# 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 1 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:

- 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 02 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, 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.

The value of the base in a binary system is

(a) 2

(b) 8

(c) 10

(d) 1

Answer:

(a) 2

Explanation:

In a binary number system, the value of the base (radix) is 2. The binary system uses only two digits for representation of numbers, therefore its base it has chosen to be 2.

Question 2.

\(4 x^{-1 / 4}\) is expressed as

(a) \(-4 x^{1 / 4}\)

(b) x^{-1}

(c) \(\frac{4}{x^{1 / 4}}\)

(d) none of these

Answer:

(c) \(\frac{4}{x^{1 / 4}}\)

Explanation:

\(4 x^{-1 / 4}=\frac{4}{x^{1 / 4}}\) [negative index rule]

Question 3.

David obtained 76, 65, 82, 67, and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks?

(a) 75

(b) 96

(c) 72

(d) 76

Answer:

(a) 75

Explanation:

Average = \(\frac{76+65+82+67+85}{5}\)

= \(\frac{375}{5}\)

= 75

Question 4.

If P = {x: \(\frac{x-8}{2}\) = 4}, then P is a

(a) empty set

(b) singleton set

(c) infinite set

(d) none of these

Answer:

(b) singleton set

Explanation:

Given, P = {x: \(\frac{x-8}{2}\) = 4}, therefore,

\(\frac{x-8}{2}\) = 4

⇒ x – 8 = 8

⇒ x = 16

Hence, P is a singleton set.

Question 5.

The 10th common term between the series 3 + 7 + 11 + ….. and 1 + 6 + 11 + ….. is

(a) 191

(b) 193

(c) 211

(d) None of these

Answer:

(a) 191

Explanation:

The first common term is 11.

Now, the next common term is obtained by adding LCM of the common difference of 4 and 5, i.e., 20.

Therefore, the 10th common term = T_{10} of the A.P. whose first term, a = 11, and common difference, d = 20.

T_{10} = a + 9d

⇒ T_{10} = 11 + 9(20)

⇒ T_{10} = 191

Question 6.

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. The number of such arrangements in which no two vowels are adjacent to each other is

(a) 360

(b) 144

(c) 72

(d) 54

Answer:

(b) 144

Explanation:

We note that there are 3 consonants and 3 vowels, first let us arrange the consonants in a row. This can be done in ^{3}P_{3} = 3! = 6 ways.

× M × C × T ×

Now, no. two vowels are adjacent to each other if they are put at places marked ×.

The 3 vowels can fill up these 4 places in ^{3}P_{3} = 4 × 3 × 2 × 1 = 24 ways.

Hence, the total number of arrangements = 6 × 24 = 144.

Question 7.

If TEMPLE is coded as VHQURL, how would you code CHURCH?

(a) EKYWIO

(b) EKUWIO

(c) EKVWIN

(d) EKYWJO

Answer:

(a) EKYWIO

Explanation:

From the above diagram, it is clear that the code for T is V, for E is H, for M is Q. It may be noticed from here that letters of TEMPLE have been replaced by new letters from the alphabet. There is a gap of one letter between T and V, a gap of two letters between E and H, a gap of three letters between M and Q, and so on in the alphabet. Therefore, coding for CHURCH is

∴ CHURCH is coded as EKYWIO.

Question 8.

If f(x) = \(x^3-\frac{1}{x^3}\), then f(x) + f(\(\frac{1}{x}\)) is equal to

(a) 2x^{3}

(b) \(\frac{2}{x^3}\)

(c) 0

(d) 1

Answer:

(c) 0

Explanation:

Question 9.

\(\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}\) is equal to

(a) \(\frac{1}{10}\)

(b) \(\frac{-1}{10}\)

(c) 1

(d) None of these

Answer:

(b) \(\frac{-1}{10}\)

Explanation:

Question 10.

If a single letter is selected at random from the word ‘PROBABILITY’, then the probability of vowels is

(a) \(\frac{1}{3}\)

(b) \(\frac{4}{11}\)

(c) \(\frac{2}{11}\)

(d) \(\frac{3}{11}\)

Answer:

(b) \(\frac{4}{11}\)

Explanation:

Total number of alphabets in the word ‘PROBABILITY’ = 11

Number of vowels = 4

∴ Required Probability = \(\frac{4}{11}\)

Question 11.

If P(A) = \(\frac{1}{2}\), P(B) = 0, then P(\(\frac{A}{B}\)) is

(a) 0

(b) \(\frac{1}{2}\)

(c) not defined

(d) 1

Answer:

(c) not defined

Explanation:

Given, P(A) = \(\frac{1}{2}\) and P(B) = 0

∵ \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)

⇒ \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{0}\) = ∞ (not defined)

Question 12.

The type of central tendency measures which divides the data set into ten equal parts is classified as

(a) percentile

(b) multiple piles of data

(c) quartile

(d) deciles

Answer:

(d) deciles

Explanation:

The nine values which divide the data set into ten equal parts are called deciles.

Question 13.

The standard deviation of some temperature data in °C is 5. If the data were converted into °F, then the variance would be

(a) 81

(b) 57

(c) 36

(d) 25

Answer:

(a) 81

Explanation:

Given, σ_{C} = 5

We know that, C = \(\frac{5}{9}\) (F – 32)

⇒ F = \(\frac{9C}{5}\) + 32

∴ \(\sigma_F=\frac{9}{5} \sigma_C\)

= \(\frac{9}{5}\) × 5

= 9

\(\sigma_F^2\) = (9)^{2} = 81

Question 14.

Relationship between annual nominal rate of interest and annual effective rate of interest, if the frequency of compounding is greater than one

(a) Effective rate > Nominal rate

(b) Effective rate < Nominal rate

(c) Effective rate = Nominal rate

(d) None of the above

Answer:

(a) Effective rate > Nominal rate

Explanation:

If interest is compounded more than once a year the effective interest rate for a year exceeds the per annum nominal interest rate.

i.e., effective rate > nominal rate

Question 15.

If ₹ 1200 is lent out at 5% per annum simple interest for 3 years, then the amount after 3 years is……

(a) ₹ 1380

(b) ₹ 1830

(c) ₹ 1300

(d) ₹ 1880

Answer:

(a) ₹ 1380

Explanation:

A = P + I

A = 1200 + Pit

= 1200 + 1200 × \(\frac{5}{100}\) × 3

= 1200(1 + 0.15)

= 1200 × 1.15

= ₹ 1380

Question 16.

The present value of an annuity is the worth of an annuity……..

(a) 10 years ago

(b) today

(c) in the future

(d) at the end

Answer:

(c) in the future

Explanation:

The present value of an annuity is the worth of an annuity in the future.

Question 17.

A point on the X-axis, which is equidistant from the point (7, 6) and (3, 4) is

(a) (0, \(\frac{15}{2}\))

(b) (\(\frac{15}{2}\), 0)

(c) (\(\frac{2}{15}\), 0)

(d) (\(\frac{8}{15}\), 0)

Answer:

(b) (\(\frac{15}{2}\), 0)

Explanation:

Let P(x, 0) be the point, which is equidistant from A(7, 6) and B(3, 4)

∴ PA = PB

⇒ PA^{2} = PB^{2}

⇒ (7 – x)^{2} + (6 – 0)^{2} = (3 – x)^{2} + (4 – 0)^{2}

⇒ 49 – 14x + x^{2} + 36 = 9 – 6x + x^{2} + 16

⇒ x = \(\frac{60}{8}=\frac{15}{2}\)

∴ The point on X-axis is (\(\frac{15}{2}\), 0).

Question 18.

The equation of the circle which touches the X-axis and whose center is (1, 2) is

(a) x^{2} + y^{2} – 2x + 4y + 1 = 0

(b) x^{2} + y^{2} + 2x – 4y + 1 = 0

(c) x^{2} + y^{2} – 2x – 4y + 1 = 0

(d) x^{2} + y^{2} – 2x – 4y – 1 = 0

Answer:

(c) x^{2} + y^{2} – 2x – 4y + 1 = 0

Explanation:

Given, that the Centre of the circle is C(1, 2).

As, the circle touches the X-axis, its radius = Perpendicular distance from center (1, 2) to the X-axis = ordinate of point C = 2.

∴ The equation of circle is (x – 1)^{2} + (y – 2)^{2} = 2^{2}

⇒ x^{2} + y^{2} – 2x – 4y + 1 = 0.

**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): log_{2} 8 = 3

Reason (R): log a^{n} = n log a and log_{a} a = 1

Answer:

(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:

log_{2} 8 = 3

L.H.S. = log_{2} 8 = log_{2} 2^{3}

= 3 log_{2} 2 [∵ log a^{n} = n log a]= 3 × 1

= 3 [∵ log_{a} a = 1]

Question 20.

Assertion (A): Range of the function f(x) = |x| is (-∞, ∞).

Reason (R): The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain.

Answer:

(d) (A) is false but (R) is true.

Explanation:

Range of |x| is [0, ∞) therefore assertion is false.

Section-B

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

Question 21.

The son of M is the father of N and the grandfather (Mother’s father) of R. S is the daughter of N and sister of B. Based on this information, how is M related to B?

Answer:

From the above chart, it can be concluded that M is the (maternal) great-grandfather or (maternal) great-grandmother of B, hence the gender of M is not specified.

Question 22.

Two-thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting a first class is 0.25 and that of a boy getting a first class is 0.28. Find the probability that a student chosen at random will get first-class marks in the subject.

OR

In an experiment of rolling a fair die, let A, B, and C be three events defined as

A: a number that is a perfect square.

B: a prime number.

C: a number which is greater than 5.

These events are mutually exclusive or exhaustive?

Answer:

Let E_{1}, E_{2}, and A be the events defined as follows:

E_{1} = A boy is chosen from the class

E_{2} = A girl is chosen from the class

A = The student gets first-class marks

Here, P(E_{1}) = \(\frac{2}{3}\), P(E_{2}) = \(\frac{1}{3}\), P(\(\frac{A}{E_1}\)) = 0.28 and P(\(\frac{A}{E_2}\)) = 0.25

Using the law of total probability, we obtain

P(A) = P(E_{1}) P(\(\frac{A}{E_1}\)) + P(E_{2}) P(\(\frac{A}{E_2}\))

= \(\frac{2}{3}\) × 0.28 + \(\frac{1}{3}\) × 0.25

= 0.27

OR

When we roll a fair die, sample space S = (1, 2, 3, 4, 5, 6}

A = (1, 4}, B = {2, 3, 5} and C = {6}

Since A ∩ B = φ, B ∩ C = φ, C ∩ A = φ, A ∩ B ∩ C = φ

∴ A, B, and C are exhaustive events.

Question 23.

The mean of 100 observations is 50 and their standard deviation is 5. Find the sum of all squares of the observations.

OR

The value of attendance of 9 students in a test were 13, 17, 20, 5, 3, 3, 18, 15 and 20. Find the first and third quartiles.

Answer:

Given, \(\bar{x}\) = 50, n = 100 and σ = 5

\(\sigma^2=\frac{\sum x_i^2}{n}-(\bar{x})^2\)

\(\frac{\sum x_i^2}{n}=\sigma^2+(\bar{x})^2\)

\(\Sigma x_i^2=n\left[\sigma^2+(\bar{x})^2\right]\)

= 100[5^{2} + (50)^{2}]= 100(25 + 2500)

= 100 × 2525

= 252500

Hence, the sum of the squares of all the observations is 252500.

OR

First, we have to arrange the data in ascending order 3, 3, 5, 13, 15, 17, 18, 20, 20

Here, n = 9

Lower quartile or first quartile, Q_{1} = \(\frac{n+1}{4}^{\text {th }}\) value

= \(\frac{10}{4}^{\text {th }}\) value

= 2.5^{th} value

∴ Q_{1} = \(\frac{2^{n d} \text { value }+3^{r d} \text { value }}{2}\)

= \(\frac{3+5}{2}\)

= 4

Upper quartile or third quartile,

Question 24.

₹ 16000 invested at 10% p.a. compounded semi-annually amounts to ₹ 18522. Find the time period of investment.

Answer:

Here, P = ₹ 16000 A_{n} = ₹ 18522

i = 10 × \(\frac{1}{2}\)% = 5% = 0.05

We have, A_{n} = P(1 + i)^{n}

⇒ 18522 = 16000(1 + 0.05)^{n}

⇒ \(\frac{18522}{16000}=\left(\frac{21}{20}\right)^n\)

⇒ \(\left(\frac{21}{20}\right)^3=\left(\frac{21}{20}\right)^n\)

Therefore, the time period of investment is three and half years i.e., 1\(\frac{1}{2}\) years.

Question 25.

A machine can be purchased for ₹ 50000. The machine will contribute ₹ 12000 per year for the next five years. Assume borrowing is 10% per annum compounded annually Determine whether the machine should be purchased or not. Given P(5, 0.10) = 3.79079.

Answer:

The present value of annual contribution

PV = C.F. P(n, i)

Here, C.F. = ₹ 12000, i = 10% = \(\frac{10}{100}\) = 0.10, n = 5

and P(5,0.10) = 3.79079

∴ PV = 12000 × 3.79079 = ₹ 45489.48

which is less than the initial cost of the machine. Therefore machine must not be purchased.

Section-C

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

Question 26.

Find the sum of the G.P. 0.15, 0.015, 0.0015,… up to 20 terms.

OR

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.

Answer:

OR

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, the remaining elements are {(-1, -1), (-1, 1) (0, -1), (0, 0), (1, 0), (1, -1), (1, 1)}

∴ Domain = {-1, 0, 1} and Range = {-1, 0, 1}

Question 27.

In a certain code ‘Ding Dong Dang’ means ‘A Hacking the enemy’, ‘Ping Pond Dong’ means ‘Enemy is retreating’, and ‘Ding Ping Mong’ means ‘Attacking and retreating’.

From the above information, answer the following questions

(i) Which code stands for Enemy?

(ii) Which code stands for Attacking?

(iii) Which code stands for Retreating?

Answer:

Let (I) Ding Dong Dang – Attacking the enemy

(II) Ping Pond Dong – Enemy is retreating

(III) Ding Ping Mong – Attacking and retreating

(i) In the I and II sentences, the common word is ‘enemy’ and the common code is ‘Dong’. Hence, ‘Dong’ stands for ‘enemy’.

(ii) In the I and III sentences, the common word is ‘Ding’, and the common code is ‘Attacking’. Hence, the code for ‘Ding’ is ‘Attacking’.

(iii) From the II and III sentences, we can see that the common word is ‘Ping’ and the common code is ‘retreating’. Hence, ‘Ping’ stands for ‘retreating’.

Question 28.

If y = \(a e^{2 x}+b e^{-x}\), then show that \(\frac{d^2 y}{d x^2}-\frac{d y}{d x}\) – 2y = 0.

Answer:

Question 29.

For a frequency distribution, Bowley’s coefficient of skewness is 1.2. If the sum of the 1st and 3rd quartiles is 200 and the median is 76, then find the value of the third quartile.

OR

If \(\bar{x}\) is the mean and σ^{2} is the variance of n observations x_{1}, x_{2}, x_{3},…, x_{n}, then prove that the mean and variance of the observations ax_{1}, ax_{2},…, ax_{n} are a\(\bar{x}\) and a^{2}σ^{2} respectively (a ≠ 0).

Answer:

Question 30.

A product is sold from Kota (Rajasthan) to Gwalior (M.P.) for ₹ 8,000 and then from Gwalior to Indore (M.P.). If the rate of tax under the GST system is 18% and the profit made by the dealer in Gwalior is ₹ 3,000, find the net GST payable by the dealer in Gwalior.

Answer:

Cost Price in Gwalior = ₹ 8,000

Profit = ₹ 3,000

Selling Price in Gwalior = ₹ (8000 + 3000) = ₹ 11,000

CGST = ₹ \(\frac{9}{100}\) × 11,000 = ₹ 990

SGST = ₹ \(\frac{9}{100}\) × 11,000 = ₹ 990

∴ Net GST paid by the dealer in Gwalior = ₹ (990 + 990) = ₹ 1980

Question 31.

If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.

OR

Find the foot of the perpendicular from the point (3, 8) to the line x + 3y = 7.

Answer:

Let LAM be the parabolic reflector such that LM is its diameter and AM is its depth.

Given that, LM = 20 cm AN = 5 cm

∴ LN = \(\frac{LM}{2}\)

= \(\frac{10}{2}\)

= 10 cm

Now, taking A as the origin, AX along the X-axis, and a line through A perpendicular to AX as the Y-axis, let the equation of parabolic reflector be y^{2} = 4ax …….(i)

The point L has coordinates (5, 10) and lies on (i).

Therefore, 10^{2} = 4a × 5

⇒ a = 5

So, the equation of reflector is y^{2} = 20x.

Its focus is at (5, 0) i.e., at point N.

Hence, the focus is at the mid-point of the given diameter.

OR

The given equation of the line is x + 3y = 7

⇒ y = \(-\frac{1}{3} x+\frac{7}{3}\)

∴ Slope of the line, m_{1} = \(-\frac{1}{3}\)

Let m_{2} be the slope of the perpendicular line.

∴ m_{1}m_{2} = -1

⇒ \(-\frac{1}{3}\) × m_{2} = -1

⇒ m_{2} = 3

∴ The equation of the perpendicular line with slope 3 and passing through (3, 8) is y – 8 = 3(x – 3)

⇒ 3x – y – 1 = 0

∴ The foot of the perpendicular is the point of intersection of the lines x + 3y – 7 = 0 and 3x – y – 1 = 0.

Solving these equations, we get x = 1, y = 2

So (1, 2) is the foot of the perpendicular.

Section-D

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

Question 32.

On what dates of March 2005 did Friday fall?

OR

A, B, and C can complete a work separately in 24,36 and 48 days respectively. They started together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed?

Answer:

First, we find the day on 01.03.2005.

01.03.2005 = (2004 years + Period from 01.01.2005 to 01.03.2005)

odd day in 1600 years = 0

odd day in 400 years = 0

4 years = (1 leap year + 3 ordinary years)

= (1 × 2 + 3 × 1) odd days

= 5 odd days

(Because leap year has 2 odd days whereas ordinary year has 1 odd day)

January + February + March = 31 days + 28 days + 1 day

= 60 days

= (8 weeks + 4 days)

= 4 odd days

Total number of odd days = 0 + 0 + 5 + 4

= 9

= 2 odd days

∴ 01.03.2005 was Tuesday. So, Friday is 04.03.2005.

Hence, Friday lies on the 4th, 11th, 18th, and 25th March, 2005.

OR

(A + B + C)’s 1 day’s work = \(\left(\frac{1}{24}+\frac{1}{36}+\frac{1}{48}\right)=\frac{13}{144}\)

Work done by (A + B + C) in 4 days = \(\frac{13}{144}\) × 4 = \(\frac{13}{36}\)

Work done by B in 3 days = \(\frac{1}{36}\) × 3 = \(\frac{1}{12}\)

Remaining work = \(\left[1-\left(\frac{13}{36}+\frac{1}{12}\right)\right]=\frac{5}{9}\)

(A + B)’s 1 day’s work = \(\frac{1}{24}+\frac{1}{36}=\frac{5}{72}\)

Now, \(\frac{5}{72}\) work is done by A and B in \(\frac{72}{5} \times \frac{5}{9}\) = 8 days

Hence, total time taken = (4 + 3 + 8) days = 15 days

Therefore, work will be completed in 15 days.

Question 33.

For any two sets A and B, prove that A ∪ B = A ∩ B ⇔ A = B.

OR

If the A.M. between p^{th} and q^{th} terms of an A.P. is equal to A.M. between r^{th} and s^{th} terms of the A.P., then show that p + q = r + s.

Answer:

Let A = B, then A ∪ B = A and A ∩ B = B

A ∪ B = A ∩ B Thus, A = B …….(i)

Conversely, let A ∪ B = A ∩ B

Let x ∈ A

x ∈ (A ∪ B) [∵ A ∪ B = A ∩ B]x ∈ (A ∩ B)

x ∈ A and x ∈ B

x ∈ B

∴ A ⊆ B ……(ii)

Now, Let y ∈ A

y ∈ A ∪ B

y ∈ A ∩ B [∵ A ∪ B = A ∩ B]y ∈ A and y ∈ B

y ∈ A

∴ B ⊆ A ……(iii)

From equations (ii) and (iii), we get A = B.

Thus, (A ∪ B) = (A ∩ B)

∴ A = B

From equations (iii) and (iv), we get

A ∪ B = A ∩ B ⇔ A = B

Hence Proved.

OR

Let a be the first term and d be the common difference of the given A.P. Then

a_{p} = p^{th} term = a + (p – 1)d

a_{q} = q^{th} term = a + (q – 1)d

a_{r} = r^{th} term = a + (r – 1)d

a_{s} = s^{th} term = a + (s – 1)d

It is given that

A.M. between ap and aq = A.M. between ar and as

⇒ \(\frac{1}{2}\)(ap + aq) = \(\frac{1}{2}\)(ar + as)

⇒ a_{p} + a_{q} = a_{r} + a_{s}

⇒ a + (p – 1)d + a + (q – 1)d = a + (r – 1)d + a + (s – 1)d

⇒ (p + q – 2)d = (r + s – 2)d

⇒ p + q = r + s

Question 34.

Compute rank correlation from the following data relating to ranks given by two judges in a contest:

Serial No. of Candidate: |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

Rank by Judge A: |
10 | 5 | 6 | 1 | 2 | 3 | 4 | 7 | 8 | 9 |

Rank by Judge B: |
5 | 6 | 9 | 2 | 8 | 7 | 3 | 4 | 10 | 1 |

Answer:

The rank correlation coefficient is given by

The lower value (almost 0) indicates hardly any agreement between the ranks given by the two Judges in the contest.

Question 35.

‘A’ earns the following income during the financial year 2018-19:

Particulars |
Amount (₹) |

(a) Interest paid by an Indian company but received in London | 2,00,000 |

(b) Pension from former employer in India, received in USA | 8,000 |

(c) Profits earned from business in Paris which is controlled in India, half of the profits being received in India | 40,000 |

(d) Income from agriculture in Bhutan and remitted to India | 10,000 |

(e) Income from property in England and received there | 8,000 |

(f) Past foreign untaxed income brought to India | 20,000 |

Determine the total income of ‘A’ for the assessment year 2019-20 if he is (i) Resident and ordinarily resident, (ii) Not ordinarily resident, and (iii) Non-resident in India.

Answer:

Section-E

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

Question 36.

A party was arranged by Rahul and a total of 25 people joined the party. Out of 25 people, 12 like to have tea, 15 like to have coffee, and 7 like to have coffee and tea.

On the basis of the above information answer the following questions:

(i) How many of them like only coffee but not tea?

(ii) How many of them like neither tea nor coffee?

(iii) If the cost of Tea per cup at the party was ₹ 35 and the cost of Coffee per cup at the party was ₹ 75 it is assumed that each person who likes only tea or only coffee has consumed two cups of tea and coffee respectively. Then what was the total cost of Tea and Coffee which was consumed in the party by such persons?

OR

How many of them like atleast one of the two drinks?

Answer:

(i) Only coffee but not tea = n(C) – n(T ∩ C)

= 15 – 7

= 8

(ii) Neither tea nor coffee = n(U) – n(T ∪ C)

= 25 – 20

= 5

(iii) Total number of person like only Tea = 5

and the total number of people who liked only Coffee = 8

So, total cost = 5 × 2 × 35 + 8 × 2 × 75 [As each person had taken two cups of tea and coffee]= 350 + 1200

= 1550

So, the total cost for tea and coffee will be ₹ 1550.

OR

Given that, n(T) = 12, n(C) = 15, n(T ∩ C) = 7

As we know that n(T ∪ C) = n(T) + n(C) – n(T ∩ C)

= 12 + 15 – 7

= 20

Therefore total person who liked atleast one of the two drinks was 20.

Question 37.

In class XI Mathematics teacher was explaining limits to students. In the case of the existence of a limit if left-hand and right-hand limits are equal and both exist then a limit exists. It is also shown below.

Left hand limit is f(a – 0) = \(\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h)\)

Similarly, right hand limit is f(a + 0) = \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h)\)

Existence of Limit: \(\lim _{x \rightarrow a} f(x)\) exists, if

(i) \(\lim _{x \rightarrow a^{-}} f(x)\) and \(\lim _{x \rightarrow a^{+}} f(x)\) both exist

(ii) \(\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)\)

On the basis of this information, the teacher asks students various questions as mentioned below.

(i) Evaluate the left-hand limit of the given function \(\begin{cases}1+x^2 ; & 0 \leq x \leq 1 \\ 2-x ; & x>1\end{cases}\) at x = 1

OR

Evaluate the right-hand limit of the given function \(\begin{cases}1+x^2 ; & 0 \leq x \leq 1 \\ 2-x ; & x>1\end{cases}\) at x = 1

(ii) Let f(x) = \(\left\{\begin{array}{l}

\cos x ; x>0 \\

x+k ; x<0

\end{array}\right.\) If limit exists find k.

(iii) Show that \(\lim _{x \rightarrow 4} \frac{|x-4|}{x-4}\) does not exist.

Answer:

Question 38.

The history of Indian playing cards begins with the circular-shaped Ganjafeh cards. The earliest mention of these comes to us from the Mughal Emperor Babur’s memoirs. In his memoir (dated June 1527), the founder of the Empire mentions having sent a set of Ganjafeh to his friend Shah Hussain in Sindh.

It’s from the late 15th century, probably made in the Burgundian Netherlands territory, and very recognizable by modern standards. There are 52 cards in four suits, with both numbered and face cards.

(i) A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is either a king or a spade.

OR

(ii) 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Answer:

Let ‘S’ be the sample space. So, n(S) = 52

Also, let ‘A’ and ‘B’ be the events of getting a king and spade respectively.

∴ n(A) = 4, n(B) = 13 and n(A ∩ B) = 1

[∵ there is only one card which is King of Spade]

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ The required probability that the card is a king or a spade is \(\frac{4}{13}\).

OR

(ii) Number of ways of drawing 4 cards from 52 cards = ^{52}C_{4}.

In a deck of 52 cards, there are 13 diamonds and 13 spades.

∴ Number of ways drawing 3 diamonds and one spade = ^{13}C_{3} × ^{13}C_{1}.

Thus, the probability of obtaining 3 diamonds and one spade = \(\frac{{ }^{13} C_3 \times{ }^{13} C_1}{{ }^{52} C_4}\).