}}{7}\) is not a rational number, given that √2 is an irrational number.

Answer:

Let us assume, to the contrary, that \(\frac{2+3 \sqrt{2}}{7}\) is rational.

That is, we can find coprime a and b (b ≠ 0) such that

\(\frac{2+3 \sqrt{2}}{7}\) = \(\frac{a}{b}\)

⇒ 2 + 3√2 = \(\frac{7 a}{b}\) ⇒ 3√2 = \(\frac{7 a}{b}\) – 2

√2 = \(\frac{7 a}{3 b}-\frac{2}{3}\) ⇒ √2 = \(\frac{7 a}{3 b}-\frac{2}{3}\)

Since a and b are integers, we get is \(\frac{7 a}{3 b}-\frac{2}{3}\) rational and so √2 is rational.

But this contradicts the fact that √2 is rational …[Given

This contradiction has arisen because of our incorrect assumption that \(\frac{2+3 \sqrt{2}}{7}\) is rational.

So, we conclude that \(\frac{2+3 \sqrt{2}}{7}\) is an irrational number.

Question 23.

In an A.P., the first term is – 4, the last term is 29 and the sum of all its terms is 150. Find its common difference.

Answer:

We have, a = – 4, a_{n} = 29, S_{n} = 150

a_{n} = 29

a + (n-1) d = 29

– 4 + (n – 1) d = 29

(n – 1) d = 29 + 4

(n – 1) d = 33

(12 – 1) d = 33 …[From (i)

11d = 33

∴ d = \(\frac{33}{11}\) = 3

S_{n} = 150

\(\frac{n}{2}\)(a + a_{n}) = 150

\(\frac{n}{2}\)(- 4 + 29) = 150

\(\frac{25 n}{2}\) = 150

n(25) = 300

n = \(\frac{300}{25}\) = 12

∴ n = 12 …. (i)

Question 25.

Prove that: 2 (sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ) + 1 = 0.

Answer:

L.H.S. 2 (sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ) + 1

= 2[(sin^{2} θ)^{3} + (cos^{2} θ)^{3}] – 3(sin^{4} θ + cos^{4} θ) + 1

= 2[(sin^{2} θ + cos^{2} θ).[(sin^{2} θ)^{2} + (cos^{2} θ)^{2} – sin^{2} θ cos^{2} θ)] – 3(sin^{4} θ + cos^{4} θ) + 1 …[∴ a^{3} + b^{3} = (a + b) (a^{2} + b^{2} – ab)

= 2[1.(sin^{4} θ + cos^{4} θ – sin^{2} θ cos^{2} θ)]

– 3(sin^{4} θ + cos^{4} θ) + 1 …[∵ sin^{2} θ + cos^{2} θ = 1

= 2(sin^{4} θ + cos^{4} θ) – 2 sin^{2} θ cos^{2} θ – 3(sin^{4} θ + cos^{4} θ) + 1

= -(sin^{4} θ + cos^{4} θ) – 2 sin^{2} θ cos^{2} θ + 1

= -(sin^{4} θ + cos^{4} θ + 2 sin^{2} θ cos^{2} θ) + 1

= -[(sin^{2} θ)^{2} + (cos^{2} θ)^{2} + 2(sin^{2} θ) (cos^{2} θ)] + 1

= -(sin^{2} θ + cos^{2} θ)2 + 1 …[∵ a^{2} + b^{2} + 2ab = (a + b)^{2}

= -(1)^{2} + 1 …[∵ sin^{2} θ + cos^{2} θ = 1

= – 1 + 1 = θ [Hence proved]