Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Maximum Marks: 80 Marks
Time Allowed : 3 Hours

General Instructions:

  1. This question paper contains five sections A, B, C, D and E. Each section is compulsory.
  2. 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.
  3. Section A: It comprises of 20 MCQs of 1 mark each.
  4. Section B: It comprises of 5 VSA type questions of 2 marks each.
  5. Section C: It comprises of 6 SA type of questions of 3 marks each.
  6. Section D: It comprises of 4 LA type of questions of 5 marks each.
  7. 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.
  8. 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

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

Question 1.
In a kilometre race, A, B and C are three participants. A can give B a start of 50 m and C a start of 69 m. What start B can allow C?
(a) 16 m
(b) 20 m
(c) 24 m
(d) 26 m [1]
Answer:
(b) 20 m

Explanation: A : B : C
= 1000 : (1000 – 50): (1000 – 69)
= 1000 : 950 : 931
In a 950 m race, B can give C a start of (950-931) m = 19 m
In a 1000 m race, B can give C a start of (\(\frac{19}{950}\) × 1000) = 20 m

Question 2.
Kartika can row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to row the boat 5.25 km upstream than to return downstream, then the speed of the stream is:
(a) 2 km/h
(b) 1.5 km/h
(c) 2.2 km /h
(d) 1.8 km/h
Answer:
(a) 2 km/h

Explanation: Let the speed of the stream be x
Speed when going upstream = 5 + x
Speed when going dowstream = 5 — x
Distance = speed × Time
So,
\(\frac{5.25}{5+x}\) + 1 = \(\frac{5.25}{5-x}\)
5.25(5 + x – 5 + x) = (25 – x2)
10.5x = 25 – x2
x2 + 10.5x – 25 = 0
(x – 2)(x + 12.5) = 0
x = 2 km/hr is the speed of the stream.

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 3.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
-1 & 1
\end{array}\right]\) and X is a 2 × 2 matrix such that X + 2A = B, then X equals
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 1 [1]
Answer:
(d) \(\left[\begin{array}{cc}
0 & 5 \\
-3 & 1
\end{array}\right]\)

Explanation: Here,
X = B – 2A
X = \(\left[\begin{array}{cc}
2-2 & 1+4 \\
-1-2 & 1-0
\end{array}\right]\)
= \(\left[\begin{array}{cc}
0 & 5 \\
-3 & 1
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{lll}
1 & -2 & 0 \\
3 & -1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
3 & 7 \\
4 & 0 \\
-1 & 2
\end{array}\right]\), then
(a) only AB is defined
(b) only BA is defined
(c) AB and BA both are defined
(d) AB and BA both are not defined
Answer:
(c) AB and BA both are defined Explanation: Here, A is of order 2 × 3 and B is of order 3 × 2
So, AB is defined and of order 2 × 2. Also, BA is defined of order 3 × 3

Question 5.
If A = \(\left[\begin{array}{lll}
1 & 2 & x \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & y \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) and AB = 13, then x + y equals:
(a) 0
(b) -1
(c) 2
(d) None of these [1]
Answer:
(a) 0

Explanation:
A = \(\left[\begin{array}{lll}
1 & 2 & x \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & y \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 6
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 7

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 6.
If S = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\), then adj A is:
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 2 [1]
Answer:
(b) \(\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)

Explanation:
S = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Then
a11 = d
a22 = a
a12 = -c
a21 = – b
∴ adj S = \(\left[\begin{array}{cc}
d & -c \\
-b & a
\end{array}\right]\) = \(\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)

Question 7.
If x = t2, y = t3, then \(\frac{\left(d^2 y\right)}{d x^2}\) is:
(a) \(\frac{3}{2}\)
(b) \(\frac{3}{4 t}\)
(c) \(\frac{1}{2 t^2}\)
(d) \(\frac{3}{2 t}\) [1]
Answer:
(b) \(\frac{3}{4 t}\)
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 8
Thus, (b) is the correct option.

Question 8.
If the supply fúnction for a commodity is p = \(\sqrt{x+9}\) and the market price is 4, then producer’s surplus is
(a) 3
(b) \(\frac{10}{3}\)
(c) 10
(d) 15 [1]
Answer:
(b) \(\frac{10}{3}\)
Explanation:
Here.
p0 = 4. This gives x0 = 7
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 9
= \(\frac{84-74}{3}\)
= \(\frac{10}{3}\)

Question 9.
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then, the value of E(X) is:
(a) \(\frac{37}{22}\)
(b) \(\frac{5}{13}\)
(c) \(\frac{1}{13}\)
(d) \(\frac{2}{13}\) [1]
Answer:
(d) \(\frac{2}{13}\)

Explanation: Let X be the random variable denoting the number of aces obtained. Then, X takes the values 0, 1, 2. The corresponding probabilities are:
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 10

Question 10.
A fair coin is tossed 99 times. Let X be the number of times head occurs. Then P(X = r) is maximum when r is equal to:
(a) 49
(b) 51
(c) 48
(d) 99 [1]
Answer:
(a) 49

Explanation: Here n = 99 and p = \(\frac{1}{2}\) ⇒ \(\frac{1}{2}\)
So, P(X = r) = 99Cr\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{99-r}\) = 99Cr is maximum.
99Cr is maximum when r = 50 or 49

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 11.
If for a Poisson variate X, P(X = k) = P(X = k + 1), then the variance of X is:
(a) k – 1
(b) k + 1
(c) k
(d) k + 2 [1]
Answer:
(b) k + 1
Explanation: Recurrence fórmula states that
P(X = k + 1) = \(\frac{\lambda}{k+1}\)P(X = k)
It is given that P(X = k + 1) = P(X = k).
So, \(\frac{\lambda}{k+1}\) = 1, i.e, λ = k + 1
⇒ Variance = λ = k + 1

Question 12.
In one sample t-test, the estimation for the population mean is:
(a) \(\frac{\bar{x}-\mu}{s / \sqrt{n}}\)
(b) \(\frac{\overline{\mathrm{X}}-\mu}{\mathrm{S} / \boldsymbol{n}}\)
(c) \(\frac{\overline{\mathrm{X}}-\mu}{\mathrm{s}^2 / \sqrt{n}}\)
(d) \(\frac{\overline{\mathrm{X}}-\mu}{\mathrm{s} / \mathrm{n}^2}\) [1]
Answer:
(a) \(\frac{\bar{X}-\mu}{S / n}\)

Question 13.
The area under the standard normal curve which lies to the left of z = -0.56 is:
(a) 0.7123
(b) 0.2877
(c) -0.2123
(d) 0.2123 [1]
Answer:
(b) 0.2877
Explanation: Here,
P(Z < -0.56) = 0.5 – P(0 < Z < 0.56)
= 0.5 – 0.2123 = 0.2877

Question 14.
A time series consists of:
(a) two components
(b) three components
(c) four components
(d) five components [1]
Answer:
(c) four components

Question 15.
The amount S of an annuity due A, when the rate; of interest is r and n is the number of years, is given by
(a) S = \(\frac{\mathrm{A}}{r}\)(1 + r)n[(1 + r)n – 1]
(b) S = \(\frac{\mathrm{A}}{r}\) (1 + r)[(1 + r)n – 1]
(c) S = Ar(1 + r)n [(1 + r)n – 1]
(d) S = Ar(1 + r)[(1 + r)n – 1] [1]
Answer:
(b) S = \(\frac{A}{r}\)(1 + r)(1 + r)n – 1]

Question 16.
The corner points of the feasible region of a LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Then the minimum value of Z = 4x + 6y occurs
(a) only at one point
(b) only at two points
(c) at finite number of points
(d) at infinite number of points [1]
Answer:
(d) at infinite number of points

Explanation: The values of Z at the corner points are: 12, 12, 24, 72 and 30
As minimum of Z occurs at two points (0, 2) and (3, 0), every point of the line segment joining these two corner points is a solution.

Question 17.
A specific characteristic of a population is called
(a) statistic
(b) mean
(c) parameter
(d) sample [1]
Answer:
(c) parameter

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 18.
The assumed hypothesis which is tested for rejection considering it to be true is called:
(a) true hypothesis
(b) alternative hypothesis
(c) simple hypothesis
(d) null hypothesis [1]
Answer:
(d) null hypothesis

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) and (IQ are true but (IQ 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): The slope of the normal to the curve y = 2x2 + 3x – 5 at x = 0 is –\(\frac{1}{3}\)
Reason (R): The Slope of the normal of the area is \(\left(\frac{d x}{d y}\right)\) [1]
Answer:
(a) Both (A) and (B) are true and (R) is the correct explanation of (A).

Explanation:
Since, y = 2x2 + 3x – 5
\(\frac{d y}{d x}\) = 4x + 3
at x = 0
\(\frac{d y}{d x}\) = 3
∴ Slope of normal = \(-\frac{1}{\left(\frac{d y}{d x}\right)}\) = \(\frac{-d x}{d y}\) = \(-\frac{1}{3}\)

Question 20.
In binomial distribution n = 200, P = 0.04. Taking poisson distribution as an approximation to the binomial distribution.
Assertion (A): Mean of poisson distribution = 8.
Reason (R): In a Poisson distribution,
P (X = 4) = \(\frac{512}{3 e^8}\) [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
Mean of p.d. = np = 8
And f(X = 4) = \(\frac{e^{-8}(8)^4}{4 !}\) = \(\frac{512}{3 e^8}\)

Section – B

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

Question 21.
A boat can row upstream at 10 km/h and downstream at 16 km/h. Find the speed of the boat in still water and the speed of the stream. [2]
Answer:
Speed of the boat in still water
= \(\frac{1}{2}\) (16 + 10) km/h
= 13 km/h
Speed of the stream = \(\frac{1}{2}\) (16 – 10) km/h
= 3 km/h

Question 22.
Four bad oranges are mixed accidently with 16 good oranges. And the probability distribution of the number of bad oranges in a draw of 2 oranges.
OR
In a binomial distribution B(n, p = \(\frac{1}{4}\)) if the probability of at least one success is greater than or equal to \(\frac{9}{10}\), then find the value of n. [2]
Answer:
Let x denote the number of bad oranges in a draw of 2 oranges drawn from the group of 16 good and 4 bad oranges. Hence, X can take values 0, 1, 2
Now, P(X = 0) = Probability of getting no bad orange
= Probability of getting 2 good oranges
= \(\frac{{ }^{16} c_{c_2}}{{ }^{20} c_{c_2}}\) = \(\frac{60}{95}\)
P(X = 1) = Probability of getting 1 good and 1 bad orange
= \(\frac{{ }^{16} c_{c_1} \times{ }^4 c_1}{{ }^{20} c_{c_2}}\) = \(\frac{32}{95}\)
P(X = 2) = Probability of getting 2 bad oranges
= \(\frac{{ }^4 c_2}{{ }^{20}{ }_{C_2}}\) = \(\frac{3}{95}\)
Thus, the required probability distribution of X is
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 11

OR
Here, p = \(\frac{1}{4}\) and q = \(\frac{3}{4}\)
It is given that P(X ≥ 1) ≥ \(\frac{9}{10}\)
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 12

Question 23.
Distinguish between a population and a sample.
OR
Suppose that a 95% confidence interval states that population mean is greater than 100 and less than 300. How would you interpret this statement? [2]
Answer:

Basis for comparison Population

Sample

Meaning Population refers to the collection of all elements possessing common characteristics that comprises universe. Sample means a subgroup of the members of population chosen for participation in the study.
Includes Each and every unit of the group Only a handful of units of population.
Characterstics Parameter Statistic

Or

The given 95% confidence interval tells that we are fairly sure that out true value lies in the range of 100 – 300.

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 24.
What amount is received the end of every six months forever, if ₹ 72,000 kept in a bank earns money is worth 8% per annum compounded half-yearly?
Answer:
We know that the present value of perpetuity of ₹R payable at the end of every period fórever at a rate of i per period is given by \(\frac{\mathrm{R}}{i}\), i.e., P = \(\frac{\mathrm{R}}{i}\)
Here, R = x, i per period = \(\frac{\mathrm{8}}{200}\) = 0.04, P = 72,000
So, P = \(\frac{\mathrm{R}}{i}\)
⇒ 72,000 = \(\frac{x}{0.04}\)
⇒ x = 2800
So, an amount of ₹ 2800 will, be received at the end of every six months.

Question 25.
Graph the following system of constraints and shade the feasible region:
y – 2x ≤ 1; x + y ≤ 3; x ≤ 2
x ≥ 0, y ≥ 0[2]
Answer:
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 13

Section – C

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

Question 26.
If a, b and c are any three unequal positive real numbers, then show that a3 + b3 + c3 ≥ 3abc
OR
If a2 + b2 = 1 and c2 + d2 = 1, then show that 1 ≥ ac + bd.
Answer:
Consider a3 + b3 + c3 – 3
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= (a + b + c)
{\(\frac{1}{2}\) = (a + b + c) {\(\frac{1}{2}\)(2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca)}
= (a + b + c) {(a – b)2 + (b – c)2 + (c – a)2}
Now, a > 0, b > 0, c > 0 and a ≠ b ≠ c.
We have (a – b)2 > 0, (b – c)2 and (c – a)2
And So, a3 + b3 + c3 – 3abc > 0, or a3 + b3 + c3 – 3abc > 0, or a3 + b3 + c3 > 3abc

OR

We have (a – c)2 ≥ 0
a2 + c2 ≥ 2ac
Similarly, (b – d)2 ≥ o
b2 + d2 ≥ 2bd
Therefore,
a2 + c2 + b2 + d2 ≥ +2ac + 2bd
1 + 1 ≥ 2ac + 2bd
1 ≥ ac + bd

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 27.
If A = \(\left[\begin{array}{ll}
p & q \\
r & s
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), then show that A2 – (p + s)A = (qr – ps)I [3]
Answer:
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 14
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 15

Question 28.
The input-output coefficient matrix for a 2-sector economy is A = \(\left[\begin{array}{cc}
0.1 & 0.5 \\
0.2 & 0.25
\end{array}\right]\). If the final demands for the two industries are 300 and 100 units, find the total output of each of the two industries. [3]
Answer:
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 16
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 17
Hence, the total output of two industries are 478.26 units and 260.87 units respectively.

Question 29.
Find the intervals in which the function f(x) = 5 + 36x + 3x2 – 2x3 is strictly increasing and strictly decreasing.
OR
If f(x) = \(\frac{1}{1-x}\), then prove that \(\int_0^a f[f\{f(x)\}] d x\) = \(\frac{a^2}{2}\) [3]
Answer:
For f(x) = 5 + 36x + 3x2 – 2x3.
f ‘(x) = 36 + 6x – 6x2 = 6(6 + x – x2)
= -6(x – 3)(x + 2)
Now, f’(x) = 0 gives -6(x – 3)(x + 2) = 0,
i.e., x = 3, -2

Sub-interval Test point Test value Sign of f’(x) Conclusion
(A) (-∞, -2) -3 f’(-3) = -54 (-) f’(x)
(B) (-2, 3) 0 f’(0) = 36 (+) f’(x) > 0
(C) (3, ∞) 4 f’(4) = -36 (-) f’(x) < 0

Hence, f(x) is strictly increasing on (-2, 3) and strictly decreasing on (-∞, -2) ∪ (3, ∞).

OR

Given f(x) = \(\frac{1}{1-x}\), we have
f{f(x)} = \(\frac{1}{1-f(x)}\) = \(\frac{1}{1-\left[\frac{1}{1-x}\right]}\)
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 18

Question 30.
4000 students appeared for an examination. The mean marks were 49 and S.D. 6. Assuming that the marks to be normally distributed, What percent of students scored more than 55 marks? [3]
Answer:
According to the problem, Z = \(\frac{X-49}{6}\)
Here, P(X > 55) = P(Z > 1)
= 0.5 – P(0 < Z < 1) = 0.5 – 0.3413 = 0.1587 So, the percent of students who scored more than 55 marks is 15.87% Further, P(X > 70) = P(Z > 3.5)
= 0.5 – P(0 < Z < 3.5)
= 0.5 – 0.4998
= 0.0002
So, the percent of students who scored grade A is 0.02%.

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 31.
A limited company intends to create a depreciation fund to replace at the end of the 20th year assets costing ₹ 5,00,000. Calculate the amount to be retained out of the profits every year if the interest rate is 5% per annum.
[Given (1.05)20 = 2.6532] [3]
Answer:
Here, S = 5,00,000, A = x(say), r = 0.05 and n = 20
So, S = \(\frac{A}{r}\left[(1+r)^n-1\right]\) gives
500000 = \(\frac{x}{0.05}\) [(1.05)20 – 1]
⇒ x = 25000 ÷ [(1.05)20 – 1]
⇒ x = 25000 ÷ 1.6532 = 15122.18
Thus, the amount to be retained out of profits is ₹15,122.18.

Section – D [20 marks]

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

Question 32.
If y = a(x + \(\sqrt{x^2-1}\))n + b(x + \(\sqrt{x^2-1}\))n, then show that (x2 – 1)\(\frac{d^2 y}{d x^2}\) + x\(\frac{d y}{d x}\) – n2y = 0.
OR
The rate at which radioactive substances decay is known to be proportional to the number of such nuclei that are present at the time in a given sample.
In a certain sample, 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. Find what percentage of the original radioactive nuclei will remain after 1000 years. [5]
Answer:
We have
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 19
Differentiating again w.r.t. x, we get
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 20
= n2y
Hence, (x2 – 1)\(\frac{d^2 y}{d x^2}\) + x\(\frac{d y}{d x}\) – n2y = 0

OR

Let x be the number of radioactive nuclei at any time t. Then,
\(\frac{d x}{d t}\) = -kx, where k is constant of proportionality.
\(\frac{d x}{d t}\) is -ive, because x decreases when t increases
⇒ \(\frac{d x}{x}\) = -k dt
Integrating both sides we have \(\int \frac{d x}{x}\) = –\(\int k d t\)
log x = -kt + C
Let x = x0, when t = 0. Then, from (1)
we have C = log x0
So, from(1), log (\(\frac{x}{x_0}\)) = -kt, or x = x0e-kt ….. (3)
At t = 100, x = 0.9x0, (3) gives us 0.9x0 = x0e-kt
e-1ook = 0.9 . i.e, k = –\(\frac{\log (0.9)}{100}\)
Thus, from (3), we have
x = xo\(e^{\frac{\log (0.9)}{100} t}\) = x0(0.9)t/100
For t = 1000, x is given by
x = x0(0.9)1000/100 = x0(0.9)10
Hence, % of remaining element = \(\frac{x}{x_0}\) × 100
= (0.9)10 × 100 = \(\frac{9^{10}}{10^8}\)
Therefore, \(\frac{9^{10}}{10^8}\)% of the radioactive elements will, remain after 1000 years.

Question 33.
Calculate 5 – yearly moving averages for the following data of the number of commercial and industrial failures in a country from 1992 to 2007:
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 3
Display the actual and trend values on the same graph, using the same axes for both.
OR
For the following data, fit a straight line trend by method of least squares. Tabulate the trend values and also,, estimate the most likely sales for the year 2007.

Year 2001 2002 2003 2004 2005 2006
Sales (₹ lakhs) 10 20 30 56 40 60

Answer:

Year Number of failures 5-yearly total
[5-yearly moving average]
1992 23
1993 26
1994 28 129[25.8]
1995 32 118[23.6]
1996 20 104[20.8]
1997 12 86[17.2]
1998 12 63[12.6]
1999 10 56[11.2]
2000 9 55[11.0]
2001 13 57[11.4]
2002 11 59[11.8]
2003 14 59[11.8]
2004 12 49[9.8]
2005 9
2006 3
2007 1

Thus, the S- yearly moving averages are 25.8, 23.6, 20.8, 17.2, 12.6, 11.2, 11.0, 11.4, 11.8, 11.8, 9.8 and 7.8
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 21

OR

Here, n = 6(Even).
Let the equation of the straight line of best fit, with the origin at the mid of the years 2003 and 2004 and units of x as \(\frac{1}{2}\) year, be y = a + bx
By the method of Least squares, the values of ‘a’ and ‘b’ are given by
a = \(\frac{\Sigma y}{n}\) and b = \(\frac{\Sigma x y}{\Sigma x^2}\) … (A)

Calculation for fitting the line of best fit

Year Value x x2 xy
2001 10 -5 25 -50
2002 20 -3 9 -60
2003 30 -1 1 -30
2004 56 1 1 56
2005 40 3 9 120
2006 60 5 25 300
Σy = 216 Σx = 0 Σx2 = 70 Σxy = 336

Using (A), we have
a = \(\frac{\sum y}{n}\) = \(\frac{216}{6}\) = 36
and b = \(\frac{\sum x y}{\sum^{x^2}}\) = \(\frac{336}{70}\) = 4.8
Hence, the required equation of the best fitted straight line is y = 36 + 4.8x

Year x Trend values (y = 36 + 4.8x)
2001 -5 36 + 4.8(-5) = 12
2002 -3 36 + 4.8(-3) = 21.6
2003 -1 36 + 4.8(-1) = 31.2
2004 1 36 + 4.8(1) = 40.8
2005 3 36 + 4.8(3) = 50.4
2006 5 36 + 4.8(5) = 60

Thus, the trend values are 12, 21.6, 31.2, 40.8, 50.4 and 60
Estimate for the year 2007 = 36 + 4.8 (7) = 69.6

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 34.
Find the purchase value of a 50,000, 6% bond, dividends payable semi-annually, redeemable at par in 10 years, if the yield rate is to be 5% compounded semi-annually. [Given (1.025)-20 = 0.61027) [5]
Answer:
Here,
F = Face value of the bond = ₹ 50,000
n = number of periodic dividend payments
= 10 × 2 = 20
i = Annual yield rate = 0.025
R = F × id = ₹ (50.000 × 0.03) = ₹ 1500
Since the bond is to be redeemed at par, C = Redemption price or Maturity value = Face value = ₹ 50,000
Let V be the purchase value of the bond. Then,
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 22
Thus, the purchase price of the bond is ₹ 53897.30

Question 35.
A dietician wished to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

Food Vitamin A Vitamin B Vitamin C
X 1 2 3
Y 2 2 1

One kg of food X costs ₹ 16 and one kg of food Y costs ₹ 20. Find the least cost of the mixture which will produce the required diet? [5]
Answer:
Let x kg of Food X, y kg of Food Y be mixed to make the mixture and Z be the cost of the mixture.

Vitamin-A Vitamin-B Vitamin-C Cost
Food X 1 2 3 ₹ 16
Food Y 2 2 1 ₹ 20
Medium requirement 10 12 8

Then, the mathematical formulation of the given LPP is
Minimise Z = 16x + 2Oy
subject to constraints
x + y ≥ 6, 3x + y ≥ 8, x, y > 0
Let us draw the graph for the system of inequalities representing constraints.
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 23
The feasible region is ABCD shown (shaded) in Fig. 10, which is unbounded.
The coordinates of the corner points of the feasible region ABCD are A(10, 0), B(2, 4), C(1, 5) and D(0, 8)
Let us evaluate the objective function Z = 16x + 20y at the corner points.

Corner points Corresponding value to Z
A(10, 0) 160
B(2, 4) 112(minimum)
C(1, 5) 116
D(0, 8) 160

Z is minimum at B(2, 4) and minimum cost of the mixture is ₹ 112.
Since the feasible region is unbounded, so we have to determine whether Z = 112 is minimum cost or not.
Now we draw the graph of the inequality 16x +20y < 112, or 4x + 5y < 28.
The open half plane of 4x + 5y < 28 and the feasible region has no common points.
So, the least cost of the mixture is 112 when 2kg of Food X and 2kg of Food Y are mixed.

Section – E [12 marks]

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

Question 36.
In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers “wrap around” when reaching a certain value, called modulus.

A familiar use of modular arithmetic is in the 12- hour clock, in which the day is divided into two 12- hour periods. If the time is 7: 00 now, then 8 hours later it will be 3: 00. Simple addition would result in 7 + 8 = 15, but clocks “wrap around” every 12 hours. Because the hour number starts over after it reaches 12, this is arithmetic modulo 12. In terms of the definition, 15 is congruent to 3 modulo 12, so “15:00″ on a 24- hour clock is displayed “3: 00″ on a 12- hour clock.

Based on the above concept of “ modulo arithmetic” answer the following questions:
(A) Evaluate 36 (mod4). [1]
(B) What is the least possible value of x for which 100 x (mod 7)? [1]
(C) Evaluate (137 + 995)( mod 12).
OR
Find the last digit of 1212. [2]
Answer:
(A) We have 32 = 9 ≡ 1 (mod4).
So, 36 = (32)3 ≡(1)3 = 1 (mod4).
(B) Here, it is given that 1oo ≡ x (mod 7).
So, (100 – x) is completely divisible by 7
⇒ the Least possible value of ‘x’ is 2, as 98 is completely divisible by 7.
(C) (137 + 995)(mod 12)
= 137 (mod 12) + 995( mod 12)
= 5(mod 12) + 11(mod 12)
= (5 + 11)(mod 12)
= 16(mod 12) 4 (mod 12)

OR

To find the last digit of 1212, we need to find 1212 (mod 10)
Since, 12 ≡2 (mad 10), 124 (mod 10) = 24
(mod 10) = 16(mod 10) = 6(mod 10)
⇒ 1212 (mod 10) = (12m )3 (mod 10) = 6
(mod 10) = 216 (mod 10) = 6
Thus, the last digit of 1212 is 6.

CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions

Question 37.
A girl designs a pizza by cutting it with a knife on a cardboard. If pizza is circular in shape which is represented by the equation x2+ y2 = 4 and edge of knife represent a straight line given by x = \(\sqrt{3 y}\)
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 4
(A) What are the coordinates of the point P? [1]
(B) The area (in sq units) of the semi-circular piece of pizza ? [1]
(C) What is the area (in sq units) of the pizza’s piece OABCO?
OR
The area (in sq units) of the pizza’s piece OABO? [2]
Answer:
(A) The point P is the point of intersection of the curve x2 + y2 = 4
and line x = \(\sqrt{3} y\). So, we have
3y2 + y2 = 4
⇒ y = 1 and hence x = \(\sqrt{3}\)
Thus, the coordinates of P are (\(\sqrt{3}\), 1)

(B) The required area = 2\(\int_0^2 \sqrt{2^2-x^2} d x\)
= 2{2.\(\frac{\pi}{2}\) – 0}
= 2π

(C) The required area = \(\int_0^2 \sqrt{2^2-x^2} d x\)
= {2.\(\frac{\pi}{2}\) – 0}
= π

OR

The required area
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 24

Question 38.
In the year 2000, Mrs. 0 Sudha took a home loan of ₹ 30,00,000 from Axis Bank at 9% per annum compounded monthly for 10 years on reducing balance method.
CBSE Sample Papers for Class 12 Applied Mathematics Set 9 with Solutions 5
{Use: (1.0075)120 = 2.4514}

Based on the above information, answer the following:
(A) What is the EMI paid by Sudha? [1]
(B) What is the outstanding amount at the end of 1st year?
OR
What is the outstanding amount at the end of 2nd year? [2]
(C) What is the total interest paid over the 10 years by Sudha? [1]
Answer:
(A) EMI = \(\frac{p \times r \times(1+r)^n}{(1+r)^n-1}\)
= \(\frac{30,00,000 \times 0.0075 \times(1.0075)^{120}}{(1.0075)^{120}-1}\)
= ₹ 38003

(B) Outstanding amount at the end of 1 year
= (P + Pi) – EMI =P(1 + i) – EMI
= 30,00,000 (1.0075) – 38003
= ₹ 29,84,497

OR

Outstanding amount at the end of 2nd year P(1 + i)2 – EMI {(1 + i) + 1}
= 30,00,000 (1.0075)2 – 38003(2.0075)
= 30,45,169 – 76291
= ₹ 29,68,878

(C) Total interest paid =
n × EMI – P = 120 × 38003 – 30,00,000
= ₹ 15,60,360