CBSE Sample Papers for Class 11 Applied Mathematics Set 6 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Applied Mathematics Set 6 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.
Add binary numbers 11011 and 10101. [1]
Solution:

Question 2.
Simplify: \(\frac{\left(81 x^4\right)^{\frac{1}{4}}}{y^{-8}}\) [1]
OR
Find the value of log 5, if given log 2 = 0.3010.
Solution:
\(\frac{\left(81 x^4\right)^{1 / 4}}{y^{-8}}\) = \(\frac{\left[(3)^4 x^4\right]^{1 / 4}}{y^{-8}}\)
= \(\frac{3 x}{y^{-8}}\) = 3xy⁸
[Using (a^m)ⁿ = a^mn and \(\frac{1}{a^{-n}}\) = aⁿ]

OR

log 5 = log \(\frac{10}{2}\) = log 10 – log 2
[Applying rule log_a \(\left(\frac{m}{n}\right)\) = log_a m = log_an]
= 1 – 0.3010 [∵ log 10 = 1 and given log 2 = 0.3010]
= 0.6990

Question 3.
Write the relationship between present and future value of an annuity regular. [1]
Solution:
resent value of an annuity regular
= \(\frac{\text { Future value of an annuity regular }}{(1+i)^n}\)
where, i = is the rate of interest per conversion period
n = no. of conversion period.

Question 4.
Co-ordinates of centroid of AABC are (1, – 1). Vertices of AABC are A(- 5, 3), B(p – 1) and C(6, q). Find p and q. [1]
Solution:
Co-ordinates of centroid of a triangle with vertices (x₁, y₁) (x₂, y₂) and (x₃, y₃) are given by
\(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
Since the given vertices are A(-5, 3), B(p, -1) and C(6, q). So the coordinates of the centroid of ∆ABC are
\(\left(\frac{-5+p+6}{3}, \frac{3-1+q}{3}\right)\) = \(\left(\frac{p+1}{3}, \frac{2+q}{3}\right)\)
Given, \(\left(\frac{p+1}{3}, \frac{2+q}{3}\right)\) = (1, 1)
⇒ \(\frac{p+1}{3}\) = 1 and \(\frac{2+q}{3}\) = -1
⇒ p = 2 and q = -5

Commonly Made Error
Some students are not able to attempt this question correctly because they don’t have the idea about centroid of triangle.

Answering Tip
If G is the centroid of a triangle whose vertices are (x₁, y₁) (x₂, y₂) and (x₃, y₃), then coordinates of its centroid is given by
G = \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

Question 5.
Find the value of [1]
¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇.
Solution:
¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇
= ¹⁵C_15-8 + ¹⁵C_15-9 – ¹⁵C₆ – ¹⁵C₇ [∵ⁿC_r = ⁿC_n-r
= ¹⁵C₇ + ¹⁵C₆ – ¹⁵C₆ – ¹⁵C₇
= 0

Question 6.
In G.P. 2\(\sqrt{2}\), 4, …………. , 128\(\sqrt{2}\), find the 4^th term from the end. [1]
OR
Find the number of terms in the A.P. 7, 10, 13, …………….. , 31.
Solution:
Given G.P. is 2\(\sqrt{2}\), 4, …………. , 128\(\sqrt{2}\)
Here, a = 2\(\sqrt{2}\), r = \(\sqrt{2}\) and l = 128\(\sqrt{2}\)
∵ n^th term from the end = l \(\left(\frac{1}{r}\right)^{n-1}\)
∴ 4^th term from the end = \(128 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)^{4-1}\)
= \(128 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)^3\)
= \(\frac{128 \sqrt{2}}{2 \sqrt{2}}\)
= 64

OR

Given, AP is 7, 10, 13, ……………. 31
Here, a = 7,1 = 31 and d = 3
Using formula,
l = a +(n – 1 )d
⇒ 31 = 7 + (n – 1)3
⇒ 31 – 7 = 3n – 3
⇒ 24 + 3 = 3n
⇒ n = 9

Question 7.
Find the range of the function f(x) = x² + 2. [1]
OR
Find the domain and range of real function/(x) = – |x|.
Solution:
f(x) = x² + 2
Let y = f(x)
y = x² + 2
x² = y – 2
or, x = \(\sqrt{y-2}\)
Clearly, x will take real values, if
y – 2 ≥ 0
y ≥ 2
∴ Range of y = [2, ∞)

OR

Given, f(x) = -|x|
Clearly, f(x) = – |z| ≤ 0, ∀ x ∈ R
∴ Domain = R
and Range = {x ∈ R : x ≤ 0} = (-∞, 0).

Question 8.
Evaluate \(\lim _{x \rightarrow 3} \frac{\sqrt{2 x+3}}{x+3}\) [1]
Solution:
\(\lim _{x \rightarrow 3} \frac{\sqrt{2 x+3}}{x+3}\) = \(\frac{\sqrt{2.3+3}}{3+3}\)
= \(\frac{\sqrt{9}}{6}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 9.
The marks of 9 students in a test were 13,17, 20, 5, 3, 3,18,15 and 20. Find the 6^th decile and P₆₂. [1]
Solution:
First we have to arrange the data in ascending order:
3, 3, 5, 13, 15, 17, 18, 20, 20
Here, n = 9
6^th decile, D₆ = \(\frac{6(9+1)}{10}\)th value
= 6^th value = 17
and
and P₆₂ = \(\frac{62(9+1)}{10}\) th value
= 6.2^th value
= 6^th value + 0.2 (7^th value – 6^th value)
= 17 + 0.2(18-17)
= 17.2

Question 10.
How many ways are there to arrange the letters of the word “GARDEN” with the vowels in alphabetical order ? [1]
Solution:
Total number of ways in which all letters of the word GARDEN can be arranged = 6! = 720
There are only two vowels in the word, A and E. When A at the first place, E can be occupy any of the remaining 5 places.
So, total arrangements = 5 × 4!
When A in the second place, E can occupy any of 4 places,
So, total arrangements = 4 × 4!
Repeat this process until A occupies the last but one place. A cannot occupy the last place.
Hence, total number of arrangement is (5 + 4 + 3 + 2 + 1) × 4!
= 15 × 24 = 360

Question 11.
Find the mean about median for the following profits (in ₹ 1000) of a firm during a week [1]
82, 56, 75, 70, 52, 80, 68
Solution:
The profits in thousand rupees is denoted by x. Arranging the values of x in an ascending order, we get
52, 56, 68, 70, 75, 80, 82
Therefore, Median, M_d = 70,
thus, median profit
= ₹ 70,000
Computation of Mean Deviation

xi	di = |xi – Md|
52	18
56	14
68	2
70	0
75	5
80	10
82	12
Total	∑di = 61

Mean Deviation = \(\frac{\Sigma d_i}{n}\) = ₹ \(\frac{61}{7}\) × 1000
= ₹ 8714.28

Commonly Made Error
Sometimes students do not convert interest rate according to conversion period and made mistake.

Answering Tip
Always convert the rate of interest according to given time period (no. of conversion periods)

Question 12.
Find the effective annual rate of interest corresponding to a nominal rate of 7% per annum payable half-yearly. [1]
Solution:
∵ Effective rate, E = (1 + i)ⁿ – 1
Here, i is 7% half yearly = \(\frac{7}{100}\) × \(\frac{1}{2}\)
n = 2
∴ E = (1 + \(\frac{7}{100}\) × \(\frac{1}{2}\))² – 1
= (1 + 0.035)² – 1
= (1.035)² – 1
= 1.0712 – 1
= 0.0712
= 7.12%
The effective interest rate is 7.12%.

Question 13.
The radii of two cylinders are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. Find the ratio of their curved surface areas. [1]
OR
X can do \(\frac{1}{3}\) of a work in 10 days, Y can do 40% of the work in 40 days and Z can do \(\frac{1}{3}\) of the work in 13 days, who will complete the work first ?
Solution:
Let the radii of the cylinders be 3z, 5z and their heights be 2y, 3y, respectively.
Ratio of their curved surface areas
= \(\frac{2 \pi \times 3 x \times 2 y}{2 \pi \times 5 x \times 3 y}\)
= \(\frac{2}{5}\) = 2 : 5
Thus, ratio of their curved surface areas is 2 : 5.

OR

Whole work will be done by X in (10 × 4) = 40 days
Whole work will be done by Y in (40 × \(\frac{100}{40}\)) = 100 days
Whole work will be done by Z in (13 × 3) = 39 days
∴ will complete the work first.

Question 14.
What do you understand by Tax ? [1]
Solution:
Tax : In order to run the government and manage the affairs of a state, money is required. So, the government impose taxes in many forms on the income of individuals and companies. It is the financial charge (fee) imposed by the government on income, commodity or activity.

Question 15.
If A = {3n + 5 : n ≤ N and n ≤ 6}, then represent set A in the roster form. [1]
OR
If (x – 2, y + 5) = (-2, \(\frac{1}{3}\)) are two equal ordered pairs, then find values of x and y.
Solution:
A = {3n + 5 : n ∈ N, n ≤ 6} = {8, 11, 14, 17, 20, 23}.

OR

Given that : (x – 2, y + 5) = (-2, 1/3)
⇒ x – 2 = -2 ⇒ x = 0 and y + 5 = \(\frac{1}{3}\)
⇒ y = \(\frac{1}{3}\) – 5 = \(\frac{-14}{3}\)

Question 16.
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then show that P\(\left(\frac{A}{B}\right)\) = 1. [1]
Solution:
GivenP(A) + P(B) – P(A and B) = P(A)
⇒ P(A) + P(B) – P(A ∩ B)= P(A)
⇒ P(B) – P(A ∩ B) = 0
⇒ P(A ∩ B) = P(B)
⇒ \(\frac{P(A \cap B)}{P(B)}\) = 1
⇒ P\(\left(\frac{A}{B}\right)\) = 1

Section – B (20 Marks)

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

Question 17.
It was Sunday on Jan 1, 2006. What was the day of the week on Jan 1, 2010 ? [2]
OR
If x^1/p = y^1/q = z^1/r and xyz = 1, then find the value of p + q + r.
Solution:
On 31^st Dec, 2005, it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (l + l+ 2 + l) = 5 days
∴ On 31^st Dec, 2009, it was Thursday.
Thus, on 1^st Jan, 2010, it was Friday.

OR

Given, x^1/p = y^1/q = z^1/r
Let, x^1/p = y^1/q = z^1/r = t
Therefore, x^1/p = t ⇒ x = t^p ………. (i)
x^1/q = t ⇒ y = t^q …………. (ii)
and x^1/r = t ⇒ z = t^r ……………… (iii)
Also, given xyz = 1
∴ From eqs (i), (ii) and (iii), we get
x.y.z = 1
⇒ t^p . t¹ . t^r = 1
⇒ t^p+q+r = 1 [using a^m aⁿ = a^m+n]
⇒ t^p+q+r = t⁰ [using a⁰ = 1]
⇒ p + q + r = 0
[using same base has same power]

Question 18.
Are sets A = {1, 2, 3, 4}, B = {x: x e N and 5 ≤ x ≤ 7} disjoint ? Why ? [2]
OR
If R₂ = {(x, y) | x and y are integers and x² + y² = 64} is a relation, then find the value of R₂.
Solution:
Yes, sets A and B are disjoint, because A ∩ B = Φ
∵ A = {1, 2, 3, 4}
and B = {5, 6, 7}
A ∩ B = {1, 2, 3, 4) ∩ (5, 6, 7} = Φ

OR

Given, R₂ = {(x, y)/x and y are integers and x² + y² = 64}
Since, 64 is the sum of squares of 0 and ± 8.
When x = 0, then y² = 64 ⇒ y = ±8
When x = 8, then y² = 64 – 8² ⇒ 64 – 64 = 0
When x = -8, then y² = 64 – (-8)² ⇒ 64 – 64 = 0
∴ R₂ = {(0, 8), (0, -8), (8, 0), (-8, 0)}

Commonly Made Error
Sometimes students forget to take 0 for x.

Answering Tip
Read and understand the question carefully.

Question 19.
In a certain code ‘Ding Dong Dang’ means ‘A Hacking the enemy’, ‘Ping Pond Dong’ means ‘Enemy is retreating’, ‘Ding Ping Mong’ means ‘Attacking and retreating’. [2]
From the above information, answer the following questions
(A) Which code stands for Enemy ?
(B) Which code stands for Attacking ?
OR
Arrange each of the following words in meaningful logical order.
(A) 1. Probation 2. Interview 3. Selection 4. Appointment 5. Advertisement 6. Application
(B) 1. Gold 2. Iron 3. Sand 4. Platinum 5. Diamond
Solution:
Given,
(I) Ding Dong Dang – Attacking the enemy
(II) Ping Pond Dong – Enemy is retreating
(III) Ding Ping Mong – Attacking and retreating
(a) In the I and II sentences, common word is ‘enemy’ and common code is ‘Dong’. Hence, ‘Dong’ stands for ‘enemy’.
(b) In the I and III sentences, common word is ‘Ding’ and common code is ‘Attacking’. Hence, code for’Ding’is’Attacking’.

OR

(a) As for a job, a person first see an advertisement, then fill application form and go for interview. If he/she is selected, got appointment letter and final stage is of probation.
Hence, the Correct arrangement of given words is 5, 6, 2, 3, 4, 1.

(b) All the given words represent substances which can be arranged in the increasing order of their cost. The least costly is sand after which comes the cost of iron, followed by gold, diamond and costliest among all is platinum. Hence, the Correct arrangement of given words is 3, 2, 1, 5, 4.

Question 20.
Find the domain of the function f(x) = \(\frac{1}{\log (4-x)}\) [2]
Solution:
Given, f(x) = \(\frac{1}{\log (4-x)}\)
To determine the domain of f(x), it should be real. Therefore log (4 – x) should be defined and should not be equal to zero.
i.e., 4 – x > 0 ⇒ x < 4 …………… (i)
Also, log (4 – x) ≠ 0 ⇒ x ≠ 3 …………… (ii)
From eqs. (i) and (ii), domain is (-∞, 3) ∪ (3, 4)

Question 21.
The value attendance of 9 students in a test were 13, 17, 20, 5, 3, 3, 18, 15 and 20. Find the first and three quartiles. [1]
Solution:
First we have to arrange the data in ascending order :
3, 3, 5, 13, 15, 17, 18, 20, 20
Here, n = 9.
Lower quartile or first quartile, Q₁ = \(\frac{n+1}{4}\) ^th value
= \(\frac{10}{4}\) ^th value
= 2.5^th value
∴ Q₁ = \(\frac{2^{\text {nd }} \text { value }+3^{\text {rd }} \text { value }}{2}\)
= \(\frac{3+5}{2}\) = 4
Upper quartile or third quartile,
Q₃ = \(\frac{3(n+1)}{4}\) ^th value
= \(\frac{30}{4}\) ^th value
∴ Q₃ = \(\frac{7^{\text {th }} \text { value }+8^{\text {th }} \text { value }}{2}\)
= \(\frac{18+20}{2}\) = 19

Question 22.
For a certain frequency distribution μ’₁ = 2, μ’₂ = 18 and μ’₃ = 100, find S_k. [2]
Solution:
Given, μ’₁ = 2, μ’₂ = 18 and μ’₃ = 100
Here, we need to calculate the central moments μ₂ and μ₃.
μ₂ = σ² – μ’₂ – (μ’₁)²
= 18 – (2)² = 18 – 4 = 14
μ₃ = μ’₃ – 3μ’₂μ’₁ + 2(μ’₁)
= 100 – 3(18)(2) + 2(2)³
= 100 – 108 + 16
= 8
Now, σ = \(\sqrt{\mathrm{n}_2}=\sqrt{14}\) = 3.74 (Approx.)
S_k = \(\frac{\mu_3}{\sigma^3}=\frac{8}{(3.74)^3}\) = 0.1529. (Approx.)

Question 23.
₹ 5000 is paid every year for three years to pay off a loan. What is the loan amount if interest rate be 14% per annum compounded annually ? [2]
Solution:
Since, C.F. = \(\frac{P V}{P(n, i)}\) where P(n, i) = \(\frac{(1+i)^n-1}{i(1+i)^n}\)
Here, C.F. = ₹ 5000
n = 3, i = 14% = 0.14
P(n, i) = \(\frac{(1+0.14)^3-1}{0.14(1+0.14)^3}\)
= \(\frac{(1.14)^3-1}{0.14(1.14)^3}\)
= \(\frac{1.4815-1}{0.14 \times 1.4815}\)
= \(\frac{0.4815}{0.2074}\) = 2.3216
Therefore, PV = C.F. P(n, i)
= 5000 × 2.3216
= ₹ 11,608

Question 24.
Mr. Kohli, a citizen of India, is an export manager of Arjun Overseas Limited, an Indian Company, since 1.5.2014. He has been regularly going to USA for export promotion. He spent the following days in U.S.A. for the last five years : [2]

Previous year ended	No. of days spent in USA
31.3.2015	317 days
31.3.2016	150 days
31.3.2017	271 days
31.3.2018	311 days
31.3.2019	294 days

Determine his residential status for assessment year 2019-20 assuming that prior to 1.5.2014 he had never travelled abroad.
Solution:
Total stays in India

2014-15	48 days
2015-16	216 days
2016-17	94 days
2017-18	54 days
2018-19	71 days

During previous year 2018-19 his stay in India is 71 days and in the four preceding years 48 + 216 + 94 + 54 = 412 days.
Resident in India (condition of 182 days for citizen not applicable as he has not gone for employment abroad but has been going out of India during the course of employment)
2017-18 – 54 days (Non- resident)
2016-17 – 94 days but more than 365 days in the 4 preceding previous year. Hence, resident
2015-16 – 216 days – resident
2014-15 – 48 days (non- resident)
Prior to 2014-15 – resident
He satisfies the first condition of being resident in at least 2 out of 10 previous year prior to relevant previous year and the 2^nd condition of being in India for 730 days or more in the 7 preceding previous years. He is “resident and ordinarily resident in India”.

Question 25.
By using slope method, show that the points P(4, 8), Q(5,12) and R(9, 28) are collinear. [2]
Solution:
The given points are P(x₁, y₁) = (4, 8), Q(x₂, y₂) = (5, 12) and R(x₃, y₃) = (9, 28).
∴ Slope of line PQ, m₁ = \(\frac{12-8}{5-4}\) = 4
and Slope of line PR, m₂ = \(\frac{28-8}{9-4}\) = 4
Since, slope of PQ = Slope of PR.
∴ The points P, Q and R are collinear.

Commonly Made Error
Some students solve this question by using area of triangle = 0 and lose their marks.

Answering Tips
Before attempting the question, carefully read the method of solving if it is mentioned.
Remember, three points P, Q, R in XY plane are collinear i.e., they lie on the same line if and only if
Slope of PQ = Slope of PR.

Question 26.
Find the equation of the circle with radius 5 units whose centre lies on X-axis and passes through the point (2, 3). [2]
Solution:
As the centre of the circle lies on the X-axis, let its centre be C(h, 0).
Since, the circle passes the point A(2, 3) and has radius 5.
∴ Radius, CA = 5
⇒ (2 – h)² + (3 – 0)² = 5²
⇒ (2 – h)² = 16
⇒ 2 – h = ± 4
⇒ h = – 2 or h = 6
∴ The centre of the circle is (- 2, 0) or (6, 0)
The equation of the circle is
When h = -2 and K = 0
(x + 2)² + (y – 0)² = 5²
i.e., x² + y² + 4x – 21 =0
When h = 6 and K = 0
or (x – 6)² + (y – 0)² = 52
or x² + y² – 12x + 11 = 0
There are two circles satisfying the given conditions.

Section – C (21 Marks)

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

Question 27.
Prove that : \(\) = 3 log₁₀ 2 [3]
Solution:
Change all the logarithms on L.H.S. to the base 10 by using the formula.

[Since, log_a mⁿ = n log_a m and log₁₀ 10 = 1]
Now,
L.H.S. = \(\frac{3 \log _{10} 2}{\log _{10} 3} \times \frac{2 \log _{10} 3}{4 \log _{10} 2} \times \frac{2 \log _{10} 2}{1}\)
= \(\frac{3 \times 2 \times 2}{4} \cdot \frac{\log _{10} 2 \times \log _{10} 3 \times \log _{10} 2}{\log _{10} 3 \times \log _{10} 2}\)
= 3 log₁₀ 2 = R.H.S.
Hence Proved

Commonly Made Error
Some students do not write the correct formula of change of base as they are confused between which term should be write in numerator or denominator.

Answering Tip
Learn all the properties of logarithm and practice as many question based on the change of base property.

Question 28.
(A) A car owner buys petrol at ₹ 7.50, ₹ 8 and ₹ 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends ₹ 4000 each year ? [3]
(B) The average of five numbers is 27. If one number is excluded, the average number becomes 25. Find the excluded number.
OR
Study the following information to answer the given questions :
(A) Eight friends A, B, C, D, E, F, G and H are seated in a circle facing centre.
(B) D is between B and G and F is between A and H.
(C) E is second to the right of A.
(a) Determine the position of A?
(b) Which of the informations statement are not required to ascertain the position of C ?
(c) Determine the position of C ?
Solution:
(a) Total quantity of petrol consumed in 3 years
= (\(\frac{4000}{7.50}\) + \(\frac{4000}{8}\) + \(\frac{4000}{8.50}\)) litres
= 4000 (\(\frac{2}{15}\) + \(\frac{1}{8}\) + \(\frac{2}{17}\))
= \([latex]\frac{76700}{51}\)[/latex] litres
Total amount spent = ₹ (3 × 4000) = ₹ 12000
∴ Average cost = ₹ \(\left(\frac{12000 \times 51}{76700}\right)\)
= ₹ \(\frac{6120}{767}\)
= ₹ 7.98
Hence, average cost of petrol = ₹ 7.98 per litre.

(b) Sum of five numbers = 27 × 5
Sum of four numbers = 25 × 4
∴ Excluded number = (27 × 5) – (25 × 4)
= 135 – 100
= 35.

OR

On the basis of the information given in the question, we have the sitting arrangement:

From figure conclude the following :

1. A is sitting left or right to C.
2. All informations given in the question are required to ascertain the position of C.
3. C is sitting between A and E.

Question 29.
Sum to n term the series 2 + 5x + 8x² + 11x³ + …………………. |x| < 1. [3]
Solution:
The given series in formed by multiplying corresponding terms of A.P. 2, 5, 8,11 and G.P. 1, x, x², x³, …………
Hence, n^th term of given arithmetico-geometric series is
[2 + (n – 1)3]. x^{n – 1} = (3n – 1)x^{n – 1}
Let S_n = 2 + 5x + 8x² + 11x³ + ……………. + (3n – 4)x^{n – 2} + (3n – 1)x^{n – 1}
x S_n = 2x + 5x² + 8x³ + …….. + ……… + (3n – 4) x^{n – 2} + (3n – 1)xⁿ
⇒ (1 – x) S_n = 2 + (3x + 3x² + 3x³ + ………. + 3x^{n – 1}) – (3n – 1) xⁿ.
= 2 + 3x \(\left(\frac{1-x^{n-1}}{1-x}\right)\) – (3n – 1)xⁿ
⇒ S_n = \(\frac{2}{1-x}+\frac{3 x\left(1-x^{n-1}\right)}{(1-x)^2}-\frac{(3 n-1) x^n}{1-x}\)

Question 30.
Find \(\lim _{x \rightarrow 1}\) f(x), where f (x) = . [3]
Solution:
L.H.L = \(\lim _{x \rightarrow 1-}\) f(x) = \(\lim _{x \rightarrow 1-}\) (x² – 1)
= (1)² – 1
= 1 – 1 = 0
and, R.H.L. = \(\lim _{x \rightarrow 1+}\) f(x) = \(\lim _{x \rightarrow 1+}\) (-x² – 1)
= – 1² – 1 = – 2
Since \(\lim _{x \rightarrow 1-}\) f(x) ≠ \(\lim _{x \rightarrow 1+}\) f(x)
So \(\lim _{x \rightarrow 1}\) f(x) does not exist.

Question 31.
Find the standard deviation from the following table : [3]

Marks	8	10	12	14	16
Frequency	4	7	8	7	4

OR
Compute Cov(x, y) for the following pair of observations :
(15, 44), (20,43), (25, 45), (30, 37), (40, 34), (50,37)
Solution:
Calculation of Arithmetic Mean

Marks x	Frequency f	Product fx
8	4	32
10	7	70
. 12	8	96
14	7	98
16	4	64
Total	∑f = 30	∑fx = 360

∴ Mean \(\bar{x}\) = \(\frac{\Sigma f x}{\Sigma f}=\frac{360}{30}\) = 12
Calculation of Standard Deviation σ(\(\bar{x}\) = 12)

Now, variance σ² = \(\frac{\Sigma f(x-\bar{x})^2}{\Sigma f}\)
= \(\frac{184}{30}\) = 6.133 (Approx.)
Thus, standard deviation, π = \(\sqrt{\text { variance }}\)
= \(\sqrt{6.133}\) = 2.476 (Approx.)
Here, \(\bar{x}\) = \(\frac{15+20+25+30+40+50}{6}\) = \(\frac{180}{6}\) = 30
and \(\bar{y}\) = \(\frac{44+43+45+37+34+37}{6}\) = \(\frac{240}{6}\) = 40
Now, construct the following table:

So, Cov(x, y) = \(\frac{1}{N} \Sigma(x-\bar{x})(y-\bar{y})\)
= \(\frac{1}{6}\) (-235) = -39.17

Question 32.
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is ₹ 1. Find the sum. [3]
Solution:
Let the sum be ₹ x. Then,
Compound Interest = A_n – P ⇒ I = P( 1 + i)ⁿ – P
Here, P = x, i = 4% = \(\frac{4}{100}\) = 0.04, n = 2
Compound Interest = x(1 + 0.04)² – x
= \(\frac{676}{625}\)x – x
= \(\frac{51}{625}\) x
Now, simple interest,
I = P i t
= x × 0.04 × 2
= \(\frac{2 x}{25}\)
Since, given difference = ₹ 1
i.e., \(\frac{51}{625}\) x – \(\frac{2 x}{25}\) = 1
⇒ x = 625

Question 33.
Five 100 W bulbs are used for 10 hours every day for 30 days. Find the cost of electricity if the rate is ₹ 4.00 per unit. [3]
Solution:
Five bulbs consume the power of 100 W each.
The total power consumed by 5 bulbs
= 5 × 100
= 500 W
= 0.5 kilowatt (1 kW = 1000 W)
Electrical energy consumed daily is,
E = P × t = 0.5 kW × 10 hours
= 5 kWh.
Thus, total electrical energy consumed for 30 days
= 5 kWh × 30 days
= 150 kWh or 150 units.
The cost of 1 unit of electricity = ₹ 4
Thus, the total cost of electricity
= ₹ 4 × 150 = ₹ 600

Section – D (15 Marks)

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

Question 34.
Let S be the sum, P be the product and R be the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = S_n.
Solution:
Let the G.E is a, ar, ar², ar³ ………….. ar^{n – 1}
Given, S = sum of n terms = a, ar, ar², ar³ ………….. ar^{n – 1}
= \(\frac{a\left(r^n-1\right)}{r-1}\) (let r > 1) ……………. (i)
and R = sum of the reciprocals of n terms

⇒ R = \(\frac{\left(r^n-1\right) r}{a r^n(r-1)}\) ………….. (ii)
and P = Product n terms
= a × ar × ar² × ar³ × ………….. × ar^{n – 1}
= a^{1+1+1………..+ n terms} r^{1+2+3+……..+(n-1) terms}
= aⁿ r^{\(\frac{n(n-1)}{2}\)} [∵ Σn = \(\frac{n(n+1)}{2}\)]
⇒ P² = a²ⁿr^n(n-1) ……………… (iii)
Now, we have to prove P²Rⁿ = Sⁿ
or P² = \(\frac{S^n}{R^n}\) or P² = \(\left(\frac{S}{R}\right)^n\)
Taking RHS = \(\left(\frac{S}{R}\right)^n=\left[\frac{a\left(r^n-1\right)}{r-1} \times \frac{a r^n(r-1)}{\left(r^n-1\right) r}\right]^n\)
[using eqs. (i) and (ii)]
= [a²rⁿr^-1]ⁿ = (a²r^n-1)ⁿ
= [a²ⁿr^n(n-1)]
= P² = LHS [from eq. (iii)]
Hence Proved.

Question 35.
The machines E₁, E₂ and E₃ in a certain factory producing electric bulbs, produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the bulbs produced by each of machines E₁ and E₂ are defective and that 5% of those produced by machine E₃ are defective. If one bulb is picked up at random from a day’s production, calculate the probability that it is defective. [5]
OR
There are three coins, one is a two headed coin (having head on both the faces), another is a biased coin that comes up heads 75% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed. If it shows head, what is probability that it was the two headed coin ?
Solution:
Let A₁ : Event that the bulb is produced by machine E₁
A₂ : Event that the bulb is produced by machine E₂
A₃ : Event that the picked up bulb is defective
Here, P(A)₁ = 50% = \(\frac{50}{100}\) = \(\frac{1}{2}\)
P(A)₂ = 25% = \(\frac{25}{100}\) = \(\frac{1}{4}\)
P(A)₃ = 25% = \(\frac{25}{100}\) = \(\frac{1}{4}\)
Also, P\(\left(\frac{A}{A_1}\right)\) = 4% = \(\frac{4}{100}\) = \(\frac{1}{25}\)
P\(\left(\frac{A}{A_2}\right)\) = 4% = \(\frac{4}{100}\) = \(\frac{1}{25}\)
and P\(\left(\frac{A}{A_3}\right)\) = 5% = \(\frac{5}{100}\) = \(\frac{1}{20}\)
∴ The probability that the picked bulb is defective,
P(A) = P(A₁) × P\(\left(\frac{A}{A_1}\right)\) + P(A₂) × P\(\left(\frac{A}{A_2}\right)\) + P(A₃) × P\(\left(\frac{A}{A_3}\right)\)
= \(\frac{1}{2}\) × \(\frac{1}{25}\) + \(\frac{1}{4}\) × \(\frac{1}{25}\) + \(\frac{1}{4}\) × \(\frac{1}{20}\)
= \(\frac{1}{50}\) + \(\frac{1}{100}\) + \(\frac{1}{80}\)
= \(\frac{8+4+5}{400}\) = \(\frac{17}{400}\) = 0.0425

OR

Let E₁ = event of selecting a two headed coin
E₂ = event of selecting a biased coin, which shows 75% times head
E₃ = event of selecting an unbiased coin
A = event that tossed coin shows head.
∴ P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
P\(\left(\frac{A}{E_1}\right)\)= P(coin showing head given that it is two headed coin)= 1
P\(\left(\frac{A}{E_2}\right)\) = P(coin showing headed given that it is a biased coin)
= \(\frac{75}{100}\) = \(\frac{3}{4}\)
P\(\left(\frac{A}{E_3}\right)\) = P(coin showing head given that it is unbiased coin)
= \(\frac{1}{2}\)
By Bayes’s theorem
P(getting two headed coin when it is known that it shows Head)

Commonly Made Error
Students forget to define the events and apply Bayes’s theorem directly and lose marks.

Answering Tip
For Bayes’ theorem and total probability theorem, event should be well defined.

Question 36.
(a) ₹ 200 is invested at the end of each month in an account paying interest 6% Toper year compounded monthly. What is the future value of this annuity after 10^th payment ? [5]
Given that (1.005)¹⁰ = 1.0511
(b) Sneha borrows ₹ 500000 to buy a house. If she pays equal instalments for 20 years and 10% interest on outstanding balance what will be the equal annual instalment ?
Given that (1.10)²⁰ = 6.7224
OR
Manjeet, Dilawar and Ravi live in the same city. Manjeet sells an article to Dilawar for ₹ 60,000 and Dilawar sells the same article to Ravi at a profit of ₹ 8,000. If all the transactions are under GST system at the rate of 12%, find :
(A) The state-government tax (SGST) paid by Dilawar.
(B) The total tax received by CGST.
(C) How much does Ravi pay for the article.
Solution:
(a) Here, C.F. = ₹ 200
n = 10
i = 6% per annum
= \(\frac{6}{12}\)% per month
= 0.005
Future value of annuity after 10 months is given by
F.V. = C.F. \(\left[\frac{(1+i)^n-1}{i}\right]\)
= 200 × \(\left[\frac{(1+0.005)^{10}-1}{0.005}\right]\)
= 200 × \(\left[\frac{(1.005)^{10}-1}{0.005}\right]\)
= 200 × \(\frac{(1.0511-1)}{0.005}\)
= 200 × \(\frac{0.0511}{0.005}\)
= 200 × 10.22
= ₹ 2044

(b) We know that,
C.E. = \(\frac{\text { Present value (P.V.) }}{\mathrm{P}(n, i)}\)
Here, P.V. = ₹ 500000
n = 20
i = 10% p.a. = 0.10
C.F. = \(\frac{500000}{P(20,0.10)}\)
∵ P(n, i) = \(\frac{(1+i)^n-1}{i(1+i)^n}\)
= \(\frac{(1+0.10)^{20}-1}{0.10(1+0.10)^{20}}\)
= \(\frac{6.7274-1}{0.10 \times 6.7274}\)
= \(\frac{5.7274}{0.67274}\)
= 8.51356
Therefore, C.F. = \(\frac{500000}{8.51356}\) = ₹ 58729.84

Commonly Made Error
Some students are confused between future and present value of an annuity.

Answering Tip
Learn all the concepts and formulae related to annuity.

OR

Given the rate of GST = 12%
∴ CGST rate= SGST rate = 6%
(a) Tax paid by Dilawar to SGST
= SGST received by Dilawar on selling price – SGST paid by him on purchasing
= \(\frac{6}{100}\) × (60,000 + 8,000) – \(\frac{6}{100}\) × 60,000
= \(\frac{6}{100}\) × 68,000 – \(\frac{6}{100}\) × 60,000
= \(\frac{6}{100}\) × (68,000 – 60,000)
= \(\frac{6}{100}\) × 8,000
= ₹480

(b) Total Tax received by CGST
= CGST paid by Manjeet + CGST paid by Dilawar
= \(\frac{6}{100}\) × 60,000 + ₹ 480
= ₹ (3600 + 480)
= ₹ 4080

(c) The amount, Ravi paid for the article
= Cost price of the article for Ravi with GST
= ₹ (68000 + \(\frac{12}{100}\) × 68000)
= ₹ (68000 + 8160)
= ₹ 76160

Section – E (8 Marks)

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

Question 37.
A labour welfare society conducted a survey on two firms, firm A and firm B in an industrial area, belonging to the same industry. Both firms have different no. of daily wages workers. An analysis 6f monthly wages paid to the workers in two firms, given in the following table : [4]

Particular	Firm A	Firm B
No. of wage earners	586	648
Mean of monthly wages	₹ 5253	₹ 5253
Variance of the distribution of wages	100	121

According to the above given information answer the following questions :
(a) Amount paid by firm A is :
(A) ₹ 30,78,258
(B) ₹ 34,03,944
(C) ₹ 37,08,258
(D) ₹ 30,43,944
Solution:
Option (A) is correct.

Explanation:
No. of wages earners = 586
Mean of monthly wages, \(\bar{x}\) = ₹ 5253
Amount paid by firm A = ₹ (586 × 5253)
= ₹ 3078258

(b) Coefficient of variation of the distribution of wages for firm A :
(A) 0.15
(B) 0.19
(C) 0.21
(D) 0.24
Solution:
Option (B) is correct

Explanation:
Variance of distribution of wages, σ² = 100
Standard deviation, σ = \(\sqrt{\sigma^2}\)
= \(\sqrt{100}\)
= 10
Coefficient of Variation = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{10}{5,253}\) × 100
= 0.19

(c) Standard deviation of the distribution of wages for firm B :
(A) 10
(B) 15
(C) 11
(D) 20
Solution:
Option (c) is correct

Explanation:
Standard deviation,
σ = \(\sqrt{\sigma^2}\)
= \(\sqrt{121}\)
= 11

(d) Amount paid by firm B is :
(A) ₹ 30,78,258
(B) ₹ 30,43,944
(C) ₹ 37,08,258
(D) ₹ 34,03,944
Solution:
Option (d) is correct.

Explanation :
No. of wage earners = 648
Mean of monthly wages, \(\bar{x}\) = ₹ 5253
Amount paid by firm B = 648 × 5253
= ₹ 3403944

Question 38.
Mohan and Mukesh go to a club to play a game of spinning arrow wheel. Mukesh spins the wheel first. The arrow is pointing at one of the numbers 1,2, 3, 4, 5, 6, 7, 8 (see figure) and these are equally likely outcomes. [1]

According to the above given information answer the following questions :
(a) The total possible outcomes are:
(A) 8
(B) 1
(C) 36
(D) 4
Solution:
Option (A) is correct

Explanation:
Since, out of 8 numbers, an arrow can point any of the number is 8 ways.
Therefore, total number of possible outcomes = 8

(b) The probability that spinning arrow will point at 8 is:
(A) 1/8
(B) 1/2
(C) 3/4
(D) 1
Solution:
Option (A) is correct

Explanation:
Favourable number of outcomes = 1
Hence P(arrow points at 8) = \(\frac{1}{8}\)

(c) The probability that spinning arrow will point at a number greater than 2 is:
(A) 1/8
(B) 1/2
(C) 3/4
(D) 1
Solution:
Option (C) is correct

Explanation:
Favourable number of outcomes = 6
Hence P(arrow points at a number > 2)
= \(\frac{6}{8}\) = \(\frac{3}{4}\)

(d) The probability that spinning arrow will point at an odd number is:
(A) 1/8
(B) 1/2
(C) 3/4
(D) 1
Solution:
Option (B) is correct

Explanation:
Favourable number of outcomes = 4
Hence P(arrow points at an odd number)
= \(\frac{4}{8}\) = \(\frac{1}{2}\)