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

Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2019 (Series: JMS/4) to familiarize themselves with the exam format and marking scheme.

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:

1. All questions are compulsory.
2. This question paper consists of 30 questions divided into four sections – A, B, C and D.
3. 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.
4. 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.
5. 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

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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]