CBSE Sample Papers for Class 11 Applied Mathematics Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Applied Mathematics Set 1 with Solutions

Time Allowed : 3 hours
Maximum Marks : 80

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 ofl 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.
Use of calculators is not permitted.

Section – A (16 Marks)

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

Question 1.
Walking at \(\frac{5}{6}\) of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey ? [1]
OR
The average annual income (in ₹) of certain agriculture workers is S and that of other workers is T. The number of agriculture workers is 11 times that of the other workers. Then find the average monthly income (in ₹) of all workers.
Solution:
5
New speed = \(\frac{5}{6}\) of the usual speed
New time taken = \(\frac{6}{5}\) of usual time
So, (\(\frac{6}{5}\) of the usual time) – (usual time) = 10 min
⇒ \(\frac{1}{5}\) of the usual time = 10 min
⇒ usual time = 50 min.
OR
1.et the number of other workers be x.
Then, number of agriculture workers = 11x
Total number of workers = 11x + x = 12x
∴ Average monthly income = \(\frac{S \times 11 x+T \times x}{12 x}\)
= \(\frac{11 S+T}{12}\)

Question 2.
Simplify: [1 – {1 – (1 – x²)^-1}^-1]^-1/2
Solution:

= \(\left[\frac{1}{x^2}\right]^{-1 / 2}\)
= (x^-2)^-1/2
= x^-2x(-1/2) = x
[Using (aⁿ)ⁿ = a^mn]

Commonly Made Error
In such type of simplification, generally students make mistakes of signs while using
rule aⁿ = \(\frac{1}{a^n}\)

Answering Tip
Be careful with signs during simplification because one wrong sign leads to incorrect answer.

Question 3.
In the given data set: 1, 2, 3, 4, 5; how many numbers are even or find the percentile ranking of even numbers ? [1]
Solution:
Here, total numbers in given data set is 5, i.e., Y = 5
Also, there are two even numbers, (2 & 4) so here the value of M is 2.
Applying formula,
PR = \(\frac{M}{Y}\) × 100 = \(\frac{2}{5}\) × 100 = 40%
Thus, there are 40% numbers in the given data set are even or we can say the percentile ranking of even number is 40%.

Question 4.
Find the value of ‘p’ so that the equation x² + y² – 2px + 4y – 12 = 0 may represent a circle of radius 5 units. [1]
Solution:
Given equation of the circle is
x² + y² – 2px + 4y – 12 = 0
⇒ (x² – 2px + p²) + (y² + 4y + 4) – p² – 4 – 12 = 0
⇒ (x – p)² + (y + 2)² = p² + 16
⇒ (x – p)² + {y – (- 2)}² = \(\left(\sqrt{p^2+16}\right)\)²
∴ The centre of the circle is (p, – 2) and radius \(\sqrt{p^2+16}\) units.
According to the question,
\(\sqrt{p^2+16}\) = 5
⇒ p² + 16 = 25
⇒ p² = 9
⇒ p = ± 3

Commonly Made Error
When converting the given equation into standard equation, some students commit algebraic error, which gives incorrect answer.

Answering Tip
Recheck the solution to avoid errors.

Question 5.
What is the next number in the given Number Series? [1]
5, 24, 94, 279, ?
OR
In a certain code, NEWYORK is written as 111, how is NEWJERSEY written in that code ?
Solution:
The given number series follows a pattern that
5 × 5 – 1 = 24
24 × 4 – 2 = 94
94 × 3 – 3 = 279
279 × 2 – 4 = 554

In NEWYORK
N = 14, E = 5, W = 23, Y = 25, O = 15, R = 18, K = 11
Total = 14 + 5 + 23 + 25 + 15 + 18 + 11 = 111
In NEWJERSEY
Total = 14 + 5 + 23 + 10 + 5 + 18 + 19 + 5 + 25 = 124

Question 6.
Arrange the following words in logical and meaningful manner. [1]
1. Leaves, 2. Branch, 3. Flower, 4. Tree 5. Fruit.
Solution:
From the above words, it is deducted that in tree, first come branches, then leaves, then flowers, then from flowers we get fruit.
Therefore the correct arrangement is

Question 7.
Find the domain of following function :
f(x) = \(\frac{1}{\sqrt{x-5}}\)
OR
Let f and g be real function be f(x) = \(\sqrt{x+4}\), x > – 4 and g(x) = \(\sqrt{x-4}\), x > 4. Then find function fg.
Solution:
Given, f(x) = \(\frac{1}{\sqrt{x-5}}\)
f(x) is defined, if x – 5 > 0 ⇒ x > 5
∴ Domain of f = (5, ∞)

fg = f(x) g(x) = \(\sqrt{x+4}\) \(\sqrt{x-4}\)
= \(\sqrt{x^2-16}\)

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

Question 9.
If x = t² and y = t³ then find \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) [1]
Solution:
Given that,
x = t² and y = t³
Then, \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 2t and \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 3t²
Thus, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{3 t^2}{2 t}=\frac{3 t}{2}\)
\(\frac{d^2 y}{d x^2}=\frac{3}{2} \frac{d t}{d x}\)
= \(\frac{3}{2} \cdot \frac{1}{2 t}=\frac{3}{4 t} .\)

Question 10.
LetP(A) = \(\frac{7}{13}\), P(B) = \(\frac{9}{13}\) and P(A ∩ B) = \(\frac{4}{13}\). Then find P\(\left(\frac{A^{\prime}}{B}\right)\). [1]
Solution:

Question 11.
Find the coordinates of focus and length of latus rectum of parabola 3y² = 8x. [1]
Solution:
Given equation of the parabola is
3y² = 8x
⇒ y² = \(\frac{8}{3}\) x
⇒ y² = 4 . \(\frac{2}{3}\)x
(on comparing it with y² = 4ax)
Here, a = \(\frac{2}{3}\)
∴ Focus = (a, 0) = (\(\frac{2}{3}\), 0)
and length of latus-rectum is 4a = \(\frac{8}{3}\)

Question 12.
Calculate range and the coefficient from the following distribution : [1]

X	10-15	15-20	20-25	25-30
frequency	4	10	16	8

Solution:
Here, lower dass limit of the first dass interval,
S = 10
and upper class limit of the last class interval,
L = 30
∴ Range = L – S = 30 – 10 = 20
Coefficient of range = \(\frac{L-S}{L+S}\) × 100
= \(\frac{30-10}{30+10}\) × 100
= 0.5 × 100 = 50

Question 13.
If P = {x : \(\frac{x -8}{2}\) = 4}, then set P is empty? [1]
OR ,
If A = (1, 2}, B = {3, 4}, then find A × (B ∩ Φ).
Solution:
Given, P = {x : \(\frac{x -8}{2}\) = 4}, therefore, \(\frac{x-8}{2}\) = 4
⇒ x – 8 = 8 ⇒ x = 16
⇒ Set P will have one element i.e., P = {16}
⇒ Set P is not a empty set, it is a singleton set.

Φ, because A × (B ∩ Φ) = A × Φ = Φ

Question 14.
What is the difference between annuity regular and annuity due ? [1]
Solution:
Each payment of an ordinary (regular) annuity belongs to the payment period preceding its date, while the payment of an annuity-due refers to a payment period following its date.

Question 15.
Find the distance between parallel lines l(x + y) + p = 0 and l(x + y) – r = 0. [1]
Solution:
Given lines are:
l(x + y) + p = 0
⇒ ly= -lx – p
⇒ y = -x – \(\frac{p}{l}\)
and l(x + y) – r = 0
⇒ ly = -lx + r
⇒ y = -x – \(\frac{r}{l}\)
∴ m = -1, c₁ = – \(\frac{p}{l}\) and c₂ = \(\frac{rp}{l}\)
∴ Distance between the lines
d = \(\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}\)
= \(\frac{\left|-\frac{p}{l}-\frac{r}{l}\right|}{\sqrt{1+(-1)^2}}=\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right|\)

Commonly Made Error
Some students get confused between formula of distance for parallel lines and perpendicular lines, so they apply wrong formula sometimes.

Answering Tip
Distance between two parallel lines y = mx + c₁ and y = mx + c₂ is given by d = \(\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}\)

Question 16.
If x, y, z are positive integers then find the value of the expression {x + y) (y + z) (z + x). [1]
OR
Which term of the sequence 25, 24 \(\frac{1}{4}\), 23 \(\frac{1}{2}\), 22 \(\frac{3}{4}\), is the first negative term ?
Solution:
We know that,
A.M. ≥ G.M.
\(\frac{x+y}{2}\) ≥ \(\sqrt{x y}\) , \(\frac{y+z}{2}\) ≥ \(\sqrt{y z}\)
and \(\frac{z+x}{2}\) ≥ \(\sqrt{z x}\)
Multiplying the three inequalities, we get
\(\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{z+x}{2}\) ≥ \(\sqrt{(x y)(y z)(z x)}\)
or, (x + y) (y + z) . (z + x) ≥ 8xyz
Hence, the value of (x + y) (y + z) (z + x) ≥ 8xyz.

The given sequence is an A.P. with common difference, d = –\(\frac{3}{4}\) and first term, a = 25.
Let the n^th term of the given A.P. be the first negative term, then
a_n < 0 ⇒ 25 + (n – 1) (-\(\frac{3}{4}\)) < 0
⇒ \(\frac{103}{4}\) – \(\frac{3n}{4}\) < 0
⇒ 103 – 3n < 0
⇒ 103 < 3n ⇒ 3n > 103
⇒ n > 34\(\frac{1}{3}\)
Since, 35 in the least natural number satisfying
n >34\(\frac{1}{3}\) ⇒ n = 35.
Hence, 35^th term of the given sequence is the first negative term.

Section – B (20 Marks)

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

Question 17.
Multiply (1011.01)₂ to (110.1)₂.
OR
Simplify:
{(x + y)^2/3 (x – y)^3/2 ÷ (\(\sqrt{x+y}\) x \(\sqrt{(x-y)^3}\))]}⁶
Solution:

Hence, (1011.01)₂ × (110.1)₂ = (1001001.001)₂

Commonly Made Error
Some students make mistakes while adding the binary numbers.

Answering Tip
It is very much important to learn and understand the rules of addition of binary numbers and do as much problems as you can to avoid mistakes.

= (x + y)^{\(\frac{1}{6}\)×6} [Using (a^m)ⁿ = a^mn]
= x + y

Question 18.
If log (0.57) = \(\overline{1} .756\), then find the value of log 57 + log (0.57)³ + log \(\sqrt{0.57}\).
Solution:
log (0.57) = \([\overline{1} .756/latex]
log 57 = 1.756
[∵ mantissa will remain same]
∴ log 57 + log(0.57)³ + log [latex]\sqrt{0.57}\)
= log 57 + 3 log (\(\frac{57}{100}\)) +log (\(\frac{57}{100}\))^{\(\frac{1}{2}\)}
[Applying rule log_n mⁿ = n log_a m]
= log 57 + 3 log 57 – 3 log 100 + \(\frac{1}{2}\) log 57 – \(\frac{1}{2}\) log 100
[Applying rule log_a \(\frac{m}{n}\) = log_a m – log_a n]
= \(\frac{9}{2}\) log 57 – \(\frac{7}{2}\) log 100
= \(\frac{9}{2}\) × 1.756 – \(\frac{7}{2}\) × 2 [∵ log₁₀ 100 = 2]
= 7.902 – 7
= 0.902

Question 19.
Write the value of \(0 . \overline{3}\). [2]
Solution:
Let x = \(0 . \overline{3}\)
x = 0.33333 ………………
x = 0.3 + 0.3 + 0.0003 + …………….
∴ a = 0.3, r = \(\frac{0.03}{0.3}\) = 0.1, n = ∞
∴ Sum of infinite terms is G.P, S = \(\frac{a}{1-r}\)
∴ x = \(\frac{0.3}{1-0.1}\) = \(\frac{0.3}{0.9}\) = \(\frac{1}{3}\)
∴ \(0 . \overline{3}=\frac{1}{3}\)

Commonly Made Error
While expanding decimal number some students were not able to expand the number in the form of G.P.

Answering Tip
When sum of G.P is asked upto infinity then the formula to be used, S_∞ = \(\frac{a}{1-r}\)

Question 20.
Using the digits 0, 1, 2, 2, 3, how many numbers greater than 2000 can be made ? [2]
OR
A committee of 7 has to be formed out of 9 boys and 4 girls. In how many ways can be this be done when the committee consists of exactly 3 girls.
Solution:
Total number of digits = 0, 1, 2, 2, 3
Total number formed by these digits = \(\frac{5 !}{2 !}=\frac{120}{2}\)
= 60
Total number formed by starting 0 = \(\frac{4 !}{2 !}=\frac{24}{2}\)
= 12
Total number formed by starting 1 = \(\frac{4 !}{2 !}=\frac{24}{2}\)
= 12
Total number formed greater than 2000
= 60 – 12 – 12 = 36 .

Commonly Made Error
Students generally don’t calculate the permutation of number formed by starting 0 and 1, they find permutation of numbers greater than 2000 directly which lengthen the solution.

Answering Tip
Learn all the concepts and formulae and practice as many questions as possible.

A committee of 7 has to be formed from 9 B and 4 Exactly 3 girls
= ⁹C₄ × ⁴C₃
=
= 72 × 7 = 504

Question 21.
Main street high school has 10 members on its football team and 14 members on its science club. 5 members at the school belong to both the football and science teams. How many students belong to only science club team or football team? [2]
Solution:
Given that,
Members in football team = n(f) = 10
Members in science club = n(s) = 14
Members in both football team and science club = n(f ∩ s) = 5
Then, members in only football team = n(s) = 10 – 5 = 5
Members in only science team = n(s) = 14 – 5 = 9
Hence, members in only football or science team
= n(f ∪ s) = n(F) + n(S) – n(f ∩ s)
= 9 + 5 – 5 = 9

Question 22.
If [x]² – 5[x] + 6 = 0, where [.] denote the greatest integer function, then find interval of x.
Solution:
Given,
[x]² – 5[x] + 6 = 0
⇒ [x]² – 2[x] – 3[x] + 6 = 0
⇒ [x]([x] – 2) – 3([x] – 2) = 0
⇒ ([x] – 3)([x] – 2) = 0
⇒ [x] = 2, 3
Hence, x ∈ [2, 3]

Question 23.
If e^y(x + 1) = 1, then solve that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\) [2]
OR
Differentiate (cos x)^x with respect to x.
Solution:
e^y(x + 1) = 1 or e^y . 1 + (x + 1) . e^y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0
or \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-1}{(x+1)}\)
\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) = \(\frac{1}{(x+1)^2}=\left(\frac{d y}{d x}\right)^2\)

Let, y = (cos x)^x
Then, y = e^{log(cos x)^x} = e^{x log cos x}
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^{x log cos x} × \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x log cos x)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (cosx)^x {log cos x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x) + x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(log cosx)}
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (cos x)^x{log cosx + x \(\frac{1}{\cos x}\)(-sinx)}
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (cos x)^x {log cosx – x tan x)}.

Question 24.
Find the mean and variance for the following data : [2]
6, 7, 10, 12, 13, 4, 8, 12.
Solution:
Here,
Mean x = \(\frac{6+7+10+12+13+4+8+12}{8}\)
= \(\frac{72}{8}\) = 9

x_i	x_i – \(\bar{x}\)	(x_i – \(\bar{x}\))²
6	-3	9
7	-2	4
10	1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9

\(\Sigma\left(x_i-\bar{x}\right)^2\) = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9
= 74
∴ Variance = \(\frac{\Sigma\left(x_i-\bar{x}\right)^2}{\Sigma f_i}=\frac{74}{8}\) = 9.25

Question 25.
Compute the compound interest on ₹ 4000 for 1 1/2 years at 10% per annum compounded half yearly. [2]
Solution:
Here, principal P = ₹ 4000. Since the interest is compounded half-yearly the number of conversion periods in 1 \(\frac{1}{2}\) years are 3. Also, the rate of interest per conversion period (6 months) is 10% × \(\frac{1}{2}\) = 5% (0.05 in decimal)
Thus, A_n(in ₹) is given by
A_n = P(1 + i)ⁿ
or, A₃ = 4000(1 + 0.05)³ = ₹ 4630.50
The compound interest is therefore ₹ (4630.50 – 4000)
= ₹ 630.50

Question 26.
A bank A offers you an (effective) annual interest rate of 6%, the bankB offers an interest rate of 1.5% per quarter. Which of these two banks offers the best return ?
Solution:
Bank B offers a quarterly rate 1.5%, the equivalent annual interest rate (or effective rate) for this interest can be obtained by the relation
(1 + i_quart)⁴ = 1 + i_ann
(1.015)⁴ = 1 + i_ann
i_ann = (1.015)⁴ – 1 = 0.06136
= 6.136%
Bank B, therefore offers a better return with (effective) annual interest rate of 6.136% than bank A.

Section – C (21 Marks)

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

Question 27.
Read the following information carefully and answer the questions given below : [3]
(A) Six flats on the floor in two facing north and south are allotted to P, Q, R, S, T and U.
(B) Q gets a north facing flat and is not next to S.
(C) S and U get diagonally opposite flats.
(D) R next to U, gets a south facing flat and T gets a north facing flat. 7
(a) Whose flat is between Q and S ?
(b) The flats of which of the other pairs than SU, is diagonally opposite to each other ?
(c) If the flat of T and P are interchanged, whose flat will be next to that of U ?
Solution:
On the basis of the given information, the position of all flats are shown as below:

(a) T is between Q and S
(b) Apart from SU, PQ are diagonally opposite to each other.
(c) After interchanging T and P, R’s position does not get affected and R is next to U.

Commonly Made Error
While doing problems related to seating arrangement students do mistakes in drawing the correct figure with the help of the statements.

Answering Tips
In question based on seating arrangement remember the following points:

Count the number of people/objects in the given question.
Draw a frame accordingly whether it is linear, circular or rectangle.
Solve sentence by sentence and then merge them.

Question 28.
To Prove : A ∩ (B – C) = (A ∩ B) – (A ∩ C)
Solution:
Let x ∈ {A ∩ (B – C)}
x ∈ A and x ∈ B – C
x ∈ A and x ∈ B and x ∉ C
(x ∈ A and x ∈ B) and (x ∈ A and x ∉ C)
x ∈ A ∩ B and x ∉ A ∩ C
x ∈ (A ∩ B) – (A ∩ C)
A ∩ (B – C) ⊆ (A ∩ B) – (A ∩ C) …………………… (i)
Again, let
y ∈ (A ∩ B) ∩n (A – C)
y ∈ A and (y ∈ B and y ∉ C)
y ∈ A and y ∈ B – C
y ∈ {A ∩ (B – C)}
(A ∩ B) – (A ∩ C) ⊆ A ∩ (B – C) ……………….. (ii)
From eqs. (i) and (ii), we get
A ∩ (B – C) = (A ∩ B) – (A ∩ C)
Hence Proved

Commonly Made Error
Some students are unable to apply appropriate properties which results in proving the answer wrong.

Answering Tip
Clarify the properties of sets and their complements properly.

Question 29.
If x = a + \(\frac{a}{r}+\frac{a}{r^2}\) + ………….. + ∞, y = b – \(\frac{b}{r}+\frac{b}{r^2}\) + ……….. + ∞ and z = c + \(\frac{c}{r^2}+\frac{c}{r^4}\) + …………. + ∞ Prove that \(\frac{x \cdot y}{z}=\frac{a b}{c}\) [3]
OR
A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if (i) they can be of any colour, (ii) two must be white and two red. (iii) they must all be of the same colour.
Solution:
Given series x = a + \(\frac{a}{r}+\frac{a}{r^2}\) …………….. + ∞ is in G.P. with common ratio \(\frac{1}{r}\).
and series y = b – \(\frac{b}{r}+\frac{b}{r^2}\) – ………… + ∞ is in G.P. with common ratio –\(\frac{1}{r}\).
and series z = c + \(\frac{c}{r^2}+\frac{c}{r^4}\) + ………… + ∞ is in G.P. with common ratio \(\frac{1}{r^2}\).
Sum of infinite terms of series
x = a + \(\frac{a}{r}+\frac{a}{r^2}\) + …………… + ∞ is x = \(\frac{a}{1-\frac{1}{r}}\)
Sum of infinite term of series y = b – \(\frac{b}{r}+\frac{b}{r^2}\) – …………… + ∞ is
y = \(\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{b}{1+\frac{1}{r}}\)
Sum of infinite terms of series z = c + \(\frac{c}{r^2}+\frac{c}{r^4}\) + …………. + ∞ is
z = \(\frac{c}{1-\frac{1}{r^2}}\)
Now, LHS = \(\frac{xy}{z}\)

Hence Proved.

Commonly Made Error
Some students find this question difficult and were not able to determine the appropriate method for evaluating it.

Answering Tip
Use the property
S_∞ = a + ar + ar² + ar3 + … ∞ term
if | r| < 1 i.e., – 1 < r < 1, then S_∞ = \(\frac{a}{1-r}\)

Total number of marbles = 6 white + 5 red = 11 marbles
(i) If they can be of any colour means we have to select 4 marbles out of 11.
∴ Required Number of ways = ¹¹C₄ = \(\frac{11 !}{7 ! 4 !}\) = 330
(ii) If two must be white, then selection will be ⁶C₂ and two must be red, then selection will be ⁵C₂
∴ Required number of ways = ⁶C₂ × ⁵C₂ = 15 × 10 = 150
(iii) If they all must be same colour, then selection of 4 white marbles out of 6 = ⁶C₄ and selection of 4
red marble out of 5 = ⁵C₄
∴ Required number of ways = ⁶C₂ + ⁵C₄ = 15 + 5 = 20

Question 30.
Find the value of k, f or which [3]

is continuous at x = 0.
Solution:
Given,

Thus, f(x) is continuous at x = 0, if k = -1

Question 31.
Find the equation of circle concentric with circle 4x² + 4y² – 12x – 16y – 21 = 0 and half its area. [3]
OR
Obtain the equation of the line passing through the intersection of lines 4x – 3y – 1 = 0 and 2x – 5y + 3 = 0 and equally inclined to the axes.
Solution:
The given equation of the circle is
4x² + 4y² – 12x – 16y – 21 = 0
⇒ x² + y² – 3x – 4y – \(\frac{21}{4}\) = 0
⇒ (x² -3x + \(\frac{9}{4}\)) + (y² – 4y + 4) – \(\frac{9}{4}\) – 4 – \(\frac{21}{4}\) = 0
⇒ (x – \(\frac{3}{2}\))² + (y – 2)² – \(\frac{15}{4}\) – 4 = 0
⇒ (x – \(\frac{3}{2}\))² + (y – 2)² = \(\left(\sqrt{\frac{23}{2}}\right)^2\)
∴ Centre = (\(\frac{3}{2}\), 2) and radius, r₁ = \(\sqrt{\frac{23}{2}}\)
Let ‘r₂’ be the radius of the concentric circle.
∴ Its equation will bes

⇒ 4x² + 4y² – 12x – 16y + 25 = 26
⇒ 2x² + 2y² – 6x – 8y + 1 = 0, is the required equation of the circle.

Commonly Made Error
Sometimes students don’t have an understanding that concentric circles have same centre and different radius.

Answering Tip
Students need to emphasise on the conditions and properties of circle.

Given lines are
4x – 3y – 1 = 0 …………… (i)
2x – 5y + 3 = 0 …………… (ii)
Now, eq. (i) – 2 × eq. (ii),
⇒ 7y – 7 = 0
⇒ y = 1
Putting the values of y in equation (i),
we get x = 1
∴ (1, 1) is the point of intersection of the lines (i) and (ii).
Now, let the equation of the line which is equally inclined to the axes be
\(\frac{x}{a}\) + \(\frac{y}{a}\) = 1
⇒ x + y = a …………… (iii)
Since, (iii) passes through (1, 1)
∴ 1 + 1 = a
⇒ a = 2
∴ Equation of the line is
x + y – 2 = 0.

Commonly Made Error
Students have difficulty in identifying that which method is going to be used to solve the problem, so they apply inappropriate formulas which leads error and they get stuck in half way of solution.

Answering Tips
Read the question carefully before starting to solve it.
Try to understand the concept and apply appropriate formulas.

Question 32.
Let Amar, Ram and Rahim be three dealers belonging to different states. Dealer Amar sells some products/services to dealer Ram for ₹ 1000 dealer Rahim at a profit of ₹ 300. Calculate the tax liability of Ram, if the rate of GST is 12%. [3]
Solution:
For dealer Amar,
Selling price = ₹ 1000 (Given)
∵ Since in case of inter-state, we get
IGST = ₹ \(\frac{12}{100}\) × 1000 = ₹ 120
For dealer Ram,
cost price = ₹ 1000
∴ selling price = ₹ (1000 + 300)
= ₹ 1300
For dealer Rahim,
Cost price = ₹ 1000
profit = ₹ 300
Input tax credit = ₹ 120
Output tax = ₹ (\(\frac{12}{100}\) ×1300)
= ₹ 156
∴ Tax liability on dealer Ram = ₹ (156 – 120)
= ₹ 36

Commonly Made Error
Calculation errors are observed. Some students found the discounted price but were unable to find the tax paid by dealer.

Answering Tip
Students should read the question carefully and try finding answers to each subpart by working out step at a time.

Question 33.
In an A.P., if the pth term is \(\frac{1}{q}\) and qth term is \(\frac{1}{p}\). Prove that the sum of first pq term is \(\frac{1}{2}\) (pq + 1).
Solution:
∵ T_n = a + (n – 1)d
Therefore, T_p = a + (p – 1)d = \(\frac{1}{q}\) (given) ………………. (i)
and a + (q – 1) = \(\frac{1}{p}\)(given) ……………….. (ii)
Subtracting Eq. (i) from Eq. (ii),
d(p – 1 – q + 1) = \(\frac{1}{q}\) – \(\frac{1}{p}\)
⇒ d(p – q) = \(\frac{p-q}{p q}\) ⇒ d = \(\frac{1}{p q}\)
Putting the value of d in Eq. (i) we get

Commonly Made Error
Sometimes students do mistake while finding first term and common difference.

Answering Tip
With an arithmetic sequence, everything depends on the value of ‘a! and ‘d’. Once you have these values you can find anything.

Section – D (15 Marks)

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

Question 34.
Experts say that the baby boom generation (Indians born between 1946 and 1960) cannot count on a company pension or social security to provide comfortable retirement, as their parents did. It is recommended that they start to save early and regularly. Mahesh, a baby boomer, has deposit ₹ 200 each month for 20 years in an account that pays interest of 7.2% compounded monthly. [5]
(a) How much will be in the account at the end of 20 years ?
(b) Mahesh believes he needs to accumulate ₹ 130,000 in the 20 years period to have enough for retirement, if he can not get higher rate to produce ₹ 130,000 in 20 years. To meet the goal, he must increase her monthly payment. What payment should he make each month ? Given that (1.006)²⁴⁰ = 4.2026.
OR
Compute the tax liability of Mr. A (aged 42), having total income of ₹ 51 lakhs for the Assessment Year 2021-22. Assume that his total income comprises of salary income. Income from house property and interest on fixed deposit.
Solution:
(i)The saving plan is an annuity with periodic payments PMT = 200, i = 7.2% monthly = \(\frac{7.2}{100} \times \frac{1}{12}\) annually = 0.006 and n = 12(20) = 240
The future value is

He will accumulate ₹ 106,252.47 in the account at the end of 20 years.

(ii) Mahesh’s goal is to accumulate ₹ 130,000 in 20 years at 7.2% compounded monthly. Therefore the future value is F.V. = ₹ 130,000, the monthly interest rate 0.072 is i = \(\frac{0.072}{12}\) = 0.006 and the number of periods is n = 12(20) = 240.
Using the sinking fund payment formula to find the payment PMT.
PMT = \(\frac{i \times \mathrm{F} . \mathrm{V}}{(1+i)^n-1}\)
= \(\frac{0.006 \times 130,000}{(1+0.006)^{240}-1}\)
= \(\frac{780}{4.2026-1}\)
= \(\frac{780}{3.2026}\)
= ₹ 243.5521
Mahesh will need payments of ₹243.55 each month for 20 years to accumulate at atleast ₹ 130,000.

Computation of tax liability of Mr. A for the A.Y. 2021-22
(a) Tax payable including surcharge on total income of ₹ 51,00,000
₹ 2,50,000 – ₹ 5,00,000 @5% ₹ 12,500
₹ 5,00,000 – ₹ 10,00,000 @20% ₹ 1,00,000
₹ 10,00,000 – ₹ 51,00,000 @30% ₹ 12,30,000
Total ₹ 13,42,500
Add: Surcharge @ 10% ₹ 1,34,250, ₹ 14,76,750
(b) Tax Payable on total income of ₹ 50 lakhs (₹ 12,500 + ₹ 1,00,000 + ₹ 12,00,000) ₹ 13,12,500
(c) Excess tax payable (a) – (b) ₹ 1,64,250
(d) Marginal Relief (₹ 1,64,250 – ₹ 1,00,000, being the amount of income in excess of ₹ 50,00,000) ₹ 64,250
(e) Tax payable (a)-(d) ₹ 14,12,500

Question 35.
Find spearman’s coefficient of rank correlation between marks obtained in History and Geography. [5]

Student	A	B	C	D	E	F	G	H	I	J
History	30	20	40	50	30	20	30	50	10	0
Geography	15	40	40	45	20	30	15	50	20	10

OR
Draw the graph of the function

Also, find its range.
Solution:
We construct the following table (giving rank 1 to the lowest):

There are 3 ties in ranks R₁ – two ties of 2 items and one tie of 3 items. Also, there are 3 ties in ranks R₂ – each of two items.

Here, f(x) = 1 + 2x, x < 0, this gives
f(-4) = 1 + 2(-4) = -7
f(-3) = 1 + 2(-3) = -5
f(- 2) = 1 + 2(- 2) = – 3
f(- 1) = 1 + 2(- 1) = – 1
f(x) = 3 + 5x, x ≥ 0
f(0) = 3 + 5(0) = 3
f(1) = 3 + 5(1) = 8
f(2) = 3 + 5(2) = 13
f(3) = 3 + 5(3) = 18
f(4) = 3 + 5(4) = 23
Now the graph of f is as shown in following figure.

Range: Let y₁ =f(x), x < 0
∴ y₁ = 1 + 2x, x < 0
∴ x = \(\frac{y_1-1}{2}\), x < 0
∵ x < 0 ⇒ y₁ – 1 < 0 = y₁ < 1
Let y₂ = f(x), x ≥ O
∴ y₂ = 3 + 5x, x ≥ 0
⇒ x = \(\frac{y_2-3}{5}\), x ≥ 0
∵ x ≥ 0 ⇒ y₂ – 3 ≥ 0 ⇒ y₂ ≥ 3
Therefore, range of f(-∞, 1), ∪ [3, ∞).

Commonly Made Error
Sometimes students get confused between the method of finding domain and range of a function.

Answering Tip
Rules of plotting graphs should be understand thoroughly.

Question 36.
A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹ 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. [5]
Solution:
(i) When no head and 4 tails appear. Let A be the event money lost = ₹ (4 × 1.50) = ₹ 6.00.
There is only one way of getting no head and 4 tails i.e, (TTTT) ⇒ n(A) = 1
n(S) = 16, since there are 16 possible outcomes
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{1}{16}\)

(ii) Let B be the event when 1 head and 3 tails appear.
∴ B = {HTTT, THTT, TTHT, TTTH}
⇒ n(B) = 4
Money lost = ₹ (3 × 1.50 – 1 × 1) = ₹ 3.50
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)

(iii) Let C be the event that 2 head and 2 tail appear.
∴ Money lost = ₹ (2 × 1.50 – 2 × 1)
= × 1
C = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
⇒ n(C) = 6
∴ P(C) = \(\frac{n(C)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)

(iv) Let D be the event that 3 head and 1 tail appear.
∴ D = {HHHT, HTHH, THHH, HHTH}
⇒ n(D) = 4
Money gained = ₹ (3 × 1 – 1 × 1.5) = ₹ 1.50
P(E) = \(\frac{n(D)}{n(S)}=\frac{4}{16}=\frac{1}{4}\)

(v) Let E be the event that all heads appear.
∴ E = {HHHH} ⇒ n(E) = 1
Money gained = ₹ (4 × 1) = ₹ 4
Also, P(E) = \(\frac{n(E)}{n(S)}=\frac{1}{16}\)

Section – E (8 Marks)

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

Question 37.
Data of all the previous cricket matches are stored to analyze the average batting score of various batsmen. The scores of a batsman in ten innings are: [4]
38, 70, 48, 34, 42, 55, 63, 46, 54, 44

On the basis of above information, answer the following questions:
(a) The median of the data is
(A) 48
(B) 46
(C) 47
(D) 51
Solution:
Option (C) is correct.

Explanation:
Arranging the data in ascending order
= 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Median= A.M. of 5th and 6th observation
= \(\frac{46+48}{2}\) = 47

(b) Mean of the given data is .
(A) 51.2
(B) 49.4
(C) 46.5
(D) 43.5
Solution:
Option (B) is correct.

Explanation:
Sum of all the scores = 494
Mean = \(\frac{494}{10}\) = 49.4

(c) What is the mean deviation about the median of the given scores?
(A) 7.6
(B) 8.6
(C) 9.2
(D) 6. 9
Solution:
Option (B) is correct.

Explanation:

x_i	\|d_i\| = \|x_i – 47\|
38	9
70	23
48	1
34	13
42	5
55	8
63	16
46	1
54	1
44	3
Total	∑\|d_i\| = 86

Mean Deviation = \(\frac{1}{n} \times \sum\left|d_i\right|=\frac{86}{10}\) = 8.6

(d) If the scores 38 and 34 are replaced by 68 and 74 what will be the mean of the data ?
(A) 52.8
(B) 45.8
(C) 58.2
(D) 56.4
Solution:
Option (D) is correct.

Explanation:
Sum of new scores = 564
New mean = \(\frac{564}{10}\) = 56.4

Question 38.
One urn contains two black balls (labelled B₁ and B₂) and one white ball. A second urn contains one black ball and two white balls (labelled W₁ and W₂). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen fro the urn. Then, a second ball is chosen at random from the same urn without replacing the first ball. [4]

Based on the above information answer the following questions :
(a) The total number of possible outcomes of the experiment is:
(A) 12
(B) 10
(C) 8
(D) 14
Solution:
Option (A) is correct.

Explanation:
It is given that one of the two urn is chosen, then a ball is randomly chosen from the urn, then a second ball is chosen at random from the same urn without replacing the first ball.
∴ Sample space S = {B₁B₂, B₁W, B₂B₁, B₂W, WB₁, WB₁, BW₂, B₁W, W₁B, W₁W₂, W₂B, W₂W₁}
∴ Total number of possible outcomes = 12

(b) The probability that two black balls are chosen is:
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
Solution:
Option (B) is correct.

Explanation:
If two black balls are chosen.
So, the favourable events are B₁B₂, B₂B₁, i.e., 2
∴ Required probability = \(\frac{2}{12}=\frac{1}{6}\)

(c) The probability that two white balls are chosen is:
(A) \(\frac{1}{2}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{5}{6}\)
Solution:
Option (B) is correct.

Explanation:
If two white balls are chosen.
So, the favourable events are W₁W₂, W₂W₁, i.e., 2
∴ Required probability = \(\frac{2}{12}=\frac{1}{6}\)

(d) The number of favourable events of the experiment, when two balls of opposite colours are chosen is:
(A) 12
(B) 10
(C) 8
(D) 14
Solution:
Option (C) is correct.

Explanation:
If two balls of opposite colours are chosen
So, the favourable events are (B₁, W), (B₂, W) WB₁, (WB₂), (BW₁), (BW₂), (W₁B₁), (W₂B) i.e, 8.