CBSE Sample Papers for Class 11 Applied Mathematics Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Applied Mathematics Set 4 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 sum of binary numbers 111 + 100. [1]
Solution:

Question 2.
Find the value of (125)^2/3 × \(\sqrt{25}\) × \(\sqrt[3]{5^3}\) × 5^1/2. [2]
OR.
Find the logarithm of 64 to the base 2 \(\sqrt{2}\).
Solution:

Question 3.
If A and B are two events such that A ⊂ B and P(B) ≠ 0, then P\(\left(\frac{A}{B}\right)\) ≥ P(A) is true or not ? [1]
Solution:
If A ⊂ B then A ∩ B = A
∴ P(A ∩ B) = P(A)
Also, P(A) < P(B)
Consider, P\(\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}\) …………… (i)
It is known that P(B) ≥ 1
∴ \(\frac{1}{P(B)}\) ≥ 1
or \(\frac{P(A)}{P(B)}\) ≥ P(A)
From eqs. (i) and (ii), we get
P\(\left(\frac{A}{B}\right)\) ≥ P(A)

Question 4.
For what value of X for which the equation 2(x² + y²) – 6x + 8y + λ = 0 represents a point circle. [1]
Solution:
Given equation of circle is
2(x² + y²) – 6x + 8y + λ = 0
or, (x² + y²) – 3x + 4y + \(\frac{\lambda}{2}\) = 0 …………. (i)
If the radius of the circle is zero, then the circle is called a point circle. Thus, eq. (1) will represents a point circle, if
\(\sqrt{\left(\frac{3}{2}\right)^2+(-2)^2-\frac{\lambda}{2}}\) = 0 [∵ r = \(\sqrt{g^2-f^2-c}\)]
⇒ \(\frac{\lambda}{2}\) = \(\frac{25}{4}\)
⇒ \(\frac{25}{2}\)

Commonly Made Error
Some students don’t understand the question due to lack of understanding of point circle.

Answering Tip
Always remember, whenever dealing with point circle, the radius is always zero.

Question 5.
There are four bus routes between A and B, and three bus routes between B and C. A man can travel round-trip in number of ways by bus from A to C via B. If he does not want to use a bus route more than once, in how many ways can he make round trip ? [1]
OR
If ⁿC₁₂ = ⁿC₈, then find the value of n.
Solution:
In the following figure, there are 4 bus routes from A to B and 3 routes from B to C. Therefore, there are 4 × 3 = 12 ways to go from A to C.

It is a round trip so that the man will travel back from C to A via B. It is restricted the man cannot use same bus routes from C to B and B to A more than once. Thus, There are 2 × 3 = 6 routes for return journey. Therefore, the required number of ways = 12 × 6 = 72.

OR

Given, ⁿC₁₂ = ⁿC₈
⇒ ⁿC₁₂ = ⁿC_n-8 [∵ ⁿC_r = ⁿC_n-r]
⇒ 12 = n – 8
⇒ n = 20

Commonly Made Error
Students solve combination separately and then put them equal which gives a lengthy solution.

Answering Tip
Apply property, ⁿC_r = ⁿC_n-r.

Question 6.
Write the Parametric equations of the parabola y² = 12x. [1]
OR
Determine the position of points (3, -4) relative to parabola y² = 9x.
Solution:
The given equation of parabola is of the form y² = 4ax
On comparing the equation, we get 4a = 12 ⇒ a = 3
Therefore, the parametric equations of the given parabola are x = 3t² and y = 6t.
[∵ The parametric equation of parabola x² = 4ay are x = 2at and y = at².]

OR

For point (3, – 4), y₁ = – 4 and x₁ = 3.
Therefore, y₁² – 9x₁ = (- 4)² – 9(3)
= 16 – 27 = – 11 < 0
Hence, the point (3, – 4) lies inside parabola y² = 9x.

Question 7.
Out of the words (A) Advertisement (B) News (C) Editor (D) Paper (v) Date, which is most appropriate to complete the sentence ‘A newspaper always has ……………… ‘. [1]
Solution:
All the given words are relevant for a newspaper, but out of these the word ‘News’ is most appropriate and remaining are not appropriate. So, the given sentence can be completed as A newspaper always has news’.

Question 8.
Find the domain of \(\sqrt{a^2-x^2}\) (a > 0). [1]
Solution:
Let f(x) = \(\sqrt{a^2-x^2}\) ; f(x) is defined.
\(\sqrt{a^2-x^2}\) ≥ 0
⇒ x² – a² ≤ 0
⇒ x² ≤ a²
⇒ -a ≤ x ≤ a
∴ Domain of f(x) = [-a, a]

Question 9.
Evaluate : \(\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}\) [1]
Solution:

Question 10.
If the variance of data is 121, then find the standard deviation of data. [1]
Solution:
We know that, S.D. = \(\sqrt{\text { variance }}\)
= \(\sqrt{121}\)
= 11

Question 11.
What percentile in 74 is the following group of data (must be in order) ? [1]
35, 65, 68, 100, 23, 69, 74, 61, 70, 95, 85, 55.
Solution:
Arrange the given data in ascending order, we get 23, 35, 55, 61, 65, 68, 69, 70, 74, 85, 95, 100
Here, total no. of observations, Y = 12
and score 74 is at the rank 9, so M = 9
Using formula,
PR = \(\frac{\mathrm{M}}{\mathrm{Y}}\) × 100 = \(\frac{9}{12}\) × 100 = 75^th percentile

Question 12.
If ₹ 1200 is lent out at 5% per annum simple interest for 3 years, then find the amount after 3 year. [1]
Solution:
A = P + I
A = 1200 + P i t
= 1200 + 1200 × \(\frac{5}{100}\) × 3
= 1200(1 + 0.15)
= 1200 × 1.15 = ₹ 1380

Question 13.
David obtained 76, 65, 82 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks? [1]
OR
January 1, 2007 was Monday. What day of the week lies on January 1, 2008 ?
Solution:
Average = \(\frac{76+65+82+67+85}{5}\)
= \(\frac{375}{5}\) = 75

OR

The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2008 will be 1 day beyond Monday. Hence, it will be Tuesday.

Question 14.
A dealer in Agra (U.P.), say Ramesh, supplies products and services worth ₹ 10,000 to Suresh a person in Lucknow (U.P.). If the rate of GST is 28% find the SGST. [1]
Solution:
Since, the GST rate is 28%,
∴ SGST rate is 14%.
i.e, SGST = 14% of ₹ 10,000
= ₹ \(\frac{14}{100}\) × 10, 000 = ₹ 1400

Question 15.
A line passes through P(1, 2) such that its intercept between the axes is bisected at P. Find the equation of line. [1]
Solution:
Equation of a line in intercept form is
\(\frac{\mathrm{x}}{\mathrm{a}}\) + \(\frac{\mathrm{y}}{\mathrm{b}}\) = 1
Given, \(\frac{\mathrm{a+0}}{\mathrm{2}}\) = 1 and \(\frac{\mathrm{0+b}}{\mathrm{2}}\) = 2
⇒ a = 2 and b = 4
∴ \(\frac{\mathrm{x}}{\mathrm{2}}\) + \(\frac{\mathrm{y}}{\mathrm{4}}\) = 1
⇒ 2x + y – 4 = 0, is the required equation of the line.

Question 16.
The collection of difficult topics in Mathematics is a set or not. Justify your answer. [1]
OR
Find the domain of the relation, R = {(x, y): x, y ∈ Z, xy = 4}.
Solution:
The collection of difficult topic in Mathematics is not a set, because the term ‘difficult topic’ is not well defined.

OR

Given, R = {(x, y): x, y ∈ Z, xy = 4}
= {(- 4, – 1), (- 2, – 2), (- 1, – 4), (1, 4), (2, 2), (4, 1)}
Therefore, domain of R = {- 4, – 2, – 1, 1, 2, 4}

Section – B (20 Marks)

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

Question 17.
Find the domain and range of the real function f(x) = \(\frac{x^2-9}{x-3}\). [2]
Solution:
Given, f(x) = \(\frac{x^2-9}{x-3}\)
Domain: Clearly,f(x) is defined for all x ∈ R except x = 3
∴ Domain of f = R – {3}
Range: Let y = f(x)
∴ y = \(\frac{x^2-9}{x-3}\) ⇒ y = x + 3
It follows from the above relation that y takes all real values except 6 when x takes values in the set R – {3}.
∴ Range of f = R – {6}

Question 18.
Find the derivative of f(e^{tan x}) w.r.t. x at x = 0. It is given that f'(l) = 5. [1]
Solution:
Given f(e^{tan x})
∴ y = f(e^{tan x})
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = f'(e^tanx) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (e^{tan x})
or \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = f'(e^tanx) e^{tan x} sec² x
Now \(\left[\frac{d y}{d x}\right]_{x=0}\) = f'(e^{tan 0}) e^{tan 0} sec² 0
= f'(e⁰)e⁰1
= f'(1)
\(\left[\frac{d y}{d x}\right]_{x=0}\) = 5. [given f'(1) = 5]

Commonly Made Error
Sometimes use of the rule for composite function as well as chain rule are forgotten by several students.

Answering Tip
Derivatives of all forms of functions require practice and revision from time to time. Sufficient time should be spent on this topic.

Question 19.
Find the unknown value in the following table : [2]

Class Interval	Frequency	Cumulative frequency
0-10	5	5
10-20	7	x1
20 – 30	x2	18
30-40	5	x3
40-50	x4	30

OR
The sum of the squares of deviations for 10 observations taken from their mean 50 is 250. Find the coefficient of variation.
Solution:
x₁ = 5 + 7 = 12
x₂ = 18 – x₁= 18 – 12 = 6
x₃ = 18 + 5 = 23
and x₄ = 30 – x₃ = 30 – 23 = 7

OR

Here, N = 10, \(\bar{x}\) = 50
and, \(\sum_{i=1}^{10}\) (x_i – 50)² = 250
⇒ \(\frac{\sum_{i=1}^{10}\left(x_i-50\right)^2}{10}\) = 25
⇒ σ² = 25
⇒ σ = 5
∴ Coefficient of variation,
C.V. = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{5}{50}\) × 100
= \(\frac{500}{50}\)
= 10
∴ 10 is the coefficient of variation.

Question 20.
For a group of 60 boys students, the mean and S.D. of statistics marks are 45 and 2 respectively. The same figures for a group of 40 girls students are 55 and 3 respectively. What is the mean and S.D. of marks if the two groups are pooled together ? [2]
Solution:
As given n₁ = 60, \(\bar{x}_1\) = 45, σ₁ = 2, n₂ = 40, \(\bar{x}_2\) = 53, σ₂ = 3
Thus, combined mean is given by

Commonly Made Error
Some students made mistake in writing the formula of combined mean and combined S.D.

Answering Tips
If there are two groups containing n₁ and n₂ observations, with \(\bar{x}_1\) and \(\bar{x}_2\) are the means and σ₁ and σ₂ are the standard deviation, then
Combined mean, \(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)
Combined standard deviation
σ = \(\sqrt{\frac{n_1 \sigma_1^2+n_2 \sigma_2^2+n_1 d_1^2+n_2 d_2^2}{n_1+n_2}}\)
where, d₁ = \(\bar{x}_1-\bar{x}\) and d₂ = \(\bar{x}_2-\bar{x}\)

Question 21.
ABC Ltd. wants to lease out an asset costing ₹ 36,000 for five years has fixed rental of ₹ 105000 per annum payable annually starting from the end of first year. Suppose rate of interest is 14% per annum compounded annually on which money can be invested by the company. Is this agreement favourable to the company ? (Given, P(5, 0.14) = 3.43308) [2]
Solution:
First we have to compute the present value of the annuity of ₹ 105000 for five years at the interest rate of 14% p.a. compounded annually.
The present value PV of the annuity is given by
= C.F. P(n, i)
Here, C.F. = 105000, i = 14% = 0.14, n = 5 and
P(n, i) = P(5, 0.14) = 3.43308
∴ PV = 105000 × 3.43308
= ₹ 360473.40
which is greater than the initial cost (₹ 36,0000) of the asset and consequently leasing is favourable to the lessor.

Question 22.
Mr. Roy need to borrow money. His neighbourhood bank charges 8% interest compounded semi-annually. An internet bank charges 7.9% interest compounded monthly. At which bank will Mr. Roy pay lesser amount of interest ? [2]
Solution:
Compare effective rates:
Neighbourhood bank,
E = (1 + \(\frac{0.08}{2}\))² – 1
= 0.0816 ≈ 8.16%
Internet bank,
E = (1 + \(\frac{0.79}{12}\))¹² – 1 = 0.08192 ≈ 8.19%
The neighbourhood bank has the lower effective rate although it has a higher nominal (actual) rate.

Question 23.
If Y = 50, SEA = 50, then “YACHT” will be equal to? [2]
OR
In a family, there are six members A, B, C, D, E and F. A and B are a married couple, A being the male member. D is the only son of C, who is the brother of A. E is the sister of D. B is the daughter-in-law of F, whose husband has died. How is E related to C ?
Solution:
The patter followed here is Y = 25 then 25 × 2 = 50 SEA, if we analyse the place value we get, SEA
= 19 + 5 + 1 then 25 × 2 = 50
Similarly for YACHT it will be, YACHT
= 25 + 1 + 3 + 8 + 20 then 57 × 2 = 114
Hence, the answer is 114.

OR

A is a male and married to B. So, A is the husband and B is the wife. C is the brother of A. D is the son of C. E. who is the sister of D will be the daughter of C. B is the daughter-in-law of F whose husband has died means F is the mother of A.
Clearly, E is the daughter of C.

Question 24.
A man starts repaying a loan as first installments of ₹ 100. If he increases the installment by ₹ 5 every month, what amount he will pay in the 30^th installment? [2]
Solution:
Given, a = 100 and d = 5
∵ T_n = a + (n – 1)d
∴ T₃₀ = 100 + (30 – 1) (5)
= 100 + 29 × 5
= 100 + 145
= 245
Therefore, he will pay ₹ 245 in the 30th instalment.

Question 25.
Justify for the sets A, B and C, (A ∩ B) ∪ C = A ∩ (B ∪ C). [2]
Solution:
Let A = {a, b}, B = {b, c} and C = {c, d}
Now, (A ∩ B) ∪ C = ({a, b} ∩ {b, c}) ∪ {c, d}
= {b} ∪ {c, d} = {b, c, d}
and A ∩ (B ∪ C) = {a, b} ∩ ({b, c} ∪ {c, d})
= {a, b} ∩ {b, c, d} = {b}
From above, it is observed that
(A ∩ B) ∪ C ≠ A ∩ (B ∪ C)
i.e., given statement is not true.

Question 26.
Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination. [2]
OR
18 mice were placed in two experimental groups and one control group with all groups equally large. In how many ways can the mice be placed into three groups ?
Solution:
One ace will be selected from four aces and four cards will be selected from (52 – 4) = 48 cards. If P is the required number of ways, then
P = C(4, 1) × C(48, 4)
= \(\frac{4 !}{1 !(4-1) !} \times \frac{48 !}{4 !(48-4) !}\)
= \(\frac{4(3 !)}{1 ! 3 !} \times \frac{48 \times 47 \times 46 \times 45 \times 44 !}{4 \times 3 \times 2 \times 1 \times 44 !}\)
= 4 × 2 × 47 × 46 × 45
= 778320 ways

OR

It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
∴ Required arrangements = \(\frac{\text { Total arrangement }}{\text { Equally likely arrangement }}=\frac{18 !}{6 ! 6 ! 6 !}\)

Section – C (21 Marks)

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

Question 27.
(A) Find the shortest distance of the point (8,1) from the circle (x + 2)² + (y – 1)² = 25. [3]
(B) Find the farthest distance of the point (1, 5) from the circle (x – l)2 + (y + l)2 = 16.
OR
Find the equation of a straight line which passes through the point (a, 0) and whose perpendicular distance from (2a, 2a) is a.
Solution:
(i) The centre of the circle (x + 2)² + (y – 1)² = 25 is C(- 2, 1) and its radius (r) is 5.

If P is the point (8, 1), then
CP = \(\sqrt{(8+2)^2+(1-1)^2}\)
= 10 units
The shortest distance of P from the circle = PA
= CP – AC
= 10 – 5
= 5 units.

(ii)The centre of the circle (x – 1)²+ (y + 1)² = 16 is C(1, – 1) and its radius (r) = 4.

If P is the point (1, 5), then
CP = \(\sqrt{(1-1)^2+(5+1)^2}\) = 6
The farthest distance of point P(1, 5) from the circle = PB
= CP + BC
= 6 + 4
= 10 units

OR

Let the straight line be Ax + By + C = 0 …………… (i)
Since, (i) passes through (a, 0), so
A . a + B . 0 + C = 0
⇒ Aa + C = 0
⇒ C = – Aa …………… (ii)
Also, the T distance of (i) from (2a, 2a) is a.

⇒ A² + B² = A² + 4AB + 4B²
⇒ – 3B² = 4AB
⇒ 4AB + 3B² = 0
⇒ B(4A + 3B) = 0
⇒ B = 0 or 4A + 3B = 0
If 4A + 3B = 0
⇒ B = –\(\frac{4 A}{3}\)
Substituting the values of B and C in eq. (i), we get
Ax – \(\frac{4}{3}\) Ay – Aa = 0
⇒ 3x – 4y – 3a = 0
∴ Equation of line is 3x – 4y – 3a = 0
Similarly, if B = 0, then substituting the values of B and C in eq. (i), we get
Ax = – 0.y + Aa
⇒ A(x – a) = 0
⇒ x – a = 0
∴ Equation of required line is x – a = 0.

Question 28.
A is 50% as efficient as B. C does half of the work done by A and B together. If C alone does the work in 40 days, then in how many days A, B and C together finish the work. [3]
Solution:
(A’s 1 day work): (B’s 1 day work) = 150 : 100 = 3: 2
Let A’s and B’s 1 day work be 3x and 2x, respectively.
Then, C’s 1 day work = \(\left(\frac{3 x+2 x}{2}\right)=\frac{5 x}{2}\)
∴ \(\frac{5 x}{2}\) = \(\frac{1}{40}\) or x = \(\left(\frac{1}{40} \times \frac{2}{5}\right)\) = \(\frac{1}{100}\)
A’s 1 day work = \(\frac{3}{100}\)
B’s 1 day work = \(\frac{1}{50}\)
C’s 1 day work = \(\frac{1}{40}\)
(A + B + C)’s 1 day work = (\(\frac{3}{100}\) + \(\frac{1}{50}\) + \(\frac{1}{40}\))
= \(\frac{15}{200}\) = \(\frac{3}{40}\)
So, A, B and C together can do the work in
= \(\frac{40}{3}\) = 13\(\frac{1}{3}\) days

Commonly Made Error
Sometimes students write incorrect relationship between efficiency and time.

Answering Tip
Learn the following relationship.
If A is a% as efficient as B, then ratio of (A’s 1 day’s work): (B’s 1 days’ work)
= (100 + a): 100

Question 29.
Two finite sets have m and n elements. The total number of subsets of first set is 56 more than the total number of subsets of the second set. Find the value of m and n. [3]
Solution:
Let A and B be two sets having m and n number of elements respectively.
Number of subsets of A = 2ⁿ
Number of subsets of B = 2ⁿ
According to questions
2^m – 2ⁿ = 56 = 8 × 7
2ⁿ(2^m-n – 1) = 2³(2³ – 1)
Therefore, n = 3 and m – n = 3
∴ m = 6

Question 30.
If a is A.M. of b and c and the two geometric means are G₁ and G₂, then prove that G₁³ + G₂³ = abc [3]
Solution:
It is given that a is the A.M. of b and c.
∴ a = \(\frac{b+c}{2}\) ⇒ b + c = 2a …………… (i)
Since G₁ and G₂ are two geometric means between b and c. Therefore, b, G₁, G₂, c is a G.P with common ratio r = \(\left(\frac{c}{b}\right)^{\frac{1}{3}}\)
∴ G₁ = br = b\(\left(\frac{c}{b}\right)^{\frac{1}{3}}\) = c^{\(\frac{1}{3}\)}b^{\(\frac{2}{3}\)}
and G₂ = br = b\(\left(\frac{c}{b}\right)^{\frac{2}{3}}\) = c^{\(\frac{1}{3}\)}b^{\(\frac{2}{3}\)}
⇒ G₁³ = b²c and G₂³ = bc²
⇒ G₁³ + G₂³ = b²c + bc²
⇒ G₁³ + G₂³ = bc(b + c)
⇒ G₁³ + G₂³ = 2abc [using (i)]

Question 31.
If a, b, c are in A.P. show that a\(\left(\frac{1}{b}+\frac{1}{c}\right)\), b\(\left(\frac{1}{c}+\frac{1}{a}\right)\) and c\(\left(\frac{1}{a}+\frac{1}{b}\right)\) are also in A.P. [1]
OR
How many different products can be obtained by multiplying two or more of the numbers 2, 5, 6, 7, 9 ?
Solution:
a, b, c are in A.P.

Commonly Made Error
Sometimes students have difficulty in assuming the terms of A.P. by which the result has been proved.

Answering Tip
Students are advised to learn all the formulae and practice as many questions as possible.

OR

The given numbers are 2, 5, 6, 7, 9.
The numbers of different product when 2 or more is taking = the number of ways of taking product of 2 numbers + number of ways of taking product of 3. numbers + numbers of ways of taking product of 4 numbers + number of ways of taking 5 together
= ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅
= \(\frac{5 !}{3 ! 2 !}+\frac{5 !}{2 ! 3 !}+\frac{5 !}{1 ! 4 !}+\frac{5 !}{0 ! 5 !}\)
= \(\frac{5.4}{2.1}+\frac{5.4}{2.1}+\frac{5}{1}\) + 1
= 10 + 10 + 5 + 1
= 26

Question 32.
The first four raw moments of a distribution are 2,136,320 and 40000. Find out coefficient of skewness and kurtosis. [3]
Solution:
Given that, μ’₁ = 2, μ’₂ = 136, μ’₃ = 320 and μ’₄ = 40000
First of all we have to calculate the first four central moments
μ₁ = μ’₁ – μ’₁ = 0 [H]
μ₂ = μ’₂ – (μ’₁)2 = 136 – (2)2 = 132 [H]
μ₃ = μ’₃ – 3m’₁ + (2m’₁)³
= 320 – 3 x 136 x 2 + 2(2)3
= 320 – 816 + 16 = – 480 [H]
μ₄ = μ’₄ – 4m’₁ μ’₃ + 6m’₂ (μ’₁)2 – 3(μ’₁)⁴
= 40000 – 4 × 2 × 320 + 6 × 136 × (2)² – 3 × (2)⁴
= 40000 – 2560 + 3264 – 48
= 40656
Skewness β₁ = \(\frac{\mu_3^2}{\mu_2^3}=\frac{(-480)^2}{(132)^3}\) = 0.10017
= 0.1002 (approx).
Kurtosis, β₂ = \(\frac{\mu_4}{\mu_2^2}=\frac{40656}{(132)^2}\) = 2.333 (approx.)
= 2.333. (approx.)
The value of β₂ is less than 3, hence the curve is platykurtic.

Question 33.
A manufacturer in a firm manufactures a machine and marks it at ₹ 80,000. He sells the machine to a wholesaler (in Gorakhpur) at a discount of 20%. The wholesaler sells the machine to a dealer (in Mathura) at a discount of 15% on the marked price. If the rate of GST 28%, find tax paid by the whole seller to central Government. [3]
Solution:
Given, Marked price = ₹ 80,000
and discount = ₹ \(\frac{20}{100}\) × 80,000
= ₹ 16000
Selling price by the manufacturer
= ₹ (80,000 – 16,000)
= ₹ 64,000
∴ Cost price of a machine by the wholesale = ₹ 64,000
and now selling price by the wholesale
= ₹ (80,000 – \(\frac{15}{100}\) × 80,000)
= ₹ (80,000 – 12000)
= ₹ 68000
∴ Tax paid by the wholesaler to the central government
= Tax on S.E – Tax on C.P
= \(\frac{28}{100}\) × 68,000 – \(\frac{28}{100}\) × 64,000
= \(\frac{28}{100}\) × (68,000 – 64,000)
= \(\frac{28}{100}\) × 4,000
= ₹ 1120

Commonly Made Error
A few calculation errors were however committed. Some candidates made calculation errors in finding the tax paid by the wholesaler. The concept that the tax charged on the additional value of a machine was not clear.

Answering Tip
Concepts of tax must be made clear to the candidates. Students should read the question carefully and try finding it at a time. This helps in avoiding errors.

Section – D (15 Marks)

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

Question 34.
It is estimated that 50% of e-mails are spam e-mails. Some software has been applied to filter these spam e-mails before they reach your inbox. A certain brand of software claims that it can detect 99% of spam e-mails, and the probability for a false positive (a non-spam email detected as spam) is 5%. Now if an email is detected as spam, then what is the probability that it is in fact a non-spam email ? [5]
Solution:
Let
A = event that an email is detected as spam,
B = event that on email is spam
B^c = event that an email is not spam
We know that,
P(B)= P(B^c) = 0.5, P(A/B) = 0.99 and P(A/B^c) = 0.05
Hence by the Bayes’ formula, we have
P(B^c/A) = \(\frac{P\left(A / B^c\right) P\left(B^c\right)}{P(A / B) P(B)+P\left(A / B^c\right) P\left(B^c\right)}\)
= \(\frac{0.05 \times 0.5}{0.05 \times 0.5+0.99 \times 0.5}\)
= \(\frac{5}{104}\)

Question 35.
(a) In what time will ₹ 8000 amount to ₹ 8820 at 10% per annum interest compounded half-yearly ? [5]
(b) Find the rate percent per annum if ₹ 200000 amount to ₹ 231525 in 1\(\frac{1}{2}\) year interest being compounded half yearly.
OR
(a) Total income of Mr. X aged 35 years resident in India is ₹ 3,35,000. Compute tax liability for AY 2019-20.
(b) Total income of Mr. X aged 35 years (NR) in India is ₹ 3,35,000. Compute tax liability for AY 2019-20.
Solution:
(a) Here, interest rate per conversion period,
i = \(\frac{10}{2}\)% = 5% = 00.05
P = ₹ 8000
and A_n = ₹ 8820
We know,
A_n = P(1 + i)ⁿ
8820 = 8000 (1 + 0.05)ⁿ
⇒ \(\frac{8820}{8000}\) = (1.05)ⁿ
⇒ 1.1025 = (1.05)ⁿ
⇒ (1.05)² = (1.05)ⁿ
⇒ n = 2
Hence, number of conversion period is 2 and the required time = \(\frac{n}{2}\) = \(\frac{n}{2}\) = 1 years

(b) Here, P = ₹ 200000
Number of conversion period (n) = 1\(\frac{1}{2}\) × 2
= \(\frac{3}{2}\) × 2 = 3
⇒ A_n = ₹ 231525
We know,
A_n = P(1 + i)ⁿ
⇒ 231525 = 200000 (1 + i)³
⇒ \(\frac{231525}{200000}\) = (1 + i)³
⇒ 1.157625 = (1 + i)³
⇒ (1.05)³ = (1 + i)³
⇒ 1 + i = 1.05
⇒ i = 0.05 = 5%
Hence, i is the rate of interest per conversion period (six months) = 5%
Thus, rate per annum = 5% × 2 = 10%

OR

(a) Since Mr. X is a resident having Total Income ₹ 3,50,000, rebate u/s 87A is available.

First ₹ 2,50,000	Nil
Next ₹ 85,000 @ 5%	₹ 4,250
Total	₹ 4,250
Rebate u/s 87A = Lower of (i) Tax payable or (ii) ₹ 2,500	(₹ 2,500)
Tax after rebate	₹ 1,750
Add: 4% HEC	₹ 70
Tax rounded off	₹ 1,820

(b) Since Mr. X is a Non- Resident, rebate u/s 87A is Not available.

First ₹ 2,50,000	Nil
Next ₹ 85,000 @ 5%	₹ 4,250
Total	₹ 4,250
Add: 4% HEC	₹ 170
Tax rounded off	₹ 4,420

Question 36.
Find the first principle the derivative of x³ – 27. [5]
OR
Calculate, the PR for the score 35 and 55 for the distribution of scores given in the following table :

Scores	10-19	20 – 29	30 – 39	40-49	50 – 59	60-69	70 – 79	80 – 89	90 – 99
f	2	4	5	10	35	20	13	8	3

Solution:

OR

Scores	f	c. f.
9.5 – 19.5	2	2
19.5 – 29.5	4	6
29.5 – 39.5	5	11
39.5 – 49.5	10	21
49.5 – 59.5	35	56
59.5 – 69.5	20	76
69.5 – 79.5	13	89
79.5 – 89.5	8	97
89.5 – 99.5	3	100
	n = 100

We use formula, PR = \(\frac{100}{N}\) + (c.f. + \(\frac{X-l}{i}\) × f)
(i) For X = 35,
X lies between the class 29.5 – 39.5
So, c.f. = 6, l = 29.5, f = 5, i = 10, N = 100
∴PR = \(\frac{100}{100}\) + (6 + \(\frac{35-29.5}{10}\) × 5
= 1 + (6 + \(\frac{5.5}{2}\))
= 1 + (6 + 2.75)
= 9.75
Thus, percentile ranking of 35 is 9.75.

(ii) For X = 55,
X lies between the class 49.5 – 59.5
So, c.f. = 21, l = 49.5, f = 35, i = 10, N = 100
∴ PR = \(\frac{100}{100}\) + (21 + \(\frac{55-49.5}{10}\) × 35
= 1 + (21 + 19.25)
= 41.25
Thus, percentile ranking of 55 is 41.25.

Section – E (8 Marks)

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

Question 37.
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. [4]

According to the above given information answer the following questions ;
(a) The position of middle most flag is:
(A) 13^th
(B) 13.5^th
(C) 14^th
(D) 12.5^th
Solution:
Option (C) is correct.

Explanation:
Since, they have 27 flags, the middle most flag is 14^th flag.

(b) How many flags are left and right to the middle flag ?
(A) 14,12
(B) 13,13
(C) 13,14
(D) 14 13
Solution:
Option (B) is correct.

Explanation:
13 flags to left of middle and 13 flags to right of middle most flag.

(c) How much distance did she cover in completing this job and returning back to collect her books ?
(A) 339 m
(B) 634 m
(C) 364m
(D) 346 m
Solution:
Option (C) is correct.

Explanation:
The total distance travelled by her is
S = 2(2 + 4 + 6 + 8 + … up to 13 terms)
= 2 × 2(1 + 2 + 3 + 4 + … up to 13 terms)
= 4 × \(\frac{13 \times 14}{2}\)
= 364 m
Thus, Ruchi need to travel total distance as 364 m

(d) The maximum distance she travelled carrying a flag is:
(A) 13 m
(B) 52 m
(C) 27 m
(D) 26 m
Solution:
Option (D) is correct.

Explanation:
The maximum distance travelled by her is 2 × 13 = 26 m.

Question 38.
A college gets affiliation of NSS and NCC. Now, the teachers announce that interested candidate can take MSS or NCC along with their regular courses.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random. [4]

According to the above given information answer the following questions:

(a) The student has opted only NCC :
(A) \(\frac{1}{2}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{19}{30}\)
(D) \(\frac{11}{30}\)
Solution:
(a) Option (A) is correct.

Explanation:
Let C and S denote the event that the student opted for NCC and NSS respectively.
Then P(C) = \(\frac{30}{60}\) = \(\frac{1}{2}\)

(b) The student opted for NCC and NSS :
(A) \(\frac{1}{2}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{19}{30}\)
(D) \(\frac{11}{30}\)
Solution:
Option (B) is correct.

Explanation:
P(C ∩ S) = \(\frac{24}{60}\) = \(\frac{2}{5}\)

(c) The student opted for NCC or NSS :
(A) \(\frac{1}{2}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{19}{30}\)
(D) \(\frac{2}{15}\)
Solution:
Option (C) is correct.

Explanation:

(d) The student opted for neither NCC nor NSS :
(A) \(\frac{1}{2}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{2}{15}\)
(D) \(\frac{11}{30}\)
Solution:
Option (D) is correct.

Explanation:
P(C’ ∩ S’)= P(C ∪ S)’ = 1 – P(C ∪ S)
= 1 – \(\frac{19}{30}\) = \(\frac{11}{30}\)