CBSE Sample Papers for Class 12 Applied Mathematics Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Mathematics Set 7 with Solutions

Maximum Marks: 80 Marks
Time Allowed : 3 Hours

General Instructions:

This question paper contains five sections A, B, C, D and E. Each section is compulsory.
Section – A carries 20 marks weightage, Section – B carries 10 marks weightage, Section – C carries 18 marks weightage, Section – D carries 20 marks weightage and Section – E carries 3 case-based with total weightage of 12 marks.
Section A: It comprises of 20 MCQs of 1 mark each.
Section B: It comprises of 5 VSA type questions of 2 marks each.
Section C: It comprises of 6 SA type of questions of 3 marks each.
Section D: It comprises of 4 LA type of questions of 5 marks each.
Section E: It has 3 case studies. Each case study comprises of 3 case-based questions, where 2 VSA type questions are of 1 mark each and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks question in each case-study.
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 of the alternatives in all such questions.

Section – A (20 marks)

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

Question 1.
If A = \(\left[\begin{array}{ll}
p & 0 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 0 \\
5 & 1
\end{array}\right]\) are two matrices, then A2 = B is true for [1]
(a) p = 1
(b) p = -1
(c) p = 4
(d) no real value of p
Solution:
(d) no real value of p

Explanation:
Here, A² = \(\left[\begin{array}{cc}
p^2 & 0 \\
p+1 & 1
\end{array}\right]\). This matrix is not equal to B for any real value of p.

Question 2.
The solution set of the inequality x + 3y ≥ 6 is: [1]
(a) half plane not containing the origin
(b) whole xy – plane except the points lying on the line x + 3y = 6
(c) open half plane containing the origin
(d) none of these
Solution:
(a) half plane not containing the origin

Question 3.
A random variable X takes the values -1, 0, 1. Its mean is 0.6 and P(X = 0) = 0.2, then P(X = 1) is: [1]
(a) 0.7
(b) 0.5
(c) 0.4
(d) 0.3
Solution:
(a) 0.7

Explanation: Let P(X = -1) = p; P(X = 1) = q
Here we are given that
⇒ P(X = -1) + P(X = 0) + P(X = 1) = 1
p + q = 1 – 0.2 = 0.8 …………………. (1)
Further, (-1) × p + (0 × 0.2) + (1 × q)
= 0.6 [Given]
⇒ – p + q = 0.6 …………… (2)
Solving (1) and (2),
we get q = 0.7, i.e., P(X = 1) = 0.7

Question 4.
If x is a real number and x ≤ 3, then [1]
(a) -x > – 3
(b) -x < – 3
(c) -x ≥ – 3
(d) -x ≤ – 3
Solution:
(c) -x ≥ – 3

Explanation:
Given x ≤ 3, we have
(-1) x ≥ (-1)3, -x ≥ -3

Question 5.
A box has 100 pens of which 10% are defective. Then, the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective is: [1]
(a) \(\frac{9^5}{10^5}\)
(b) \(\frac{7}{5} \cdot \frac{9^5}{10^5}\)
(c) \(\frac{7}{5} \cdot \frac{9^4}{10^4}\)
(d) \(\frac{1}{2} \cdot \frac{9^5}{10^5}\)
Solution:
(c) \(\frac{7}{5} \cdot \frac{9^4}{10^4}\)

Explanation :
Let X be the random variable denoting the number of defective pens in the draw of 5 pens one by one with replacement
Here, p = P(a defective pen) = \(\frac{10}{100}\) = \(\frac{1}{10}\). So, q = \(\frac{1}{10}\)
We need to determine P(X ≤ 1)
Now, P(X ≤ 1) = P(X = 0) + P(X = 1)
= ⁵C₀ \(\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^5\) + ⁵C₁ \(\left(\frac{1}{10}\right)^1\left(\frac{9}{10}\right)^4\)
= \(\left(\frac{9}{10}+\frac{1}{2}\right)\left(\frac{9}{10}\right)^4=\frac{7}{5} \cdot \frac{9^4}{10^4}\)

Question 6.
The amount S of an ordinary annuity A, when the rate of interest is r and n is the number of years, is given by [1]
(a) S = \(\frac{A}{r}\) [(1 + r)ⁿ – 1]
(b) S = \(\frac{A}{r}\) [(1 + r)ⁿ – 1]
(c) S = Ar[(1 + r)ⁿ – 1]
(d) S = Ar[(1 + r)ⁿ – 1]
Solution:
(a) S = \(\frac{A}{r}\) [(1 + r)ⁿ – 1]

Question 7.
The order of the differential equation x²\(\frac{\mathrm{d}^{2} \mathrm{~x}}{\mathrm{dx}^{2}}\) – 3 \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = 0 is: [1]
(a) 2
(b) 1
(c) 0
(d) not defined
Solution:
(a) 2

Explanation:
The order of highest order derivative \(\frac{\mathrm{d}^{y} \mathrm{~s}}{\mathrm{dx}^{2}}\) is 2.

Question 8.
Present value of perpetuity of ₹ R payable at the end of every period forever at a rate of / per period is given by [1]
(a) R^j
(b) 0
(c) R + \(\frac{\mathrm{R}}{\mathrm{i}}\)
(d) \(\frac{\mathrm{R}}{\mathrm{i}}\)
Solution:
(d) \(\frac{\mathrm{R}}{\mathrm{i}}\)

Question 9.
Let Z be a standard normal variate. If P (Z < z) = 0.7967, then the value of z is: [1]
(a) 0.73
(b) 0.83
(c) 0.64
(d) 0.30
Solution:
(b) 0.83

Explanation: Here it is given that
P(Z < z) = 0.7967
P(Z < 0) + P(0 < Z < z) = 0.7967
P(0 < Z < z) = 0.2967
z = 0.83 [From the table]

Question 10.
The mean E(X) of the numbers obtained on throwing a die having written 1 on three faces , 2 on two faces and 5 on one face, is [1]
(a) 1
(b) 2
(c) 5
(d) \(\frac{8}{3}\)
Solution:
(b) 2

Explanation: Let X be the random variable denoting the number obtained on throwing a die.
Then, X takes the values 1, 2 and 5 . The corresponding probabilities are:
P(X = 1) = \(\frac{3}{6}\) or \(\frac{1}{2}\) ; P(X = 2) = \(\frac{2}{6}\) or \(\frac{1}{3}\) ;
P(X = 5) = \(\frac{1}{6}\)
So, the mean of X is
E(X) = 1 × \(\frac{1}{2}\) + 2 × \(\frac{1}{3}\) + 5 × \(\frac{1}{6}\) = 2

Question 11.
A man can row at 8 km/hr in still water. If the velocity of current is 2 km/hr and it takes him 4 hours to row to a place and return back, how for is the place? [1]
(a) 12 km
(b) 15 km
(c) 8 km
(d) 2 km
Solution:
(b) 15 km
Explanation: Downstream speed
= 8 + 2 = 10 km/hr
upstream speed = 8 – 2 = 6 km/hr
Let the distance be D
ATQ,
\(\frac{D}{10}\) + \(\frac{D}{6}\) = 4
\(\frac{3D+5D}{30}\) = 4
8D = 120
D = 15 km
∴ Distance of the place = 15 km.

Question 12.
If y = Ae^5x + Be^-5x, then its second derivative y₂ (or \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\)) is : [1]
(a) 25y
(b) 5y
(c) -25y
(d) 15y
Solution:
(a) 25y

Explanation:
Given y = Ae^5x + Be^-5x,
we have
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 5Ae^5x – 5 Be^-5x;
\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) = 25Ae^5x + 25 Be^-5x
= 25y

Question 13.
A random variable X follows Poisson distribution with parameter λ. If P(X = 0) = 0.2, then P(X = 1) equals.
(Given that ln 0.2 = -1.61) [1]
(a) 0.261
(b) 0.322
(c) 0.378
(d) 0.400
Solution:
(b) 0.322

Explanation:
For a P.D.
we have P(X = r) = \(\frac{\lambda^r}{r !}\).e^-λ
P(X = 0) = \(\frac{\lambda^0}{0 !}\) . e^-λ
⇒ e^-λ = 0.2
⇒ λ = – ln 0.2
= – (- 1.61) = 1.61 ………….. (1)
Also, we have P(X = r + 1) = \(\frac{\lambda}{r+1}\) P(X = r)
P(X = 1) = \(\frac{\lambda}{r+1}\) P(X = 0)
= λ × 0.2 = 0.322 {Using (1)}

Question 14.
\(\int\)(4e^3x + 1) dx is equal to : [1]
(a) \(\) – x + C
(b) 12e^3x + x + C
(c) 12e^3x – x + C
(d) \(\frac{2}{3}\) x^3/2 + x + C
Solution:
(d) \(\frac{2}{3}\) x^3/2 + x + C

Explanation:
\(\int\)(4e^3x + 1) dx = \(\int\)(4e^3x + 1) dx
= \(\frac{4 e^{3 x}}{3}\) x + C

Question 15.
The corner points of the feasible region of a LPP are (0, 0), (300, 0), (180, 120) and (0, 240). Then, the maximum value of Z = 2x + 3 y is: [1]
(a) 600
(b) 720
(c) 800
(d) 900
Solution:
(b) 720

Explanation:
Here,

Corner point	Corresponding value of Z
(0, 0)	2(0) + 3(0) = 0
(300, 0)	2(300) + 3(0) = 600
(180, 120)	2(180) + 3(120) = 720

The maximum value of Z is 720.

Question 16.
The point on the parabola y² = 8x for which the abscissa and ordinate change at the same rate, is: [1]
(a) (2, 4)
(b) (2, -4)
(c) (8, 8)
(d) (1, 2\(\sqrt{2}\))
Solution:
(a) (2, 4)

Explanation: Let y denote the ordinate and x denote the abscissa of the point at instant t.
According to the problem
\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = \(\frac{\mathrm{dx}}{\mathrm{dt}}\) ……………. (1)
Now, y² = 8x gives
2y\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 8\(\frac{\mathrm{dx}}{\mathrm{dt}}\)
⇒ 2y = 8 {Using (1)}
⇒ y = 4
Putting y = 4 in the equation of the parabola we have x = 2.
So, the required point is (2, 4).

Question 17.
If -2x + 5 > – 13, then [1]
(a) x > 9
(b) x < 9
(c) x ≥ – 9
(d) x ≤ – 9
Solution:
(b) x < 9 Explanation: -2x + 5 > – 13 implies -2x > – 18, which further gives 2x < 18, or x < 9

Question 18.
The feasible region of an LPP is shown in the figure. If Z = 3x – 4y, then Z_min = …………. [1]
(A) 0
(B) -16
(C) – 30
(D) – 46

Solution:
(d) – 46

Explanation:
The Line through (0, 0) and (6,12) is y = 2x.
The line through (0,4) and parallel to this line is y = 2x + 4
∴ The corner points are (0, 0), (6, 12), (6, 16) and (0, 4)
Hence, Z_min = -46

Assertion-Reason Questions

Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A)t 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): Feasible region is the set of points which satisfy all of the given constraints. [1]
Reason (R): The optimal value of the objective function is attained at the points n x-axis only.
Solution:
(c) (A) is true but (R) is false.

Explanation:
Because optimal value can be attained anywhere in the feasible region, not on the x-axis.

Question 20.
Assertion (A): A r.v. X can take values 8, 2 and P (X = 0) = P(X – 1) = P and E(x) = E (x²), then the value of p is \(\frac{1}{2}\). [1]
Reason (R): E (X) = Σx × p(x).
Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
We know that,
P (x = 0) + P (x = 1) + P(x = 2) = 1
P(X = 2) = 1 – 2 p
Thus, E(x) = (0 × P) + (1 × P) + (2 × (1 – 2P))
= 2 – 3P
E(x²) = (0 × P) + (1 + P) + (4 × (1 – 2P))
= 4 – 7P
Now, E(x) = E (x²)
2 – 3p = 4 – 7p
⇒ P = \(\frac{1}{2}\)

Section – B (10 Marks)

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

Question 21.
A man can row a boat in still water at a speed of 4.5 km/h. In a river flowing at the rate of 1.5 km/h, he starts his journey from point downstream and comes back to the same point. What is his average speed for the total journey? [2]
OR
Solve : \(\frac{4 x}{5}-\frac{2 x}{7}\) ≤ 1 + \(\frac{1}{2}\)
Solution:
Upstream speed = (4.5 – 1.5) km/hr
⇒ 3 km/h r
Downstream speed = (4.5 + 1.5) km/hr
⇒ 6 km/hr
Let speed covered by upstream and downstream be 6 km
Time taken to cover 6 km upstream distance = (6/3) hrs
⇒ 2 hrs
Time taken to cover 6 km dowstream distance = (6/6) hrs .
⇒ 1 hr
Total distance = (6 + 6) km
⇒ 12 km
Total time = (2 + 1) hrs
⇒ 3 hrs
Required Average speed = (12/3) km/hr
⇒ 4 km/hr
∴ The average speed during the total journey is 4 km/hr.

We have \(\frac{4 x}{5}-\frac{2 x}{7}\) ≤ 1 + \(\frac{1}{2}\)
⇒ \(\frac{18 x}{35}\) ≤ \(\frac{3}{2}\)
⇒ 12x ≤ 35, or x \(\frac{35}{12}\)
Thus, solution set is (-∞, \(\frac{35}{12}\)]

Question 22.
Find the present value of a sequence of payments of ₹ 360 made at the end of every six months continuing forever, if money is worth 6% p.a. compounded half-yearly. [2]
Solution:
We know that the present value of perpetuity of ₹ R payable at the end of every period forever at a rate of i per period is given by \(\frac{\mathrm{R}}{i}\), i.e.,
P_∞ = \(\frac{\mathrm{R}}{i}\)
Here, R = 360, i per second = \(\frac{6}{200}\) = 0.03
So, P_∞ = \(\frac{\mathrm{R}}{i}\) = \(\frac{360}{0.03}\) = ₹ 12,000
So, the required present value is ₹ 12,000.

Question 23.
Calculate Fisher’s Ideal Index and Dorbish and Bowley’s index numbers, if Laspeyre’s Index and Paasche’s Index numbers are 201 and 192 respectively. [2]
Solution:
We are given
\(\frac{\sum p_c q_b}{\sum p_b q_b}\) = 2.01
and \(\frac{\sum p_c q_c}{\sum p_b q_c}\) = 1.92
So, Dorbish and Bowley’s index number
= \(\frac{1}{2}\left[\frac{\sum p_c q_b}{\sum p_b q_b}+\frac{\sum p_c q_c}{\sum p_b q_c}\right]\) × 100
= \(\frac{1}{2}\) [ 2.01 + 1.92] × 100 = 196.5
and Fisher’s Ideal Index number
= \(\sqrt{\frac{\sum p_c q_b}{\sum p_b q_b} \times \frac{\sum p_c q_c}{\sum p_b q_c}}\) × 100
= \(\sqrt{2.01 \times 1.92}\) × 100
= \(\sqrt{3.8592}\) × 100
= 1.965 × 100 = 196.45

Question 24.
Arjun borrowed a home loan amount of ₹ 50,00,000 from a bank at an interest rate of 12% per annum for 30 years. Find his EMI [Given : (1.01)^-360 = 0.0278] [2]
Solution:
If a loan of amount P is borrowed at flat rate r per rupee per month for a period of n months.
Then, EMI = P × r × \(\frac{1}{1-(1+r)^{-n}}\)
Here, P = 50,00,000,
r per rupee per month = \(\frac{12}{12000}\) = 0.01
n = 30 × 12
= 360 months
So. required EMI,= P × r × \(\frac{1}{1-(1+r)^{-n}}\)
= ₹ {50,000 × 0.01 × \(\frac{1}{1-0.0278}\)
= ₹ 50,000 × \(\frac{1}{0.9722}\) = 51,430 (Approx.)
Thus, the Thus, required EMI is ₹ 51, 430

Question 25.
Using Cramer’s rule, solve the system of equations: [2]
x – y = 3
2x + 3y = 11
OR
Construct a matrix A= [aij]_{3 × 2} whose elements a_ij are given by
(A) a_ij= \(\frac{1}{2}\) |-3i + J|
(B) a_ij = \(\frac{1}{2}\) (i – 2 j)²
Solution:
Here,
D = \(\left|\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right|\) = 5 ;
D_x = \(\left|\begin{array}{cc}
3 & -1 \\
11 & 3
\end{array}\right|\) = 20 ;
D_y = \(\left|\begin{array}{cc}
1 & 3 \\
2 & 11
\end{array}\right|\) = 5
Since D = 5 ≠ 0. the system of equations has a unique solution given by
x = \(\frac{D_x}{D}=\frac{20}{5}\) = 4,
y = \(\frac{D_y}{D}=\frac{5}{5}\) = 1
Hence, x = 4, y = 1

(A) It is given that a_ij = \(\frac{1}{2}\)|-3i + j|
So. a₁₁ = \(\frac{1}{2}\)|-3 + 1|
= 1 ; a₁₂ \(\frac{1}{2}\)|-3 + 2| = \(\frac{1}{2}\) ;
a₂₁ = \(\frac{1}{2}\)|-6 + 1| = \(\frac{5}{2}\) ; a₂₂ = \(\frac{1}{2}\)|-6 + 2| = 2;
a₃₁ = \(\frac{1}{2}\)|-9 + 1| = 4 ; a₃₂ = \(\frac{1}{2}\)|-9 + 2| = \(\frac{7}{2}\)
Thus, A = \(\left[\begin{array}{ll}
1 & \frac{1}{2} \\
\frac{5}{2} & 2 \\
4 & \frac{7}{2}
\end{array}\right]\)

(B) It is given that. a_ij = \(\frac{1}{2}\)(i – 2j)²
So, a₁₁ = \(\frac{1}{2}\)(1 – 2)² = \(\frac{1}{2}\);
a₁₂ = \(\frac{1}{2}\)(1 – 4)² = \(\frac{3}{2}\)
a₂₁ = \(\frac{1}{2}\)(2 – 2)² = 0
a₂₂ = \(\frac{1}{2}\)(2 – 4)² = 1
a₃₁ = \(\frac{1}{2}\)(3 – 2)² = \(\frac{1}{2}\)
a₃₂ = \(\frac{1}{2}\)(3 – 4)² = \(\frac{1}{2}\)
Thus, A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{3}{2} \\
0 & 1 \\
\frac{1}{2} & \frac{1}{2}
\end{array}\right]\)

Section – C (18 marks)

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

Question 26.
If a, b and c are any three distinct real numbers, then show that (a + b) (b + c)(c + a) ≥ 8 abc. [3]
Solution:
We have, (\(\sqrt{a}\), \(\sqrt{b}\))² > o
a + b > 2 \(\sqrt{ab}\)
Similarly, b + c > 2 \(\sqrt{bc}\) and c + a > 2 \(\sqrt{ca}\)
Multiplying all the three, we have
(a + b) (b + c) (c + a) > (2\(\sqrt{ab}\))(2\(\sqrt{bc}\)) (2\(\sqrt{ca}\))
or (a + b)(b + c)(c + a) > 8 abc

Question 27.
Divide 15 into two parts such that the square of one part multiplied with the cube of other part is maximum. [3]
OR
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Solution:
Let the two parts of 15 be x and (15 – x).
Let P = x² (15 – x)³
⇒ \(\frac{\mathrm{dP}}{\mathrm{dx}}\) = 2x(15 – x)³ + x² [3(15 – x)² (-1)]
= x(15 – x)² (30 – 5x)
Now, \(\frac{\mathrm{dP}}{\mathrm{dx}}\) = 0 gives x(15 – x)³ (30 – 5x) = 0 dx
This gives x = 0, 6 and 15
Rejecting 0 and 15, we have only one critical point 6.
\(\frac{\mathrm{d}^{2} \mathrm{~P}}{\mathrm{dx}^{2}}\) = x(15 – x)² (-5) + (15 – x)² (30 – 5x) dx² – 2x (15 – x)(30-5x)
Now, at x = 6, \(\frac{\mathrm{d}^{2} \mathrm{~P}}{\mathrm{dx}^{2}}\) = 6(15 – 6)² (-5) + (15 – 6)² (30 – 30) – 2x (15 – 6)(30 – 30)
= -2430 < 0
So, p is maximum when x = 6.
Thus, the two parts of 15 are 6 and 9.

Here, the slope of the chord joining the points
(2, 0) and (4, 4) is \(\frac{4-0}{4-2}\), i.e., 2 4-2
So, the slope of the tangent to the curve is 2.
Now, y = (x – 2)² gives \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2(x – 2) dx
Equating it to 2, we get x = 3.
Hence, the required point on the curve is x = 3,
y = (3 – 2)² = 1, i.e., (3, 1)

Question 28.
The length of a rectangle is twice the breadth. If the minimum perimeter of the rectangle is 120 cm, then find the possible values of the breadth. [3]
Solution:
Let x denote the breadth of the rectangle. Then, its length is 2x.
According to the problem, 2(2x + x) ≥ 120
⇒ 6x ≥ 120x ≥ 20
Thus the possible values of breadth are 20 cm or more.

Question 29.
Write an example of [3]
(A) a row matrix
(B) a column matrix
(C) a row matrix which is also a column matrix
(D) a diagonal matrix which is not a scalar matrix
(E) an identity matrix of order 2.
Solution:
(A) \(\left[\begin{array}{lll}
1 & 2 & -3
\end{array}\right]\)
(B) \(\left[\begin{array}{l}
2 \\
0 \\
3
\end{array}\right]\)
(C) \([4]\)
(D) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right]\)
(E) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(F) \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 30.
A machine costs the company ₹ 1,00,000 and its effective life is estimated to be 12 years. If the scrap realises ₹ 5000 only, what amount should be retained out of profits at the end of each year to accumulate at a compound interest of 5% per annum?
[Given (1.05)¹² = 1.797) [3]
OR
A machine costs a company ₹ 52,000 and its effective life is estimated to be 25 years. A sinking fund is created for replacing the machine by a new model at the end of its life time, when its scrap realises a sum of ₹ 2500 only. The price of the new model is estimated to be 25% more than the price of the present one. Find what amount should be set aside at the end of each year out of the profits for the sinking fund , if it accumulates at 3.5% per annum compounded annually. [Use (1.035)²⁵ = 2.3632]
Solution:
If ₹ A be the sum of money set aside periodically, S is the amount accumulated after n periods and r is the rate of interest per period, then S is given by
S = \(\frac{\mathrm{A}}{r}\) [(1 + r)ⁿ – 1]
Thus, periodic payment is given by A = \(\frac{S r}{(1+r)^n-1}\)
Here, S = 1,00,000 – 5000 = 95,000; r = 0.05 and n = 12
So, A = \(\frac{95,000 \times 0.05}{(1.05)^{12}-1}\) = \(\frac{47500}{1.797-1}\) = 5959.85
Thus, an amount of ₹ 5959.85 should be retained out of profits at the end of each year.

Let ₹ A be the amount set aside each year.
Here, S = Amount needed to purchase the machine after 25 years
= ₹ (52,000 + 25% of 52,000) – ₹ 2500
= ₹ 62,500
r = 3.5% = 0.035 and n = 25 years
Now, A = \(\frac{S r}{(1+r)^n-1}\) gives
A = \(\frac{(0.035)(62400)}{(1.035)^{25}-1}\)
= \(\frac{21875}{1.3682}\) = 1604.68
Thus, ₹ 1604.68 are to be set aside each year out of the profits for the sinking fund.

Question 31.
Find: \(\int \frac{3 x+1}{(x-1)^2(x+3)}\) dx [3]
Solution:

⇒ 3x + 1A(x – 1)(x + 3) + B(x + 3) + C (x -1)²
Comparing the coefficients of x², x and the constatnt, we have
A + C = 0; 2A + B – 2 C = 3 ; -3A + 3B + C = 1
Solving these equations, we get A = \(\frac{1}{2}\), B = 1, C = –\(\frac{1}{2}\)
Thus,

Section – D (20 Marks)

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

Question 32.
A wire of length 36 m is to be cut into two pieces. One of the pieces is to be made a square and the other into a circle. What would be the lengths of the two parts, so that the combined area of the square and the circle is minimum? [5]
OR
Find \(\int\){log log (logx) + \(\frac{1}{(\log \log x)^2}\)} dx
Solution:
Let the length of a side of the square be x metres and the radius of the circle be y metres. Then,
4x + 2πy = 36
⇒ y = \(\frac{18-2 x}{\pi}\) …………….. (1)
Let A denote the’ combined area of the square and the circle, Then,
A = x² + π\(\left(\frac{18-2 x}{\pi}\right)^2\) = x² + \(\frac{1}{\pi}\) (18 – 2x)²
⇒ \(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 2x – \(\frac{4}{\pi}\) (18 – 2x) and \(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\) = 2 + \(\frac{8}{\pi}\) > 0 for all x
The critical points of A are given by \(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 0
∴ \(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 0 implies 2x – \(\frac{4}{\pi}\) (18 – 2x) = 0, which gives x = \(\frac{36}{\pi+4}\)
Clearly, \(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\) at x = \(\frac{36}{\pi+4}\) is + ive
So, A is minimum when x = \(\frac{36}{\pi+4}\). Putting x = \(\frac{36}{\pi+4}\) in (1), y = \(\frac{18}{\pi+4}\)
Thus, the lengths of two pieces of the 36 m long wire are 4x and 2πy, i.e., \(\frac{144}{\pi+4}\) metres and \(\frac{36 \pi}{\pi+4}\)

Let I = \(\int\left\{\log \log (\log x)+\frac{1}{(\log \log x)^2}\right\} d x\)
Put log x = t so that x = e^t and dx = e^t dt
Then

Question 33.
Two types of drugs were used to control the high blood pressure on 6 and 8 patients and decrease in systolic blood pressure ( upper limit of b.p.) were as below: [5]

Drug A:	12	18	30	15	7	14
Drug B:	15	16	12	10	21	25	28	17

Is there any significant difference in the efficiency of the two types of drugs? (Given t₁₂(0.05) = 2.18)
Solution:
The null hypothesis and alternative hypothesis can be written as:
H_o : There is no significant difference in the efficiency of two drugs, i.e., µ₁ = µ₂
H₁ : There is a significant difference in the efficiency of two drugs, i.e., µ₁ ≠ µ₂
Here, n₁ = 6 and n₂ = 8
\(\overline{\mathrm{X}}\) = \(\frac{12+18+30+15+7+14}{6}\)
= \(\frac{96}{6}\) = 16
\(\overline{\mathrm{Y}}\) = \(\frac{15+16+12+10+21+25+28+17}{8}\)
= \(\frac{144}{8}\) = 18
For the calculation of S², we have the following table :

Thus,
S² = \(\frac{1}{n_1+n_2-2}]\) [Σ(X – \(\overline{\mathrm{X}}\))² + Σ(Y – \(\overline{\mathrm{Y}}\))²]
= \(\frac{1}{12}\) [302 + 272] = 47.83
⇒ S = \(\sqrt{47.83}\) = 6.91
Now, test statistics is given by
t = \(\frac{\overline{\mathrm{X}}-\overline{\mathrm{Y}}}{\mathrm{S} \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\) = \(\frac{16-18}{(6.91) \sqrt{\frac{1}{6}+\frac{1}{8}}}\)
= \(\frac{-2}{(6.91) \times(0.54)}\) = \(\frac{-2}{3.731}\) = -0.536
⇒ |t| = 0.536
The table value of t at α =0.05 and (6 + 8 – 2)
= 12 d.f. is 2.18
Conclusion: Since |t| < t_α, the null hypothesis is accepted, i.e., there is no significant difference in the efficiency of two types of drugs.

Question 34.
A bond whose par value is ₹ 1000 bears a coupon rate of 12% and has a maturity period of 3 years. The required rate of return on the bond is 10%. What is the present value of the bond? (Given (1.1)^-3 = 0.751) [5]
Solution:
Here,
F = Face value of the bond = 51000
n = number of periodic divident payments = 3
i = Annual yield rate = 0.1
R = F × i_d = ₹ (1000 × 0.12) = ₹ 120
Since the bond is to be redeemed at par,
C = Redemption price or Maturity value = Face value = ₹ 1000
Let V be the purchase value of the bond. Then,
V = R \(\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)^-n
= ₹ {120 \(\left[\frac{1-(1.1)^{-3}}{0.1}\right]\) + 1000(1.1)^-3}
= ₹ {1200 [1 – (1.1)^-3] + 1000 (1.1)^-3)} .
= ₹ {1200 [1 – 0.751] + 1000 (0.751)}
= ₹ {298.80 + 751}, or ₹ 1049.80
Hence, the present value of the bond is ₹ 1049.80

Question 35.
A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ₹ 50 per kg to purchase Food T and ₹ 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture. [5]
OR
A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as ₹ 10,500 and ₹ 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further, no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?
Solution:
Let x packages of Food I, y packages of Food II be mixed and Z be the cost of the mixture.

	Food I	Food II	Requirement
Vitamin A (units/kg)	2	1	8
Vitamin C (units/kg)	1	2	10
Cost (5/ kg)	50	70

Then, the mathematical formulation of the given LPP is
Minimise Z = 50x + 70y
subject to constraints
2x + y ≥ 8, 4x + 2y ≥ 10, x, y ≥ 0
Let us draw the graph for the system of ine¬qualities representing constraints.

The feasible region is ABC shown (shaded) in Fig. 1, which is unbounded.
The coordinates of the corner points of the feasible region ABC are A( 0, 8), B(2, 4) and C (10, 0).
Let us evaluate the objective function Z = 50x + 70y at the corner points.

Corner point	Corresponding value of Z
A(0, 8)	560
B(2, 4)	380 (minimum)
C(10, 0)	500

Z is minimum at B(2, 4) and minimum cost of the mixture is ₹ 380.
Thus, the optimal mixing strategy for the dietician would be to mix 2 kg of Food I and 4 kg of food II for arriving at the minimum cost of the mixture ₹ 380

OR
Let x hectares of land be allocated to crop X and y hectare to crop Y.
Then, the mathematical formulation of the given LPP is
Maximise Z = 10500x + 9000y
subject to constraints
2x + y ≤ 50,2x + y ≤ 80, x, y > 0
Let us draw the graph for the system of inequalities representing constraints.

The feasible region is OABC shown (shaded) in Fig. 2, which is bounded.
The coordinates of the corner points of the feasible region OABC are 0(0, 0), A(40, 0), B(30, 20) and C (0, 50).
Let us evaluate the objective function Z = 10500 x + 9000 y at the corner points.

Corner point	Corresponding value of Z
O(0, 0)	0
A(40, 0)	420000
B (30, 20)	495000 (maximum)
C (0, 50)	450000

Z is maximum at B(30, 20) and the maximum profit is ₹ 495000
Thus, the society will get the maximum profit of ₹ 495000 by allocating 30 hectares for crop X and 20 hectares to crop Y.

Section – E (12 Maks)

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

Question 36.
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for cooperation and helping others and some others ( say z) for supervising the workers to keep the colony neat and clean.

(i) The sum of all awardees is 12.
(ii) Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33.
(iii) The sum of the number of awardees for honesty and supervision is twice the number of awardees for cooperation andhelping others.
Based on this information, answer the following questions:
(A) What is the group of equations for the statements (i), (ii) and (iii) [1]
(B) Write the above group of equation in matrix form AX = B: [1]
(C) Find adj (A): [2]
OR
Find |A| is:
Solution:
(A) x + y + z = 12 ; 2x + 3(y + z) = 33;
x + z = 2y

(B) \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 3 \\
1 & -2 & 1
\end{array}\right]\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] \cdot=\left[\begin{array}{c}
12 \\
33 \\
0
\end{array}\right]\)

A₃₃ = (-1)³⁺³ \(\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right]\) = 1
∴ Adj (A) = \(\left[\begin{array}{ccc}
9 & -3 & 0 \\
1 & 0 & -1 \\
-7 & 3 & 1
\end{array}\right]\)

|A| = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 3 \\
1 & -2 & 1
\end{array}\right]\)
= 1(3 + 6) – 1(2 – 3) + 1(-4 – 3) = 3

Question 37.
The hourly wages of 500 workers are normally distributed around a mean of ₹ 80 and with a standard deviation of ₹ 10.

Based on this information, answer the following questions:
(A) Find the number of workers whose hourly wages are between ₹ 65 and ₹ 95? [1]
(B) What is t he number of workers whose hourly wages are more than ₹ 95? [1]
(C) How many workers gets exactly ₹ 100? [2]
OR
What is lowest hourly wages of the 50 highest paid workers?
Solution:
Here, Z = \(\frac{X-80}{10}\). So, we have

X:	65	75	93	100
Y:	-1.5	-0.5	1.3	130

(A) P( 65 < X < 95) = P(-1.5 < Z < 1.5)
= 2 P(0 < Z < 1.5) = 2 × 0.4332 = 0.8664 So, required number of workers is = 500 × 0.8664 = 433.2 – 433 (B) P(X > 95) = P(Z > 1.5)
= 0.5 – P(0 < Z < 1.3)
= 0.5 – 0.4032
= 0.0968
So, required number of workers is =
500 × 0.0968 = 48.4 ~ 48

(C) P( 99 < X < 101) = P(1.9 < Z < 2.1)
= P(0 < Z < 2.1) – P(0 < Z < 1.9)
= 0.4821 – 0.4713
= 0.0108
So, required number of workers is = 500 × 0.0108
= 5.4 5

Proportion of the 50 highest paid workmen = 0.10
So, we need to determine X = x₁ such that P(X > x₁) = 0.1
When X = x₁, Z = \(\frac{x_1-80}{10}\) = z₁ (say)
Then, P(Z > z₁) = 0.1
⇒ P( 0 < Z < z₁) = 0.5 – 0.1 = 0.4
⇒ z₁ = 1.28
⇒ x₁ = 10 × 1.28 + 80
= 92.8

Question 38.
The table given below contains the prices (per unit) of few selected items

Commodity	Weight	Year 2005	Year 2010	Year 2015
A	70	15	20	24
B	15	45	60	70
C	40	24	10	16
D	30	3	6	10

Based on the above data, answer the following questions:
(A) What is the percentage relative price index for the year 2010 taking 2005 as the base year? [1]
(B) What is the percentage relative price index for the year 2015 taking 2010 as the base year? [1]
(C) What is the price index for the year 2010 taking 2005 as the base year using weighted average of price relatives? [2]
OR
What is the price index for the year 2015 taking 2005 as the base year using weighted average of price relatives?
Solution:
(A) The percentage relative price index for the year 2010 taking 2005 as the base year
= \(\frac{90}{84}\) × 100 = 107.14

(B) The percentage relative price index for the year 2015 taking 2005 as the base year
= \(\frac{110}{84}\) × 100 = 131

(C) The price index for the year 2010 taking 2005 as the base year using weighted
average of price relatives: \(\frac{\Sigma P W}{\Sigma W}\)
= \((133.33 \times 70)+(133.33 \times 15)+ \frac{(41.66 \times 40)+(200 \times 30)}{155}\)
= \(\frac{18999.45}{155}\) = 122.57

The price index for the year 2015 taking 2005 as the base year using weighted
average of price relatives: \(\frac{\Sigma P W}{\Sigma W}\)
= \((160 \times 70)+(155.56 \times 15)+ \frac{(66.67 \times 40)+(333.33 \times 30)}{155}\)
= \(\frac{26200.10}{155}\) = 169