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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
∴ kx2 – 2kx + 6 = 0, k ≠ 0
For two equal roots …[Given
b2 – 4ac = 0, a = k, b = -2k, c = 6
⇒ (-2k)2 – 4(k) (6) = 0
⇒ 4k2 – 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 = \(\frac{31-36}{2}\) = \(\frac{-5}{2}\)
an = -47 …[Given
an = [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 22 × 53 × 32 × 17.
Answer:
1st 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.

IInd 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
22 × 53 × 32 × 17
= 22 × 52 × 5 × 32 × 17
= (2 × 5)2 × 5 × 32 × 17
= (10)2 × 5 × 32 × 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 7th 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, an = 204
Now, an = 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
a3 = 16
a + 2d = 16
a + 2(6) – 16 ……[From (i)
a + 12 = 16
a = 4

a7 = a5 + 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 = 2n = 23 = 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,
a1 = k, b1 = 2, c1 = 3,
a2 = 3, b2 = 6, c2 = 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
⇒ x2 – 8x – 1280 = 0
⇒ x2 – 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) = x2 – 4x + 3, find the value of (α4β2 + α2β4).
Answer:
We have, f(x) = x2 – 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 = \(\frac{c}{a}\) = \(\frac{3}{1}\)
α4β2 + α2β4
⇒ = α2β22 + β2)
⇒ = (αβ)2[(α + β)2 – 2αβ)]
= (3)2 (42 – 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) = a2 – b2
= [(sin2 θ + cos2 θ + 2 sin θ cos θ – 1]. sec θ cosec θ
= (1 + 2 sin θ cos θ – 1). sec θ cosec θ …..[∵ sin2 θ + cos2 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
∴ AC2 = AB2 + BC2 …[Pythagoras’ Theorem]
∴ AC2 = 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 = \(\frac{360^{\circ}-\theta}{360^{\circ}}\)πr2
= \(\frac{300}{360}\) × \(\frac{22}{7}\) × 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 cm2

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 f0
20 – 25 20 f1
25 – 30 8 f2
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, f1 = 20, f0 = 7, f2 = 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 m2. If the difference of their perimeter is 64 m, find the sides of the square.
Answer:

⇒ 2ax + bx + 2x2 = -ab
⇒ 2x2 + 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 + y2 = 640
256 + y2 + 32y + y2 = 640
2y2 + 32y – 384 = 0
y2 + 16y – 192 = 0 …[divide by 2)
y2 + 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.
Sp = Sq …..[Given
\(\frac{p}{2}\)[2a + (p – 1)d] = \(\frac{q}{2}\)[2a + (q – 1)d]
…[ Sn = \(\frac{n}{2}\)(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 (p2 – p – q2 + q) = 0
⇒ 2a (p – q) + d (p2 – q2 – 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, Sp+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: sin8 θ – cos8 θ = (1 – 2 cos2 θ) (1 – 2 sin2 θ cos2 θ).
Answer:
L.H.S. sin8 θ – cos8 θ
= (sin4 θ)2 – (cos4 θ)2
= (sin4 θ – cos4 θ) (sin4 θ + cos4 θ) …[∵ a2 – b2 = (a – b) (a + b)]
= [(sin2 θ)2 – (cos2 θ)2] (sin2 θ . sin2 θ + cos2 θ . cos2 θ)
= (sin2 θ – cos2 θ) (sin2 θ + cos2 θ) (sin2 θ .
(1 – cos2 θ) + cos2 θ (1 – sin2 θ)) …. [∵ sin2 θ = 1 – cos2 θ, cos2 θ = 1 – sin2 θ, sin2 θ + cos2 θ = 1
= (1 – cos2 θ – cos2 θ) . 1 . (sin2 θ – sin2 θ cos2 θ + cos2 θ – sin2 θ cos2 θ)
– (1 – 2 cos2 θ) (sin2 θ + cos2 θ – 2 sin2 θ cos2 θ)
– (1 – 2 cos2 θ) (1 – 2 sin2 θ cos2 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 (fi) Class Mark (xi) fixi
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
Σfi = 50 Σfixi = 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 4x2 – 12x – k = 0 have no real roots?
Answer:
We have, 4x2 – 12x – k = 0
Here, a = 4, b = – 12, c = -k
For No Real Roots,
D < 0
∴ b2 – 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 m2. 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,
x2 + y2 = 157 …(i) …[∵ Area of square = side2
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 + y2 = 157
⇒ 289 + y2 – 34y + y2 = 157
⇒ 2y2 – 34y + 132 = 0
⇒ y2 – 17y + 66 = 0 …[divided by 2)
⇒ y2 – 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)
= [x2 – (sum of zeores) x + (product of zeroes)]
= [(x2 – (- 1)x + (-20)]
= (x2 + 1x – 20)
Zeroes of quadratic polynomial,
(x2 + 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 m2.
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,

⇒ x2 + 250x = 2 (375000)
⇒ x2 + 250x = 750000
⇒ x2 + 250x – 750000 = 0
⇒ x2 + 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 m2 …[Given
xy = 200 …[Area of Rectangle = (side)2
⇒ x (30 – x) = 200 …[From (i)
⇒ 30x – x2 = 200
⇒ x2 – 30x + 200 = 0
⇒ x2 – 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
nth term, an = x
Sn = 1800 …[Given
⇒ \(\frac{n}{2}\) [2a + (n – 1) d] = 1800
⇒ \(\frac{n}{2}\) [2(2) + (n – 1) 4] = 1800
⇒ n(4 + 4n – 4) = 3600
⇒ 4n2 = 3600
⇒ n2 = \(\frac{3600}{4}\) = 900
n = \(\sqrt{900}\) = ± 30
∴ n = 30 …[∵ number of terms cannot be negative
Now, nth term,
an = 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 = m2 – m2 sin θ
sin θ + m2 sin θ = m2 – 1
(1 + m2) sin θ = m2 – 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
an = 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, an = 29, Sn = 150
an = 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

Sn = 150
\(\frac{n}{2}\)(a + an) = 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 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
Answer:
L.H.S. 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1
= 2[(sin2 θ)3 + (cos2 θ)3] – 3(sin4 θ + cos4 θ) + 1
= 2[(sin2 θ + cos2 θ).[(sin2 θ)2 + (cos2 θ)2 – sin2 θ cos2 θ)] – 3(sin4 θ + cos4 θ) + 1 …[∴ a3 + b3 = (a + b) (a2 + b2 – ab)
= 2[1.(sin4 θ + cos4 θ – sin2 θ cos2 θ)]
– 3(sin4 θ + cos4 θ) + 1 …[∵ sin2 θ + cos2 θ = 1
= 2(sin4 θ + cos4 θ) – 2 sin2 θ cos2 θ – 3(sin4 θ + cos4 θ) + 1
= -(sin4 θ + cos4 θ) – 2 sin2 θ cos2 θ + 1
= -(sin4 θ + cos4 θ + 2 sin2 θ cos2 θ) + 1
= -[(sin2 θ)2 + (cos2 θ)2 + 2(sin2 θ) (cos2 θ)] + 1
= -(sin2 θ + cos2 θ)2 + 1 …[∵ a2 + b2 + 2ab = (a + b)2
= -(1)2 + 1 …[∵ sin2 θ + cos2 θ = 1
= – 1 + 1 = θ [Hence proved]


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