Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 3 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 11 Applied Mathematics Set 3 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 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.
- 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 decimal equivalent of the octal number (648)_{8}. [1]

OR

Simplify : \(\frac{4}{(32)^{1 / 5}}\)

Solution:

Octal to Decimal conversion is obtained by multiplying 8 to the power of base index along with the value at that index portion.

The decimal equivalent of the octal number (648)_{8} is:

Given digits 6 4 8

Position Number 2 1 0

Positional value 8^{2} 8^{1} 8^{0}

Decimal Number : 6 × 8^{2} + 4 × 8^{1} + 8 × 8^{0}

= 6 × 64 + 4 × 8 + 8 × 1

= 384 + 32 + 8

= 424

Hence, (648)_{8} = (424)_{10}

OR

\(\frac{4}{(32)^{1 / 5}}\) = \(\frac{4}{\left(2^5\right)^{1 / 5}}\)

= \(\frac{4}{(2)^{(5 \times 1 / 5)}}\) [∵ (a^{m})^{n} = a^{mn}]

= \(\frac{4}{2}\) = 2

Question 2.

Given, log 2 = 0.3010 and log 3 = 0.4771, then find the value of log 6. [1]

Solution:

log 6 = log (2 × 3)

= log 2 + log 3 = 0.3010 + 0.4771

= 0.7781

Question 3.

If P(A) = 0.4, P(B) = 0.8 and \(P\left(\frac{B}{A}\right)\) = 0.6, then find the P(A ∪ B). [1]

Solution:

Given, P(A) = 0.4, P(B) = 0.8 and \(P\left(\frac{B}{A}\right)\) = 0.6

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

⇒ 0.64 = \(\frac{P(A \cap B)}{0.4}\)

⇒ P(A ∩ B) = 0.6 × 0.4 = 0.24

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.4 + 0.8 – 0.24

= 1.2 + 0.24 = 0.96

Question 4.

If A = (1, 5},B = {2,6} and C = {2,4}, then A × (B ∩ C). [1]

OR

If R = {(x, y): x, y ∈ N, x + 2y = 21}, then find range of R.

Solution:

Given that, A = {1, 5}, B = {2, 6} and C = {2, 4}

Then, B ∩ C = {2} and A × (B ∩ C)= {1, 5} × {2} = {(1, 2), (5, 2)}.

OR

Given, x + 2y = 21 ⇒ x= 21 – 2y

When y = 1, x = 21 – 2 × 1 = 19

When y=2, x = 21 – 2 × 2= 17

When y = 3, x = 21 – 2 × 3 = 15

………………

………………..

When y = 10, x = 21 – 2 × 10 = 1

For other values of y ∈ N, we do not get x ∈ N.

Range of R = {1, 2, 3 ,…, 10}

Question 5.

A circle of radius r is in the second quadrant. If the circle touches both the axes, then find the equation of the circle. [1]

Solution:

Since, the circle is in the second quadrant, its centre will also be in the second quadrant. Also, the circle touches the axes, so that coordinates of its centre will be (- r, r).

Hence, the required equation of the circle is

(x + r)^{2} + (y – r)^{2} = r^{2}

or, (x^{2} + 2rx + r^{2}) + (y^{2} – 2ry + r^{2}) = r^{2}

or, x^{2} + y^{2} + 2rx – 2ry + r^{2} = 0

Question 6.

Find the length of the latus-rectum of the parabola y^{2} + 8x – 2y + 17 = 0. [1]

OR

If V and S are respectively the vertex and focus of the parabola y^{2} + 6y + 2x + 5 = 0, then find SV.

Solution:

Given equation of parabola

y^{2} + 8x – 2y + 17 = 0

⇒ (y – 1)^{2} + 8x + 16 = 0

⇒ (y – 1)^{2} = – 8(x + 2)

Let X = x + 2 and Y = y – 1

Y^{2} = – 8x

Comparing with Y^{2} = – 4aX, we get

4a = 8, which is the length of latus-rectum.

OR

Given equation of parabola is

y^{2} + 6y + 2x + 5 = 0

⇒ y^{2} + 6y = – 2x – 5

⇒ (y + 3)^{2} = – 2x + 4

⇒ (y + 3)^{2} = -2(x – 2)

⇒ (y + 3)^{2} = -4.\(\frac{1}{2}\)( x – 2)

⇒ Y^{2} = – 4 . A . X

Where, Y = y + 3, X = x – 2, A = \(\frac{1}{2}\)

Since, SV is the distance between vertex and focus

= A = \(\frac{1}{2}\)

Question 7.

Arrange the following words in a logical and meaningful order. [1]

1. Travel, 2. Destination, 3. Payment, 4. Berth/seat number, 5. Reservation, 6. Availability of berth/seat for reservation.

Solution:

From the given words, it is very clear that in order to perform a journey. First destination is defined, secondly availability of berth is known, which follows payment for reservation. As a result berth is allotted and travelling is performed.

Hence, the correct, logical and meaningful arrangement of the given words is 2, 6, 3, 5, 4, 1.

Question 8.

If f(x) = x^{3}, then find the value of \(\frac{f(5)-f(1)}{5-1}\). [1]

Solution:

Given, f(x) = x^{3}

At x = 5, f(5) = 5^{3} = 125

At x = 1, f(1) = 1^{3} = 1

∴ \(\frac{f(5)-f(1)}{5-1}\) = \(\frac{125-1}{5-1}\)

= \(\frac{124}{4}\) = 31

Question 9.

Evaluate : \(\lim _{x \rightarrow 1 / 2} \frac{4 x^2-1}{2 x-1}\) [1]

Solution:

Question 10.

Pointing to a person, a man said to a woman, “His mother is the only daughter of your father.” How was the woman related to the person ? [1]

Solution:

Daughter of your father – Your sister. So, the person’s mother is woman’s sister or the woman is person’s aunt.

Question 11.

Find the third quartile of the data set 33, 25, 42, 25, 31, 37, 46, 29, 38. [1]

Solution:

On arranging the given data in ascending order, we get 25, 25, 29, 31, 33, 37, 38, 42, 46

Question 12.

Find the compound interest on a sum of ₹ 25,000 after 3 years at the rate of 12%. [1]

Solution:

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

Here, P = 25000, i = 12 %

= \(\frac{12}{100}\) = 0.12, n = 3

∴ A_{n} = 25000 (1 + 0.12)^{3}

= 25000 × (1.12)^{3}

= 25000 × 1.4049

= 35,123.20

∴ Interest, I = A_{n} – P

= 35123.20 – 25000

= ₹ 10123.20.

Question 13.

The average of 20 numbers is zero. Of them, at most, how many may be greater than zero ? [1]

OR

How many times in a day, are the hands of a dock in straight line but opposite in direction ?

Solution:

Average of 20 numbers = 0

∴ Sum of 20 numbers = (0 × 20) = 0

It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

OR

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours (Because between 5 and 7, they point in opposite directions at 6 O’clock only). So, in a day, the hands point in the opposite directions 22 times.

Question 14.

Define IGST. [1]

Solution:

IGST (Integrated Goods and Services Tax) : InterState supplies of taxable goods and/or services are subject to Integrated Goods and Services Tax (IGST). IGST is the total sum of CGST and SGST/UTGST and is levied by Centre on all inter-State supplies.

Question 15.

Find the odd one out. [1]

2, 5, 10, 17, 26, 37, 50, 64

Solution:

(1 × 1) + 1, (2 × 2) + 1, (3 × 3) + 1, (4 × 4) + 1, (5 × 5) + 1, (6 × 6) + 1, (7 × 7) + 1, (8 × 8) + 1

But, 64 is out of pattern.

Question 16.

In how many ways can 4 red, 3 yellow and 2 green discs can be arranged in a row if discs of the same colour are indistinguishable ? [1]

OR

If ^{n}P_{r} = 720 and ^{n}C_{r} =120, find r.

Solution:

In total, we have 4 + 3 + 2 = 9 discs of which 4 are of one kind (red), 3 of another kind (yellow) and 2 of another kind (green).

The number of ways of arranging these discs in a row are

= \(\frac{9 !}{4 ! 3 ! 2 !}\)

= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 6 \times 2}\)

= 1260

OR

We know that,

^{n}C_{r} = \(\frac{{ }^n P_r}{r !}\)

⇒ 120 = \(\frac{720}{r !}\)

⇒ r! = \(\frac{720}{120}\) ⇒ r! = 6

⇒ r! = 3! ⇒ r = 3.

Section – B (20 Marks)

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

Question 17.

Find the domain of function : [2]

f(x) = \(\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}\)

Solution:

Given, f(x) = \(\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}\)

Clearly, f(x) is defined when 4 – x ≥ 0 and x^{2} – 1 > 0

4 – x > 0

x ≤ 4

and x^{2} – 1 > 0

(x – 1) (x + 1) ≥ 0

Hence, Domain of f is (- ∞, – 1) ∪ (1, 4]

Question 18.

Find the value of k for which the function : [2]

f(x) = \(\frac{x^2+3 x-10}{x-2}\), x ≠ 2

k, x = 2

is continuous at x = 2.

Solution:

\(\lim _{x \rightarrow 2}\) f(x) = f(2) = k

\(\lim _{x \rightarrow 2} \frac{(x+5)(x-2)}{x-2}\) = k (x ≠ 2)

∴ k = 7.

Question 19.

Find the mean deviation about the mean of the distribution. [2]

Size | 20 | 21 | 22 | 23 | 24 |

Frequency | 6 | 4 | 5 | 1 | 4 |

OR

Let f be the subset of Z × Z defined by /f = {(ab, a + b): a,b Z}. Is/a function from Z to Z ? Justify your answer.

Solution:

x_{i} |
f_{i} |
f_{i}x_{i} |
|x_{i}–\(\bar{x}\)| |
f_{i} |x_{i} – \(\bar{x}\)| |

20 | 6 | 120 | 1.65 | 9.90 |

21 | 4 | 84 | 0.65 | 2.60 |

22 | 5 | 110 | 0.35 | 1.75 |

23 | 1 | 23 | 1.35 | 1.35 |

24 | 4 | 96 | 2.35 | 9.40 |

20 | 433 | 25 |

∴ \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{433}{20}\) = 21.65

∴ Required mean deviation about mean (\(\bar{x}\)) i.e.,

M.D. \((\bar{x})\) = \(\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}\) = \(\frac{25}{20}\) = 1.25

Commonly Made Error

Calculation errors are found in distribution table sometimes students apply wrong formula for mean deviation.

Answering Tip

For grouped data where x_{1}, x_{2}, ………………. x_{n} be n distinct values and their corresponding frequencies are f_{1}, f_{2}, ………………. f_{n} respectively. The mean deviation about mean is given by \(\frac{\sum_{i=1}^n f_i\left|x_i-\bar{x}\right|}{\sum_{i=1}^n f_i}\)

OR

Given, f = {(ab, a + b): a, b ∈ Z}

Taking a = b = 1, we have (ab, a + b) = (1, 2) ∈ f

Taking a = b = -1, we have (ab, a + b) = (1, -2) ∈ f

⇒ f-image of 1 is not unique

Hence, f is not a function.

Question 20.

For a distribution Karl Pearson’s coefficient of skewness is 0.64, standard deviation is 13 and mean is 59.2. Find mode and median. [2]

Solution:

Given, S_{k} = 0.64, s = 13 and Mean \((\bar{x})\) = 59.2 We know that,

S_{k} = \(\frac{\text { Mean – Mode }}{\sigma}\)

⇒ 0.64 = \(\frac{59.20-\text { Mode }}{13}\)

⇒ Mode = 59.20 – 8.32 = 50.88

⇒ Mode = 3 Median – 2 Mean

⇒ 50.88 = 3(Median) – 2(59.2)

⇒ Median = \(\frac{50.88+118.4}{3}\)

= \(\frac{169.28}{3}\) = 56.42. (approx.)

Question 21.

₹ 16000 invested at 10% p.a. compounded semi-annually amounts to ₹ 18522. Find the time period of investment. [2]

Solution:

Here, P = ₹ 16000

A_{n} = ₹ 18522

i = 10 × \(\frac{1}{2}\) % = 5% = 0.05 2

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

⇒ 18522 = 160000 (1 + 0.05)^{n}

⇒ \(\frac{18522}{16000}\) = (1.05)^{n}

⇒ 1.157625 = (1.05)^{n}

⇒ (1.05)^{3} = (1.05)^{n}

⇒ n = 3

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

Question 22.

Give four examples of regular annuity.

Solution:

Examples of regular annuity are :

(a) Depositing in a saving account with regular period of time of a fixed payment.

(b) Monthly insurance payments.

(c) Pension payments.

(d) Monthly rent of a house.

Question 23.

In a group of persons travelling in a bus, 6 persons can speak Tamil, 15 can speak Hindi and 6 can speak Gujarati. In that group, none can speak any other language. If 2 persons in the group can speak two languages and one person can speak all the three languages, then how many persons are there in the group ? [2]

OR

The A.M. and G.M. between two positive numbers are 10 and 8 respectively, find the numbers.

Solution:

Let us assume the two persons who can speak two languages speak Hindi and Tamil.

The third person then speaks all the three languages.

Tamil – Number of persons who can speak is 6. Only Tamil 6 – 2 – 1 = 3

Hindi – Number of persons who can speak is 15. Only Hindi 15 – 2 – 1 = 12

Gujarati – Number of persons who can speak is 6. Only Gujarati 6 – 1 = 5

Thus the number of persons who can speak only one language is 3 + 12 + 5 = 20

Number of persons who can speak two languages = 2

Number of person who can speak all the languages = 1

Total number of persons = 23

OR

Let the two positive numbers be a and b. Then A.M.

(given)

= \(\frac{a+b}{2}\) = 10

⇒ a + b = 20 …………..(i)

and G.M. = \(\sqrt{ab}\) = 8 (given)

⇒ ab = 64

⇒ a(20 – a) = 64 [using (i)]

⇒ a^{2} – 20a + 64 = 0

⇒ (a – 4) (a – 16) = 0

⇒ a = 4, 16

Taking a = 4, we get b = 20 – 4 = 16

Taking a = 16, we get b = 20 – 16 = 4

Thus, the two numbers are 4, 16, or 16, 4.

Question 24.

In a certain code language, if SUGAR is coded as PKLTN and TEA is coded as QGT, then find code of GREAT in the same language. [2]

Solution:

Here, we can see S is substituted for P, U for K, G for L, A for T, R for N, In the word TEA, T is substituted for Q, E for G and A for T. We find that coding of A in both words is T, Hence, substitution of letters in coded word takes place in the same order as in basic word. Thus, GREAT has been coded as

Commonly Made Error

Sometimes students ignore the direct substitution and they are trying to find out some pattern in the given question. In this way, they complicate the simple question.

Answering Tip

When letters of a word are substituted for new letters and are placed in the coded word at the same time at same position as in the basic word, then this method of substitution is known as direct substitution.

Question 25.

If (2a + b, a – b) = (8, 3), find a and b. [2]

Solution:

We know that, two ordered pairs are equal, if their corresponding first and second elements are equal

We have, (2a + b, a – b) = (8, 3)

∴ 2a + b = 8 …………… (i)

and a – b = 3 …………… (ii)

From eq.(ii), we get b = a – 3 ………………. (iii)

In substituting, b = a – 3, in eq. (1), we get

⇒ 2a + (a – 3) = 8

⇒ 3a = 11

⇒ a = \(\frac{11}{3}\)

Again, on substituting a = \(\frac{11}{3}\) in eq. (iii), we get

b = \(\frac{11}{3}\) – 3

⇒ b = \(\frac{2}{3}\)

∴ a = \(\frac{11}{3}\) and b = \(\frac{2}{3}\)

Question 26.

Let A = {1, 2} and B = {3, 4}. Write A × B. How many subset will A x B have? List them. [2]

OR

Find the number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line.

Solution:

Given, A = {1, 2} and B = {3, 4}

⇒ A × B = {1, 2} × {3, 4}

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

⇒ n(A × B) = 4

Total subsets of A × B = 24 = 16

Subsets of A × B are :

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)} {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

OR

Total number of triangles formed by 12 points taking 3 at a time = ^{12}C_{3}

But any three points selected from given seven collinear points does not from a triangle.

Number of ways of selecting three points from seven collinear points = ^{7}C_{3}

Required number of triangles

= ^{12}C_{3} – ^{7}C_{3}

= 220 – 35

= 185

Section – C (21 Marks)

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

Question 27.

Let A = {9, 10, 11, 12, 13} and let f : A → N defined by f(n) = the highest prime factor of n. Find the range of f . [3]

Solution:

Given, A = {9, 10, 11, 12, 13} and f : A → N and

f(n) = the highest prime factor of ‘n’

For n = 9, 9 = 1 × 3 × 3

⇒ Highest prime factor of 9 = 3.

For n = 10, 10 = 1 × 2 × 5

⇒ Highest prime factor of 10 = 5

For n = 11, 11 = 1 × 11

⇒ Highest prime factor of 11 = 11

For n = 12, 12 = 1 × 2 × 2 × 3

⇒ Highest prime factor of 12 = 3

For n = 13, 13 = 1 × 13

⇒ Highest prime factor of 13 = 13

∴ f(n) = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13)}

⇒ Range = {3, 5, 11, 13}

Question 28.

Differentiate f(x) = x^{5} e^{x} + x^{3} log x – 2^{x} with respect to x. [3]

Solution:

Given,

f (x) = x^{5} e^{x} + x^{3} log x – 2^{x}

∴ f'(x) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x^{5} e^{x} + x^{3} log x – 2^{x})

= \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x^{5}e^{x}) + \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x^{3} log x) – \(\frac{\mathrm{d}}{\mathrm{dx}}\) 2^{x}

= e^{x} \(\frac{\mathrm{d}}{\mathrm{dx}}\) x^{5} + x^{5} \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^{x} + log x \(\frac{\mathrm{d}}{\mathrm{dx}}\) x^{3} + x^{3} \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x = 2^{x} log_{e} 2

= e^{x}.5x^{4} + x^{5} . e^{x} + log x . 3x^{2} + x^{3} . \(\frac{1}{2}\) 2^{5} log_{e}2

= 5x^{4} e^{x} + e^{x} x^{5} + 3 log x. x^{2} + x^{2} – 2^{x} log_{e}2

= (5x^{4} + x^{5}) e^{x} + (3 log x + 1) x^{2} – 2^{x} log_{e} 2.

Question 29.

Whatever be 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. [3]

OR

Find the equation of a straight line which makes acute angle with positive direction of X-axis, passes through point (-5, 0) and is at a perpendicular distance of 3 units from the origin.

Solution:

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 eqs. (iii) and (iv), we get

α^{2}(cos^{2} t + sin^{2} t) + β^{2}(sin^{2} t + cos^{2} t) = a^{2} + b^{2}

⇒ α^{2} + β^{2} = a^{2} + b^{2}

[∵ sin^{2} θ + cos^{2} θ = 1]

Therefore, the locus of the point P(α, β) is

x^{2} + y^{2} = a^{2} + b^{2}

which represents a circle with centre (0, 0) and radius

= \(\sqrt{a^2+b^2}\)

OR

Let ‘α’ be the acute angle made by the line with positive x-axis.

∴ Equation of the line is

x cos α + y sin α = 3 ……………. (i)

Since the line passes through (- 5, 0),

So, – 5 cos α + 0 . sin α = 3

⇒ cos α = –\(\frac{3}{5}\)

Therefore, sin α = \(\sqrt{1-\left(-\frac{3}{5}\right)^2}\)

= \(\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)

From (i), x\(\left(-\frac{3}{5}\right)\) + y\(\left(\frac{4}{5}\right)\) = 3

⇒ 3x – 4y + 15 = 0 is the required equation of line.

Question 30.

(a) Prove that the last day of a century cannot be Tuesday or Thursday or Saturday. [3]

(b) How many times does the 29th day of the month occur in 400 consecutive years ?

Solution:

(a) 100 years contain 5 odd days.

∴ Last day of 1^{st} century is Friday.

200 years contain (5 × 2) = 10 days = 3 odd days

∴ Last day of 2^{nd} century is Wednesday.

300 years contain (5 × 3) = 15 days = 1 odd day

∴ Last day of 3^{rd} century is Monday.

400 years contain 0 odd day.

∴ Last day of 4^{th} century is Sunday.

This cycle is repeated.

∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

Hence Proved.

(b) In 400 consecutive years, there are 97 leap years. Hence, in 400 consecutive years, February has the 29^{th} day 97 times and the remaining 11 months have the 29 th day 400 × 11 = 4400 times.

Hence, 29^{th} day of the month occurs (4400 + 97) = 4497 times

Question 31.

In a survey it was found that 21 persons liked product A, 26 liked product B, and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find : [3]

(a) The number of people who liked at least one product.

(b) The number of people who liked product C only.

Solution:

Given, n(A) = 21, n(B) = 26, n(C) = 29

n(A ∩ B)= 14, n(C ∩ A) = 12

n(B ∩ C)= 14, n(A ∩ B ∩ C) = 8

(a) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

= 21 + 26 + 29 – 14 – 12 – 14 + 8

= 44

(b) n(C only) = n(C) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)

= 29 – 12 – 14 + 8

= 11

Question 32.

If the 4^{th}, 10^{th} and 16^{th} terms of G.P. are x, y and z respectively, prove that x, y, z are in G.P. [3]

OR

Find the sum of n terms of the sequence 8, 88,888, 8888 ……………..

Solution:

Given, T_{4} = x ⇒ ax^{4-1} = x ⇒ ar^{3} = x ………….. (i)

T_{10} = y ⇒ ar^{10-1} = y ⇒ ar^{9} = y ……………… (ii)

T_{16} = z ⇒ ar^{16-1} = z ⇒ ar^{15} = y ……………. (iii)

Now, multiplying Eq. (i) by Eq. (iii),

⇒ ar^{3} × ar^{15} = x × z

⇒ a^{2}r^{3+15} = xz

⇒ a^{2}r^{18} = xz

⇒ (ar^{9})^{2} = xz

⇒ y^{2} = xz

^ y2 = xz [From Eqn. (ii)]

Therefore, x, y, z are in G.P

OR

Let S = 8, 88, 888, 8888 + + n terms

⇒ S = 8 (1 + 11 + 111 + 1111 + …. + n terms)

= \(\frac{8}{9}\) [9 + 99 + 999 + 9999 + …. + n terms]

= \(\frac{8}{9}\) [(10 -1) + (100 -1) + (1000 -1) + …. + n terms ]

= \(\frac{1}{2}\) [10 + 100 + 1000 + ……………. + n terms – (1 + 1 + 1 + …………. + n terms)]

Question 33.

From 6 different novels and 3 different dictionaries, 4 novels and a dictionary is to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then find the number of such arrangements. [3]

Solution:

The number of ways in which 4 novels can be selected from 6 different novels = ^{6}C,sub>4 = 15

The number of ways in which 1 dictionary can be selected out of 3 different dictionaries = ^{3}C_{1} = 3

Now, it is given that dictionary is always in the middle.

Hence, the arrangement look like M M D M M

∴ Dictionary is always in the middle as the novels are different, hence they can be arranged in 4! ways.

Hence, total number of such arrangement

= 15 × 3 × 4!

= 15 × 3 × 4 × 3 × 2 × 1

= 1080

Commonly Made Error

Students commit error in finding total number of arrangements.

Answering Tip

To solve the problem like this, you have to revise properties of combination and revise the concept thoroughly.

Section – D (15 Marks)

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

Question 34.

A factory has three machines A, B and C producing 1500, 2500 and 3000 bulbs per day, respectively. Machine A produces 1.5% defective bulbs, machine B produces 2% defective bulbs and machine C produces 2.5% defective bulbs. At the end of the day, a bulb is drawn at random and is found to be defective. What is the probability that the defective bulb has been produced by machine B ? [5]

Solution:

Given that, total daily production of bulbs

= 1500 + 2500 + 3000

= 7000 bulbs.

Let A, B and C be the event if drawing a bulb produced by machines A, B and C respectively.

∴ P(A) = \(\frac{1500}{7000}\) = \(\frac{3}{14}\)

P(B) = \(\frac{2500}{7000}\) = \(\frac{5}{14}\)

and P(C) = \(\frac{3000}{7000}\) = \(\frac{3}{7}\)

Let F be the event of producing a defective bulb

Commonly Made Error

Students forget to define the events and apply Bayes’ theorem directly and lose marks.

Answering Tip

For Bayes’ theorem and total probability law events should be well defined.

Question 35.

(a) A sum of money doubles itself in 4 years compound interest. It will amount to 8 times itself at the same rate of interest in how many years? [5]

(b) Compound interest on a sum of money for 2 years at 4% per annum is ₹ 2448. Find simple interest on the same sum of money at the same rate of interest for 2 years.

OR

Compute the taxable value of the perquisite in respect of medical facilities availed of by X from his employer in the following situations :

(a) The employer reimburses the following medical expenses :

(i) Treatment of X by his family physician ₹ 8,400

(ii) Treatment of Mrs. X in a private nursing home ₹ 7,200

(iii) Treatment of X’s mother (dependent upon him) ₹ 2,400 by a private doctor

(iv) Treatment of X’s brother (not dependent upon him) ₹ 800

(v) Treatment of X’s grandfather (dependent upon him) ₹ 3,000

(b) The employer reimburses an insurance premium of ₹ 6,000 paid by X under a health insurance scheme on the life of X and his wife.

(c) The employer maintains a hospital for the employees where they and their family members are provided free treatment The expenses on treatment of X and his family members during the previous year 2018-19 were as under :

S.No. | Particulars | Amount (₹) |

(i) | treatment of X’s major son (dependent upon him) | 4,400 |

(ii) | treatment of X | 10,400 |

(iii) | treatment of X’s uncle | 9,200 |

(iv) | treatment of Mrs. X | 16,000 |

(v) | treatment of X’s widowed sister (dependent upon him) | 8,200 |

(vi) | treatment of X’s handicapped nephew | 5,000 |

(d) Expenses on cancer treatment of married daughter of X at Tata Memorial Hospital, Mumbai paid by the employer ₹ 1,00,000 and reimbursement of expenses for medical treatment of himself amounting to ₹ 40,000.

Solution:

(a) Let Principal = P

Rate = i% = \(\frac{\mathrm{i}}{\mathrm{100}}\)

t = 4 years

∴ Amount = 2P

By comparing both sides,

t = 12 years

(b) Time (t) = 2 years

Rate, i = 4% = 0.04

Effective interest rate of compound interest 4% of 2 years

= (1 + i)^{n} – 1

= (1 + 0.04)^{2} – 1

= (1.04)^{2} – 1

= 1.0816 – 1

= 0.0816 ≈ 8.16%

Effective rate of interest of simple interest for 2 years = 8%

According to the question,

8.16% of sum = ₹ 2448

1% of sum = ₹ \(\frac{2448}{8.16}\)

8% of sum = ₹ \(\frac{2448}{8.16}\) × 8

= ₹ 2400

Commonly Made Error

The questions on compound and simple interest is usually formula based, but many students make calculations errors while simplifications.

Answering Tips

- Always double check your calculation to remove errors and not to lose marks due to simplification errors.
- Practice adequate question in order to make your calculation speed faster.

OR

S.No. | Particulars | Expenses Taxable |

(i) | Treatment of X | 8,400 |

(ii) | Treatment of Mrs. X | 7,200 |

(iii) | Treatment of X’s mother | 2,400 |

(iv) | Treatment of X’s brother | 800 |

(v) | Treatment of X’s grandfather Less. Exempt | 3,000 |

21,800 |

Hence ₹ 21,800 shall be taxable perquisite w.e.f. assessment year 2019-20, medical reimbursement shall be fully taxable.

(b) Reimbursement of insurance premium on the health of the employee and his family members is a tax-free

perquisite. Hence nothing is taxable.

(c) The expenses of medical treatment of the employee and his family members in a hospital maintained by the

employer are tax-free. Therefore, expenses on treatment of X, X’s major son, X’s widowed sister and Mrs. X are not taxable. Only the following expenses are taxable :

S.No. | Particulars | Amount (₹) |

(i) | Treatment of X’s uncle | 9,200 |

(ii) | Treatment of X’s handicapped nephew | 5,000 |

Taxable perquisite | 14,200 |

(d) Expenses on medical treatment of the employee/family members in respect of prescribed diseases, in any hospital approved by the Chief Commissioner of Income- tax, are tax-free. In this case as cancer is a prescribed disease and Tata Memorial Hospital, Mumbai is approved by Chief Commissioner of Income-tax, there is no taxable perquisite. However, ? 40,000 reimbursement of medical expenses, shall be taxable perquisite in this case.

Question 36.

If x^{y} – y^{x} = a^{b}, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) [5]

OR

Find the correlation coefficient between the heights of husbands and wives based on the following data (given in inches) and interpret the result.

Couple | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

Height of husband | 76 | 75 | 75 | 72 | 72 | 71 | 71 | 70 | 68 | 68 | 68 | 68 | 67 | 67 | 62 |

Height of wife | 71 | 70 | 70 | 67 | 71 | 65 | 65 | 67 | 64 | 65 | 65 | 66 | 63 | 65 | 61 |

Solution:

Given, x^{y} – y^{x} = ab

⇒ e^{y logx} + e^{x logy} = e^{b loga}

On differentiating both sides w.r.t. x, we get

⇒ {x^{y} log x + xy^{x-1}} \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + {yx^{y-1} + y^{x} log y} = 0

⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – \(\left\{\frac{y x^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}\right\}\)

Commonly Made Error

Some students take logarithm directly without separating the equation.

Answering Tip

log (a + b) ≠ log a + log b. So, separate the function and find the derivatives.

OR

We use assumed mean A = 70, B = 66 and use the formula

Now, using (i),

which is a strong positive correlation. This shows that tall men usually marry tall women and short men marry short women (called assertive matching)

Section – E (8 Marks)

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

Question 37.

In order to increase the framing business, a farmer decided to purchases a tractor. As the cost of new tractor is very high and due to the lack of money, he decided to buy a used tractor. The cost of used tractor is ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalment of ₹ 500 plus 12% interest on the unpaid amount. [4]

(a) Interest paid by farmer on 1^{st} instalment is:

(A) ₹ 720

(B) ₹ 660

(C) ₹ 600

(D) ₹ 700

Solution:

Option (A) is correct.

Explanation:

Total cost = ₹ 12000

Down payment = ₹ 6000

Balance Amount = ₹ 6000

Interest of 1st instalment = \(\frac{6000 \times 12 \times 1}{100}\)

= ₹ 720 (∵ I = \(\frac{P \times R \times T}{100}\))

(b) Interest paid by farmer on 2^{nd} instalment is:

(A) ₹ 720

(B) ₹ 660

(C) ₹ 600

(D) ₹ 700

Solution:

Option (B) is correct.

Explanation:

Unpaid amount = 6000 – 500

= ₹ 5500

Interest of II ^{nd} instalment = \(\frac{5500 \times 12 \times 1}{100}\)

= ₹ 660

(c) Interest paid by farmer on 3^{rd} instalment is:

(A) ₹ 720

(B) ₹ 660

(C) ₹ 600

(D) ₹ 700

Solution:

Option (C) is correct.

Explanation:

Unpaid amount = 5500 – 500

= ₹ 5000

Interest on III^{rd} instalment = \(\frac{5000 \times 12 \times 1}{100}\)

= ₹ 600

(d) Total interest paid by farmer is :

(A) ₹ 4680

(B) ₹ 12000

(C) ₹ 4860

(D) ₹ 16680

Solution:

(d) Option (A) is correct.

Explanation:

Total interest paid by him = 720 + 660 + 600 + … + 12 terms which is an

A.P with a = 720, d = 660 – 720 = – 60

Therefore, total interest = \(\frac{12}{2}\) [2 × 720 + (12 – 1) (- 60)]

= 6[1440 – 11 × 60]

= 6[1440 × 660]

= 6 × 780 = ₹ 4680

Question 38.

A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15, 0.20, 0.31, 0.26 and 0.80. Find the probabilities that a particular surgery will be rated : [4]

Solution:

Let E_{1}, E_{2}, E_{3}, E_{4} and E_{5} be the event that surgeries are rated as very complex, complex, routine, simple or very simple, respectively.

P(E_{1}) = 0.15,

P(E_{2}) = 0.20,

P(E_{3}) = 0.31,

P(E_{4}) = 0.26,

P(E_{5}) = 0.08

(a) Complex or very complex :

(A) 0.35

(B) 0.77

(C) 0.51

(D) 0.57

Solution:

Option (A) is correct.

Explanation:

P (complex or very complex)

= P(E_{1} or E_{2})

= P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

= 0.15 + 0.20 – P(E_{1} ∪ E_{2}) = 0

⇒ P(E_{1} ∪ E_{2}) = 0.35

(b) Neither very complex nor very simple :

(A) 0.35

(B) 0.77

(C) 0.51

(D) 0.57

Solution:

Option (B) is correct.

Explanation:

P (neither very complex nor very simple),

P (E’_{1} ∩ E’_{5}) = P (E_{1} ∪ E_{5})’

= 1 – P(E_{1} ∪ E_{5})

= 1 – [P(E_{1}) + P(E_{5})]

= 1 – (0.15 + 0.08)

= 1 – 0.23

= 0.77

(c) Routine or complex :

(A) 0.35

(B) 0.77

(C) 0.51

(D) 0.57

Solution:

Option (C) is correct.

Explanation:

P (routine or complex)

= P(E_{3} ∪ E_{2})

= P(E_{3}) + P(E_{2})

= 0.31 + 0.20

= 0.51

(d) Routine or simple :

(A) 0.35

(B) 0.77

(C) 0.51

(D) 0.57

Solution:

Option (D) is correct.

Explanation:

P (routine or simple) = P(E_{3} ∪ E_{4})

= P(E_{3}) + P(E_{4})

= 0.31 + 0.26

= 0.57