Article

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:

  • This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there is some internal choice in some questions.
  • Section A has 18 MCQs and 2 Assertion Reason-based questions of 1 mark each.
  • Section B has 5 Very Short Answer (VSA) questions of 2 marks each.
  • Section C has 6 Short Answer (SA) questions of 3 marks each.
  • Section D has 4 Long Answer (LA) questions of 5 marks each.
  • Section E has 3 source-based/case-based/passage-based/integrated units of assessment (4 marks each) with sub-parts.
  • Internal Choice is provided in 2 questions in Section B, 2 questions in Section C, and 2 Questions in Section D. You have to attempt only one alternative in all such questions.

Section-A

(All Questions are compulsory. No internal choice is provided in this section)

Question 1.
Which of the following is the correct representation of a binary number?
(a) (124)2
(b) 110
(c) (110)2
(d) (000)2
Answer:
(d) (000)2
Explanation:
The binary numbers should comprise only two digits 0 and 1. Also, for the base, the value should be 2 and it should be written as a subscript enclosing the entire number.
Hence, the fourth option i.e., (000)2 gives the correct representation.

Question 2.
January 1, 2008, was Tuesday. What day of the week lied on Jan 1, 2009.
(a) Monday
(b) Wednesday
(c) Thursday
(d) Sunday
Answer:
(c) Thursday
Explanation:
The year 2008 was a leap year. So, it had 2 odd days.
1st day of the year 2008 was Tuesday. (Given)
So, 1st day of the year 2009 was 2 days beyond Tuesday.
Hence, it would be Thursday.

Question 3.
A man can do a piece of work in 5 days, but with the help of his son, he can do it in 3 days. At what time can the son do it alone?
(a) 6\(\frac{1}{2}\) days
(b) 7 days
(c) 7\(\frac{1}{2}\) days
(d) 8 days
Answer:
(c) 7\(\frac{1}{2}\) days
Explanation:
Son’s 1 day’s work = \(\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2}{15}\)
∴ The son alone can do the work in \(\frac{15}{2}\) = 7\(\frac{1}{2}\) days.

Question 4.
The set {x : x ∈ R, -3 ≤ x ≤ 7} as an interval can be written as
(a) [-3, -7]
(b) [3, 7]
(c) (-3, 7)
(d) [-3, 7)
Answer:
(d) [-3, 7)
Explanation:
Because the set of all real numbers lying between a and b and including the number a is said to form a closed open interval, denoted by [a, b).
It is written as [a, b) = {x: x ∈ R, a ≤ x ≤ b}.

Question 5.
If the nth term of an A.P. is 6n – 7, then its 50th term is:
(a) 300
(b) 297
(c) 293
(d) None of these
Answer:
(c) 293
Explanation:
Given, Tn = 6n – 7
∴ T50 = 6(50) – 7
= 300 – 7
= 293

Question 6.
If the sum of the first n terms of an A.P. is 2n2 + 7n, then its nth term is:
(a) 2n + 7
(b) 4n + 7
(c) 2n + 5
(d) 4n + 5
Answer:
(d) 4n + 5
Explanation:
Given, Sn = 2n2 + 7n
We know that, an = Sn – Sn-1
∴ an = 2n2 + 7n – {2(n – 1)2 + 7(n – 1)}
= 2n2 + 7n – {2(n2 – 2n + 1) + 7n – 7}
= 2n2 + 7n – (2n2 – 4n + 2 + 7n – 7)
= 2n2 + 7n – (2n2 + 3n – 5)
= 2n2 + 7n – 2n2 – 3n + 5
= 4n + 5

Question 7.
If LINGER is 123456 and FORCE is 56789, then FIERCE is
(a) 345667
(b) 456678
(c) 345677
(d) 556789
Answer:
(b) 456678
Explanation:
L → 1 F → 5
I → 2 O → 6
N → 3 R → 7
G → 4 C → 8
E → 5 E → 9
R → 6
Letters R and E are common to LINGER and FORCE and numbers 6 and 5 are common to their respective coding. Therefore, 5 and 6 will stand for R and E but not respectively.
Now, since the word FIERCE has a double E, hence either 5 or 6 will appear twice in its coding.
Besides, the coding of FIERCE will have two numbers from the coding of FORCE for letters F and C, and one from the coding of LINGER for letter I.
But from the given option we will consider 4 among 1, 2, 3, 4. Hence, the coding of FIERCE will be 456678.

Question 8.
Domain of \(\sqrt{a^2-x^2}\) (a > 0) is
(a) (-a, a)
(b) [-a, a]
(c) [0, a]
(d) (-a, 0]
Answer:
(b) [-a, a]
Explanation:
Let f(x) = \(\sqrt{a^2-x^2}\); f(x) is defined
\(\sqrt{a^2-x^2}\) ≥ 0
⇒ x2 – a2 ≤ 0
⇒ x2 ≤ a2
⇒ x ≤ ±a
⇒ -a ≤ x ≤ a
∴ The domain of f(x) = [-a, a].

Question 9.
\(\lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x}\) is equal to:
(a) n
(b) 1
(c) -n
(d) 0
Answer:
(a) n
Explanation:

Question 10.
If A and B are events such that P(A) = 0.2, P(B) = 0.4 and P(A ∪ B) = 0.5, the value of P(\(\frac{A}{B}\)) is
(a) 0.1
(b) 0.25
(c) 0.5
(d) 0.08
Answer:
(b) 0.25
Explanation:
Given, P(A) = 0.2, P(B) = 0.4 and P(A ∪ B) = 0.5
We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∩ B) = 0.2 + 0.4 – 0.5
⇒ P(A ∩ B) = 0.1
Now, P(\(\frac{A}{B}\)) = \(\frac{P(A \cap B)}{P(B)}\)
= \(\frac{0.1}{0.4}\)
= 0.25

Question 11.
If A and B are two events associated with a random experiment such that P(A) = 0.3, P(B) = 0.2, and P(A ∩ B) = 0.1, then the value of P(A ∩ \(\bar{B}\)) is _________
(a) 0
(b) 0.2
(c) 0.5
(d) 1
Answer:
(b) 0.2
Explanation:
Given, P(A) = 0.3, P(B) = 0.2 and P(A ∩ B) = 0.1
We know that, P(A ∩ \(\bar{B}\)) = P(A) – P(A ∩ B)
⇒ P(A ∩ \(\bar{B}\)) = 0.3 – 0.1
⇒ P(A ∩ \(\bar{B}\)) = 0.2

Question 12.
The third quartile of the data set 33, 25, 42, 25, 31, 37, 46, 29, 38 is:
(a) 30
(b) 40
(c) 45
(d) 55
Answer:
(b) 40
Explanation:
On arranging the given data in ascending order, we get
25, 25, 29, 31, 33, 37, 38, 42, 46

Question 13.
Dispersion measures
(a) The scatteredness of a set of observations
(b) The concentration of a set of observations
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer:
(a) The scatteredness of a set of observations
Explanation:
The number that describes the variability or scatteredness of a set of observations, is called a measure of dispersion.

Question 14.
SGST is applicable when
(a) Goods are sold within a state
(b) Goods are sold from one GST dealer to a customer
(c) Goods are sold by a GST dealer to another GST dealer
(d) Interstate supply
Answer:
(a) Goods are sold within a state
Explanation:
Integrated Goods and Services Tax is applicable when goods are sold within a state.

Question 15.
Which of the below are GST’s advantages?
1. Establishment of a single national market
2. Strengthening the ‘Make in India’ program
3. Lessening the burden of conformity on taxpayers
4. Increase in government revenue
5. Dual taxation, as well as double taxation, are eliminated.
(a) 1, 3, 4, and 5 are the first, third, fourth, and fifth numbers.
(b) 2, 3, 4 & 5
(c) 1, 2, 4 and 5
(d) 1, 2, 3, 4 and 5
Answer:
(c) 1, 2, 4 and 5

Question 16.
Mr. X started his business on 1st Sept ’18 and does not have any other source of income. His first previous (financial) year will start from _________
(a) 1st April 2019
(b) 1st April 2018
(c) 1st September 2019
(d) 1st September 2018
Answer:
(d) 1st September 2018
Explanation:
1st Sept 2018, because Mr. X started his business on 1st Sept ’18, so his previous (financial) year will start from 1st Sept 2018.

Question 17.
The value of x for which the points (x, -1), (2, 1), and (4, 5) are collinear is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Explanation:
Let the given point be A(x, -1), B(2, 1) and C(4, 5)
Since, A, B, and C are collinear, so
Slope of AB = Slope of BC
Slope of AB = \(\frac{1+1}{2-x}=\frac{2}{2-x}\)
Slope of BC = \(\frac{5-1}{4-2}\) = 2
2 = \(\frac{2}{2-x}\)
⇒ 4 – 2x = 2
⇒ x = 1

Question 18.
For specifying a straight line, how many geometric parameters should be known?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Explanation:
Different forms of straight lines are:
(i) Slope intercept form, y = mx + c;
parameters = 2 i.e., m and c
(ii) Intercept form, \(\frac{x}{a}+\frac{y}{b}\) = 1;
parameters = 2 i.e., a and b.
(iii) Normal form, x cos α + y sin α = p;
parameters = 2 i.e., α and p.
Hence, two parameters should be known.

Direction (Q.19 & Q.20): In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.

(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 19.
Assertion (A): If the first moment of a distribution is 2 about the value 2, then the mean is 4.
Reason (R): Mean, \(\bar{x}=\mu_1^{\prime}+\mathrm{A}\)
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
Given, \(\mu_1^{\prime}\) = 2 and A = 2.
∵ \(\bar{x}=\mu_1^{\prime}+\mathrm{A}\)
∴ \(\bar{x}\) = 2 + 2 = 4

Question 20.
Assertion (A): The value of \(\left\{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2\right\} \div\left\{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3\right\}=1\)
Reason (R): The value of \(\left\{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2\right\} \div\left\{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3\right\}=0\)
Answer:
(c) (A) is true but (R) is false.
Explanation:
\(\frac{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2}{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3}=\frac{3^6 \times 4^6 \times 5^6}{3^6 \times 4^6 \times 5^6}\) = 1 [Using (am)n = amn]

Section-B

(All Questions are compulsory. In case of internal Choice, attempt any one question only)

Question 21.
In this question, a few statements, followed by four conclusions numbered I, II, III, and IV are given. You have to consider every given statement as true, even if it does not confirm the well-known facts. Read the conclusions and then decide which of the conclusions can be logically derived.
Statements:
(i) Some books are copies.
(ii) All copies are green.
(iii) Some green are yellow.
Conclusions:
I. All copies are yellow.
II. Some yellow are green.
III. Some copies are yellow.
IV. All green are copies.
Answer:
Conclusions I and II are mediate inferences drawn from statements (ii) and (iii). Since the middle term ‘green’ is not distributed, hence both conclusions do not follow from statements. Conclusion II is an immediate inference drawn from statement (iii). Also, conclusions III and IV do not follow. Hence, only conclusion II follows.

Question 22.
An experiment involves tossing two coins and recording them in the following events:
A: no tail
B: exactly one tail
C: at least one tail
write the sets representing events
(i) A and C
(ii) A but not B.
OR
If P(E) = \(\frac{7}{13}\), P(F) = \(\frac{9}{13}\) and P(E ∩ F) = \(\frac{4}{13}\), then evaluate
(i) \(P\left(\frac{\bar{E}}{F}\right)\)
(ii) \(P\left(\frac{\bar{E}}{\bar{F}}\right)\)
Answer:
When we toss two coins, the sample space is
S = {HH, HT, TH, TT}
A = {HH}, B = {HT, TH}, C = {HT, TH, TT}
(i) A and C = A ∩ C = φ
(ii) A but not B = A – B = {HH}
OR
Given, P(E) = \(\frac{7}{13}\), P(F) = \(\frac{9}{13}\) and P(E ∩ F) = \(\frac{4}{13}\)

Question 23.
For several towns, the coefficient of rank correlation between the people living below the poverty line and the increase in population is 0.50. If the sum of squares of the differences in rank awarded to these factors is 82.50, find the number of towns.
OR
Compute Sk if
(i) Mean = 108, Mode = 99, σ = 5
(ii) σ = 4, Mean = 20.5, Mode = 22
Also, mention the type of skewness.
Answer:

Question 24.
You invest ₹ 3000 in a two-year investment that pays you 12% per annum. Calculate the future value of the investment. [Given (1.12)2 = (1.2544)]
Answer:
We know that
F.V. = C.F.(1 + i)n
Here, C.F. = ₹ 3000
i = 12% = \(\frac{12}{100}\) = 0.12
n = time period = 2 years
∴ F.V. = 3000(1 + 0.12)2
= 3000(1.12)2
= 3000 × 1.2544
= ₹ 3763.20

Question 25.
What is the difference between annuity regular and annuity due? Write four applications of ordinary (regular) annuity.
Answer:
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.
Applications of regular annuities are

  • Leasing
  • Capital Expenditure (Investment Decision)
  • Valuation of Bond
  • Sinking Fund

Section-C

(All Questions are compulsory. In case of internal Choice, attempt any one question only)

Question 26.
For any set A and B, show that (A – B) = (A ∩ B’)
OR
In a survey of 450 people, it was found that 110 play cricket, 160 play tennis, and 70 play both cricket as well tennis. How many play neither cricket nor tennis?
Answer:
Let x ∈ (A – B), then
⇒ x ∈ A and x ∉ B
⇒ x ∈ A and x ∈ B’
⇒ x ∈ A ∩ B’ ……(i)
∴ (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)’ …..(ii)
From (i) and (ii), we have
(A – B) = (A ∩ B’)
OR
Let C and T denote the students who play cricket and tennis, respectively.
Given, n(C) = 110, n(T) = 160, n(C ∩ T) = 70, n(U) = 450
Using identity, n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
= 110 + 160 – 70
= 200
∴ No. of students play neither cricket nor tennis = n(U) – n(C ∩ T)
= 450 – 200
= 250

Question 27.
(i) 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?
(ii) If A stands for ‘+’ B stands for ‘-‘, and C stands for ‘×’, then what is the value of (10 C 4) (A) (4 C 4) B6?
Answer:
(i) 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 the common code for clean is ‘#’.
Similarly from statements II and III, the common word is ‘neat’ and the common code for it is ‘@’.
Now, from statement II, we can conclude that the code for ‘and’ is ‘D’.
(ii) Given, codes for letters 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, the value of (10 C 4)(A)(4 C 4) B6 is 50.

Question 28.
Let f(x) = \(\left\{\begin{array}{lll}
x^2, & \text { when } & 0 \leq x \leq 2 \\
2 x, & \text { when } & 2 \leq x \leq 5
\end{array}\right.\)
g(x) = \(\left\{\begin{array}{lll}
x^2, & \text { when } & 0 \leq x \leq 3 \\
2 x, & \text { when } & 3 \leq x \leq 5
\end{array}\right.\)
Show that f is a function, while g is not a function.
Answer:
The relation f is defined as
f(x) = \(\left\{\begin{array}{lll}
x^2, & \text { when } & 0 \leq x \leq 2 \\
2 x, & \text { when } & 2 \leq x \leq 5
\end{array}\right.\)
It is observed that for
0 ≤ x ≤ 2, f(x) = x2 and 2 ≤ x ≤ 5, f(x) = 2x
Also, at x = 2, f(x) = 22 = 4 or f(x) = 2 × 2 = 4
i.e., at x = 2, f(x) = 4.
Therefore, for 0 ≤ x ≤ 5, the images of f(x) are unique.
Thus, the given relation ‘f’ is a function.
The relation g is defined as g(x) = \(\begin{cases}x^2, & 0 \leq x \leq 3 \\ 2 x, & 3 \leq x \leq 5\end{cases}\)
It can be observed that for x = 3, g(x) = 32 = 9 and g(x) = 2 × 3 = 6
Hence, element 3 of the domain of relation ‘g’ corresponds to two different images i.e., 9 and 6.
Hence, this relation is not a function.

Question 29.
Find the value of a and b from the following data:

Marks0-55-1010-1515-2020-25Total
No. of Students7a2530b100

Given that the third decile is 11.
Answer:
We construct the cumulative frequency table as under:

Marks0-55-1010-1515-2020-25
No. of Students7a2530b
Cumulative Frequency77 + a32 + a62 + a62 + a + b

As the total number of students is 100,
62 + a + b = 100
⇒ a + b = 38 …….(i)
Third decile, D3 is \(\frac{3}{10}\) of 100th i.e., 30th value.
Given third decile is 11, so it lies in class 10 -15.
Using formula, D3 = \(l+\frac{\frac{3}{10} n-c}{f} \times i\)
⇒ 11 = \(10+\frac{30-(7+a)}{25} \times 5\)
⇒ 1 = \(\frac{23-a}{5}\)
⇒ 5 = 23 – a
⇒ a = 18
From equation (i),
b = 38 – a
⇒ b = 38 – 18
⇒ b = 20
Hence, a = 18 and b = 20.

Question 30.
The difference between simple and compound interest compounded annually on a certain sum of money for 2 years at 4% per annum is ₹ 1. Find the sum.
Answer:
Let the sum be ₹ x. Then,
Compound Interest = An – P
⇒ l = 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 = Pit
= 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 31.
Whatever the value of t, prove that the locus of the point of intersection of the lines x cos t + y sin t = a and x sin t – y cos t = b is a circle.
OR
Find the vertex, axis, and focus of the parabola y2 – 8y – x + 19 = 0.
Answer:
The given lines are
x cos t + y sin t = a …….(i)
and x sin t – y cos t = b …….(ii)
Let P(α, β) be the point of intersection of given lines, then
α cos t + β sin t = a ……(iii)
α sin t – β cos t = b ……..(iv)
To eliminate the parameter t, on squaring and adding equations (iii) and (iv), we get
α2 (cos2t + sin2t) + β2 (sin2t + cos2t) = a2 + b2
⇒ α2 + β2 = a2 + b2 [∵ sin2θ + cos2θ = 1]
Therefore, the locus of the point P(α, β) is x2 + y2 = a2 + b2
Which represents, a circle with centre (0, 0) and radius = \(\sqrt{a^2+b^2}\)
OR
The given equation of parabola is y2 – 8y – x + 19 = 0
⇒ y2 – 8y = x – 19
⇒ y2 – 8y + 16 = x – 19 + 16
⇒ (y – 4)2 = x – 3 …….(i)
Shifting the origin to the point (3, 4) without rotating the axes and denoting the new coordinates concerning these new axes by X and Y, we have
x = X + 3 and y = Y + 4 ……(ii)
Using these relations, equation (i) reduces to
Y2 = X ……(iii)
This is of the form Y2 = 4aX.
On comparing, we get 4a = 1
⇒ a = \(\frac{1}{4}\)
Vertex: The coordinates of the vertex concerning the new axes are (X = 0, Y = 0)
So, the coordinates of the vertex w.r.t. to the old axes are (3, 4) [Putting X = 0, Y = 0 in equation (iii)]
Axis: The equation of the parabola [equation (iii)] w.r.t. the new axes are (X = a, Y = 0).
So, the equation of the axis w.r.t. the old axes is y = 4 [Putting Y = 0 in equation (ii)]
Focus: The coordinates of the focus w.r.t. the new axes are (X = a, Y = 0) i.e., (X = \(\frac{1}{4}\), Y = 0).
So, the coordinates of the focus w.r.t. the old axes are (\(\frac{13}{4}\), 4) [Putting X = \(\frac{1}{4}\) and Y = 0 in equation (ii)]

Section-D

(This section comprises long answer type questions (LA) of 5 marks each)

Question 32.
Five persons are sitting in a row. One of the two persons at the end is intelligent, and the other is fair. A fat person is sitting to the right of a weak person. A tall person is sitting to the left of the fair person and the weak person is sitting between the intelligent and the fat person.
(i) Tall person is at which place counting from right?
(ii) A person to the left of a weak person possesses which characteristic?
(iii) Which person is sitting at the center?
OR
If a = log24 12, b = log36 24 and c = log48 36, then prove that 1 + abc = 2bc.
Answer:
The first information given in the question that one of the two persons at extreme ends is intelligent and the other one is fair suggests two figures as shown below in fig (1) and fig (2)

Information that a tall person is sitting to the left of a fair person rules out the possibility of Fig (1) as no person in Fig (1) can sit to the left of a fair person. Therefore, only fig (2) shows the correct positions of intelligent and fair persons. Now, the rest of the information regarding the position of the other person can easily be inserted. The final ranking of their sitting arrangement is shown in Fig (3).

It is clear from Fig 3,
(i) The tall person is in second place counting from right.
(ii) A person left to a weak person possesses intelligence.
(iii) Fat person is sitting at the center.
OR

Question 33.
A committee of 7 has to be formed out of 9 boys and 4 girls. In how many ways can this be done when the committee consists of
(i) exactly 3 girls
(ii) At most 3 girls?
Answer:
A committee of 7 has to be formed from 9 boys and 4 girls.
(i) exactly 3 girls = 9C4 × 4C3
= \(\frac{\underline{9}}{\lfloor 4\lfloor 5} \times \frac{\lfloor 4}{\lfloor 3 ! 1}\)
= 72 × 7
= 504
(ii) at most 3 girls
(a) No girl and 7 boys
(b) 1 girl and 6 boys
(c) 2 girls 5 boys
(d) 3 girls and 4 boys
∴ The committee consists of at most 3 girls = 4C0 × 9C7 + 4C1 × 9C6 + 4C2 × 9C5 + 4C3 × 9C4
= 36 + 336 + 1296 + 504
= 2172

Question 34.
Draw the graph of the function |x – 2|.
OR
If f(x) = \(\left\{\begin{array}{lc}
m x^2+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^3+m, & x>1
\end{array}\right.\). For what integers m and n do both \(\lim _{x \rightarrow 0} f(x)\) and \(\lim _{x \rightarrow 1} f(x)\) exist?
Answer:
Clearly, y = |x – 2| = \(\left\{\begin{array}{cc}
x-2, & x-2 \geq 0 \\
-(x-2), & x-2<0
\end{array}\right.\)
= \(\begin{cases}x-2, & x \geq 2 \\ 2-x, & x<2\end{cases}\)
We know that a linear equation in x and y represents a line for drawing a line, we need only two points for y = x – 2.

So, plot the points P(2, 0), Q(4, 2) and join PQ to get the graph of y = x – 2
for y = 2 – x:

Plot the points R(1, 1), S(0, 2) and join RS to get the graph of y = 2 – x

OR

Question 35.
Calculate the electricity bill amount for 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, and a washing machine of 400 W for 3 h are used per day. The cost per unit is ₹ 1.80.
Answer:
Electric energy consumed per day by 4 bulbs = 4 × 40 × 5 = 800 Wh
Electric energy consumed per day by 4 tube lights = 4 × 60 × 5 = 1200 Wh
Electric energy consumed per day by TV = 100 × 6 = 600 Wh
Electric energy consumed per day by washing machine = 400 × 3 = 1200 Wh
∴ Total electric energy consumed per day by all electric appliances = (800 + 1200 + 600 + 1200) Wh
= 3800 Wh
= 3.8 kWh
= 3.8 units
Total electric energy consumed in 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-E

(This section comprises 3 source-based questions (Case Studies) of 4 marks each)

Question 36.
Vijeta and Rohini are playing cards. Total number of playing cards are 52 in number.

The following events happen while playing the game of choosing a card out of 52 cards.
Answer the following questions accordingly:
(i) Vijeta draws a card, what is the probability that the drawn card is either red or king?
(ii) Vijeta drew the card again, now what is the probability that the card drawn is red?
OR
Rohini draws the card again, now what is the probability that the card is an ace of red colour?
(iii) Vijeta draws two cards successively with replacement and finds the probability distribution of several aces when one ace is there in selection.
Answer:
(i) Favourable events are:
Total red cards = 26 (includes two kings)
Total cards = 26 + Two black kings = 28
Required Probability = \(\frac{28}{52}=\frac{7}{13}\)
(ii) Total number of red cards = 26
Required Probability = \(\frac{26}{52}=\frac{1}{2}\)
OR
Total number of aces = 2
Required probability = \(\frac{2}{52}=\frac{1}{26}\)
(iii) When one ace is there, then X = 1
Required conditions:
(a) ace is selected first and then no ace
(b) no ace is selected first and an ace is selected
P(X = 1) = \(\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52}\)
⇒ P(X = 1) = \(\frac{24}{169}\)

Question 37.
Data from 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 38, 70, 48, 34, 42, 55, 63, 46, 54, 44.

Based on above information answer the following questions:
(i) Find the median of the data.
(ii) Find the mean of the given data.
(iii) What is the mean deviation from the median of the given scores?
OR
If the scores 38 and 34 are replaced by 68 and 74 what will be the mean of the data?
Answer:
(i) 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
(ii) Sum of all the scores = 494
Mean = \(\frac{494}{10}\) = 49.4
(iii)

Mean Deviation = \(\frac{1}{n} \times \sum\left|d_i\right|\)
= \(\frac{86}{10}\)
= 8.6
OR
Sum of new scores = 494 – 38 – 34 + 68 + 74 = 564
New mean = \(\frac{564}{10}\) = 56.4

Question 38.
Reena and Lilly, students of class XI, were discussing the topic ‘Geometric Progression’. They prepared the notes of their discussion which are as follows:
General Term of a G.P.
Consider a G.P. with the first non-zero term ‘a’ and common ratio ‘r’, then G.P. can be written as a, ar, ar2, ar3, ……. and so on.
Then, the general term or nth term of G.P. is given by Tn = an = arn-1
Let term ‘l’ of a G.P. is the same as the nth term and is given by l = arn-1
The sum of the first n terms of a G.P.
If a and r are the first term and common ratio of a G.P., respectively, then the sum of n terms of this G.P. is given by
Sn = \(\frac{a\left(1-r^n\right)}{1-r}\), when r < 1 Sn = \(\frac{a\left(r^n-1\right)}{r-1}\), when r > 1
and Sn = na, when r = 1
The sum of an infinite G.P.
The sum of an infinite G.P. with first term ‘a’ and common ratio ‘r’ (-1 < r < 1 i.e., |r| < 1) is given by
S (or S) = \(\frac{a}{1-r}\)
(i) Find the sum of n terms of the series \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \ldots .\) Also, find the sum to infinity of the given series.
OR
(ii) Find the sum of the series 1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + …….
Answer:
(i) The given series is an infinite G.P. with the first term,
a = 1 and common ratio, r = \(\frac{1}{3}\)
Hence, the sum of the first n terms is

OR
(ii) Let Sn = 1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + …… to n terms
Hence, (1 – x) Sn = (1 – x) + (1 – x)(1 + x) + (1 – x)(1 + x + x2) + (1 – x)(1 + x + x2 + x3) + ….. to n terms
⇒ (1 – x) Sn = (1 – x) + (1 – x2) + (1 – x4) …… to n terms
⇒ (1 – x) Sn = n – (x + x2 + x3 + x4 + ……) to n terms
⇒ (1 – x) Sn = \(n-\frac{x\left(1-x^n\right)}{(1-x)}\)
Hence, Sn = \(\frac{n}{1-x}-\frac{x\left(1-x^n\right)}{(1-x)^2}\)


Show More
यौगिक किसे कहते हैं? परिभाषा, प्रकार और विशेषताएं | Yogik Kise Kahate Hain Circuit Breaker Kya Hai Ohm ka Niyam Power Factor Kya hai Basic Electrical in Hindi Interview Questions In Hindi