Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 9 are designed as per the revised syllabus.
CBSE Sample Papers for Class 11 Applied Mathematics Set 9 with Solutions
Maximum Marks : 80
Time Allowed : 3 hours
General Instructions :
 All the questions are compulsory.
 The question paper consists of 38 questions divided into 5 sections A, B, C, D and E.
 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.
 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, tivo 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.
 Use of calculators is not permitted.
Section – A
All questions are compulsory. In case of internal choices attempt any one.
Question 1.
Subtract 1101101 – 11011. [1]
Answer:
Question 2.
Find value of
x^{ab} × x^{bc} × x^{ca}
OR
If log x + log y = log (x + y), then find the relation between x and y. [1]
Answer:
x^{a – b} × x^{b – c} × x^{c – a} = x^{a – b + b – c + c – a}
OR
Given,
log x + log y = log (x + y)
⇒ log (x . y) = log (x + y)
[Applying rule log_{a}(mn) = log_{a}m + log_{a}n
⇒ xy = x + y
⇒ y(x – 1) = x
⇒ xy – y = x
⇒ y(x – 1) = x
⇒ y = \(\frac{x}{x1}\)
Question 3.
What is the range and its coefficient for the following distribution of weight ?
Weights in kgs  5054  5559  6064  6569  7074 
No. of students  12  18  23  10  3 
[1]
Answer:
The lowest class boundary is 49.50 kg and the highest class boundary is 74.50 kg (after changing the given interval in continuous)
Thus, L = 74.50 kg and S = 49.50 kg
∴ Range = 74.50 – 49.50 = 25 kg
Also, coefficient of range
= \(\frac{LS}{L+S}\) × 100
= \(\frac{74.5049.50}{74.50+49.50}\) × 100
= \(\frac{25}{124}\) × 100 = 0.2016 × 100
= 20.16
Question 4.
Find the distance between the points (3, – 2) and ( 1, – 1). [1]
Answer:
Let, the point be A(x_{1}, y_{1}) = (3, 2) and B(x_{2}, y_{2}) = (1, 1).
Then, the distance
AB = \(\sqrt{\left(x_1x_2\right)^2+\left(y_1y_2\right)^2}\) = \(\sqrt{(3+1)^2+(2+1)^2}\)
= \(\sqrt{(4)^2+(1)^2}\)
= \(\sqrt{17}\) units
Question 5.
Find the 13^{th} and 14^{th} terms of the sequence defined by
a_{n} = {n^{2}, when n is even
n^{2} + 1, when n is odd
OR
Which term of the G.P. 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\) is \(\frac{1}{1024}\) ? [1]
Answer:
As 13 is odd, a_{13} = n^{2} + 1 = (13)^{2} + 1 = 169 + 1 = 170 and as 14 is even, a_{14} = n^{2} = (14)^{2} = 196
OR
Given G.P. is 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\) ……
First term a = 2
and common ratio = \(\frac{1}{2}\)
Let n^{th} term of G.P. be \(\frac{1}{1024}\)
∴ T_{n} = ar^{n1}
⇒ \(\frac{1}{1024}\) = 2\(\left(\frac{1}{2}\right)^{n1}\)
⇒ \(\frac{1}{2048}\) = \(\left(\frac{1}{2}\right)^{n1}\)
⇒ \(\left(\frac{1}{2}\right)^{11}\) = \(\left(\frac{1}{2}\right)^{n1}\)
∴ n – 1 = 11
or n = 12
Commonly Made Error
Sometimes students apply wrong formula for n^{th} term of G.P. Sometimes they write ar^{n} in place of ar^{n1} and get incorrect answer.
Answering Tip
It is necessary to revise formula of n^{th} term of G.P because it is used in almost each questions.
Question 6.
In an examination there are three multiple choice questions and each question has 4 choices. Find the number of ways in which a student can fail to get all answer correct is.
OR
In how many ways a committee consisting of 3 men and 2 women, can be chosen from 7 men and 5 women ? [1]
Answer:
There are three multiple choice questions, each has four possible answers. Therefore, the total number of possible answers will be 4 × 4 × 4 = 64. Out of these possible answers only one will be correct and hence number of ways in which a student can fail to get correct answer is 64 – 1 = 63.
OR
Out of the 7 men, 3 men can be chosen in ^{7}C_{3} ways and out of the 5 women, 2 women can be chosen in ^{5}C_{2} ways. Hence, the committee can be chosen in 7C_{3} × 5C_{2} = 350 ways.
Question 7.
Out of the Lactometer, Barometer, Diameter, Hydrometer and Thermometer, which is odd one out. [1]
Answer:
Diameter is odd one out, because all the other are apparatus to measure something.
Question 8.
Introducing Asha to guests, Gopal said, “Her father is the only son of my father.” How is Asha related to Gopal? [1]
Answer:
The only son of Gopal’s father is the Gopal himself. This means that Gopal is the father of Asha. Hence, Asha is the daughter of Gopal.
Question 9.
Find the range of function f(x) = 1 + 3 cos 2x. [1]
Answer:
Given, f(x) = 1 + 3 cos 2x
We know that, – 1 ≤ cos 2x ≤ 1
⇒ 3 ≤ 3 cos 2x ≤ 3
⇒ 1 – 3 ≤ 1 + 3 cos 2x ≤ 1 + 3
⇒ – 2 ≤ 1 + 3 cos 2x ≤ 4
⇒ Range of f(x) = [ 2, 4]
Question 10.
Determine the value of the constant ‘k’
f(x) = {\(\frac{k}{x}\), if x < 0 is continuous at x = 0 [1]
3, if x ≥ 0
Answer:
Since f is continuous at x = 0,
Question 11.
What is the probability that a given twodigit number is divisible by 15 ? [1]
Answer:
No. of two digits numbers = 90
⇒ n(S) = 90.
Let A’ be the event of getting a number divisible by 15.
⇒ A = {15, 30, 45, 60, 75, 90}
⇒ n(A) = 6
∴ p(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{90}\) = \(\frac{1}{15}\)
∴ The probability of getting a number divisible by 15
Commonly Made Error
Some students are not able to determine the values of event A because sample has large number of values.
Answering Tip
It has been observed that the experimental probability of an event approaches to its theoretical probability, if the number of trials of an experiment is very large.
Question 12.
^{n}P_{r} = 720 and ^{n}C_{r} = 120, find r. [1]
Answer:
We know that,
^{n}C_{r} = \(\frac{{ }^n P_r}{r !}\)
⇒ 120 = \(\frac{720}{r !}\)
⇒ r! = \(\frac{720}{120}\) ⇒ r! = 6 ⇒ r! = 3! ⇒ r = 3
Question 13.
Find the average of first 40 natural numbers.
OR
How many times are the hands of a dock at right angles in a day ? [1]
Answer:
Sum of first n natural numbers = \(\frac{n(n+1)}{2}\)
sum of first 40 natural numbers = \(\frac{40 \times 41}{2}\)
= 820
∴ Required average = \(\frac{820}{40}\) = 20.5
Commonly Made Error
Sometimes students forgot to apply the formula of sum of n natural numbers and they used classical approach of average i.e, they write all first 40 natural numbers and sum up, which makes the question complex.
Answering Tip
Always use formula of sum of first n natural numbers:
s = \(\frac{n(n+1)}{2}\)
OR
In 12 hours, they are right angles 22 times.
∴ In 24 hours, they are at right angles 44 times.
Question 14.
Interest obtained on a sum of ₹ 5000 for 3 years is ₹ 1500. Find the rate percent. [1]
Answer:
Let rate is i%
We have,
I = P i t
Here, I = 1500, t = 3, P = 5000 and i% = \(\frac{i}{100}\)
∴ 1500 = 5000 × \(\frac{i}{100}\) × 3
i = \(\frac{1500 \times 100}{5000 \times 3}\) = 10%
Thus, rate of interest is 10%.
Question 15.
Specify with the reason, whether the following acts can be considered as tax planning or tax management “Mr. P deposit ₹ 1 lakh in PPF account so as to reduce his total income from ₹ 6 lakh to ₹ 5 lakh.” [1]
Answer:
The investment of ₹ 1 lakh in PPF account so as to reduce his total income from ₹ 6 lakh to ₹ 5 lakh is considered as Tax Planning because the same is carried out within the framework of law by availing the deductions permitted by law and thereby minimising the tax liability.
Question 16.
List all of the proper subsets of {0,1, 2, 3}.
OR
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {x, y} : the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster from. [1]
Answer:
Proper subsets of the given set are:
ϕ, {0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}, {0, 1, 2}, {0, 1, 3}, {0, 2, 3}, {1, 2, 3}.
OR
Here Given, R = {(x, y): the difference between x and y is odd, x ∈ A, y ∈ B} or, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5,4), (5, 6)}.
Section – B
All questions are compulsory. In case of internal choices attempt any one.
Question 17.
If lx + my = 1 touches the circle x^{2} + y^{2} = a^{2} then prove that the point (l, m) lies on the circle x^{2} + y^{2} = a^{2}.
OR
Find the coordinates of points on parabola y^{2} = 8x, whose focal distance is 4. [2]
Answer:
Since, the line lx + my – 1 =0 touches the circle x^{2} + y^{2} = a^{2} (radius = a), so the perpendicular distance of the centre (0, 0) from the line is
r = \(\frac{0 . 1+0 . m1}{\sqrt{l^2+m^2}}\)
⇒ a = \(\frac{1}{\sqrt{l^2+m^2}}\)
⇒ \(\sqrt{l^2+m^2}\) = \(\frac{1}{a}\)
⇒ l^{2} + m^{2} = a^{2}
Since, (l, m) satisfies x^{2} + y^{2} = circle x^{2} + y^{2} = a^{2}.
OR
Given equation of the parabola is y^{2} = 8x
i.e., y^{2} = 4.2.x
∴ a = 2, focus, S = (2, 0)
Let P(x, y) be any point on the parabola, then PS = 4
i.e., \(\sqrt{(x2)^2+(y0)^2}\) = 4
⇒ x^{2} – 4x + 4 + 8x = 16
⇒ x^{2} + 4x – 12 = 0
⇒ (x + 6)(x – 2) = 0
∴ x = – 6 or x = 2
when x = 6, y^{2} = 16 ⇒ y = ±4
when x = 2, y^{2} = 48 ⇒ y = \(\sqrt{48}\), not possible
Hence, the coordinates of point P are (2, 4) and (2,4).
Question 18.
Find the equation of the line passing through (1, 2) and parallel to the line y = 3x – 1. [2]
Answer:
Given line: y = 3x – 1
i.e., 3x – y – 1 = 0
Any line parallel to it is written as
3x – y + λ = 0
Since, this line passes through (1, 2), we get 3(1) – 2 + λ = 0
⇒ λ = – 1
Hence, 3x – y – 1 is the equation of required line, which is the equation of given line itself.
Question 19.
Mr. Raghav aged 26 years and a resident in India, has a total income of ₹ 4,40,000, comprising his salary income and interest on bank fixed deposit. Compute his tax liability for A.Y. 202021. [2]
Answer:
Computation of tax liability of Mr. Raghav for A.Y. 202021
Particulars  ₹ 
Tax on total income of ₹ 4,40,000  
Tax @5% of ₹ 1,90,000 (₹ 4,40,000 – ₹ 2,50,000)  9,500 
Less: Rebate u/s 87A (Since total income < ₹ 5,00,000)  9,500 
Tax Liability  Nil 
Question 20.
For the given set of data, calculate the percentile rank and quartile rank for observation 6.
5, 6, 6, 9, 11, 13, 15, 18, 20.
OR
For a frequency distribution the Bowley’s coefficient of skewness is 1.2. If the sum of 1st and 3^{rd} quartile is 200 and median is 76, then find the value of third quartile. [2]
Answer:
The given data is already arranged in ascending order, so there is no need to arrange the data in ascending order.
Here, we can see that observation 6 comes under the repeated rank
So, here Y = 9, R = 2 and M = 2
(Here we choose the 6 which is at the farthest side, so, here we take 6 which is at 3^{rd} rank, since, 6 is at 3^{rd} rank, therefore the ranks below to 6 are 2 i.e., the value of M = 2)
We apply formula,
PR = \(\left[\frac{M+(0.5 \times R)}{\Upsilon}\right]\) × 100
= \(\frac{2+(0.5 \times 2)}{9}\) × 100
= \(\frac{3}{9}\) × 100 = 0.33 × 10 = 33%
Now, locating this percentile rank on quartile scale, we observe that, it lies between Q_{1} and Q_{2}. Therefore, quartile ranking for score 6 is Q_{2}.
Commonly Made Error
Some students are confused whether they use formula PR = \(\frac{M}{Y}\) × 100 or PR = \(\frac{M+(0.5 \times R)}{Y}\) × 100
Answering Tips
* Whenever you found the repeating rank or position in the given data set, then we apply formula \(\frac{M+(0.5 \times R)}{\Upsilon}\) × 100
otherwise, we use the formula PR = \(\frac{M}{Y}\) × 100.
* Also, do not make mistake in choosing value of M and R. R is the number of times the rank is repeating for the given observation or scores where as M is the number of rank below to the chosen rank.
* When there is a repetition, then choose the score or observation which at farthest place.
OR
We have given, S_{k} = 1.2, Q_{1} + Q_{2} = 200, Q_{2} = 76
∵ S_{k} = \(\frac{Q_3+Q_12 Q_2}{Q_3Q_1}\)
⇒ 1.2 = \(\frac{(2002 \times 76)}{Q_3Q_1}\)
⇒ Q_{3} – Q_{1} = \(\frac{48}{1.2}\)
⇒ Q_{3} – Q_{1} = 40
From eq (i), we get
Q_{3} – (200 – Q_{3}) = 40
⇒ 2Q_{3} = 240
⇒ Q_{3} = 120.
Question 21.
Let A = (All prime numbers less than 10} and B = {all odd number less than 10}. Find A – (A ∩ B)). [2]
Answer:
Here, A = (2, 3, 5, 7} and B = {1, 3, 5, 7, 9}
A ∩ B = {2, 3, 5, 7} ∩ {1, 3, 5, 7, 9} = {3, 5, 7}
A – (A ∩ B) = {2, 3, 5,7} – {3, 5, 7} = {2}
Question 22.
Differentiate \(\frac{x}{\sin x}[latex] with respect to x.
OR
Evaluate [latex]\lim _{x \rightarrow 1} \frac{x^{15}1}{x^{10}1}\) [2]
Answer:
Let,
f(x) = \(\frac{x}{\sin x}\)
∴ f'(x) = \(\frac{d}{d x}\left(\frac{x}{\sin x}\right)\)
= cosec x – x cot x cosec x
= (1 – x cot x). cosec x.
Commonly Made Error
Some students make arithmetic errors while applying quotient rule which leads to incorrect answer.
Answering Tip
Differentiation rules for different functions and terms need attention.
OR
Question 23.
Find the domain of the function f(x) = \(\frac{1}{\sqrt{\left[x^2\right][x]6}}\) [2]
Answer:
Given, f(x) = \(\frac{1}{\sqrt{\left[x^2\right][x]6}}\)
Clearly, f(x) is defined when [x]^{2} – [x] – 6 > 0
[x]2 – 3[x] + 2 [x] – 6 > 0
([x] – 3) ([x] + 2) > 0
[x] > 3 and [x] < – 2
x ≥ 3 + 1 and x < – 2
x ≥ 4 and x < – 2
∴ Domain of f = ( ∞, 2), ∪ [4, ∞).
Question 24.
If ‘men are very busy’ means, ‘1234’, ‘Busy person need encouragement’ means ‘4567’ ‘encouragement is very important’ means ‘3589’ and ‘Important persons are rare means, ‘2680’, then
(A) What is the code for ‘encouragement’.
(B) According to given code, what is the code of ‘Men need encouragement’ ? [2]
Answer:
Given statements and their codes are as follows:
(I) Men are very busy 1 2 3 4
(II) Busy persons need encouragement – 4 5 6 7
(III) Encouragement is very important – 3 5 8 9
(IV) Important persons are rare – 2 6 8 0
(i) In second (ii) and third (iii) sentences, common word is ‘encouragement’ and common code no. is ‘5’ . Hence, number 5 stands for encouragement.
(ii) In first (i) and fourth (iv) sentences, the common word is ‘are’, hence ‘are’ stands for ‘2’ from first (i) and third (iii) sentences, ‘very’ stand for ‘3’ from first (i) and second (ii) sentences, ‘busy’ stands for ‘4’. Hence, we get from first sentence ‘Men stand for ‘1’ similarly ‘needs’ stand for ‘7’. From this, we conclude that ‘Men need encouragement’ will be coded as ‘1 7 5’.
Question 25.
Find the sum of the series: 5 + 13 + 21 + ….. + 181.
Answer:
The terms of the given series form an A.P with first term, a = 5 and common difference, d = 8. Let there be n terms in the given series. Then, a_{n} = 181
⇒ a + (n – 1)d = 181
⇒ 5 + (n – 1)8 = 181
⇒ 8n = 184 ⇒ n = 23
Since, S_{23} = \(\frac{23}{2}\)(5 + 181)
= 2139 [∵ a_{n} = l = 181]
Question 26.
Simplify 2x^{1/2}3x^{1} if x = 4. [2]
Answer:
We have, 2x^{1/2}3x^{1}
= 6.x^{1/2}3x^{1} [Using a^{m} . a^{n} = a^{m + n}]
= 6.x^{1/2} [Using a^{n} = \(\frac{1}{a^n}\)]
= \(\frac{6}{x^{1 / 2}}\) [Using a^{n} = \(\frac{1}{a^n}\)]
Now at x = 4
\(\frac{6}{(4)^{1 / 2}}\) = \(\frac{6}{\left(2^2\right)^{1 / 2}}\) = \(\frac{6}{2^{2 \times \frac{1}{2}}}\) = \(\frac{6}{2}\) = 3
Section – C
All questions are compulsory. In case of internal choices attempt any one.
Question 27.
Convert the following binary numbers to their decimal representation.
(A) (101010)_{2}
(B) (111.10101)_{2} [3]
Answer:
(i)Given, binary number is 101010
Decimal Number 1 × 2^{5} + 0 × 2^{4} + 1 × 2^{3} + 0 × 2^{2} + 1 × 2^{1} + 0 × 2^{0}
= 32 + 0 + 8 + 0 + 2 + 0
= 42
Thus, (101010)_{2} = (42)_{10}
(ii)
Decimal Value 1 × 2^{2}1 × 2^{1} × 2^{0} . 1 × 2^{1} 0 × 2^{2} 1 × 2^{3}0 × 2^{4}1 × 2^{5}
= 4 + 2 + 1.0.5 + 0 + 0.125 + 0 + 0.03125
= 7.65625
Thus, (111. 10101)_{2} = (7.65625)_{10}
Question 28.
A man who went out between 5 and 6 and returned between 6 and 7 found that the hands of the watch had exactly changed place. When did he go out ?
OR
A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B complete the remaining work in 23 days. Find the number of days after which A left the work. [3]
Answer:
Between 5 and 6 to 6 and 7, hands will change place after crossing each other one time, i.e., they together will make 1 + 1 = 2 complete revolutions.
Hour hand will move through 2 × \(\frac{60}{13}\) or \(\frac{120}{13}\) minute divisions
Between 5 and 6 → \(\frac{120}{13}\) minute divisions.
At 5, minute hand is 25 minute division behind the hour hand.
Hence, it will have to gain 25 + \(\frac{120}{13}\) minute divisions on the hourhand = \(\frac{445}{13}\) minute divisions on the hour hand.
The minute hand gains \(\frac{445}{13}\) minute divisions in \(\frac{445}{13}\) × \(\frac{12}{11}\) minutes
= \(\frac{5340}{143}\)
= 37\(\frac{49}{143}\) minurtes
∴ The required time of departure is 37\(\frac{49}{143}\) min past 5.
OR
(A + B)’s 1 day’s work = \(\left(\frac{1}{45}+\frac{1}{40}\right)\) = \(\frac{17}{360}\)
Work done by Bin 23 days = \(\left(\frac{1}{40} \times 23\right)\) = \(\frac{23}{40}\)
Remaining work = \(\left(1\frac{23}{40}\right)\) = \(\frac{17}{40}\)
∵ \(\frac{17}{360}\) work was done by (A + B) in 1 day
∵ \(\frac{17}{40}\) work done by (A + B) in \(\left(1 \times \frac{360}{17} \times \frac{17}{40}\right)\)
= 9 days
Therefore, A left after 9 days.
Question 29.
Using the letters of the word ‘ARRANGEMENT’ how many different words (using all letters at a time) can be made such that both A, both E, both R and both N occur together.
OR
How many different products can be obtained by multiplying two or more of the numbers 2, 5, 6, 7,9? [3]
Answer:
There are 11 letters in the word ‘ARRANGEMENT’ out of which 2A’s, 2E’s and 2N’s and 2R’s
Considering both A, both E, both R and both N together, 8 letters should be counted as 4.
So, there are total 7 letters (AA EE RR NN G M T)
These 7 letters can be arranged in 7! ways
Hence, total ways = 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040
OR
The given numbers are 2, 5, 6, 7, 9.
The numbers of different products when 2 or more is taking = the number of ways of taking product of 2 numbers + number of ways of taking product of 3 numbers + numbers of ways of taking product of 4 numbers + number of ways of taking 5 together
= ^{5}C_{2} + ^{5}C_{3} + ^{5}C_{4} + ^{5}C_{5}
Question 30.
Draw the graph of the function
Also, find its range. [3]
Answer:
Given
The given function is called the signum function and can be written as
The domain of f = R and the range of f = {1, 0, 1}
Question 31.
If y = \(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\), prove that (2xy)\(\frac{d y}{d x}\) = \(\frac{x}{a}\frac{a}{x}\) [3]
Answer:
Given,
Question 32.
The heights (to nearest cm) of 60 students of a certain school are given in the following frequency distribution table:
Height(in cm)  151  152  153  154  155  156  157 
No. of students  6  4  11  9  16  12  2 
Find the
(A) median
(B) lower quartile
(C) upper quartile. [3]
Answer:
The given variates are already in ascending order. We construct the cumulative frequency table as under:
Variate(Height)  Frequency (No. of students)  Cumulative frequency 
151  6  6 
152  4  10 
153  11  21 
154  9  30 
155  16  46 
156  12  58 
157  2  60 
(i) Median = \(\frac{n+1}{2}\)th item
(ii) Q_{1} = \(\frac{n+1}{4}\)th value
= \(\frac{60+1}{4}\)th value
= 15.25th value
= 153 cm (as both 15th and 16th value are 153)
(iii) Q_{3} = \(\frac{3(n+1)}{4}\)th value
= \(\frac{3(60+1)}{4}\)th value
= 45.75th value
= 155 cm (as both 45th and 46th values are 155)
Question 33.
₹ 2000 is invested at annual rate of interest of 10%. What is the amount after two years if compounding is done (A) Annually, (B) semiannually, (C) Quarterly.
Answer:
Here, P = ₹ 2000, i = 10% = \(\frac{10}{100}\) = 0.1
(a) Since, interest is compounded yearly
n =2
Since, A_{n} = P(1 + i)^{n}
∴ A_{2} = 2000(1 + 0.1)^{2}
= 2000(1.1)^{2}
= 2000 × 1.21
= ₹ 2420
(b)For semiannual compounding
n = 2 × 2 = 4
Since, i = \(\frac{0.1}{2}\) = 0.05
∴ A = 2000 (1 + 0.5)^{4}
= 2000 × (1.05)^{4}
= 2000 × 1.2155
= ₹ 2431
(c) For quarterly compounding
n = 4 × 2 = 8
Since, i = \(\frac{0.1}{4}\) = o.o25
∴ A_{4} = 2000 × (1 + 0.025)^{8}
= 2000 × (1.025)^{8}
= 2000 × 1.2184
= ₹ 2436.80
Section – D
All question are compulsory. In case of internal choices attempt any one.
Question 34.
X, who is a person with disability, submits the following information.
S.No.  Particulars  Amount (₹) 
(A)  Salary (per annum)  3,05,000 
(B)  Rent received (per month)  2,000 
(C)  Dividend from Cooperative Society  3,000 
(D)  Interest on Bank Deposits  7,000 
(E)  Interest on Government securities  5,000 
(F)  Winnings from Lotteries  2,000

(G)  NSC (VIII Issue) purchased during the year  15,000 
(H)  Deposit under PPF Scheme  35,000 
He earned a longterm capital gain of ₹ 78,000 on sale of shares during the year. Compute
(a) the taxable income;
(b) the tax payable for the assessment year 202122. [5]
Answer:
Note: Income from winning of lotteries cannot be shifted to other income to claim full exemption of ₹ 250,000.
Question 35.
A girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3, or 4, she tosses a coin two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3, or 4 with the die ?
OR
If p^{th}, q^{th} and r^{th} terms of an A.P. and G.P, are both a, b and c respectively, then show that
a^{bc}.b^{ca}.c^{ab}= 1. [5]
Answer:
When a die is thrown,
Sample space = {1, 2, 3, 4, 5, 6}
let E_{1} : Girl gets 5 or 6
E_{2} : Girl gets 1, 2, 3 or 4
Then, P(E_{1}) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
and P(E_{2}) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
When she get 5 or 6, she throws a coin three times.
Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)
Out of these 8 outcomes, exactly two heads are obtained in the 3 outcomes i.e., HHH, HHT, THH. If A denotes the event of getting exactly two heads, then
\(P\left(\frac{A}{E_1}\right)\) = \(\frac{3}{8}\)
Again, when she gets 1, 2, 3 or 4 she throws a coin two times.
Sample space { HH, HT, TH, TT}
∴ \(P\left(\frac{A}{E_2}\right)\) = \(\frac{1}{4}\)
Thus, by Bayes’ theorem, we have probability that girl gets exactly two heads, when she threw 1, 2, 3 or 4 with the die
OR
Let A, d are the first term and common difference of A.P and x, R are the first term and common ratio of G.P., respectively.
According to the given condition,
A + (p – 1)d = a ….. (i)
A + (q – 1)d = b …(ii)
A + (r – 1)d = c ……(iii)
and
a = xR^{p1} ..(iv)
b = xR^{q1} …(vi)
On subtracting Eq. (ii) from Eq. (i). we get
d(p – 1 – q + 1) = a – b
⇒ a – b = d(p – q) (vii)
On subtracting Eq. (iii) from Eq. (ii), we get
d(q – 1 – r + 1) = b – c
b – c = d(q – r) …(viii)
On subtracting Eq. (i) from Eq. (iii), we get
d(r – 1 – p + 1) = c – a
c – a = d(r – p) .. (ix)
Now, we have to prove a^{bc}b^{ca}c^{ab} = 1
Taking LHS = a^{bc}b^{ca}c^{ab}
Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),
Hence proved
Question 36.
Let the sum of n, 2n, 3n terms of an A.P. be S_{1}, S_{2} and S_{3} respectively. Show that S_{3} = 3 (S_{2} – S_{1}).
OR
Find the first principle derivative of \(\frac{a x+b}{c x+d}\). [5]
Answer:
Let a and d be the first term and common difference respectively of an A.P
Given, S_{1} = Sum of n terms
= \(\frac{n}{2}\)[2a + (n – 1)d] ….. (i)
S_{2} = Sum of 2n terms
= \(\frac{2 n}{2}\)[2a + (2n – 1)d] …(ii)
and S_{3} = Sum of 3n terms
= \(\frac{3 n}{2}\)(3n – 1)d] ….. (iii)
Now, we have to prove S_{3} = 3(S_{2} – S_{1})
Now, RHS = 3(S_{2} – S_{1})
= S_{3} = LHS [from Eq.(iii) ]
∵ LHS = RHS
OR
Let, f(x) = \(\frac{a x+b}{c x+d}\)
Section – E
Both the Case study based questions are compulsory. Each Sub parts carries 1 mark.
Question 37.
In Sweta’s school a fate is organized on the Holi. There are many games and stalls in the school fair. Sweta wanted to play a game. The game is based on coin tossing. The entry fee of game is Rs. 5. The game consist of a tossing a coin 3 times. If one or two heads show. Sweta gets her entry fee back. If she throws 3 heads, she receives double entry fees. Otherwise she will lose. For tossing a coin three times, find the following: [4]
(a) What is the total number of outcomes for this game?
(A) 8
(B) 2
(C) 6
(D) none of these
Answer:
Option (A) is correct.
Explanation: Since, one coin is tossed 3 times so total number of favourable outcomes = 2^{3} = 8.
which are (HHH), (HHT), (HTH), (THH) and (replacing H → T and T → H) (TTT), (TTH), (THT), (HTT)
(b) Probability that she loses the entry fee:
(A) 1
(B) \(\frac{1}{8}\)
(C) \(\frac{3}{4}\)
(D) none of these
Answer:
Option (B) is correct.
Explanation: Losing the game means getting no head
Number of favourable outcomes of getting no head
= F(E) =.1
So, P(Losing the entry fee), i.e., P(E) = \(\frac{F(E)}{T(E)}\) = \(\frac{1}{8}\)
(c) Probability that she gets double entry fee:
(A) 1
(B) \(\frac{1}{8}\)
(C) \(\frac{3}{4}\)
(D) none of these
Answer:
Option (B) is correct.
Explanation:
Gets double entry fee back by getting 3 heads
Number of favourable outcomes of getting 3 heads,
i.e., F(E) = 1
∴ P(getting double entry fee) P(E) = \(\frac{F(E)}{T(E)}\) = \(\frac{1}{8}\)
(d) Probability that she lust gets her entry fee:
(A) 1
(B) \(\frac{1}{8}\)
(C) \(\frac{3}{4}\)
(D) none of these
Answer:
Option (C) is correct.
Explanation: Just gets her entry fees back by getting either one or two heads. Number of favourable outcomes of getting either one or two heads, i.e., F(E) = 1
∴ P(just getting entry fee) i.e.,
P(E) = \(\frac{F(E)}{T(E)}\) = \(\frac{6}{8}\)
= \(\frac{3}{4}\)
Question 38.
In modern school before coming to a pre – board, teachers of class X and XII planned to take surprise tests. Teachers of class XII conducted this surprise test of Mathematics, Physics and Chemistry on Monday. The total strength of class is 70, but on that day the strength was only 50.
The mean and standard deviation of marks obtained by 50 students of the class in three subjects are given below:
Subject  Mathematics  Physics  Chemistry 
Mean  42  32  40.9 
Standard  12  15  20 
[4]
(a) What is the coefficient of variation of marks obtained by the students in Mathematics ?
(A) 28.57
(B) 46.87
(C) 48.89
(D) none of these
Answer:
Correct Option (A)
Explanation: Standard deviation of Mathematics = 12
The coefficient of variation (C.V.) is given by
CV (in Mathematics) = \(\frac{12}{42}\) × 100 = 28.57
(b) What is the coefficient of variation of marks obtained by the students in Physics ?
(A) 28.57
(B) 46.87
(C) 48.89
(D) none of these
Answer:
Option (B) is correct.
Explanation: Standard deviation of Physics 15
C.V(in Physics) = \(\frac{15}{32}\) × 100 = 46.87
(c) What is the coefficient of variation of marks obtained by the students in Chemistry ?
(A) 28.57
(B) 46.87
(C) 48.89
(D) none of these
Answer:
Option (C) is correct.
Explanation:
Standard deviation of,Chemistry = 20
CV (in Chemistry) = \(\frac{20}{40.9}\) × 100 = 48.89
(d) Which of the three subjects show the highest variability ?
(A) Mathematics
(B) Physics
(C) Chemistry
(D) none of these
Answer:
Option (C) is correct.
Explanation: The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry.