CBSE Sample Papers for Class 11 Applied Mathematics Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Applied Mathematics Set 5 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.
Find the input hexadecimal representation of 1110. [1]
Solution:
In hexadecimal number 1110 = 14, which is represented by the alphabet E.

Question 2.
Simplify: 2^1/2.4^3/4 [1]
OR
Find the value of log \(\left(\frac{1}{81}\right)\) to the base 9.
Solution:
2^{\(\frac{1}{2}\)} . 4^{\(\frac{3}{4}\)} = 2^{\(\frac{1}{2}\)} . (2²)^{\(\frac{3}{4}\)}
= 2^{\(\frac{1}{2}\)} . 2^{2 × \(\frac{3}{4}\)} [Using (a^m)ⁿ = a^mn]
= 2^{\(\frac{1}{2}\)} . 2^{\(\frac{3}{2}\)}
= 2^{\(\frac{1}{2}\)+\(\frac{3}{2}\)} = 2^{\(\frac{4}{2}\)} = 2² = 4
[Using a^m × aⁿ = a^m+n]

OR

log₉ \(\left(\frac{1}{81}\right)\) = log₉ log \(\left(\frac{1}{9^2}\right)\) = log₉(9)^-2
= -2 log₉ 9 = -2 × 1 = -2
[Applying rules log_a (m)ⁿ = n log_n m and log_a a = 1]

Question 3.
Mr. A is running a business from 1993 onwards. Determine the previous year for the assessment year 2020-21. [1]
Solution:
The previous year will be 1.4.2019 to 31.3.2020.

Question 4.
Why r is preferred to covariance as a measure of association ? [1]
Solution:
Although correlation coefficient is similar to the covariance in a manner that both measure the degree of linear relationship between two variables, but the former is generally preferred to covariance due to the following reasons :
(a) The value of the correlation coefficient (r) lies between -1 and 1 i.e., -1 ≤ r ≤ 1
(b) The correlation coefficient is scale free.

Question 5.
A gentleman has 6 friends to invite. In how many ways can he send invitation card to them if he has three servants to carry the cards ? [1]
OR
Find the number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men.
Solution:
Given, that a gentleman has 6 friends to invite but he send invitations to them if he has 3 servants.
He can send invitations to one of his friends in 3 different ways,
Similarly, he send invitations to all 6 friends if he has 3 servants = 3 × 3 × 3 × 3 × 3 × 3 = 36
= 729.

OR

Given,
Number of men = 4
Number of women = 6
It is given that committee includes at least two men and exactly twice as many women as men.
So, we can select either 2 men and 4 women or 3 men and 6 women.
∴ Total number of committees formed
= ⁴C₂ × ⁶C₄ + ⁴C₃ × ⁶C₆
= 6 × 15 + 4 × 1
= 94

Question 6.
Find the minimum value of the expression 3^x + 3^1-x, x ∈ R. [1]
OR
Find the value of 5^{\(\frac{1}{2}\)}.5^{\(\frac{1}{4}\)}.5^{\(\frac{1}{8}\)} up to infinity is.
Solution:
We know that, A.M. ≥ G.M. for positive numbers.

Question 7.
If f(x) = x³ – \(\frac{1}{x^3}\) then find value of f(x) + f\(\left(\frac{1}{x}\right)\).
Solution:

Question 8.
If the function f(x) = is given to be continuous at x = 1, then the value of k.
Solution:

Question 9.
From the words, Discernment, Perception, Penetration and Insinuation, which is odd one out? [1]
Solution:
Discernment means the ability of judge something will.
Perception means an opinion or the ability to understand something.
Penetration means having the power to enter or pierce deeply.
Insinuation means to introduce or work in; to suggest, without being direct, that something unpleasant is true.
Hence, the correct answer is “Penetration”.

Question 10.
For the data : 5, 24, 36,12, 20 and 8, find value of D_s. [1]
Solution:
Arranging the given data in ascending order, we get 5, 8, 12, 20, 24, 36
D₅ = \(\frac{5(6+1)}{10}\) th value
= 3.5^th value
= 3^rd value + \(\frac{1}{2}\) (4^th – 3^rd value)
= 12 + \(\frac{1}{2}\) (20 – 12)
= 12 + 4 = 16

Question 11.
Find the least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled ? [1]
Solution:
According to question,
P(1 + \(\frac{20}{100}\))ⁿ > 2P
⇒ \(\left(\frac{6}{5}\right)^n\) > 2
Now, \(\left(\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}\right)\) > 2
So, n = 4 years

Question 12.
What does the P stands for in this formula ? [1]
Present value = P\(\left[\frac{1-(1+i)^{-n}}{i}\right]\)
Solution:
Present value = cash flow × \(\frac{\left[(1+i)^n-1\right]}{i(1+i)^n}\)

Question 13.
A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x ? [1]
OR
The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of ₹ 800 per sq. metre.
Solution:
Clearly, We have
\(\frac{3+11+7+9+15+13+8+19+17+21+14+x}{12}\) = 12
or, 137 + x = 144
⇒ x = 144 – 137 = 7

OR
Area of floor = (5.5 × 3.75) m²
= 20.625 m²
Cost of paving = ₹ (800 × 20.625)²
= ₹ 16,500.

Question 14.
Find the equation of the line for which p = 2, sin α = \(\frac{4}{5}\) [1]
Solution:
Equation of a line in normal form is
x cos α + y sin α = p
Given, p = 2, sin α = \(\frac{4}{5}\)
⇒ cos α = \(\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}\) [∵ sin²θ + cos²θ = 1]
⇒ x . \(\frac{3}{5}\) + y . \(\frac{4}{5}\) = 2
⇒ 3x + 4y – 10 = 0, is the required equation of the line.

Question 15.
Find the parametric equations of the circle x² + y² – 2x + 4y – 4 = 0. [1]
Solution:
We know that, the parametric equations of the circle
(x – h)² + (y – k)² = r² are
x = h + r cos t and y = k + r sin t, 0 ≤ t ≤ 2π
The given equation of circle is
x² + y² – 2x + 4y – 4 = 0
It can be written as
(x² – 2x + 1) + (y² + 4y + 4) = 1 + 4 + 4
or, (x – 1)² + (y + 2)² = 32, which is comparable with
(x – h)² + (y – k)² = r².
Here, h = 1, k = – 2 and r = 3
Therefore, parametric equations of the given circle are :
x = 1 + 3 cos t and y = -2 + 3 sin t, 0 ≤ t ≤ 2π

Question 16.
If A = {0, {0, 1}}, then cardinal number of set P(A) is. [1]
OR
Let A = {3, 9, 27, 81}, then the set builder form of set A.
Solution:
P(A) = {Φ, {0}, {{0, 1}}, {0, {0, 1}}}
Therefore, n (P(A)) = 4

OR

In set builder form :
B = {x : x = 3ⁿ, n ∈ N and 1 ≤ n ≤ 4}

Section – B (20 Marks)

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

Question 17.
Let f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x – 7. [2]
(a) For what real numbers x,f(x) = g(x) ?
(b) For what real numbers x,f(x) < g(x) ?
OR
Find the domain and range of the real functions
f(x) = \(\frac{|x-1|}{x-1}\)
Solution:
Given, f(x) = 2x + 1 and g(x) = 4x – 7
(a) ∵ f(x) = g(x)
∴ 2x + 1 = 4x – 7
2x = 8
x = 4

(b) ∵ f(x) < g(x)
∴ 2x + 1 < 4x – 7
2x – 4x + 1 < 4x – 7 – 4x
– 2x + 1 < – 7
– 2x < – 7 – 1
– 2x < – 8 \(\frac{-2 x}{-2}\) > \(\frac{-8}{-2}\)
x > 4.

OR

Given, f(x) = \(\frac{|x-1|}{x-1}\)
Domain : Clearly, f(x) is defined for all x → R except x = 1
∴ Domain of f = R – {1}
Range : Now, f(x) = \(\frac{x-1}{x-1}\) = 1, when x > 1
and f(x) = \(\frac{-(x-1)}{x-1}\) = – 1, when x < 1
∴ Range of f = {-1, 1]

Question 18.
Evaluate the left hand and right hand limits of the following function at x = 2. Does \(\lim _{x \rightarrow 2}\) f(x) exist ? [2]
f(x) = 2x + 3, if x ≤2 x + 5, if x > 2
OR
If f(x) = \(\frac{x-4}{2 \sqrt{x}}\) then find the value of f'(1).
Solution:
L.H.L = \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{-}}\) (2x + 3)
= 4 + 3
= 7
and R.H.L. = \(\lim _{x \rightarrow 2^{+}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) (x + 5)
= 2 + 5
= 7
Since \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 7
∴ \(\lim _{x \rightarrow 2}\) f(x) exists and is equal to 7.

Commonly Made Error
Several students make errors because of confusion between left hand limit and right hand limit.

Answering Tip
Understand the difference between left hand limit and right hand limit and its method to solve with appropriate sign convention.

OR

Question 19.
Find the mean deviation about median and coefficient of deviation for the following series : Marks : 17,18, 25, 20, 26, 28, 38, 30, 22 [2]
Solution:
Arranging the marks in ascending order, the obtained is :
17, 18, 20, 22, 25, 26, 28, 30, 38
Here, number of terms n = 9 is odd, so that
Median, M_d = Values of \(\left(\frac{9+1}{2}\right)^{\text {th }}\) term
= Value of 5^th term
= 25
Hence, Mean deviation

Also, coefficient of deviation
= \(\frac{\text { Mean deviation }}{\text { Median }}\) × 100
= \(\frac{5}{25}\) × 100 = 20

Question 20.
For some symmetrical distribution, Q₁ = 36 and Q₃ = 63. Using Bowely’s measures of skewness, find the median. [2]
Solution:
For symmetrical distribution, skewness = 0
Given, Q₁ = 36 and Q₃ = 63
Since, S_k = \(\frac{Q_3+Q_1-2 Q_2}{Q_3-Q_1}\)
⇒ 0 = \(\frac{63+36-2 Q_2}{63-36}\)
⇒ 2Q₂ = 99
⇒ Q₂ = \(\frac{99}{2}\) = 49.5
Q₂ = median = 49.5

Question 21.
₹ 1000 is invested every 3-months at 4.8% compounded quarterly. How much will the annuity be worth in 2 years ? [1]
OR
A dealer is in Jhansi buys some articles worth ₹ 8,000. If the rate of GST is 18%, find how much will the dealer pay for the articles bought.
Solution:

Hence, in two years the annuity worth will be ₹ 8344.18

OR

In a case of Intra-state transaction,
Cost of an article = ₹ 8,000
and GST rate = 18%
∴ CGST = 9% of ₹ 8,000
= ₹ \(\frac{9}{100}\) × 8,000
= ₹ 720
SGST = 9% of ₹ 8,000
= ₹ \(\frac{9}{100}\) × 8,000
= ₹ 720
and IGST = ₹ 0
∴ The dealer paid = ₹ (8000 + 720 + 720 + 0)
= ₹ 9440

Question 22.
The cartesian product A × A has 9 elements among which are found (-1, 0) and (0, 1). Find the set A and the remaining elements of A × A. [2]
Solution:
Given, (-1, 0) ∈ A × A and (0, 1) ∈ A × A
⇒ A = {- 1, 0, 1}
∴ A × A = {- 1, 0, 1} × {- 1, 0, 1}
= {(- 1, – 1), (- 1, 0), (- 1, 1), (0, – 1), (0, 0), (0, 1), (1, – 1), (1, 0), (1, 1)}
Thus, remaining elements are {(- 1, – 1), (- 1, 1) (0, – 1), (0, 0), (1, 0), (1, – 1), (1, 1)}

Question 23.
(a) In a certain code language $ # * means ‘Shirt is clean’, @ D # means ‘clean and neat’ and @ ? means ‘neat boy’, then what is the code for ‘and’ in this language ? [2]
(b) If A stands for ‘+’ B stand for C stands for ‘x’, then what is the value of (10 C 4) (A) (4 C 4) B6 ?
Solution:
(a) Given code statements are :
(I) $ # * ‘Shirt is clean’
(II) @ D # ‘Clean and neat’
(III) @ ? ‘neat boy’
Here, we can observe that from statements I and II, the word ‘clean’ is common and common code for clean is ‘#’. Similarly from statement II and III, the common word is ‘neat’ and common code for it is ‘@’. Now, from statement II, we can conclude that code for ‘and’ is ‘D .

(b) Given, codes for letter are :
A → + (addition)
B → – (subtraction)
C → × (multiplication)
Hence,(10 C 4) (A) (4 C 4) B6
= (10 × 4) + (4 × 4) – 6
= 40 + 16 – 6
= 56 – 6
= 50
Thus, value of (10 C 4) (A) (4 C 4) B6 is 50.

Question 24.
For any set A and B, show that (A – B) = (A ∩ B’) [2]
Solution:
Let x ∈ (A – B), then
x ∈ A and x ∉ B
x ∈ A and x ∈ B’
x ∈ A ∩ B’
(A – B) ⊆ A ∩ B
Again, Let y ∈ (A ∩ B’), then
y ∈ (A ∩ B’)
y ∈ A and y ∈ B’
y ∈ A and y ∉ B
y ∈ (A – B)
∴ (A ∩ B’) ⊆ (A – B)
From (i) and (ii), we have (A – B) = (A ∩ B’)

Question 25.
A clock is set right at 5 a.m, The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4^th day ? [2]
Solution:
Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours
Now, 23 hrs 44 min of this clock
= 24 hours of correct clock
∴ \(\frac{356}{15}\) hrs of this clock = 24 hrs of correct clock
89 hrs of this clock clock = (24 × \(\frac{15}{356}\) × 89) hrs of correct clock
= 90 hrs of correct clock
So, the correct time is 11 p.m.

Question 26.
From a class of 40 students, in how many ways can five students be chosen for an excursion party. [2]
Solution:
From a class of 40 students, five students can be chosen for an excursion party in ⁴⁰C₅ ways
= \(\frac{40 !}{5 !(40-5) !}\)
= \(\frac{40 !}{5 ! 35 !}\)
= \(\frac{40.39 \cdot 38 \cdot 37 \cdot 36 \cdot 35 !}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 35 !}\)
= 6,58,008

Section – C (21 Marks)

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

Question 27.
Find the equation of the circle which passes through (1, -2) and (4, – 3) and whose centre lies on the line 3x + 4y = 7. [3]
OR
Prove that the product of the lengths of the perpendicular drawn from the points (\(\sqrt{a^2-b^2}\), 0) and (-\(\sqrt{a^2-b^2}\), 0)
to the line \(\frac{\mathrm{x}}{\mathrm{a}}\) cos0 + \(\frac{\mathrm{y}}{\mathrm{b}}\) sin 9 = 1 is b².
Solution:
Let the required equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ……………. (i)
Since, the circle passes through the points (1, -2) and (4, – 3),
we have 1 + 4 + 2g – 4f + c = 0
or, 2g – 4f + c = – 5 ……………… (ii)
and 16 + 9 + 8g – 6f + c = 0
⇒ 8g – 6f + c = – 25 …………(iii)
Subtracting eq. (iii) from eq. (ii), we get
– 6g + 2f = 20
⇒ 3g – f = – 10 …………….. (iv)
Again, the centre of the circle (- g, – f) lies on the line
3x + 4y = 7
Therefore, – 3g – 4f = 7 ………….. (v)
Solving eq. (iv) and eq(v), we get
⇒ f = \(\frac{3}{5}\) and g = –\(\frac{47}{15}\)
Now, from eq. (ii), we get
\(\frac{-94}{15}\) – \(\frac{12}{5}\) + c = -5
or, c = \(\frac{55}{15}\)
Substituting the values of g, f and c in eq. (i), the required equation of the circle is :
or, 15x² + 15y² – 94x + 18y + 55 = 0.

OR

Let P₁ and P₂ be the perpendicular distances of (\(\sqrt{a^2-b^2}\),0) and (-\(\sqrt{a^2-b^2}\),0) from \(\frac{\mathrm{x}}{\mathrm{a}}\) cos θ + \(\frac{\mathrm{y}}{\mathrm{b}}\) sin θ = 1, then we have
P₁ = \(\left|\frac{\frac{\sqrt{a^2-b^2}}{a} \cos \theta-1}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}\right|\), P₂ = \(\left|\frac{\frac{-\sqrt{a^2-b^2}}{a} \cos \theta-1}{\sqrt{\frac{\cos ^2 \theta}{a^2}+\frac{\sin ^2 \theta}{b^2}}}\right|\)


= b²
Hence Proved

Question 28.
Two cyclists start from the same place to ride in the same direction. A starts at noon at 8 km/hr and B at 1.30 p.m. at 10 km/h. How far will A have ridden before he is overtaken by B ? Find also at what times A and B will be 5 km apart. [3]
OR
Two cyclists start from the same place to ride in the same direction. A starts at noon at 8 km/hr and B at 1.30 p.m. at 10 km/h. How far will A have ridden before he is overtaken by B ? Find also at what times A and B will be 5 km apart.
Solution:
If A rides for X hours before he is overtaken, then B rides for (X – 1.5) hrs
⇒ 8X = 10 (X – 1.5)
⇒ X = 7.5
⇒ A will have ridden 8 X 7.5 km or 60 km. [1]
For the second part, if the required number of hours after noon = Y, then
8Y = 10(Y – 1.5) ± 5
⇒ Y = 10 or 5 according as B is ahead or behind A
⇒ The required times are 5 p.m. and 10 p.m. [1]

OR

If A rides for X hours before he is overtaken, then B rides for (X – 1.5) hrs
⇒ 8X = 10 (X – 1.5)
⇒ X = 7.5
⇒ A will have ridden 8 X 7.5 km or 60 km.
For the second part, if the required number of hours after noon = Y, then
8X = 10(X – 1.5) ± 5
⇒ X = 10 or 5 according as B is ahead or behind A
⇒ The required times are 5 p.m. and 10 p.m.

Question 29.
Let I be the set of all integers. A relation R on I, such that xRy holds iff (x – y) is divisible by 5, x ∈ I, y ∈ I, i.e., R ={(x, y): x ∈ I, y ∈ I, x – y is divisible by 5}. Prove that it is an equivalence relation. [3]
Solution:
Here we observe that,
(a) For each x ∈ I, x – x = 0 and 0 is divisible by 5. Thus, ∀ x ∈ I, we have x R x. Therefore, R is reflexive.

(b) Suppose x R y; then x – y is divisible by 5 and hence (y – x) = – ( x – y) is also divisible by 5. Thus, x R y ⇒ y R x. Therefore, R is symmetric.

(c) Suppose x R y and y R z; then (x – y) and (y – z) are both divisible by 5. Hence 5 is also a divisor of (x – y) + (y – z), i.e., 5 is also a divisor of (x – z). Thus, x R y and y R z
⇒ x R z.
Therefore, R is transitive.
Since, R is reflexive, symmetric and transitive, therefore R is an equivalence

Question 30.
Find the sum of the series [3]
1 + (1 + x) + (1 + x + x²) + (1 + x + x² + x³) + ……
Solution:
Let
S_n = 1 + (1 + x) + (1 + x + x²) + (1 + x + x² + x³) + …… to n terms
Hence, (1 – x)S_n = (1 – x) + (1 – x) (1 + x) + (1 – x) (1 + x + x²) + (1 – x) (1 + x + x² + x³) + … to n terms
or, (1 – x)S_n = (1 – x) + (1 – x²) + (1 – x³) … to n terms
= n – (x + x ²+ x³ + x⁴ + …) to n terms
= n – \(\frac{x\left(1-x^n\right)}{(1-x)}\)
Here, S_n = \(\frac{n}{1-x}-\frac{x\left(1-x^n\right)}{(1-x)^2}\)

Question 31.
In the following table some capital alphabets are written in a row, below them their coding have been given. Now in questions a particular word has been coded in a particular manner using codes as given below the capital letter. You have to understand the pattern of coding and answer the following questions. [3]

A

B

c

D

E

F

G

H

I

J

K

L

M

N

o

P

Q

R

S

T

u

V

W

X

Y

Z

u

a

2

V

b

w

3

t

4

x

s

y

5

z

6

c

d

8

7

e

r

h

9

i

P

q

(a) If MAIDEN is coded as yux2b6, then what is the code for FIBRES?
(b) If MAY is coded as yzqaiq, then what is the code for TIE ? ‘
(c) If LONDON is coded as 5c62z5, then what is the code for EUROPE ?
Solution:
(a) In the word MAIDEN, for the 2^nd and 5^th letter, codes have been used from just below the letters, for the 1st and 4th letter, codes have been used from one position back and for 3rd and 6th letter codes have been used one position ahead as their position in table.
Therefore, the word FIBRES can be coded as b42dbe.

(b) For the each of the letters of the word MAY, adjacent codes have been used. Therefore, the word TIE can be coded as 7rtxvw.

(c) In word LONDON, first three letters have been coded one position in forward direction and last three position in backward direction. Therefore, word EUROPE can be coded as wh7z6v.

Question 32.
Find the quartile deviation for the marks obtained by 10 students : [3]
56, 48, 65, 35, 42, 75, 82, 60, 55
Solution:
After arranging the marks in an ascending order of magnitude, we get
35, 42, 48, 50, 55, 56, 60, 65, 75, 82
Firt quartile (Q₁) = \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= \(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= 2.75^th observation
= 2^nd observation + 0.75 × (difference between the third and the 2^nd observation)
= 42 + 0.75 × (48 – 42)
= 46.50
Third quartile (Q₃) = \(\frac{3(n+1)}{4}\) th observation
= \(3\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= 8.25^th observation
= 65 + 0.25 × (difference between 9^th and 8^th observation)
= 65 + 0.25 × 10
= 65 + 2.5 = 67.50
Quartile Deviation = \(\frac{Q_3-Q_1}{2}\)
= \(\frac{67.50-46.50}{2}\)
= 10.50

Question 33.
Calculate the electricity bill amount for a month of April, if 4 bulbs of 40 W for 5 h, 4 tube lights of 60 W for 5 h, a TV of 100 W for 6 h, a washing machine of 400 W for 3 h are used per day. The cost per unit is ₹ 1.80. [3]
Solution:
Electric energy consumed per day by 4 bulbs = 4 × 40 × 5 = 800 W
Electric energy consumed per day by 4 tube lights = 4 × 60 × 5 = 1200 W
Electric energy consumed per day by TV = 100 × 6 = 600 W
Electric energy consumed per day by washing machine = 400 × 3 =1200 W
∴ Total electric energy consumed by all electric appliances = (800+ 1200 + 600 + 1200) W
= 3800 W
= 3.8 kW = 3.8 units
Total electric energy consumed in the month of April (30 days) = 3.8 × 30 = 114 units
Cost of one unit = ₹ 1.80
Cost of 114 units = 114 × ₹ 1.80 = ₹ 205.20
∴ Electricity bill amount = ₹ 205.20

Section – D (15 Marks)

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

Question 34.
In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves into another section. On a given day, one of the three people A, B and C carries out this task. A has 45% chance, B has 35% chance and C has 20% chance of doing the task. The probability that A, B and C will take more than the allotted time is \(\frac{1}{6}\), \(\frac{1}{10}\) and \(\frac{1}{20}\) respectively- If it is found that the time taken is more than the allotted time, what is the probability that A has done the task ? [5]
Solution:
Let E₁, E₂ and E₃ denote the events of carrying out the task by A, B and C, respectively.
Let H denote the event of taking more time.
Then, P(E₁) = 0.45, P(E₂) = 0.35 and P(E₃) = 0.20
Also,

Commonly Made Error
Many students do Bayes’ Theorem correctly but took the probabilities of A, B, C as 45, 35, and 20, instead of percentages. Some candidates do not ] implement the theorem correctly.

Answering Tip
Be clear that probabilities are ratios and not numbers. Learn conditional probability concept which helps the student to understand the advance concept of Bayes’ Theorem.

Question 35.
Differentiate x^sinx + (sin x)^{cos x} with respect to x. [5]
OR
Examine whether there is any correlation between age and blindness on the basis of the following data :

Age in years	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of Persons : (in thousands)	90	120	140	100	80	60	40	20
No. of blind persons :	10	15	18	20	15	12	10	06

Solution:
Let y = x^{sin x} + (sin x)^{cos x}
= u + v
Let u = x^{sin x} and v = (sin x)^{cos x}
log u = sin x log x
and log v = cos x log sin x
Differentiate with respect to x,

OR

Let us denote the mid-value of age in years as x and the number of blind persons per lakh as y. Then, to compute coefficient of correlation between x and y, we construct the following table :

The correlation coefficient between age and blindness is given by

= \(\frac{1090}{1136.268}\)
= 0.959 ≈ 0.96
which exhibits a very high degree of positive correlation between age and blindness.

Commonly Made Error
Some students made mistake while doing arithmetic calculations.

Answering Tip
Since, the calculation of r is sometimes tedious, so take precaution while simplification and also recheck to avoid mistakes.

Question 36.
Compute the tax liability of Mr. A (aged 42), having total income of ₹ 1,01,00,000 for AY 2019-20. [5]
OR
A person opened an account on April, 2019 with a deposit of ₹ 800. The account paid 6% interest compounded quarterly. On October 1, 2019 he closed the account and added enough additional money to invest in a 6 month time-deposit for ₹ 1000, earning 6% compounded monthly.
(a) How much addition amount did the person invest on October 1 ?
(b) What was the maturity value of his time deposit on April 1 2002 ?
(c) How much total interest was earned ?
Given that (i + if is 1.03022500 for i = 1 \(\frac{1}{2}\), n = 2 and (1 + i)ⁿ; is 1.03037751 for i = \(\frac{1}{2}\) %, n = 6.
Solution:
Computation of tax liability of Mr. A for AY 2021-22.

1. Tax payable including surcharge on total income of ₹ 1,01,00,000 (Here the total income exceeds ₹ 10,00,000 then tax rate is 1,12,500 +30% of the amount by which the total income exceeds ₹ 10,00,000 Therefore, ₹ 1,12,500 + ₹ (1,01,00,000 – 10,00000) × 30% = ₹ 1,12,500 + ₹ 91,00000 × 30% = ₹ 1,12,500 + ₹ 27,30,000 = ₹ 2,842,500 Add : Surcharge @ 15% ₹ 426,375)	₹ 32,68,875
2. Tax Payable on total income of ₹ 1 crore (Here the total income exceeds ‘ 10,00,000 then tax rate is 1,12,500 +30% of the amount by which the total income exceeds ₹ 10,00,000 Therefore, ₹ 1,12,500 + ₹ (1,00,00,000 – 10,00000) × 30% = ₹ 1,12,500 + ₹ 90,00000 × 30% = ₹ 1,12,500 + ₹ 27,00,000 = ₹ 2,812,500 Add : Surcharge @ 10% ₹ 281250)	₹ 30,93,750
3. Excess tax payable = [(1)-(2)] = ₹ 32,68,875 – ₹ 30,93,750	₹ 1,75,125
4. Marginal Relief = [(₹ 1,75,125 – ₹ 1,00,000) Income in excess of ₹ 50,00,000]	₹ 75,125

Tax Payable = ₹ 32,68,875 – ₹ 75,125 + 4 %HEC = ₹ 31,93,750 + 4%
HEC = 33,21,500.

OR

(a) The initial investment earned for April- June and July-September quarter i.e, two quarters.
In this case i = \(\frac{6}{4}\) = 1 \(\frac{1}{2}\) % = 0.015,
n = \(\frac{6}{12}\) × 4 = 2
and compounded amount
= 800 (1 + 0.015)²
= 800 × 1.03022500 = ₹ 824.18
The additional amount invested
= ₹ (1000 – 824.18) = ₹ 175.82

(b) In this case the time-deposit earned interest compounded monthly for six-month
Here, i = \(\frac{6}{12}\) = \(\frac{1}{2}\) % = 0.005,
n = 6
and P = ₹ 1000
Maturity value = 1000 (1 + 0.005)⁶
= 1000 × 1.03037751
= ₹ 1030.38

(c) Total interest earned
= ₹ (824.18 – 800) + (1030.38 – 1000)
= ₹ (24.18 + 30.38)
= ₹ 54.56

Section – E (8 Marks)

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

Question 37.
In a modern school construction work for five new classes is going on. The school committee has also decided to construct a small terrace at a football playground. This small terrace at football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. [4]
Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m.
(see figure below).

According to the above given information answer the following questions :

(a) The volume of the required concrete for step 1 is:
(A) \(\frac{25}{4}\)
(B) \(\frac{50}{4}\)
(C) \(\frac{75}{4}\)
(D) \(\frac{150}{4}\)
Solution:
Option (A) is correct.

Explanation:
For the 1^st step, the volume of required concrete is
\(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 = \(\frac{25}{4}\)

(b) The volume of the required concrete for step 2 is:
(A) \(\frac{25}{4}\)
(B) \(\frac{50}{4}\)
(C) \(\frac{75}{4}\)
(D) \(\frac{150}{4}\)
Solution:
Option (B) is correct.

Explanation:
For the 2nd step, the volume of required concrete is
\(\frac{2}{4}\) × \(\frac{1}{2}\) × 50 = \(\frac{50}{4}\)

(c) The volume of the required concrete for step 3 is:
(A) \(\frac{25}{4}\)
(B) \(\frac{50}{4}\)
(C) \(\frac{75}{4}\)
(D) \(\frac{150}{4}\)
Solution:
Option (C) is correct.

Explanation:
For the 3rd step, the volume of required concrete is
\(\frac{3}{4}\) × \(\frac{1}{2}\) × 50 = \(\frac{75}{4}\)

(d) The given problem is based on which mathematical concept ?
(A) AT
(B) GP
(C) Both AP and GP
(D) None of these
Solution:
Option (A) is correct.

Explanation:
The number involved in this case is in AP in which
a = \(\frac{25}{4}\), d = \(\frac{25}{4}\), n = 15

Question 38.
Mr, Roy thew a party for his new venture. The party was in club. His friends were enjoying the party. Some of his friends were playing cards. A man named Ramesh draw a card from a well-shuffled deck of cards. [4]

According to the above given information answer the following questions :
(a) The probability of getting a kind of red colour is:
(A) \(\frac{1}{26}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{3}{26}\)
(D) \(\frac{1}{52}\)
Solution:
Option (A) is correct.

Explanation:
Since, total number of possible outcomes = 52 are two suits of red cards, i.e., diamond and heart. Each suit contains one kind. Favourable outcomes = 2
Hence, P(a king of red colour) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(b) The probability of getting a face card is:
(A) \(\frac{1}{26}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{3}{26}\)
(D) \(\frac{1}{52}\)
Solution:
Option (B) is correct.

Explanation:
There are 12 face cards in a pack. Favourable outcomes = 12
Hence P(a face card) = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(c) The probability of getting a jack of heart is :
(A) \(\frac{1}{26}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Solution:
Option (C) is correct.

Explanation:
There is only one jack of heart.
Favourable outcome = 1
Hence, P(the jack of hearts) = \(\frac{1}{52}\)

(d) The probability of getting a spade card is:
(A) \(\frac{1}{26}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Solution:
Option (D) is correct.

Explanation:
There are 13 cards of spade.
Favourable outcomes = 13
Hence, P(a spade) = \(\frac{13}{52}\) = \(\frac{1}{4}\)