Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 2 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 2 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
- Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
- Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
- Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
- Section E has 3 sourced based/Case Based/passage based/integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
- All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.
Section – A (20 Marks)
Question 1.
The zeroes of the quadratic polynomial x2 + kx + k, where k > 0: 1
(a) are both positive
(b) are both negative
(c) are always equal
(d) are always unequal
Answer:
(b) are both negative
Explanation: Let α and β be the zeroes of x2 + kx + k. Then, α + β = – k and αβ = k.
This is possible only when α and β both are negative.
Question 2.
If p, 2p – 1, 2p + 1 are three consecutive terms of an A.P., then the value of ‘p’ is: 1
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(a) 3
Explanation: Since p, 2p – 1 and 2p + 1 are three consecutive terms of an A.P.,
2(2p – 1) = p + (2p + 1)
⇒ 4p – 2 = 3p + 1
⇒ P = 3
Question 3.
In which quadrant does the point (-1, -2) lie? 1
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III
Explanation: Third quadrant since, both ‘x’ and ‘y’ are negative.
Question 4.
If radii of two concentric circles are 4 cm and 5 cm, then the length of the chord to one circle which is tangent to the other circle is: 1
(a) 6 cm
(b) 5 cm
(c) 8 cm
(d) 9 cm
Answer:
(a) 6 cm
Explanation:
We know tangent is perpendicular to radius at the point of contact.
∴ OP ⊥ AB
Also, AB is a chord to outer circle and perpendicular from centre to the chord bisects it.
∴ AP = BP = \(\frac{1}{2}\) AB
In right ΔOPA,
AP = \(\sqrt{O A^2-O P^2}\)
= \(\sqrt{5^2-4^2}\)
= \(\sqrt{25-16}\)
= √9
= 3 cm
Thus, AB = 2 × AP
= 6 cm
Question 5.
The tangent of a circle makes angle with radius at point of contact is: 1
(a) 60°
(b) 30°
(c) 90°
(d) 45°
Answer:
(c) 90°
Explanation: Tangent at any point of a circle is perpendicuLar to the radius at the point of contact. So, the tangent makes a right angle with the radius at the point of contact.
Question 6.
In the given figure, D and E are two points lying on side AB, such that AD = BE. If DP || BC and EQ||AC then: 1
(a) PQ || AB
(b) PQ = AB
(c) PQ || CD
(d) PQ = AC
Answer:
(a) PQ || AB
Explanation: In ΔABC, we have DP || BC and EQ || AC
∴ By BPT theorem,
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AP}}{\mathrm{PC}}\)
and \(\frac{B E}{E A}=\frac{B Q}{Q C}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{BQ}}{\mathrm{QC}}\)
[∵ EA = ED + DA = ED + BE = BD]
⇒ \(\frac{\mathrm{AP}}{\mathrm{PC}}=\frac{\mathrm{BQ}}{\mathrm{QC}}\)
∴ PQ || AB [By converse of BPT theorem]
Question 7.
Consider the following frequency distribution. 1
Class Interval | Frequency |
0 – 10 | 3 |
10 – 20 | 9 |
20 – 30 | 15 |
30 – 40 | 30 |
40 – 50 | 18 |
50 – 60 | 5 |
The modal class is:
(a) 10 – 20
(b) 20 – 30
(c) 30 – 40
(d) 40 – 50
Answer:
(c) 30 – 40
Explanation: The class 30 – 40 has maximum frequency. So, the modal class 30 – 40.
Question 8.
in the distribution given below, the frequency of the class succeeding the modal class is: 1
Marks | Frequency |
Below 10 | 3 |
Below 20 | 12 |
Below 30 | 27 |
Below 40 | 57 |
Below 50 | 75 |
Below 60 | 80 |
(a) 15
(b) 5
(c) 9
(d) 18
Answer:
(d) 18
Explanation:
Marks | Frequency |
0 – 10 | 3 |
10 – 20 | 9 |
20 – 30 | 15 |
30 – 40 | 30 |
40 – 50 | 18 |
50 – 60 | 5 |
Here, maximum frequency is 30, which belongs to class 30 – 40. So, the modal class is 30 – 40.
The class succeeding the modal class is 40 – 50 with frequency 18.
Question 9.
tan2 θ sin2 θ is: 1
(a) tan2 θ – sin2 θ
(b) tan2 θ + sin2 θ
(c) \(\frac{\tan ^2 \theta}{\sin ^2 \theta}\)
(d) sin2 θ cot2 θ
Answer:
(a) tan2 θ – sin2 θ
Explanation:
tan2 θ – sin2 θ = \(\frac{\sin ^2 \theta \sin ^2 \theta}{\cos ^2 \theta}\)
sin2 θ(\(\frac{1-\cos ^2 \theta}{\cos ^2 \theta}\))
= \(\frac{\sin ^2 \theta}{\cos ^2 \theta}\) – sin2 θ
= tan2 θ – sin2 θ
Question 10.
In the given figure, x is: 1
(a) \(\frac{a b}{a+b}\)
(b) \(\frac{a c}{b+c}\)
(c) \(\frac{b c}{b+c}\)
(d) \(\frac{a c}{a+c}\)
Answer:
(b) \(\frac{a c}{b+c}\)
Explanation: In ΔKPN and ΔKLM,
∠K is common and we have
∠KNP = ∠KML = 46
Thus by A – A criterion of similarity
ΔKNP ~ ΔKML
Thus \(\frac{\mathrm{KN}}{\mathrm{KM}}=\frac{\mathrm{NP}}{\mathrm{ML}}\)
\(\frac{c}{b+c}=\frac{x}{a}\)
x = \(\frac{a c}{b+c}\)
Question 11.
if α and β are the zeros of the polynomial p(x) = x2 – px + q, then the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is: 1
(a) p
(b) \(\frac{-p}{q}\)
(c) pq
(d) \(\frac{p}{q}\)
Answer:
(d) \(\frac{p}{q}\)
Explanation: Since α and β are the zeros of
p(x) = x2 – px + q
∴ α + β = p and αβ = q
Now, \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{p}{q}\)
Question 12.
In figure XY ||QR and \(\frac{P X}{X Q}=\frac{P Y}{Y R}=\frac{1}{2}\), then the value of XY is: 1
(a) \(\frac{1}{3}\)QR
(b) \(\frac{1}{3}\)XQ
(c) \(\frac{1}{3}\)PQ
(d) \(\frac{1}{3}\)PR
Answer:
(a) \(\frac{1}{3}\)QR
Explanation:
Question 13.
The discriminant of the quadratic equation (p + 3)x2 – (5 – p)x + 1 = 0 is: 1
(a) (p + 13)
(b) (p + 1)
(c) (p + 2)
(d) (p – 13) (p – 1)
Answer:
(d) (p – 13) (p – 1)
Explanation: The given quadratic equation is
(p + 3)x2 – (5 – p)x +1 = 0.
Discriminant = b2 – 4ac
= [-(5 – p)]2 – 4(p + 3)(1)
= 25 + p2 – 10p – 4p – 12
= p2 – 14p + 13
= p2 – 13p – p + 13
= p (p – 13) – 1 (p – 13)
= (p – 13) (p – 1)
Question 14.
In a ΔABC, DE || BC with D on AB and E on AC. If \(\frac{A D}{D B}\), then \(\frac{B C}{D E}\) is: 1
(a) \(\frac{5}{2}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{5}{3}\)
(d) \(\frac{2}{3}\)
Answer:
(a) \(\frac{5}{2}\)
Explanation: Since DE is parallel to BC.
∴ ΔADE ~ ΔAMBC
\(\frac{A B}{A D}=\frac{5}{2}\)
\(\frac{A D}{A B}=\frac{2}{5}\)
So, by eqn. (i)
\(\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{5}{2}\)
Question 15.
The roots of the quadratic equation 2x2 – 3x – 5 = 0 are: 1
(a) both equal
(b) opposite integers
(c) rational and unequal
(d) not real
Answer:
(c) rational and unequal
Explanation: We have,
2x2 – 3x – 5 =0
Here, a = 2, b = – 3 and c = – 5.
∴ D = b2 – 4ac
= (- 3)2 – 4(2)(- 5)
= 9 + 40
= 49 > 0
and 49 is a perfect square also.
Thus, given equation has rational and unequal roots.
Question 16.
The ratio of LCM and HCF of the least composite and the least prime numbers is: 1
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 1 : 3
Answer:
(b) 2 : 1
Explanation: Least composite number is 4 and the least prime number is 2.
LCM (4, 2): HCF (4, 2) = 4 : 2 = 2 : 1
Question 17.
The 8th term from the end of A.P.: – 12, – 7, – 2, …………, 68 is: 1
(a) 33
(b) 35
(c) 30
(d) 36
Answer:
(a) 33
Q Explanation: Write the given A.P in reverse order, then series is,
68, 63,……… – 2,- 7,- 12
Then, a = 68 and d = – 5
a8 = a + (8 – 1) d
= 68 + 7 (- 5)
= 33
Question 18.
In the given figure, ΔABC ~ ΔPQR. The value of x is: 1
(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 5 cm
Answer:
(c) 3 cm
Explanation: Since, ΔABC ~ ΔPQR
Then, \(\frac{A B}{P Q}=\frac{B C}{R Q}=\frac{A C}{P R}\)
⇒ \(\frac{6}{4.5}=\frac{4}{x}=\frac{5}{3.75}\)
⇒ x = \(\frac{4 \times 4.5}{6}\)
= 3 cm
Question 19.
Assertion (A): The length of the tangent is 7 cm in a circle of radius 3 cm and a point’s distance from the centre of the circle is 5 cm.
Reason (R): (hypotenuse)2 = (base)2 + (height)2 1
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer:
(d) Assertion (A) is false but reason (R) is true.
Explanation:
A tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, ∠OBA = 90° and ΔOBA is a right¬angled triangle.
By Pythagoras theorem,
OA2 = OB2 + AB2
52 = 32 + AB2
AB2 = 25 – 9
AB2 = 16
AB = 4 cm
Question 20.
Assertion (A): 2x2 – 4x + 3 = 0 is a quadratic equation.
Reason (R): All polynomials of degree n, when n is a whole number can be treated as quadratic equation. 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.
Explanation: Assertion (A) is true but Reason (R) is false as only polynomials of degree 2, can be treated as quadratic equations.
Section – B (10 Marks)
Question 21.
Show that 3 + √5 is an irrational number, assume that √5 is an irrational number. 2
Answer:
If possible, let us assume that 3 + √5 be a , rational number. So, there exists positive integers a and b such that, 3 + √5 = \(\frac{a}{b}\), where a and b are integers having no common factor other than 1 and b ≠ 0.
⇒ √5 = \(\frac{a}{b}\) – 3
√5 = \(\frac{a-3 b}{b}\)
Since, \(\frac{a-3 b}{b}\) is a rational number, √5 is a rational number which is a contradiction to the fact that “ √5 is irrational”.
Hence, 3 + √5 is irrational.
Question 22.
Evaluate: (sin4 60° + sec4 30°) – 2 (cos2 45° – sin2 90°)
OR
Prove that \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ 2
Answer:
(sin4 60° + sec4 30°) – 2(cos2 45° – sin2 90°)
OR
L.H.S = \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\)
Question 23.
A dice is thrown twice. Find the probability that 5 will not come up either time.
OR
Two dice are thrown simultaneously and the outcomes are noted. Find the probability that:
(A) doublets are obtained
(B) sum of numbers on the two dice is 5. 2
Answer:
Total number of possible outcomes = 36
Number of favourable outcomes = 36 – 11
= 25
[Excluding (1, 5), (5, 1), (2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4), (5, 5), (5, 6), (6, 5)]
SO required probability = \(\frac{25}{36}\)
OR
On throwing two dice together, total number of outcomes = 36
(A) Favourable outcomes = {(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.
⇒ Number of favourable outcomes = 6 6 1
∴ P (a doublet) = \(\frac{6}{36}\) i.e., \(\frac{1}{6}\)
(B) Favourable outcomes = {(1, 4), (2, 3), (3, 2), (4, 1)}.
⇒ Number of favourable outcomes = 4
∴ P (sum of 5 on two numbers)
= \(\frac{4}{36}\) i.e., \(\frac{1}{9}\)
Question 24.
Find a point on x-axis which is equidistant from A (-3, 4) and B (7, 6). 2
Answer:
Let P (x, 0) be a point on x-axis, equidistant from A (-3, 4) and B (7, 6), i.e.,
AP = BP or AP2 = BP2
i.e. (- 3 – x)2 + (4 – 0)2 = (7 – x)2 + (6 – 0)2
i.e. 9 + x2 + 6x + 16 = 49 + x2 – 14x + 36
⇒ 6x + 25 = – 14x + 85
⇒ 20x = 60
⇒ x = 3
Thus, the required point is (3, 0).
Question 25.
In the given figure, find the area of the shaded region, [use π = 3.14].
Answer:
Diameter of the circle = Diagonal AC of the rectangle ABCD
∴ AC = \(\sqrt{A D^2+D C^2}=\sqrt{6^2+8^2}\)
[by using Pythagoras theorem]
= \(\sqrt{36+64}\)
= \(\sqrt{100}\)
= 10 cm
Radius of the circle (r)
= \(\frac{A C}{2}\)
= \(\frac{1}{2}\) × 10 cm
= 5 cm
Area of the rectangle ABCD
= AB × BC
= (8 × 6) cm2
= 48 cm2
∴ Area of circle with radius 5 cm
= πr2
= 3.14 × (5)2
= 3.14 × 25 cm2
= 78.50 cm2
Hence, Area of shaded region
= Area of circle – Area of rectangle ABCD
= 78.50 cm2 – 48 cm2
= 30.50 cm2
Section – C (18 Marks)
Question 26.
The sum of the squares of two consecutive multiples of 7 is 637. Find the two multiples.
OR
The length and breadth of a rectangle are (3x + 1) cm and (2x – 1) cm respectively. If the area of the rectangle is 144 sq. cm, then find the value of x. 3
Answer:
Let 7x and 7(x + 1) be two consecutive multiples of 7.
Then,
(7x)2 + [7 (x + 1)]2 = 637
⇒ 49x2 + 49 (x2 + 2x + 1) = 637
⇒ 98x2 + 98x + 49 = 637
⇒ 98x2 + 98x – 588 = 0
⇒ x2 + x – 6 = 0
⇒ (x + 3)(x – 2) = 0
⇒ x + 3 = 0 or x – 2 = 0
⇒ x = -3 or x = 2
Thus, two consecutive multiples of 7 are 14 and 21, or – 21 and – 14.
OR.
Given, length of a rectangle = (3x + 1) cm
Breadth of a rectangLe = (2x – 1) cm
Area of a rectangle = 144 sq. cm
ATQ,
Area of rectangle = l × b
⇒ (3x + 1) (2x – 1) = 144
⇒ 6x2 + 2x – 3x – 1 = 144
⇒ 6x2 – x – 1 = 144
6x2 – x – 145 = 0
Solving by using quadratic formula,
Question 27.
The figure drawn on the graph paper shows a ΔABC with vertices A (-4, 1), B (-4, 6) and C (8, 1).
(A) Show that the Length of BC is 13.
(B) Find AB. 3
Answer:
Question 28.
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
OR
In the given figure, PT is a tangent and PAB is a secant. If PT = 6 cm and AB = 5 cm then find the length of PA. 3
Answer:
Let, AB and CD are two tangents to a circle and AB || CD. Tangent BD subtends ∠BOD at ‘the centre.
To prove: ∠BOD = 90°
Construction: Join OP, OQ and OR.
Proof: Here, OP ⊥ BD
[∵ tangent at any point of a circLe is perpendicular to the radius through the point of contact]
In right anged ΔOQB and ΔOPB.
BQ = BP
[∵ the lengths of tangents drawn from an external point are equal]
OQ = OP [radii]
OB = OB [common]
∴ ΔOQB = ΔOPB [b SSS congruency]
Then ∠1 = ∠2 [by CPCT] …….. (i)
Similarly, in right angled ΔOPD and ΔORD,
∠3 = ∠4 …….. (ii)
∴ ∠BOD = ∠1 + ∠3
= \(\frac{1}{2}\)(2∠1 + 2∠3)
= \(\frac{1}{2}\)(∠1 + ∠1 + ∠3 + ∠3)
= \(\frac{1}{2}\)(∠1 + ∠2 + ∠3 + ∠4)
[from Eqs. (i) and (ii)]
= \(\frac{1}{2}\)(180°) = 90°
[∵ QR is a straight line, therefore ∠1 + ∠2 + ∠3 + ∠4 = 180°]
Hence, proved.
OR
Given, PT and PAB be a tangent and a secant, respectively of a circle.
Also, PT = 6 cm and AB = 5 cm Join OT, OA and OP. Draw OC ⊥ AB.
Let radius of circle be r
∴ OT ⊥ PT
[∵ tangent is perpendicular to the radius at the point of contact]
In right angled ΔOTP,
(OP)2 = (PT)2 + (OT)2
[by Pythagoras theorem]
⇒ (OP)2 = 62 + r2
⇒ (OP)2 – r2 = 36
⇒ (OP)2 – (OA)2 = 36 [∵ OA = OT = r] …….. (i)
Also, in right angled ΔOCA
(OA)2 = (OC)2 + (AC)2
[by Pythagoras theorem]
On putting the value of OA2 in Eq. (i), we get
(OP)2 – (OC2 + AC2) = 36
⇒ (OP)2 – (OC)2 – (AC)2 = 36
⇒ (PC)2 – (AC)2 = 36 [∵ In ΔOCP, (OP)2 – (OC)2 = (PC)2]
⇒ (PC – AC)(PC + AC) = 36 [a2 – b2 = (a – b)(a + b)]
⇒ AP (PC + BC) = 36
⇒ AP(PB) = 36
⇒ AP(AP + AB) = 36
⇒ AP(AP + 5) = 36 [∵ AB = 5 cm]
⇒ AP2 + 5AP -36 = 0
⇒ (AP + 9)(AP – 4) = 0
⇒ AP = 4 or – 9
Hense AP = 4 cm [∵ side cannot be negative]
Question 29.
The math teacher puts a clock in front of the children in the class and ask them if the minute hand of a clock is long. Find the area of the face of the clock described by the minute hand between 7 am and 7 : 15 am. 3
Answer:
We know that a minute hand sweeps by 6° angle in one minute.
So, it will sweep by an angle of 90° in 15 minutes.
So, area swept by the minute-hand in 15 minutes
= \(\frac{90^{\circ}}{360^{\circ}}\) × π(2)2 sq.cm
= π sq. cm, or \(\frac{22}{7}\) sq. cm.
Question 30.
If a + b = c + d where a, b, c, d are rational numbers, then prove that either a = c and b = d or b and d are squares of rational numbers. 3
Answer:
Let a = c, then
a + √b = c + √d
⇒ √b = √d
⇒ b = d
So let a ≠ c, then, there exist a positive rational number x, such that a = c + x.
Now, a + √b = c + √d
⇒ c + x + √b = c + √d [∵ a = c+ x]
⇒ x + √b = √d ………. (i)
Squaring on both site, we get
⇒ (x + √b)2 = (√d)2
⇒ x + 2√b x + b = d
⇒ √b = \(\frac{d-x^2-b}{2 x}\)
⇒ √b is rational [∵ d, x, b are rationals, … \(\frac{d-x^2-b}{2 x}\) is rational]
⇒ b is a square of a rational number From (i), we have
√d = x + √b
⇒ √d is rational
⇒ d is the square of a rational number
Hence, either a = c and b = d or b and d are the squares of rational numbers.
Question 31.
Manoj has a bag containing 15 white balls and some black balls, if the probability of drawing a black ball from the bag is thrice that of a white ball, find the number of black balls in the bag. 3
Answer:
Let the bag contains ‘x’ number of black balls. Then, the total number of balls in the bag is (x + 15).
It is given that,
P(a black ball) = 3 × P(a white ball)
⇒ \(\frac{x}{x+15}\) = 3 × \(\frac{15}{x+15}\)
⇒ x = 45
Thus, there are 45 black balls in the bag.
Section – D (20 Marks)
Question 32.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(A) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
(B) (sin A + cosec A)2 + (cos A + sec A)2
= 7 + tan2A + cot2A 5
Answer:
= 5 + cosec2A + sec2A
= 5 + 1 + cot2 A + 1 + tan2A
[∵ cosec2θ = 1 + cot2θ, sec2θ = 1 + tan2θ]
= 7 + tan2A + cot2A
= R.H.S.
Question 33.
A vessel contains mixture of 24 l milk and 6 l water and a second vessel contains a mixture of 15 l milk and 10 l water. How much mixture of milk and water should be taken from the first and the second vessel separately and kept in a third vessel so that the third vessel may contain a mixture of 25 l milk and 10 l water?
OR
Solve algebraically: 4x + 3y = 14 and 3x – 4y = 23. 5
Answer:
Let x L of mixture be taken from 1st vessel and y l of the mixture be taken from 2nd vessel and kept in 3rd vessel so that (x + y) l of the mixture in third vessel may contains 25 l of milk and of 10 l of water.
A mixture of x l from 1st vessel contains \(\frac{24}{30}\) x l = \(\frac{4}{5}\) x l of milk and \(\frac{x}{5}\) l of water. And a mixture of y l from 2nd vessel contains \(\frac{3 y}{5}\) of milk and \(\frac{2 y}{5}\) l of water.
∴ \(\frac{4}{5}\)x + \(\frac{3}{5}\)y = 25
⇒ 4x + 3y = 125 ………. (i)
and \(\frac{x}{5}\) + \(\frac{2}{5}\)y = 10
⇒ x + 2y = 50 ………… (ii)
Multiplying (ii) by 4 and then subtracting (i) from it, we get
4x + 8y – 4x – 3y = 200 – 125
⇒ 5y = 75
⇒ y = 15
From (ii),
x + 2 × 15 = 50
⇒ x = 50 – 30
⇒ x = 20
OR
Given, 4x + 3y = 14 ……….. (i)
and 3x – 4y = 23 …………. (ii)
Adding (i) and (ii), we get
7x – y = 37
∴ y = 7x – 37 ………… (iii)
Substituting the value of y in equation (i), we get
4x + 3 (7x – 37) = 14
⇒ 4x + 21x – 111 = 14
⇒ 25x = 125
⇒ x = 5
Putting x = 5 in eq. (iii), we get
y = 7 (5) – 37
= 35 – 37 = – 2
∴ The solution is x = 5 and y = – 2.
Question 34.
In one corner of the drawing room, a flower basket is kept inside the glass, lies on the table. The basket is designed in such a way that every one pleases to see it.
The shape of flower basket is hemisphere with ratio 60 cm and upper shape is conical with height 120 cm from the bottom surface.
(A) Find the capacity of the glass.
(B) Find the volume of the cone. 5
Answer:
(A) Height of the cylinder, h = 180 cm = 1.8 m [∵ 1 cm = \(\frac{1}{100}\)m]
Radius of the cylinder, r = 60 cm = 0.6 m
∴ Capacity of glass = Volume of cylinder
= πr2h
= \(\frac{22}{7}\) × 0.6 × 0.6 × 1.8
= \(\frac{14.256}{7}\)m3
(B) Height and radius of cone are
h1 = 120 cm = 1.2 m
r1 = 60 cm = 0.6 m
Volume of cone = \(\frac{1}{3}\)πr122h1
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × (0.6)2 × (1.2)
= \(\frac{22}{21}\) × 0.432
= 0.45 m3
Question 35.
Determine the median for the following frequency distribution:
Class | Frequency |
100 – 120 | 12 |
120 – 140 | 14 |
140 – 160 | 8 |
160 – 180 | 6 |
180 – 200 | 10 |
OR
Find the mean and modal marks of students from distribution: the following frequency
Marks | Number of students |
0 – 10 | 3 |
10 – 20 | 5 |
20 – 30 | 7 |
30 – 40 | 10 |
40 – 50 | 12 |
50 – 60 | 15 |
60 – 70 | 12 |
70 – 80 | 6 |
80 – 90 | 2 |
90 – 100 | 8 |
Answer:
Class | Frequency | Cumulative frequency |
100 – 120 | 12 | 12 |
120 – 140 | 14 | 26 |
140 – 160 | 8 | 34 |
160 – 180 | 6 | 40 |
180 – 200 | 10 | 50 |
Here, N = 50, \(\frac{N}{2}\) = 25
Cumulative frequency just greater than 25 is 26, which beLongs to class 120 – 140.
Hence, the median class is 120 – 140
For this class,
l = 120, f= 14, cf= 12, h = 70 , N
Thus, median = l + \(\frac{\frac{N}{2}-c f}{f}\) × h
= 120 + \(\frac{25-12}{14}\) × 20
= 120 + 18.57
= 138.57
Thus, median of the given data is 138.57.
OR
Here, the modal class is 50-60 [class with max. frequency]
For this class,
l = 50,h= 10, f1 = 15, f0 = 12, f2 = 12
So, mode = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h
= 50 + \(\frac{15-12}{30-12-12}\) × 10
= 50 + \(\frac{30}{6}\)
= 55
Calculation of Mean:
Section – E (12 Marks)
Question 36.
CASE STUDY 1
Mr. Naik is a paramilitary Intelligence Corps officer who is tasked with planning a coup on the enemy at a certain date. Currently 1 he is inspecting the area standing on top of the cliff. Agent Vinod is on a hot air balloon , in the sky. When Mr. Naik looks down below the cliff towards the sea, he has Ajay and Maran in boats positioned to get a good vantage point.
The main goal is to scope out the range and angles at which they should train their soldiers.
Based on the above information, answer the following questions:
(A) Write a pair of‘angle of elevation’ and ‘angle of depression’. 1
(B) If the vertical height of the balloon from the top of the cliff is 12 m and ∠b = 30°, then find the distance between the Naik and Vinod. 1
(C) Ajay’s boat is 25 m away from the base of the cliff. If ∠d = 30°. What is the height of the cliff?
(Use √3 = 1.73).
OR
If the height of the cliff is 30 m, ∠c = 45° and ∠d = 30°, then find the horizontal distance between the two
boats (use √3 = 1.73). 2
Answer:
(A) A pair of ‘angle of elevation’ is ∠b°, ∠e° and one pair of angle of depression is ∠c° and ∠d°
(B) Then, sin 30°
= \(\frac{\text { Vertical height }}{\text { Distance between Naik and Vinod }}\)
⇒ \(\frac{1}{2}=\frac{12}{D_{\text {Nand } V}}\)
⇒ Distance = 24 m
(C) Here, ∠d° = ∠f° = 30°
And ∠d = ∠f = 30°
∴ tan 30° = \(\frac{\text { height of cliff }}{\text { Distance of Ajay’s boat }}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{30}{D_A}\)
⇒ DA = 3√3
∴ Distance between boats
= 30√3 – 30
= 30(√3 – 1)
= 30(1.73 – 1)
= 30 × 0.73
= 21.9 m
Question 37.
CASE STUDY 2
in the backyard of house, Shikha has some empty space in the shape of a ΔPQR. She decided to make a garden. She divided the whole space into three parts by making boundaries AB and CD using bricks to grow flowers and vegetables where AB // CD // QR as shown in figure.
Based on the above information, answer the following questions:
(A) Find the length of AB. 1
(B) Find the length of CD and \(\frac{P B}{B D}\).
OR
Find \(\frac{P Q}{R Q}\) and \(\frac{A Q}{A P}\). 2
(C) What is the area of whole empty land? 1
Answer:
(A) In ΔPAB and ΔPQR,
∠P = ∠P [Common]
∠A = ∠Q [Corresponding angles]
By AA similarity criterion, ΔPAB ~ ΔPQR
∴ \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{PA}}{\mathrm{PQ}}\)
⇒ \(\frac{\mathrm{AB}}{12}=\frac{6}{24}\)
⇒ AB = 3m
(B) Similarly, ΔPCD and ΔPQR are similar.
∴ \(\frac{P C}{P Q}=\frac{C D}{Q R}\)
⇒ \(\frac{14}{24}=\frac{C D}{12}\)
⇒ CD = 7 m
In ΔPCD, AB || CD
∴ \(\frac{\mathrm{PA}}{\mathrm{AC}}=\frac{\mathrm{PB}}{\mathrm{BD}}\) [By BPT theorem]
⇒ \(\frac{6}{8}=\frac{P B}{B D}\)
⇒ \(\frac{\mathrm{PB}}{\mathrm{BD}}=\frac{3}{4}\)
OR
We have,
PQ = 6 + 8 + 10 = 24 m
RQ = 12 m
∴ \(\frac{P Q}{R Q}=\frac{24}{12}\)
= 2
AQ = 8 + 10
= 18 m
AP = 6 m
∴ \(\frac{\mathrm{AQ}}{\mathrm{AP}}=\frac{18}{6}\)
= 3
(C) Area of whole empty land
\(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 12 × 15
= 90 m2
Question 38.
CASE STUDY 3
While playing in garden, Sahiba saw a honeycomb and asked her mother what is that. She replied that it’s a honeycomb made by honey bees to store honey. Also, she told her that the shape of the honeycomb formed is parabolic. The mathematical representation of the honeycomb structure is shown in the graph.
Based on the above information, answer the following questions:
(A) What will be the expression of the polynomial represented by the graph? 1
(B) Find the value of the polynomial represented by the graph when x = 6 and – 6. 1
(C) Find the sum of zeroes of the obtained polynomial in question (A). Also, find the value of the expression at x = 5.
OR
If the sum of zeroes of polynomial at2 + 5t + 3a is equal to their product, then find the value of a. Also find the values of the expression at t = – 3. 2
Answer:
(A) Since the graph of the polynomial cuts the x-axis at (- 6, 0) and (6,0). So, the zeroes of polynomial are – 6 and 6.
∴ Required polynomial is
p(x) = x2 – (- 6 + 6)x + (- 6)(6)
= x2 – 36
(B) From the graph, it is clear that P(6) = 0 and P(- 6) = 0.
(C) Let P(x) = x2 – 36. Then, coefficient of x
Sum of zeroes = \(\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)
= – \(\frac{0}{1}\) = 0
P(5) = 52 – 36
= -11
OR
The given polynomial is at2 + 5t + 3a
Given, sum of zeroes = product of zeroes
⇒ \(\frac{-5}{a}=\frac{3 a}{a}\)
⇒ a = \(\frac{-5}{3}\)
At t = – 3,
(\(\frac{-5}{3}\))(-3)2 + 5(-3) + 3(\(\frac{-5}{3}\)) = -35