CBSE Sample Papers for Class 11 Applied Mathematics Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Applied Mathematics Set 2 with Solutions

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions:

1. All the questions are compulsory.
2. The question paper consists of 38 questions divided into 5 sections A, B, C, D and E.
3. Section A comprises of 16 questions of 1 mark each. Section B comprises of 10 questions of 2 marks each. Section C comprises of 7 questions of 3 marks each. Section D comprises of 3 questions of 5 marks each. Section E comprises of 2 questions of 4 marks each.
4. There is no overall choice. However, an internal choice has been provided in five questions of 1 mark each, three questions of 2 marks each, two questions of 3 marks each, and two question of 5 marks each. You have to attempt only one of the alternatives in all such questions.
5. Use of calculators is not permitted.

Section – A (16 Marks)

All questions are compulsory. In case of internal choices attempt any one.

Question 1.
Divide (1010)₂ by (10)₂ [1]
OR
Simplify : (x^a y^-b)³ (x³y²)^-a
Solution:

∴ 1010 ÷ 10 = 101

OR

x^a y^-b)³ (x³y²)^-a
= (x^a)³ . (y^-b)³ . (x³)^-a . (y²)^-a
= x^3a . y^-3b . x^-3a . y^-2a [Using (a^m)ⁿ = a^mn]
= x^3a-3a . y^-2a-3b
= x⁰ . y^-(2a+3b) [Using a^m . aⁿ = a^m+n]
= 1. y^-(2a+3b) [Using a⁰ = 1]
= \(\frac{1}{y^{2 a+3 b}}\) [Using a^-r = \(\frac{1}{a^r}\)]

Question 2.
Of the three numbers, second is twice the first and is also thrice the third. If the average of the three numbers is 44, find the largest number. [1]
Solution:
Let the third number be x.
Then, second number = 3x
first number = \(\frac{3x}{2}\)
∴ x + 3x + \(\frac{3x}{2}\) = 44 × 3
or, \(\frac{11}{2}\)x = 44 × 3
⇒ x = 24
So, largest number = 2^nd number = 3 × 24 = 72.

Question 3.
Write four applications or ordinary (regular) annuity. [1]
Solution:
Applications of regular annuity are:

1. Leasing
2. Capital expenditure (investment decision)
3. Valuation of Bond
4. Sinking fund

Question 4.
Two dice are thrown. If it is known that the sum of numbers on the die was less than 6, then find the probability of getting a sum 3. [1]
Solution:
Let,
E₁ = Event that sum of numbers on the dice was less than 6
and E₂ = Event that sum of numbers on the dice is 3.
∴ E₁ = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
⇒ n(E₁) = 10
and E₂ = {(1, 2), (2, 1)}
⇒ n(E₁) = 2
∴ Required Probability = \(\frac{n\left(E_2\right)}{n\left(E_1\right)}=\frac{2}{10}=\frac{1}{5}\) = 0.2

Question 5.
In a certain code, COMPUTER is written as PMOCRETU, how is DECIPHER written in that code ? [1]
OR
Find the odd one out.
396, 462, 572, 427, 671, 264.
Solution:
Here,

OR

Here the given series is 396, 462, 572, 427, 671, 264. In all the terms, the middle digit is the sum of first and third digit except 427.

Question 6.
Pointing of a man in a photograph a man said to a woman, “His mother is the only daughter of your father” The woman is the related to of the man in the photograph. [1]
Solution:
From the information given in the question, it is clear that, the only daughter of woman’s Father is the woman herself, and hence the man in the photograph is her son. Therefore, the woman is the mother of the man in photograph.

Question 7.
If f(x) = ax + b, where a and b are integers, f(-1) = – 5 and f(3) = 3, then find the values of a and b. [1]
OR
If A and B are finite sets such that n(A) = 5 and n(B) = 7, then find the number of function from A to B.
Solution:
We have, f(x) = ax +b
∴ f(- 1) = a(- 1) + b
⇒ – 5 = – a + b ………………… (i)
Also, f(3) = a (3) + b
⇒ 3 = 3a + b …………… (ii)
Solve eqs. (i) and (ii) to get the values of a and b.

OR

Given, n(A) = 5 and n(B) = 7
We know that, if n(A) = p and n(B) = q, then number of functions from A to B = q^p.
∴ Number of function from A to B = 7⁵

Commonly Made Error
Some students are confused between the formulae of number of relations and number of functions of two sets.

Answering Tip
Learn the formula of number of functions of two sets. If n(A) = p and n(B) = q, then their number of function = q^p.

Question 8.
Evaluate : \(\lim _{x \rightarrow \pi} \frac{\tan x}{x-\pi}\)
Solution:
Given, \(\lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi}\)
= \(\lim _{\pi-x \rightarrow 0} \frac{\sin (\pi-x)}{-(\pi-x)}\) = -1
[∵ \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) = 1 and π – x → 0 ⇒ x → π]

Question 9.
If y = \(\sqrt{x}\) + \(\frac{1}{\sqrt{x}}\), then find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at x = 1.
Solution:
Given, y = \(\sqrt{x}\) + \(\frac{1}{\sqrt{x}}\)
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{3 / 2}}\)
\(\left(\frac{d y}{d x}\right)_{\mathrm{at} x=1}\) = \(\frac{1}{2}\) – \(\frac{1}{2}\) =0

Question 10.
Find the 13th and 14th terms of the sequence defined by [1]
a_n = n², when n is even
a_n = n² + 1, when n is odd
Solution:
As 13 is odd, a₁₃ = n² + 1 = (13)² + 1 = 169 + 1 = 170
and as 14 is even, a₁₄ = n² = (14)²= 196

Question 11.
Following are the wages of 8 workers expressed in ₹ : 82, 96, 52, 75, 65, 70, 50, 70. Find the range and its coefficient. [1]
Solution:
The largest and smallest wages are L = ₹ 96 and S = ₹ 50
Thus, range = ₹ 96 – ₹ 50 = ₹ 46 and coefficient of range
= \(\frac{L-S}{L+S}\) × 100
= \(\frac{96-50}{90+50}\) × 100
= \(\frac{46}{146}\) × 100 = 0.315 × 100
= 31.5

Question 12.
If the nominal rate of interest is 10% per annum and there is quarterly compounding, then find the effective rate of interest. [1]
Solution:
We know that,
effective rate, E = (1 + i)ⁿ – 1
Here, i = 10% quarterly
= \(\frac{10}{100}\) × \(\frac{1}{4}\) = \(\frac{0.1}{4}\) = 0.025
n = 4
E = (1 + 0.025)⁴ – 1
= (1.025)⁴ – 1
= 1.1038 -1
= 0.1038
= 10.38%

Question 13.
Write the set {x : x ∈ R, – 3 ≤ x < 7} as an interval. [1]
OR
Find the domain of the relation, R = {(x, y): x, y ∈ Z, xy = 4}.
Solution:
The set of all real numbers lying between a and b and including the number a is said to form a closed open interval, denoted by [a, b). It is written as [a, b) = {x: x ∈ R, a ≤ x < b}.

OR

Explanation: Given, R = {(x, y): x, y ∈ Z, xy = 4}
= {(- 4, – 1), (- 2, – 2), (- 1, – 4), (1, 4), (2, 2), (4, 1)}
Therefore, domain of R = {- 4, – 2, – 1, 1, 2, 4}

Question 14.
Find the equation of the parabola whose focus is (6, 0) and the directrix is x = 0. [1]
Solution:
Let P(x, y) be any point on the parabola. Join the point P to the focus S(6, 0) and draw PM perpendicular from P upon the directrix x = 0 (y-axis). Then by the definition of parabola, we have
SP = PM
⇒ \(\sqrt{(x-6)^2+(y-0)^2}\) = x
⇒ (x – 6)² + y² = x²
⇒ x² + 36 – 12x + y² = x²
⇒ y² = 12x – 36

Question 15.
If the line \(\frac{\mathrm{x}}{\mathrm{a}}\) + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 1 passes through the points (2, – 3) and (4, – 5), find the value of a and b. [1]
Solution:
Equation of a line in intercept form is
\(\frac{\mathrm{x}}{\mathrm{a}}\) + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 1 ……………… (i)
Since, the line passes through (2, -3) and (4, -5).
∴ \(\frac{\mathrm{2}}{\mathrm{a}}\) + \(\frac{\mathrm{3}}{\mathrm{b}}\) = 1
⇒ 2p – 3q = 1 [Let \(\frac{\mathrm{1}}{\mathrm{a}}\) = p, \(\frac{\mathrm{1}}{\mathrm{b}}\) = q] …………… (ii)
and \(\frac{\mathrm{4}}{\mathrm{a}}\) – \(\frac{\mathrm{5}}{\mathrm{b}}\) = 1
⇒ 4p – 5q = 1 ……………… (iii)
From equations (ii) and (iii).
p = -1, q = -1
∴ a = -1, b = -1

Question 16.
All the letters of the word ‘EAMCOT’ are arranged in different possible ways. The number of such arrangements in which no two vowels are adjacent to each other. [1]
OR
Find the number of signals that can be sent by 6 flags of different colours taking one or more at a time.
Solution:
We note that, there are 3 consonants and 3 vowels, first let us arrange the consonants in a row. This can be done in ³P₃ = 3! = 6 ways.
× M × C × T ×
Now, no. two vowels are adjacent to each other if they put at places marked X. The 3 vowels can fill up these 4 places in ⁴P₃ = 4 × 3 × 2 × 1 = 24 ways. Hence, total number of arrangements = 6 × 24 = 144.

OR

Number of signals using one flag = ⁶P₁ = 6
Number of signals using two flags = ⁶P₂ = 30
Number of signals using three flags = ⁶P₃ = 120
Number of signals using four flags = ⁶P₄ = 360
Number of signals using five flags = ⁶P₅ = 720
Number of signals using six flags = ⁶P₆ = 720
Therefore, the total number of signals using one or more flags at a time
= 6 + 30 + 120 + 360 + 720 + 720 (using addition principle)
= 1956

Section – B (20 Marks)

All questions are compulsory. In case of internal choices attempt any one.

Question 17.
Convert (0.675)₁₀ to binary. [2]
OR
Convert (10101100.010111)₂ to hexadecimal.
Solution:

Since, the fraction part (0.400) is the repeating value in the calculation, the multiplication is stopped. Now, we have to write the integral part from top to bottom to get binary number for the given fraction part.
Therefore, (0.675)₁₀ = (0.1010110)₂

OR

Making perfect group of 4 bits
Writing hexadecimal A C . 5 C
symbol for each group
Therefore, (10101100.010111)₂ = (AC.5C)₁₆

Commonly Made Error
Some student made error while doing grouping of binary numbers.

Answering Tip
Always remember that in grouping of fractional binary number, make groups of fractional part starting from left to right where as groups of integral part starting from right to left.

Question 18.
log₁₀ 2 = x and log₁₀ 3 = y, then find the value of log₁₀60. [2]
Solution:
18. Given, log₁₀ 2 = x and log₁₀ 3 = y
We know that,
log_a (mn) = log_a m + log_a n
Therefore, log₁₀ (60) = log₁₀ (6 × 10)
= log₁₀ 6 + log₁₀ 10
= log (2 × 3) + 1 [∵ log₁₀ 10 = 1]
= log₁₀ 2 + log₁₀ 3 + 1
= x + y + 1
Hence, value of log₁₀ 60 = x + y + 1.

Commonly Made Error
Sometimes students get confused with the answer of this question because they think logarithm always has a numerical value.

Answering Tip
In such type of questions, where the values log10 2 and log10 3 are x and y, respectively, we will get the answer in terms of variables like x and y.

Question 19.
A = {1, 2, 3, 4, 5}, S = {{x, y): x ∈ A, y ∈ A}, then find the ordered which satisfy the conditions given below : [1]
(a) x + y < 5 (b) x + y > 8
Solution:
Given, A = {1, 2, 3, 4, 5}
and S = {(x, y): x ∈ A, y ∈ A}

1. The set of ordered pairs satisfying x + y < 5 is {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}.
2. The set of ordered pairs satisfying x + y > 8 is {(4, 5), (5, 4), (5, 5)}

Question 20.
How many words, can be formed out of the letters of the word ‘OBEDIENCE’ so that vowels and consonants occur together ? [2]
OR
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
The word ‘OBEDIENCE’ has 5 vowels – three E’s, one O and one I; it has four different consonants: B, D, N, C.
Considering 5 vowels as a block and 4 consonants as another block. The two blocks can be arranged in 2! ways.
Now, with the block of vowels, 5 vowels can be arranged in \(\frac{5 !}{3 !}\) ways. Also, within the blocks consonants, 4 different consonants can be arranged in 4! ways.
By the multiplication principle of counting, the required number of words formed:
= 2! × \(\frac{5 !}{3 !}\) × 4!
= 2 × 5 × 4 × 4 × 3 × 2 × 1
= 960

OR

The number of ways of selecting 3 red balls out of 6 red balls = ⁶C₃
The number of ways of selecting 3 white balls out of 5 white balls = ⁵C₃
The number of ways of selecting 3 blue balls out of 5 blue balls = ⁵C₃
The number of ways of selecting 3 red balls of each colour = ⁶C₃ × ⁵C₃ × ⁵C₃ = ⁶C₃ × ⁵C₂ × ⁵C₂
= \(\frac{6 \times 5 \times 4}{1 \times 2 \times 3} \times \frac{5 \times 4}{1 \times 2} \times \frac{5 \times 4}{1 \times 2}\)
= 20 × 10 × 10
= 2000

Question 21.
Out of 120 students in a school, 5% can play all the three games Cricket, Chess and Carroms. If so happens that the number of players who can play any and only two games is 30. The number of students who can play the Cricket alone is 40. What is the total number of those who can play Chess alone or Carroms alone ? [2]
Solution:

Given U = 120
5% of 120 = 6
Therefore, students who can play Chess alone or Carroms alone = 120 – (30 + 40 + 6) = 44

Question 22.
If f(x) = x – \(\frac{\mathrm{1}}{\mathrm{x}}\), prove that [2]
[f(x)]³ = f(x³) + 3f(\(\frac{\mathrm{1}}{\mathrm{x}}\))
OR
Find the range of the function f(x) = \(\sqrt{x^2+4}\).
Solution:
We have,

[f(x)]³ = (x³ – \(\frac{1}{x^3}\)) – 3(x – \(\frac{1}{x}\))
[f(x)]³ = f(x³) – 3f(x)
[f(x)]³ = f(x³) + 3 f(\(\frac{1}{x}\)) (∵ f(x) = – f(\(\frac{1}{x}\)))

OR

Range of f : Let y = f(x)
y = \(\sqrt{x^2+4}\)
x = \(\sqrt{y^2+4}\)
x is defined, y² – 4 ≥ 0 ⇒ (y + 2) (y – 2) ≥ 0
∴ Range of f = (-∞, -2] ∪ [2, ∞).

Question 23.
Find derivative of function y = (x² + x + 1)⁴ with respect to x. [2]
Solution:
Given, y = (x² + x + 1)
Put (x² + x + 1) = u
Then, y = u⁴ and u = (x² + x + 1)
∴ \(\frac{\mathrm{dy}}{\mathrm{du}}\) = 4u³ and \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2x +1
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{dy}}{\mathrm{du}}\) × \(\frac{\mathrm{du}}{\mathrm{dx}}\)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4u³ × (2x + 1)
= 4(x² + x + 1)³(2x + 1)

Question 24.
Find the standard deviation for the following : [2]
20, 22, 27, 30, 31, 32, 35, 40, 45, 48
Solution:
Here, \(\bar{x}\) = \(\frac{20+22+27+30+31+32+35+40+45+48}{10}\)
\(\bar{x}\) = \(\frac{330}{10}\) = 33

Variable x	Deviation x – \(\bar{x}\)	(Deviation)2 = (x – \(\bar{x}\))2
20	– 13	169
22	-11	121
27	-6	36
30	-3	9
31	-2	4
32	-1	1
35	2	4
40	7	49
45	12	144
48	15	225
n = 10		∑(x – \(\bar{x}\))2 = 762

Hence,
Standard Deviation = \(\sqrt{\frac{\Sigma(x-\bar{x})^2}{n}}\)
= \(\sqrt{\frac{762}{10}}\) = \(\sqrt{76.2}\) = 8.73

Question 25.
What is the monthly equivalent interest rate to a quarterly interest 2.5% ? [2]
Solution:
We have to find imon knowing that fquart = 2.5%.
According to the relation of rate equivalence, the equality
(1 + i_quart)⁴ = (1 + i_mon)¹² must be satisfied
∴ (1 + i_mon)¹² = (1 + \(\frac{2.5}{100}\))⁴
[∵ give i_quart = 2.5% = \(\frac{2.5}{100}\)]
⇒ (1 + i_mon)¹² = (1.025)⁴
⇒ [(1 + i_mon)¹²]^1/4 = [(1.025)⁴]^1/4
⇒ 1 + i_mon = (1.025)^1/3
⇒ i_mon = (1.025)^1/3 – 1
⇒ i_mon = 1.008265 – 1
⇒ i_mon = 0.008265 ≈ 0.8265%

Question 26.
While computing rank correlation coefficient between profits and investment for 10 years of a firm, the difference in rank for a year was taken as 7 instead of 5 by mistake and the value of rank correlation coefficient was computed as 0.80. What would be the correct value of rank correlation coefficient after rectifying the mistake ? [2]
Solution:
We are given that n = 10
r_s = 0.80 and the wrong d = 7 should be replaced by 5.

Corrected Σd² = 33 – 7² + 5² = 9
Hence, rectified value of rank correlation coefficient
= 1 – \(\frac{6 \times 9}{10 \times\left(10^2-1\right)}\) = 0.95

Section – C (21 Marks)

All questions are compulsory. In case of internal choices attempt any one.

Question 27.
If the 1^st, 2^nd and last terms of an A.P. are a, b and c, respectively then find the sum of all terms of the A.P. [3]
OR
The ratio of the sum of n terms of two A.P.’s is (7n – 1): (3n + 11). Find the ratio of their 10th terms.
Solution:
Given A.P is a, b, … c
First term (A) = a
Common difference (d) = b – a
Last term i.e., n^th term (A_n) = c
Using formula, A_n = A + (n – 1)d, we get
c = a + (n – 1) (b – a)
⇒ c – a = (n – 1) (b – a)
⇒ n – 1 = \(\frac{c-a}{b-a}\)
⇒ n = \(\frac{c-a}{b-a}\) + 1
⇒ n = \(\frac{b+c-2 a}{b-a}\) ……………….. (i)
Now, using formula S_n = \(\frac{n}{2}\) (A + A_n) for the sum of n terms, we get
S_n = \(\frac{n}{2}\) [a + c]
= \(\frac{(b+c-2 a)(a+c)}{2(b-a)}\) using (i)
= \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)

Commonly Made Error
Generally students commit arithmetic error while calculation.

Answering Tip
Sum of n terms when first and last term are given, S_n = \(\frac{n}{2}\) (A + A_n), where
A = first term and A_n = last term.

OR

Let a₁, a₂ be the first terms and d₁, d₂ the common differences of the two given A.P.’s. Then, the sums of their n terms are given by

∴ Ratio of the 10^th terms of the two A.P.’s = \(\frac{33}{17}\)

Question 28.
A machine with useful life of seven years costs ₹ 10,000 while another machine with useful life of five years costs ₹ 8000. The first machine saves labour expenses of ₹ 1900 annually and the second one saves labour expenses of ₹ 2200 annually. Determine preferred course of action. Assume cost of borrowing as 10% compounded per annum. Given, P(7, 0.10) = 4.86842, P(5, 0.10) = 3.79079 [3]
Solution:
The present value of annual cost saving for the first machine = C.F. × P(n, i)
Here, C.F. = ₹ 1900, P(n, i) = P(7, 0.10) = 4.86842
n = 7 and i = 10% = \(\frac{10}{100}\) = 0.10
∴ present value = 1900 × P(7, 0.10)
= 1900 × 4.86842
= 9249.99 = ₹ 9250
Cost of machine being ₹ 10,000 it costs more by ₹ 750 than it saves in terms of labour cost.
The present value of annual cost saving of second machine
= C.F × P(n, i)
= 2000 × P(5, 0.10)
= 2000 × 3.79079 = ₹ 8339.74
Cost of the second machine being ₹ 8000 effective saving in labour cost is ₹ 339.74. Hence, the second machine is preferable.

Question 29.
Show that the perpendicular drawn from the point (4, 1) on the line segment joining (6, 5) and (2, -1) divide it internally in the ratio 8 : 5. [3]
OR
Prove that the circles
x² + y² + 2ax + c = 0 and x² + y² + 2by + c = 0 will touch each other, if \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}\)
Solution:
Let the points be A(4, 1), B(6, 5) and C(2, – 1) AP ⊥ BC is drawn

Slope of line BC,
m = \(\frac{-1-5}{2-6}\) = \(\frac{3}{2}\)
Since AP⊥ BC
∴ m’ = –\(\frac{1}{m}\) = –\(\frac{2}{3}\)
∴ Equation of line AP is
y – 1 = –\(\frac{2}{3}\) (x – 4)
⇒ 2x + 3y – 11 = 0 ……………. (i)
Let P divide BC in the ratio m:n
∴ Coordinates of P are \(\left(\frac{2 m+6 n}{m+n}, \frac{-m+5 n}{m+n}\right)\)
Since P lies on AP, so (i)
⇒ 2 . \(\frac{2 m+6 n}{m+n}\) + 3 . \(\left(\frac{-m+5 n}{m+n}\right)\) – 11 = 0
⇒ 4m + 12n – 3m + 15n – 11m – 11n = 0
⇒ -10m + 16n = 0
⇒ \(\frac{m}{n}=\frac{16}{10}\)
⇒ m : n = 8 : 5
Hence Proved

OR

Centre of the first circle is C₁(- a, 0) and its radius
= \(\sqrt{a^2-c}\)
Centre of the second circle is C₂(0, – b) and its radius
= \(\sqrt{b^2-c}\)
The given circles will touch each other if the distance between their centres is equal to the sum or difference of their radii.
i.e., C₁C₂ = r₁ ± r₂
⇒ \(\sqrt{(-a-0)^2+(0+b)^2}=\sqrt{a^2-c} \pm \sqrt{b^2-c}\)
⇒ a² + b² = (a² – c) + (b² – c) ± 2 \(\sqrt{a^2-c} \cdot \sqrt{b^2-c}\)
⇒ 2c = ± 2 \(\sqrt{\left(a^2-c\right)\left(b^2-c\right)}\)
⇒ c² = (a² – c)(b² – c)
[on squaring both sides]
⇒ c² = a²b² – c(a² + b²) + c²
⇒ c(a² + b²) = a²b²
⇒ \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}\)
Hence Proved.

Question 30.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right)\) = ? [3]
OR
Find the value of ‘k’ if \(\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}\)
Solution:

= \(\frac{\left(\sqrt{1+0^3}+\sqrt{1+0}\right)}{(0+1)\left(\sqrt{1+0^2}+\sqrt{1+0}\right)}\)
= \(\frac{2}{1 \times 2}\)
= 1

OR

Commonly Made Error
Some students make arithmetic errors in their work when solving a limit for an unknown value.

Answering Tip
Students need to revise standard limits formula and practice more such problems bases on it.

Question 31.
Study the following statement A, B, C, D and E and answer the following questions : [3]
(A) 1, 5, 9 means : ‘you better go’
(B) 1, 6, 7 means : ‘better come here’
(C) 5, 6, 7 means : ‘you come here’
(D) 1, 5, 6 means : ‘better you here’
(E) 3, 7, 9 means : ‘come and go’
(a) How many minimum number of statements are necessary to find the code no. of ‘better’ ?
(b) Which numeral means ‘go’ ?
(c) Which numeral means ‘you’ ?
Solution:

1. In statements A and B, A and D & B and D common word is ‘better’ and in all the combinations common code is 1. Hence, code no. of better is 1 and minimum number of required statements are 2 (two).
2. In statements A and E common word is ‘go’ and common code no. is ‘9’. Hence, numeral ‘9’ means ‘go’.
3. In statements A and C common words is ‘you’ and common code no. is ‘5’. Hence, numeral ‘5’ means ‘you’.

Question 32.
In a survey of 450 people, it was found that 110 play cricket, 160 play tennis and 70 play both cricket as well tennis. How many play neither cricket nor tennis ? [3]
Solution:
Let C and T denotes the students who play cricket and tennis, respectively.
Given, n(C) = 110, n(T) = 160, n(C ∩ T) = 70, n(U) = 450.
Using identity,
n(C ∪ T) = n(C) + n(T) — n(C ∩ T)
= 110 + 160 – 70
= 200
No. of students play neither cricket nor tennis
= n(U) – n(C ∩ T)
= 450 – 200
= 250

Question 33.
Study the following information to answer the given questions [3]
(a) Eight persons E, F, G, H, I, J, K and L are seated around a square table – two on each side. 0>) There are three lady members and they are not seated next to each other.
(c) J is between L and F. ‘
(d) G is between I and F.
(e) H, a lady member, is second to the left of J.
(f) F, a male, is seated opposite to E, a lady member.
(g) There is a lady member between F and I
(i) Identify the three lady members.
(«) Determine the gender of J.
(iii) Who is seating between E and H ?
Solution:
On the basis of the given information given in the question the positions of all the eight persons sitting around the square table is given below.

(i) It is clear from the figure that the three lady members are H, E and G.
(ii) It is clear that J is a male member.
(iii) K is sitting between E and H.

Section – D (15 Marks)

All questions are compulsory. In case of internal choices attempt any one.

Question 34.
(a) A man purchased a house valued at ₹ 300000. He paid ? 200000 at the time of purchased and agreed to pay the balance with the interest at 12% per annum compounded half yearly in 20 equal half yearly instalments. If first instalment is paid after six months from the date of purchase than find the amount of each instalment. [5]
[Given that (1.06)²⁰ = 3.2071]
(b) A person invests ₹ 500 at the end of each year with a bank which pays interest at 10% p.a. C.I. annually, rind the amount standing to his credit one year after he has made his yearly investment for the 12^th time. [Given that (1.1)¹² = 3.1348]
OR
From the following information, compute the tax liability of R, for the assessment year 2019-20.

	₹
Business income	4,60,000
Receipt from sale of trees of spontaneous growth	1,00,000
Agricultural income	60,000

Solution:
(a) Here, Present value
= 300000 – 200000 = ₹ 100000
i = 12% half yearly
= \(\frac{12}{100} \times \frac{1}{2}\) = 0.06
n = 20


= 100000 × 0.0871845
= 8718.45
= ₹ 8719 (approx.)
Hence ₹ 8719 is the amount of each instalment.

(b)

Amount after one year i.e., after 12^th instalment
= F.V × (1 + i)
= 10692.14 × (1.1)
= ₹ 11761.2

Commonly Made Error
Some times there are calculation mistakes while simplification.

Answering Tip
When computing the present value, you may have to round the principal value up to the nearest hundred to ensure that much money in the given amount of time. Always check you solution.

OR

Computation of tax liability of R


Note: Receipt from sale of trees which are of spontaneous growth is a capital receipt and not taxable under the head capital gain as trees of spontaneous growth can be said to be self generated and their cost of acquisition cannot be determined.

Question 35.
Three coins are tossed simultaneously. Consider the event E ‘three heads or three tails/ F ‘at least two heads’ and G ‘at least two heads’. Of the pairs (E, F), (E, G) and (F, G), which are independent ? Which are dependent ? [5]
Solution:
The sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT THT, TTH, TTT}
∴ n(S) = 8
Here, E = {HHH, TTT}; n(E) = 2
F = {HHH, HHT, HTH, THH}, n(F) = 4
and G = {HHT, HTH, THH, HTT, THT, TTH, TTT}
n(G) = 7
Also, E ∩ F ={HHH}, ∩ n(E ∩ F) = 1
E ∩ G = {TTT}, n(E ∩ G) = 1
F ∩ G = {HHT, HTH, THH}, n(F ∩ G) = 3


Thus, P(E ∩ F) = P(E) . P(F)
P(E ∩ G) ≠ P(E) . P(G)
P(F ∩ G) ≠ P(F) . P(G)
Hence, the events (E and F) are independent, and the events (E and G) and (F and G) are dependent.

Commonly Made Error
Sometimes students get confused between mutually exclusive events and independent events.

Answering Tips

• Two events are mutually exclusive if they can’t both happen. In case of mutually exclusive events P(E₁ ∩ E₂) = f.
• Independent events are events where knowledge of the probability of one doesn’t change the probability of the other.
• In case of two independent events
  P(E₁ ∩ E₂) = P(E₁). P(E₂)

Question 36.
Ten competitors in a beauty contest are ranked by three judges of the following order : [5]

1st judge	1	5	4	8	9	6	10	7	3	2
2nd judge	4	8	7	6	5	9	10	3	2	1
3rd judge	6	7	8	1	5	10	9	2	3	4

Use Spearman’s rank correlation coefficient to test which pair of judges has nearest approach to common tasks in beauty.
OR
If p, q, r are in G.P. and the equations px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root, then show that \(\frac{d}{p}, \frac{e}{q}, \frac{f}{r}\) are in A.P.
Solution:
Construct the following table:

Spearman’s correlation of rank coefficient between ranking given by:
1^st and 2^nd judges,

Since, r₂₃ is maximum, we conclude that 2^nd and 3^rd judges have the nearest approach to common beauty test.

OR

It is given that p, q, r are in G.P
∴ q² = pr
Now, px² + 2qx + r = 0
⇒ px² + 2x . \(\sqrt{p r}\) + r = 0
⇒ (x . \(\sqrt{p}\) + \(\sqrt{r}\))² = 0
⇒ \(\sqrt{p} x+\sqrt{r}\) = 0
⇒ x = –\(\sqrt{\frac{r}{p}}\)
It is given that the equations px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root and the equation px² + 2qx + r = 0 has equal roots equal to – \(\sqrt{\frac{r}{p}}\).

Commonly Made Error
Students have notation errors such as writing p² = qr instead of q² = pr. Incorrect signs are given for term of roots of equations. Arithmetic errors are numerous.

Answering Tip
It is necessary to identify the given questions, understand and proceed to prove that result.

Section – E (4 Marks)

Both the Case study-based questions are compulsory. Each Sub-parts carries 1 mark.

Question 37.
In a hockey match, both same A and B scored same number of goals up to the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team, whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, the answer the following questions : [4]

(a) The probability that team A wins in a first throw is:
(A) \(\frac{1}{6}\)
(B) \(\frac{5}{6}\)
(C) \(\frac{5}{6}\) × \(\frac{1}{6}\)
(D) \(\left(\frac{5}{6}\right)^2\) × \(\frac{1}{6}\)
Solution:
Option (A) is correct.

Explanation:
Let S denote the success (getting a ‘6’)
and F denotes the failure (not getting a ‘6’)
Thus P(S) = \(\frac{1}{6}\), P(E) = \(\frac{5}{6}\)
P(A wins in the first throw) = P(S) = \(\frac{1}{6}\)

(b) The probability that team A wins in a third throw is:
(A) \(\frac{1}{6}\)
(B) \(\frac{5}{6}\)
(C) \(\frac{5}{6}\) × \(\frac{1}{6}\)
(D) \(\left(\frac{5}{6}\right)^2\) × \(\frac{1}{6}\)
Solution:
Option (D) is correct.

Explanation:
A gets the third throw, when the first throw by A and second throw by B result into failures.
Therefore, P(A wins in the 3^rd throw)
= P(FFS)
= P(F)P(F)P(S)
= \(\frac{5}{6}\) × \(\frac{5}{6}\) × \(\frac{1}{6}\)
= \(\left(\frac{5}{6}\right)^2\) × \(\frac{1}{6}\)

(c) The probability that team A wins in a fifth throw is:
(A) \(\frac{1}{6}\)
(B) \(\left(\frac{5}{6}\right)^4\) × \(\frac{1}{6}\)
(C) \(\frac{5}{6}\) × \(\frac{1}{6}\)
(D) \(\left(\frac{5}{6}\right)^2\) × \(\frac{1}{6}\)
Solution:
Option (B) is correct.

Explanation:
P(A wins in the 5^th throw)
= P(FFFFS)
= \(\left(\frac{5}{6}\right)^4\) × \(\frac{1}{6}\)

(d) The probability that team A wins is:
(A) \(\frac{1}{6}\)
(B) \(\frac{5}{6}\)
(C) \(\frac{6}{11}\)
(D) \(\frac{5}{11}\)
Solution:
Option (C) is correct.

Explanation: P(A wins)
= \(\frac{1}{6}\) + (\(\frac{5}{6}\))² (\(\frac{1}{6}\)) + (\(\frac{5}{6}\)⁴ (\(\frac{1}{6}\)) + ………………
(sum GP S_n = \(\frac{a}{1-r}\)) a = \(\frac{1}{6}\), r = \(\frac{25}{36}\)
S_n = \(\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}\)

Question 38.
Five friends Rahul, Chetan, Ravi, Sunil and Pramod were playing in a ground, where they sit in a row in a straight line. On the basis of this, answer the following questions:

(a) In how many ways these five students can sit in a row ?
(A) 140
(B) 110
(C) 130
(D) 120
Solution:
Option (D) is correct.
Explanation:
Total number of ways = 5! = 120

(b) Total number of sitting arrangements if Rahul and Chetan sit together:
(A) 34
(B) 24
(C) 48
(D) 27
Solution:
Option (C) is correct.

Explanation:
Two position are fixed for Rahul and Chetan therefore considering it as one unit, total students left = 3 + 1 = 4
Total possible arrangement = 4! × 2! = 48

(c) What are the possible arrangements if Ravi and Sunil sits at the extreme positions?
(A) 12
(B) 20
(C) 16
(D) 14
Solution:
Option (A) is correct.

Explanation:
Total possible arrangements = 3! × 2! = 12

(d) What are the possible orders if Pramod is sitting in the middle ?
(A) 16
(B) 18
(C) 36
(D) 24
Solution:
Option (D) is correct.

Explanation:
Total possible arrangements = 4! = 24