Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 6 with Solutions

Time Allowed: 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-ll (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 Case 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
  • Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
The distance of the point (7, -8) from the origin is: [1]
(a) \(\sqrt{112}\)
(b) \(\sqrt{115}\)
(c) \(\sqrt{113}\)
(d) 96
Answer:
(c) \(\sqrt{113}\)

Explanation:
Distance of (7, – 8) from origin
= \(\sqrt{(7-0)^2+(-8-0)^2}\)
= \(\sqrt{49+64}\)
= \(\sqrt{113}\) units

Question 2.
The ratio in which the line segment joining the points ( -1, 7) and ( 4, -3) is divided by the point (1, 3) is: [1]
(a) 2 : 3
(b) 3 : 2
(c) 1: 4
(d) 2 : 5
Answer:
(a) 2:3

Explanation:
Let P(1, 3) divide AB in the ratio k : 1
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 1
Then, P(1, 3) = P\(\left(\frac{4 k-1}{k+1}, \frac{-3 k+7}{k+1}\right)\)
⇒ \(\frac{4 k-1}{k+1}\) = 1 and \(\frac{-3 k+7}{k+1}\) = 3
3k = 2 and -6k = -4
⇒ k = \(\frac{2}{3}\)
Thus,the ratio is 2: 3

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 3.
For the following distribution, [1]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 2
The sum of the lower limits of the median class and the modal class is:
(a) 20
(b) 35
(c) 30
(d) 25
Answer:
(d) 25

Explanation:

Frequency Cumulative Frequency
0-5 10 10
5-10 15 25
10-15 12 37
15-20 20 57
20-25 9 66

Here, the modal class is 15 – 20 and the median class is 10 – 15.
So, the sum of the two lower limits = 25

Question 4.
An equilateral triangle ABC is inscribed in a circle with centre O. The measure of ∠BOC is: [1]
(a) 120°
(b) 130°
(c) 60°
(d) 45°
Answer:
(a) 120°

Explanation: As ΔABC is equilateral
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 3
∴ ∠A = 60°
⇒ ∠BOC = 2 × ∠A = 120°

Question 5.
The value of 4 tan2A – 4 sec2A is: [1]
(a) -3
(b) 2
(c) -4
(d) 4
Answer:
(c) -4

Explanation:
4 tan2A – 4 sec2A
= 4 tan2A – 4 (1 + tan2A)
= – 4.

Question 6.
What is the perimeter of triangle with vertices (0, 0) (1, 0) and (0,1)? [1]
(a) (1 + √2) units
(b) (3 + √2) units
(c) √2 units
(d) (2 + √2) units
Answer:
(d) (2 + √2) units

Explanation:
Perimeter of Δ AOB
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 4
= \(\overline{O A}+\overline{A B}+\overline{O B}\)
= 1 + √2 + 1
= (2 + √2) units.

Question 7.
If 2 cos 3θ = √3 (0° ≤ θ ≤ 90°), then the value of θ is: [1]
(a) 10°
(b) 40°
(c) 20°
(d) 60°
Answer:
(a) 10°

Explanation: Given, 2 cos 3θ = √3
⇒ cos 3θ = \(\frac{\sqrt{3}}{2}\) = cos 30°
⇒ 3θ = 30°
⇒ θ = 10°

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 8.
If the area of a sector of a circle of radius 2 cm is 7t sq m, then what is the central angle of the sector ? [1]
(a) 90°
(b) 45°
(c) 30°
(d) 60°
Answer:
(a) 90°

Explanation:
Let the central angle be θ. Then, area of the sector
= [\(\frac{\theta}{360}\) × π(2)2] sq.cm

Equating it to π sq cm, we have,
\(\frac{\theta}{360}\) × π × 4 = π
⇒ θ = 90°

Question 9.
If 18, a, b, – 3 are in A.P., then the value of a + b is: [1]
(a) 15
(b) 20
(c) 25
(d) 30
Answer:
(a) 15

Explanation:
Since 18, a, b, – 3 are is A.P, so their common difference will be same.
a – 18 = b – a = – 3 – b
So, a – 18 = – 3 – b
⇒ a + b = 18 – 3
= 15

Question 10.
In ΔABC, D and E are prints on the sides AB and AC respectively, such that DE ∥ BC
If AD = 2.5 cm, BD = 3 cm and AE = 3.75 cm, then the value of AC is: [1]
(a) 8 cm
(b) 9.1 cm
(c) 8.25 cm
(d) 9 cm
Answer:
(c) 8.25 cm

Explanation:
Since DE ∥ BC
So, by B.P.T, we have
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 5

Question 11.
In the given figure, if ∠MPQ = 40°, then ∠OPM is: [1]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 6
(a) 30°
(b) 20°
(c) 50°
(d) 60°
Answer:
(c) 50°

Explanation:
In the figure,
∠OPQ = 90° [∵ Tangent ⊥ Radius]
⇒ ∠OPM = 90° – ∠MPQ
= 90° – 40°
= 50°

Question 12.
A box contains 7 red balls and 6 blue balls. A ball is drawn at random from the box. The probability that this drawn is a red or blue ball is: [1]
(a) 2
(b) 4
(c) 1
(d) 3
Answer:
(c) 1

Explanation:
P (a red or blue ball)
= \(\frac{7+6}{13}=\frac{13}{13}\) = 1

Question 13.
The empirical relationship between mean, median and mode is: [1]
(a) Mode + Mean = Median
(b) 3 Median = Mode – Median
(c) Mode = Median + 2 Mean
(d) 3 Median = Mode + 2 Mean
Answer:
(d) 3 Median = Mode + 2 Mean

Explanation:
The emperical relation between the measures of central tendency which is given by
Mode = 3 Median – 2 Mean

Question 14.
If the product of the zeros of the polynomial ax2 – 6x – 12 is 4, then the value of ‘a’ is: [1]
(a) 3
(b) 2
(c) 4
(d) -3
Answer:
(d)-3

Explanation:
Given equation: ax2 – 6x – 12
Product of zeros = \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)
⇒ \(\frac{(-12)}{a}\) = 4
a = – 3

Question 15.
The value of x, in the adjoining figure, if DE ∥ BC, is: [1]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 7
(a) 8
(b) 9
(c) 10
(d) 11
Answer:
(a) 8

Explanation:
In ΔABC, DE ∥ BC
\(\)
(By Thales theorem)
\(\)
(2x – 1) (x – 1) = (2x + 5) (x – 3)
⇒ 2x2 – 2x – x + 1 = 2x2 + 5x – 6x – 15
⇒ 2x = 16
⇒ x = 8

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 16.
What is value of x + y, if ΔABC and ΔPQR are similar ? [1]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 8
(a) 12.8 cm
(b) 14.3 cm
(c) 12.5 cm
(d) 14 cm
Answer:
(b) 14.3 cm

Explanation:
As, ΔABC and ΔPQR are similar
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 9
⇒ y = 12.8
∴ x + y = 1.5 + 12.8
= 14.3 cm

Question 17.
A quadratic polynomial whose zeros are – 7 and 5 is: [1]
(a) x2 + 2x – 35
(b) x2 – 2x + 35
(c) x2 + 3x – 25
(d) x2 – 3x – 35
Answer:
(a) x2 + 2x – 35

Explanation:
Sum of zeroes = – 7 + 5 = -2
Product of zeroes =-7 x 5 = -35
A quadratic polynomial with sum and product of zeroes is given as,
x2 – (sum of zeroes) x + Product of zeroes.
⇒ x2 – (- 2) x – 35
⇒ x2 + 2x – 35

Question 18.
The value for x and y:
x + y = 2 and 2x – y = 1 is: [1]
(a) 2
(b) 1
(c) 4
(d) 3
Answer:
(b) 1

Explanation:
On adding both the equations, we get
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 10
⇒ x = 1
Then, y = 2 – x = 2 – 1 = 1
⇒ x = y = 1 is the required solution.

Direction for questions 19 and 20: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(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.

Question 19.
Assertion (A) : In rhomus the diagonals are 20 cm and 10√6 cm in length; then side length of rhombus is 15.8 cm.
Reason (R) : The sum of the sides of a rhomus is equal to the sum of the squares of its diagonals. [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
The diagonals of a rhombus bisect each other at right angles.
∴ In ΔAOD, using pythagoras theorem,
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 11
AD2 = OA2 + OD2
= 102 + (5√6)2
= 100 + 150
= 250
= 15.8 cm

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 20.
Assertion (A) : sec2θ = 1 + tan2θ is a trigonometric identity.
Reason (R) : An equation involving trigonometric ratios of an angle is called trigonometric identity, which is true for all values of the angles involved. [1]
Answer:
(a) Both assertion (A) and reason (R) are correct and reason (R) is correct explanation of assertion (A).

Explanation:
Here, sec2θ = 1 + tan2θ
Put, θ = 45°
sec245° = (√2)2 = 2
1 + tan2θ = 1 + (tan 45°)
= 1 + 1 = 2
Thus, sec θ = 1 + tan2θ

SECTION – B (10 Marks)
(Section – B consists of 5 questions of 2 marks each)

Question 21.
Write any two irrational numbers whose product is a rational number. [2]
Answer:
Consider two irrationals as, 5 – 2√2 and 5 + 2√2
Here,
(5 – 2√2)(5 + 2√2) = 52 – (2√2)2
= 25 – 8 = 17 (a rational number)

Question 22.
If the zeros of the polynomial x3 – 3x2 + x + 1 are a – b, a and a + b, then find the values of a and b. [2]
Answer:
As (a – b), a and (a + b) are zeroes of x3 – 3x2 + x + 1, we have:
a – b + a + a + b = 3 [∵ Sum of zeroes = \(-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)]
⇒ 3a = 3, or a = 1 …(i)
Also, a (a – b) + a (a + b) + (a – b) {a + b) = 1 [∵ Sum of product of zeroes = \(-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)
⇒ 3a2 – b2 = 1 …….(ii)
(a – b) a (a + b) = -1 [∵ Product of Zeroes = \(-\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)]
⇒ a(a2 – b2) = -1 …(iii)
From (i) and (ii), we have b = ± √2
Thus, a = 1, b = ± √2

Question 23.
The product of A’s age 5 years ago with his age 9 years later is 15. Find A’s present age. [2]
OR
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in the same two variables such that the geometrical representation of the pair of equations so formed is:
(A) parallel lines
(B) coincident lines
Answer:
Let A’s present age (in years) be x. Then,
(x – 5) (x + 9) = 15 •
⇒ x2 + 4x – 45 = 15
⇒ x2 + 4x – 60 = 0
⇒ x2 + 10x – 6x – 60 = 0
⇒ x (x + 10) – 6 (x + 10) = 0
⇒ (x + 10) (x – 6) = 0
⇒ x – 6 = 0 (∵ x+ 10 ≠ 0)
⇒ x = 6
Thus, A’s present age is 6 years.

OR
(A) For parallel lines, we must have equation ax + by + c = 0
which must satisfy \(\frac{2}{a}=\frac{3}{b} \neq \frac{-8}{c}\)
So, we can write the required equation as
2x + 3y – 2 = 0

(B) For coincident lines, we must have equation
ax + by + c = 0
which must satisfy \(\frac{2}{a}=\frac{3}{b}=\frac{-8}{c}\)
So, we can write the required equation as
4x + 6y – 16 = 0

Question 24.
A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 × 16 m. Find the area of the field, that a cow can graze. [2]
OR
Prove that: (tan θ + 2) (2 tan θ + 1) = 5 tan θ + 2 sec2θ
Answer:
Shaded area can be grazed by the cow tied at the corner B. .
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 12

= 154 sq cm.
OR
(tan 0 + 2) (2 tan 0 + 1)
= 2 tan2 0 + 4 tan 0 + tan 0 + 2 = 2 tan2 0 + 5 tan 0 + 2 = 2 (tan2 0 + 1) + 5 tan 0 = 2 sec2 0 + 5 tan 0

Question 25.
Determine the mean of the following data : [2]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 13
Answer:

X f fx
2 5 10
4 6 24
3 8 24
7 12 84
9 10 90
5 7 35

Here, Σf = 48 and Σfx = 267
We know,
mean = \(\frac{\Sigma f x}{\Sigma f}=\frac{267}{48}\) = 5.5625

SECTION – C (20 Marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
Prove that √3 is an irrational number. [3]
OR
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Let us suppose that √3 is a rational number.
Then √3 can be written in the form \(\frac{p}{q}\) where p, q are co-prime i.e., they do not have common factor other than 1.
Now, √3 = \(\frac{p}{q}\)
⇒ 3 = \(\frac{p^2}{q^2}\)[squaring both sides]
⇒ p2 = 3q2
⇒ 3 divides p2
⇒ 3 divides p
⇒ 3 is factor of p.

∴ Let p = 3m.
⇒ P2 = 3 q2
⇒ (3m)2 = 3q2
⇒ 3m2 = q2
⇒ 3 divides q2
⇒ 3 divides q
It means, 3 is a factor of both p and p and q cannot have any common other than 1.
It means, our assumption is wrong.
Hence, √3 is an irrational number.
OR
Number are of two types – prime and composite
Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78 = 13 × 13 × 6
The given expression has 6 and 13 as its factors.
Therefore, it s a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 cannot be factorized further
Therefore, the given expression has 5 and 1009 as its factors.
Hence, it is a composite number.

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 27.
Two friends Reena and Sunita applied for the post of Computer Engineer in two different companies and got selected. Reena has been offered a job with a starting monthly salary of ₹ 48000, with an annual increment of ₹1400in her salary. Sunita has been offered a job with a starting monthly salary of ₹ 40000, with an annual increment of ₹ 1800 in her salary.
(A) Determine their monthly salaries for the 13th year. [1]
(B) Find each of them’s total salary of 13 years. [1]
(C) Who will get more salary in 13 years? And how much more? [1]
Answer:
Reena’s yearly amount (in ₹) of monthly salary.
48000, 49400, 50800, ………………..
It is A.P. with a = 48000 and d = 1400.
Sunita’s yearly amount (in ₹) of monthly salary.
40000, 41800, 43600, ………………..
It is A.P. with a’ = 40000 and d’ = 1800.

(A) So, Reena’s 13th year monthly salary = a + 12d
= 48000 + 12 x 1400
= ₹ 64800
Sunita’s 13th year monthly salary = a + 12d
= 40000 + 12 x 1800
= ₹ 61600

(B) Reena’s total salary of 13 years
= \(\frac{13}{2}\)[a1 + a13] x 12
= 78 [48000 + 64800]
= ₹ 8,7,98,400
Similarly, Sunita’s total salary of 13 years
= \(\frac{13}{2}\) [40000 + 61600] x 12
= ₹ 79,24,800

(C) Reena will get more than Sunita by ₹ (8798400 – 7924800) i.e., ₹ 8,73,600.

Question 28.
Find the roots of the equation
\(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\) (x ≠ -4, 7)
OR
Find a fraction which becomes \(\frac{1}{2}\) when the denominator is increased by 4, and \(\frac{1}{8}\) when the numerator is decreased by 5. [3]
Answer:
The given equation is:
\(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\)
⇒ \(\frac{x-7-x-4}{(x+4)(x-7)}=\frac{11}{30}\)
⇒ \(\frac{x-7-x-4}{(x+4)(x-7)}=\frac{11}{30}\)
⇒ (x + 4)(x – 7) = 30
⇒ x2 + 4x – 7x – 28 = – 30
or x2 – 3x + 2 = 0
or (x – 2) (x – 1) = 0
⇒ x – 2 = 0, x – 1 = 0
⇒ x = 2,1
So, the two roots are x = 2 and x = 1.
OR
Let the required fraction be \(\frac{p}{q}\).
Then,
\(\frac{p}{q+4}=\frac{1}{2}\)
⇒ 2p – q – 4 = 0 …(i)
\(\frac{p-5}{q}=\frac{1}{8}\)
8p – q – 40 = 0 …(ii)
Subtracting the first equation from the second equation, we get
6p -36 = 0
⇒ p = 6.
Substituting this value p = 6 in either equations, we get
q = 8
Thus, the required fraction is \(\frac{6}{8}\).

Question 29.
In the figure, a square OABC is inscribed in a quadrant OPBQ.
If OA = 20 cm, find the area of the shaded region. (Use π = 3.14) [3]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 14
Answer:
As, OA = 20 cm
∴ Diagonal
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 15
So, radius of the quadrant = 20√2 cm
Area of the quadrant
= \(\frac{\pi}{4}\)(20√2)2 sq cm
= 200π sq cm Area of the square
= (20)2 sq cm, i.e. 400 sq cm

Area of the shaded region
= (200π – 400) sq cm
= (628 – 400) sq cm
= 228 sq cm

Question 30.
Given that ΔPQR is similar to ΔBAR. Find: [3]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 16
(A) the value of x;
(B) the value of y;
Answer:
(A) Here,
∠PRQ = 54° (vertically opposite angles)
Now, In APQR,
∠PQR = 180° – (68° + 54°) = 58°
∵ ΔPQR – ΔBAR
∠Q = ∠A
⇒ x = 58°

(B) Again, ΔPQR – ΔBAR,
∴ \(\frac{P Q}{B A}=\frac{Q R}{A R}\)
⇒ \(\frac{2 y}{y+2}=\frac{y+3}{y}\)
⇒ 2 y2 = y2 + 5y + 6
⇒ y2 – 5y – 6 = 0
⇒ (y – 6) (y + 1) = 0
y = 6 (∵ y ≠ -1)

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 31.
The annual rainfall record of a city for 66 days is given below in the table: [3]
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 17
Calculate the median rainfall, using the formula.
Answer:
The cumulative frequency table for the given data is:

Ranifall (in cm) Frequency Cumulative frequency
0-10 22 22
10-20 10 32
20-30 8 40
30-40 15 55
40-50 5 60
50-60 6 66

Here, N = 66
So, \(\frac{N}{2}\) = 33
A cumulative frequency just greater than 33 is 40, which belongs to class 20 – 30.
So, the median class is 20 – 30.
For this class,
l = 20, cf = 32, f = 8, \(\frac{N}{2}\) = 33 and h = 10
So, Median = l + \(\frac{\frac{\mathrm{N}}{2}-c f}{f}\) × h
= 20 + \(\frac{33-32}{8}\) × 10
= 20 + \(\frac{1}{8}\) × 10
= 21.25
Thus the median rain fall (in cm) is 21.25

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

Question 32.
Draw the graphs of the equations:
4x – y = 4 and 4x + y = 12
Hence, determine the vertices of the triangle formed by the lines representing these equations and the x – axis. Shade the triangular region so formed.
OR
Solve 2x + 3y = 11 and 2x-4y = -24 and hence find the value of‘m’ for which y = mx + 3. [5]
Answer:
Table of values of
4x – y = 4
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 18

Table of Values of
4x + y = 12
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 19
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 20
From the graph, we find the vertices A, B, C of Δ ABC as A(2, 4), B(l, 0), C(3, 0).
Also, triangular region ABC is shaded.
OR
2x + 3y = 11 …(i)
2x – 4y = -24 …(ii)
Using equation (ii), we can say that
2x = – 24 + 4y
⇒ x = -12 + 2y
Putting this in equation (i), we get
2(-12 + 2y) + 3y = 11
⇒ -24 + 4y + 3y = 11
⇒ 7y = 35
⇒ y = 5

Putting value of y in equation (i), we get
2x + 3(5) = 11
⇒ 2x + 15 = 11
⇒ 2x = 11 – 15 = -4
⇒ x = -2
Therefore, x = -2 and y = 5
Putting values of x and y in y = mx + 3, we get
5 = m(-2) + 3
⇒ 5 = -2m + 3
⇒ -2m = 2
⇒ m = -1

Question 33.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Using the above result, prove the following:
If a circle touches all the four sides of a quadrilateral ABCD, prove that:
AB + CD = BC + DA. [5]
Answer:
I-Part:
We are given a circle with centre 0, a point A lying outside the circle and two tangents AX and AY on the circle from the point A.
We need to prove that
AX = AY
Join OX, OY and AO.
We know, tangent is perpendicular to radius, at the point of contact.
∠AXO = ∠AYO = 90°
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 21
Now, in right triangles AXO and AYO we have
AO = AO (common)
XO = YO (radii of the same cirde)
∠AXO ≅ ∠AYO (each 90°)
Therefore, by R.H.S. congruence criterion,
ΔAXO ≅ ΔAYO
⇒ AX = BY

IInd-Part:
Here, a circle touches all the four sides of quadrilateral ABCD.
From the figure, using the above result we have,
AS = AP, BP = BQ, CQ = CR
and DS = DR
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 22
Now, AB + CD = (AP + BP) + (CR + DR)
= (AP + DR) + (BP + CR)
= (AS + DS) + (BQ + CQ)
= AD + BC

Question 34.
Prove that:
\(\frac{\cot \theta+{cosec} \theta-1}{\cot \theta-{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}\)
OR
Prove that \(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\) = 1 + tan A + cot A. [5]
Answer:
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 23
OR
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 24

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 35.
Two stations are located at a distance of ‘a’ and ‘b’ from the foot of a leaning tower that leans in the direction of the north. If a and p be the elevations of the top of the tower from these stations, show that the inclination 6 to the horizontal is given by cot θ = \(\frac{b \cot \alpha-a \cot \beta}{b-a}\). [5]
Answer:
Let, the height of the tower DE = h
Distance between first station to foot of tower AD = a + x
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 25
Distance between second station to foot of tower BD = b + x
Distance between C and D = x
Given, a and p are angles of elevation of two station to the top of the tower.
i.e., ΔDAE = α, ΔDBE = β, αDCE = 0
In ΔCDE,
cot θ = \(\frac{x}{h}\) …(i)

In ΔBDE,
cot b = \(\frac{b+x}{h}\)
⇒ b + x = h cot β

In ΔADE
cot α = \(\frac{a+x}{h}\)
⇒ a + x = h cot α

Multiply ‘b’ on both sides
ba + bx = bh cot α …(iii)

Subtract (iii) from (ii)
(b – a)x = h (b cot β – a cot β)
\(\frac{x}{h}=\frac{b \cot \alpha-a \cot \beta}{b-a}\)
cot θ = \(\frac{b \cot \alpha-a \cot \beta}{b-a}\) [from (i)]
Hence, proved

SECTION – E (12 marks)
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory)

Question 36.
To make the learning process more interesting, creative and innovative, Amayra’s class teacher brings clay in the classroom, to teach the topic. Surface Areas and Volumes. With clay, she forms a cylinder of radius 6 cm and height 8 cm. Then she moulds the cylinder into a sphere and asks some questions to students.
(A) Find the radius of the sphere so formed.
OR
Find the total surface area of the cylinder. [2]
(B) What is the volume of the sphere so formed? [1]
(C) Find the ratio of the volume of sphere to the volume of cylinder. [1]
Answer:
(A) Since, volume of sphere = volume of cylinder
⇒ \(\frac{4}{3}\)R3 = πr2h, where R, r are the radii of sphere and cylinder respectively.
⇒ R3 = \(\frac{6 \times 6 \times 8 \times 3}{4}\) = (6)3
R = 6 cm
∴ Radius of sphere = 6 cm
Total surface area of the cylinder = 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 6(6 + 8)
= 2 × 2 × \(\frac{22}{7}\) × 6 × 14 = 528 cm2

(B) Volume of sphere = \(\frac{4}{3}\)πR3
= \(\frac{4}{3} \times \frac{22}{7}\) × 6 × 6 × 6 = 905.14 cm3

(C) Volume of sphere = Volume of cylinder
∴ Required ratio = 1:1

Question 37.
A linguist is performing a statistical analysis of word frequency distributions as part of her quantitative stylistics to understand the measurable aspects of lexical structure. She picks a random newspaper sentence (structure of which is shown below) that has 20 words in it.
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 26
The number of letters in each word is counted and the table below shows the frequency distribution:
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 27
On the basis of the above information, answer the following questions:
(A) A word is chosen a random from the whole sentence. What is the probability that it has 4 letters? [1]
(B) A word is chosen at random from those with an odd number of letters. What is the probability that it has 7 letters? [1]
(C) First person chooses a word at random from the whole sentence, Another person chooses a word at random from the whole sentence. What is the probability that one person chooses a 2-letter word and the other chooses a 6-letter word?
OR
Find the mean number of letters in the whole sentence. [2]
Answer:
(A) No. of four letter words = 5
Total number of words = 20
∴ P(4 letter words) = \(\frac{5}{20}=\frac{1}{4}\)

(B) No. of words with 7 letters =2
Total no. of words with odd letters = 9
∴P(word has 7 letter) = \(\frac{2}{9}\)

(C) If first person chooses a 2-letter word then second person chooses a 6-letter word or vice-versa.
∴ Required probability
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 28

CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions

Question 38.
Resident Welfare Association (RWA) of a M2K Society in Azadpur have put up three electric poles A, B and C in a society’s common park near Tower A. Despite these three poles, some parts of the park are still in dark.
So, RWA decides to have one more electric pole D in the park.
CBSE Sample Papers for Class 10 Maths Basic Set 6 with Solutions 29
On the basis of the above information, answer the following questions:
(A) What is the distance of the pole B form the corner O of the park? [1]
(B) Find the position of the fourth pole D so that four points A, B, C and D form a parallelogram.
OR
Find the distance between poles A and C. [2]
(C) Find the distance between poles B and D. [1]
Answer:
Coordinates of B are (6. 6) Distance from origin
= \(\sqrt{(6-0)^2+(6-0)^2}\)
= \(\sqrt{(6-0)^2+(6-0)^2}\)
= \(\sqrt{72}\) units

(B) If ABCD forms a parallelogram, then the diagonals bisects each other.
Mid point of AC = \(\left(\frac{2+5}{2}, \frac{7+4}{2}\right)\) = (3.5, 5.5)
Now, mid-point of diagonal, BD will be same. Let, the coordinates of D be (x, y)
Then,
\(\frac{6+x}{2}\) = 3.5 and \(\frac{6+y}{2}\) = 5.5
Coordinates of A are (2, 7)
Coordinates of C are (5, 4)
Distance of AC
= \(\sqrt{(5-2)^2+(4-7)^2}\)
= \(\sqrt{9+9}\)
= \(\sqrt{18}\) units

(C) Coordinates of B(6, 6)
Coordinates of D(1, 5)
Distance between BD
= \(\sqrt{(6-1)^2+(6-5)^2}\)
= \(\sqrt{5^2+1^2}\)
= \(\sqrt{25+1}\)
= \(\sqrt{26}\) units