Class XII Session 2024-25

Subject - Applied Mathematics

Sample Question Paper - 9

Time Allowed: 3 hours Maximum Marks: 80

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
1. If A is a matrix of order 3 and |A| = 8, then |adj A| = [1]

a) 1 b) 23

c) 26 d) 2

2. A grain wholeseller visits the granary market. While going around to make a good purchase, he takes a handful [1]
of rice from random sacks of rice, in order to inspect the quality of farmers produce. The handful rice taken from
a sack of rice for quality inspection is a

a) Statistic b) Sample

c) Parameter d) Population
3. The present value of a sequence of payment of ₹1000 made at the end of every 6 months and continuing forever, [1]
if money is worth 8% per annum compounded semi-annually is

a) 2500 b) 15,000

c) 1000 d) 25,000
4. Linear programming of linear functions deals with: [1]

Page 1 of 21
 a) Minimizing b) Optimizing

c) Maximizing d) Normalizing

5. If A = [
−3 x
] and A = A', then [1]
y 5

a) x = 5, y = -3 b) x = -3, y = 5

c) x = y d) x = 1, y = 2
6. A dice is thrown twice, the probability of occurring of 5 atleast once is [1]

a) b)
5 35

12 36

c) 11

36
d) 7

12

7. If m is the mean of Poisson distribution, then P(r = 0) is given by: [1]

a) e b) m-e

c) em d) e-m

8. Integrating factor of the differential equation (1 − y 2

)
dx

dy
+ xy = ay is [1]

a) 1
b) 1

2
y −1
√1−y 2

c) 1
d) 1

2
1−y
√y 2 −1

9. If in a 600 m race, A can beat B by 50 m and in a 500 m race, B can beat C by 60 m. Then, in a 400 m race, A [1]
will beat C by:

a) 70 m b) 77 m 1

c) 77 m d) 81.33 m
2 3 4
∣2
∣
2 2 ∣
∣
[1]
10. The value of ∣ 2 3
2
4
2
5
∣ is
∣ ∣
4 5 6
∣2 2 2 ∣

a) 29 b) 26

c) 213 d) 0

11. A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol [1]
and now the percentage of alcohol is found to be 26%. The quantity of whisky replaced is

a) b)
2 2
part part
5 3

c) 3

5
part d) 1

3
part

12. If x ∈ R, |x| ≥ -7, then [1]

a) x ∈ [-7, 7] b) x ∈ (−∞ , -7) ∪ [7, ∞)

c) x ∈ R d) x ∈ (−∞ , -7) ∪ (7, ∞)

13. A tank has a leak that would empty it in 10 hours. A tap is turned on which delivers 4 litre a minute into the tank [1]
and now it emptied in 12 hours. The capacity of the tank is

a) 1800 litres b) 648 litres

c) 1440 litres d) 1200 litres

Page 2 of 21
14. Solution set of inequations x − 2y ≥ 0, 2x − y ≤ −2, x ≥ 0, y ≥ 0 is: [1]

a) First quadrant b) Empty

c) Closed halfplane d) Infinite

15. Corner points of the feasible region for an LPP are: (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let z = 4x + 6y the [1]
objective function. The minimum value of z occurs at

a) any point on the line segment joining the b) the mid-point of the line segment joining the
points (0, 2) and (3, 0) points (0, 2) and (3, 0) only

c) (3, 0) only d) (0, 2) only

16. A specific characteristic of a sample is known as a [1]

a) parameter b) variance

c) statistic d) population
17. The value of ∫ 1

x+x log x
dx is [1]

a) log (1 + log x) b) x + log x

c) x log (1 + log x) d) 1 + log x

18. For the given values 15, 23, 28, 36, 41, 46, the 3-yearly moving averages are: [1]

a) 22, 29, 35, 41 b) 24, 29, 35, 41

c) 24, 28, 35, 41 d) 22, 28, 35, 41

19. Assertion (A): If A = [

3 −2
] and I = [
1 0
] , then the value of k such that A2 = kA - 2I, is -1. [1]
4 −2 0 1

Reason (R): If A and B are square matrices of same order, then (A + B)(A + B) is equal to A2 + AB + BA + B2.

a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.

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

20. Let f(x) = x4 - 2x2 + 5 be defined on [-2, 2]. [1]

Assertion (A): The range of f(x) is [2, 13].

Reason (R): The greatest value of f is attained at x = 2.

a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.

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

Section B
21. Calculate five yearly moving averages of the number of students who have studied in a school given below: [2]

Year 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002

No. of students 442 427 467 502 512 515 520 527 515 541

22. A simple interest of ₹ 1000 is paid on a certain sum of money at 10% p.a for 4 years. Find the sum. [2]
OR
Jyoti buys a car for which she makes down payment of ₹3,50,000 and the balance is to be paid in 3 years by monthly
installments of ₹34,000 each. If the financer charges interest at the rate of 12% per annum and uses flat rate method,

Page 3 of 21
 find the actual price of the car.
[2]
1

23. By using property of definite integrals, evaluate ∫ |2x − 1|dx

0

2 −1 0 4 [2]
24. If A = [ ] and B = [ ] , find 3A2 - 2B + I
3 2 −1 7

OR
∣x+ 1 x− 1 ∣ ∣4 −1 ∣
Find the values of x, if ∣ ∣ = ∣ ∣
∣x− 3 x+ 2 ∣ ∣1 3 ∣

25. In what ratio must a grocer mix two varieties of tea worth ₹ 60 per kg and ₹ 65 per kg so that by selling the [2]
mixture at ₹ 68.20 per kg may gain 10%?
Section C
26. Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, [3]
approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will
take for one-half of the original amount of radium to decompose? [Given loge 0.989 = 0.01106 and loge 2 =

0.6931]
OR

Solve: (x2 + 1) + 2xy - 4x2 = 0 subject to the initial condition y(0) = 0.

dy

dx

27. Find the purchase price of a ₹600, 8% bond, dividends payable semi-annually redeemable at par in 5 years, if the [3]
yield rate is to be 8% compounded semi-annually.
28. A company suffers a loss of ₹1,000 if its product does not sell at all. Marginal revenue and Marginal cost [3]
functions for the product are given by MR = 50 - 4x and MC = -10 + x respectively. Determine the total profit
function, break-even points and the profit maximization level of output
29. The income of a group of 10,000 persons was found to be normally distributed with mean ₹ 750 p.m. and [3]
standard deviation ₹ 50. Show that of this group about 95% had income exceeding ₹ 668 and only 5% had
income exceeding ₹ 832. What was the lowest income among the richest 100?
OR
In a binomial distribution the sum and product of the mean and the variance are 25

3
and 50

3
respectively. Find the
distribution.
30. Given below are the consumer price index numbers (CPI) of the industrial workers. [3]

Year 2014 2015 2016 2017 2018 2019 2020

Index number 145 140 150 190 200 220 230

Find the best fitted trend line by the method of least squares and tabulate the trend values.
31. A random sample of 17 values from a normal population has a mean of 105 cm and the sum of the squares of [3]

deviations from this mean is 1225 cm2. Is the assumption of a mean of 110 cm for the normal population
reasonable? Test under 5% and 1% levels of significance. Also, obtain the 95% and 99% confidence limits.
(Given t16 (0.05) = 2.12 and t16 (0.01) = 2.921)

Section D
32. A manufacturer has three machines installed in his factory. Machines I and II are capable of being operated for at [5]
most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each
requiring the use of three machines. The number of hours required for producing one unit each of the items on
the three machines is given in the following table:

Item Number of hours required by the machine

Page 4 of 21
 I II III

A 1 2 1

B 2 1 5

He makes a profit of ₹6.00 on item A and ₹4.00 on item B. Assuming that he can sell all that he produces, how
many of each item should he produce so as to maximize his profit? Determine his maximum profit. Formulate
this LPP mathematically and then solve it.
OR
A farm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. In view of the need to
ensure certain nutrient constituents (call them X, Y and Z), it is necessary to buy two additional products, say, A and
B. One unit of product A contains 36 units of X, 3 units of Y, and 20 units of Z. One unit of product B contains 6
units of X, 12 units of Y and 10 units of Z. The minimum requirement of X, Y and Z is 108 units, 36 units and 100
units respectively. Product A costs ₹ 20 per unit and product B costs ₹ 40 per unit. Formulate the above as a linear
programming problem to minimize the total cost, and solve the problem by using graphical method.
33. Show that the solution set of the following linear in equations is an unbounded set: x + y ≥ 9, 3x + y ≥ 12, x ≥ [5]
0, y ≥ 0
34. Find the probability distribution of the number of green balls drawn when 3 balls are awn, one by one, without [5]
replacement from a bag containing 3 green and 5 white balls.
OR
A random variable X has the following probability distribution:

X 0 1 2 3 4 5 6 7

P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k

Find each of the following:

i. k
ii. P(X < 6)
iii. P(X ≥ 6)
iv. P(0 < X < 5)
35. A loan of ₹ 400000 at the interest rate of 6.75 % p.a. compounded monthly is to be amortized by equal payments [5]
at the end of each month for 10 years. Find
i. the size of each monthly payment.
ii. the principal outstanding at the beginning of 61st month.
iii. the interest paid in 61st payment.
iv. the principal contained in 61st payment.
v. total interest paid.

Given (1.005625)120 = 1.9603, (1.005625)60 = 1.4001)

Section E
36. Read the text carefully and answer the questions: [4]
A factory owner wants to construct a tank with rectangular base and rectangular sides, open at the top, so that its
depth is 2 m and capacity is 8 m3. The building of the tank costs ₹280 per square metre for the base and ₹180

Page 5 of 21
 per square metre for the sides.

(a) If the length and the breadth of the rectangular base of the tank are x metres and y metres respectively,
then find a relation between x and y.
(b) If C (in ₹) is the cost of construction of the tank, then find C as a function of x.
(c) Find the value of x for which the cost of construction of the tank is least.
OR
Find the least cost of construction of the tank.
37. Read the text carefully and answer the questions: [4]
An equated monthly installment (EMI) is a set monthly payment provided by a borrower to a creditor on a set
day, each month. EMIs apply to both interest and principal each month, and the loan is paid off in full over some
years.
How is EMI calculated?
There are two ways in which EMI can be calculated. These methods are:
The flat rate method: When the loan amount is progressively being repaid, each interest charge is computed
using the original principal amount in the flat rate method.
The reducing balance method: The reducing balance technique, compared to the flat rate method,
determines the interest payment according to the outstanding principal.
Example:
A loan of ₹250000 at the interest rate of 6% p.a. compounded monthly is to be amortized by equal payments at
the end of each month for 5 years.
(Given (1.005)60 = 1.3489, (1.005)21 = 1.1104)
(a) Find the size of each monthly payment.
(b) Find the principal outstanding at beginning of 40th month.
(c) Find interest paid in 40th payment.
OR
Find principal contained in 40th payment.
38. Find the inverse of the matrix : [4]
−1 1 2
⎡ ⎤

A = ⎢ 3 −1 1⎥
⎣ ⎦
−1 3 4

and hence show that AA-1 = 1.

OR
Using matrix method, solve the following system of equations for x, y and z :
x-y+z=4
2x + y - 3z = 0
x+y+z=2

Page 6 of 21
 Solution
Section A
1.
(c) 26
Explanation: |A| = d
|adj A| = |A|n - 1
Here, n = 3, |A| = 8
|adj A| = 82
|adj A| = (23)2 = 26
2.
(b) Sample
Explanation: Sample
3.
(d) 25,000
Explanation: The given annuity is a perpetuity.
cash f low
present value of perpetuity = Interest rate

Here, cash flow = ₹1000

8/2
interest rate = 100

= 4

100
= 0.04
1000
So, present value = 0.04

= ₹25,000
4.
(b) Optimizing
Explanation: Optimizing
5.
(c) x = y
−3 x −3 y
Explanation: A = [ ]⇒ A = [
′
]
y 5 x 5

∵ A = A' ⇒ x = y
∴ Option (x = y) is the correct answer.

6.
11
(c) 36

Explanation: Here, n = 2, p = 1

6
,q =
5

6
2

P(X ≥ 1) = 1 - P(0) = 1 - 2
C0 (
5

6
) = 1 −
25

36
=
11

36

7.
(d) e-m
Explanation: Given is the mean of a Passion distribution.
Let's assume that a discrete random Variable.
x follow Poission distribution for over an infinite number of trials and a small finite probability of success then
x −λ

PMF of this random variable x is P (x) = λ e

x!

Here λ is rate parameter and is equal to the mean

i.e. λ = μ
So posission distribution with its mean 'm'
x −m
m e
P (x) =
x!

Page 7 of 21
 0 −m
m e
∴ P (0) =
0!

⇒ P(0) = e-m
8. (a) 1

√1−y 2

dx y ay
Explanation: dy
+ 2
x= 2
, which is linear in x.
1−y 1−y
y
∫ dy 1 2
− log(1− y )
I.F. = e 1−y 2
=e 2 = 1

2
√1−y

9.
(d) 81.33 m
Explanation: When A cover 600 m, B cover 550 m
When B cover 500 m, C cover 440 m
When B cover 400 m, C cover = × 400 = 88 × 4 = 352 m
440

500

In a 400 m race,
B beat C by = 400 - 352 = 48 m
When A cover 400 m, B cover = 550

600
× 400 =
1100

3
m = 366.67 m
In a 400 m, race A beat B by = 400 - 366.67 = 33.33 m
∴ In a 400 m race, A beat C by = 48 + 33.33 = 81.33 m
10.
(d) 0
Explanation: Taking 22, 23 and 24 common from R1, R2 and R3 respectively, we get
2
∣1 2 2 ∣
∣ ∣
22⋅ 23 ⋅ 24 ∣1 2 2
2
∣ (Operate C2 → 1

2
C2)
∣ ∣
2
∣1 2 2 ∣
2
∣1 1 2 ∣
∣ ∣
= 29 ⋅ 2 ∣1 1 2
2
∣ = 210 × 0 = 0 (∵ C1 and C2 are same)
∣ ∣
2
∣1 1 2 ∣

∴ Option (d) is the correct answer.

11.
(b) 2

3
part

Explanation:

40 26
−

So, ratio 100

26
100

19
=
14

7
=
2

1
.
−
100 100

∴ The quantity of whisky replaced by 19% alcohol = 2

2+1
i.e.
2

3
part
12.
(c) x ∈ R
Explanation: x ∈ R
13.
(c) 1440 litres
Explanation: Let' say the capacity of the cistern is x litres
x
so it is leaking at 10
litres per hour
Tap fills in 4 litres a min i.e. 60 × 4 = 240 litres per hour
x
Now, with tap turned on, the water leakage per hour is ( - 240) 10
x
It takes, 12 hours to be emptied now, so per hour leakage is 12

so x

10
− 240 = x

12

on solving for x,
x = 14400

Page 8 of 21
14.
(b) Empty
Explanation: There will be no common region.
15. (a) any point on the line segment joining the points (0, 2) and (3, 0)
Explanation: Here the objective function is given by:
F = 4x + 6y
Corner points Z = 4x + 6y

(0, 2) 12...(Min.)

(3, 0) 12...(Min.)

(6, 0) 24

(6, 8) 72...(Max.)

(0, 5) 30
Hence, it is clear that the minimum value occurs at any point on the line joining the points (0, 2) and (3, 0)
16.
(c) statistic
Explanation: statistic
17. (a) log (1 + log x)
Explanation: I = ∫ 1

x+x log x
dx

dx
I = ∫
x(1+log x)

Put 1 + log x = t
1
⇒ dx = dt
x
1
I = ∫ dt
t

⇒ I = log |t| + C
I = log (1 + log x) + C
18. (a) 22, 29, 35, 41
Explanation: 22, 29, 35, 41
19.
(d) A is false but R is true.
Explanation: Assertion: Given, A2 = kA - 2I
⇒ AA = kA - 2I

3 −2 3 −2 3 −2 1 0
⇒ [ ][ ] = k[ ]− 2[ ]
4 −2 4 −2 4 −2 0 1

9 − 8 −6 + 4 3k −2k 2 0
⇒ [ ] = [ ]− [ ]
12 − 8 −8 + 4 4k −2k 0 2

1 −2 3k − 2 −2k
⇒ [ ]= [ ]
4 −4 4k −2k − 2

By definition of equality of matrix, the given matrices are equal and their corresponding elements are equal.
Now, comparing the corresponding elements, we get
3k - 2 = 1 ⇒ k = 1
⇒ -2k = -2 = k = 1

⇒ 4k = 4 ⇒ k = 1

⇒ -4 = -2A - 2 ⇒ k = 1

Hence, k = 1
Reason: We have,
(A + B)(A + B) = A(A + B) + B(A + B)
= A2 + AB + BA + B2
20.
(d) A is false but R is true.
Explanation: f(x) = x4 - 2x2 + 5 ⇒ f'(x) = 4x3 - 4x

Page 9 of 21
 ⇒ f'(x) = 4x (x - 1) (x + 1)
⇒ f'(x) = 4x (x2 - 1
⇒ f'(x) = 4x (x -1) (x + 1).

For critical points, f'(x) = 0 ⇒ x = 0, -1, 1.

Now, f(-2) = (-2)4 - 2(-2)2 + 5 = 16 - 8 + 5 = 13
f(2) = 24 - 2(2)2 + 5 = 16 - 8 + 5 = 13
f(-1) = (-1)4 - 2(-1)2 + 5 = 1 - 2 + 5 = 4
f(0) = 0 - 2 × 0 + 5 = 5
f(1) = 14 - 2(1)2 + 5 = 4
So, the range of f is [4, 13]
∴ Assertion is false.

Also, f attains it maximum value at x = - 2 and x = 2

∴ Reason is true.

Section B
21. Calculation of 5-year moving averages:

22. Let P be ₹ x, S.I. = ₹ 1000, r = 10% p.a., n = 4 years

x×10×4
∴
100
= 1000
⇒ x = 2500
∴ Sum = ₹ 2500
OR
Let the amount of loan be ₹P
EMI = ₹34000, i = = 0.01, n = 3 × 12 = 36
12

12×100
P+Pni
EMI = n
P(1+36×0.01) P×1.36
⇒ 34000 = 36
= 36
34000×36
⇒ P= 1.36
⇒ P = 900,000
∴ Price of the car = down payment + loan
= ₹350000 + ₹900000 = ₹1250000
1

1 2 1

23. ∫ |2x − 1|dx = ∫ |2x − 1|dx + ∫ |2x − 1|dx

0 0 1

2 1

= ∫ −(2x − 1)dx + ∫ (2x − 1)dx

0 1

2
1
1
2 2 2
= [− (x − x)] + [(x − x)] 1
0
2

1 1 1 1 1 1 1
= −( − ) − 0 + 0 − ( − ) = + ( ) =
4 2 4 2 4 4 2

Page 10 of 21
 2 −1 0 4
24. Given, A = [ ] and B = [ ]
3 2 −1 7

3A2 - 2B + I
2 −1 2 −1 0 4 1 0
= 3[ ][ ] -2[ ] +[ ]
3 2 3 2 −1 7 0 1

4 − 3 −2 − 2 0 8 1 0
= 3[ ] -[ ] +[ ]
6 + 6 −3 + 4 −2 14 0 1

1 −4 0 8 1 0
=3[ ] -[ ] +[ ]
12 1 −2 14 0 1

3 −12 0 8 1 0
=[ ] -[ ] +[ ]
36 3 −2 14 0 1

3 − 0 + 1 −12 − 8 + 0
=[ ]
36 + 2 + 0 3 − 14 + 1

4 −20
=[ ]
38 −10

OR
∣x + 1 x − 1∣ ∣4 −1 ∣
We have ∣ ∣ = ∣ ∣
∣x − 3 x + 2∣ ∣1 3 ∣

⇒ (x + 1)(x + 2) - (x - 1)(x - 3) = 4 × 3 - 1 × (-1)

⇒ (x2 + 2x + x + 2) - (x2 - 3x - x + 3) = 12 + 1
⇒ -2x -1 = 13

⇒ -2x = 14

⇒ x = -7
25. We have,
S.P. of mixture = ₹ 68.20 and Gain = 10%
∴ C.P. =
SP

Gain
1+
100

68.20 682
⇒ C.P. = ( 10
) =( 11
) = ₹ 62
1+
100

The allegation grid is as given below:

By using the rule of allegation, we obtain

Tea of first kind: Tea of second kind = 3 : 2
Hence, the grocer must mix in the ratio 3 : 2
Section C
26. Let A be the quantity of radium present at time t and A0 be the initial quantity of radium.
Therefore, we have,
dA
∝ A
dt
dA

dt
= - 2A
dA

A
= - 2dt
dA
∫ = −λtdt
A

log A = −λt + c ...(i)

Now, A = A0 when t = 0
log A0 = 0 + c
c = log A0

Page 11 of 21
 Put value of c in equation
log A = −λ t + log A0

log(
A0
A
) = −λt ...(ii)
Given that, In 25 years, bacteria decomposes 1.1 %, so
A = (100 - 1.1)% = 98.996 %= 0.989 A0, t = 25
Therefore, (ii) gives,
0.989A0
log(
A0
) = - 25λ
log (0.989) = −25λ
λ = − log (0.989)
1

25

Now, equation (ii) becomes,

log(
A0
A
) ={ 1

25
log (0.989)} t
1
Now A = 2
A0

log(
2A
A
) =
1

25
log (0.989) t
− log 2×25
=t
log(0.989)

−
0.6931×25

0.01106
=t
t = 1567 years
Required time = 1567 years
OR
The given differential equation can be written as
dy 2

dx
+
2x

2
y =
4x

2
...(i)
1+x 1+x

dy
This is a linear differential equation of the form dx
+ Py = Q, where
2

P= 2x

2
and Q = 4x

2
1+x 1+x
2x
∫

= 1 + x2
2 2

I.F. = e ∫ P dx (1+x )dx log(1+ x )

∴ = e = e

Multiplying both sides of (i) by I.F. = (1 + x2), we get

(1 + x2) + 2xy = 4x2
dy

dx

Integrating both sides with respect to x, we get

y(1 + x2) = ∫ 4x2 dx + C [Using: y(I.F.) = ∫ Q (I.F.) dx + C]
y (1 + x2) =
3

⇒
4x

3
+ C ...(ii)
It is given that y = 0, when x = 0. Putting x = 0 and y = 0 in (i), we get
0=0+C⇒C=0
3

Substituting C = 0 in (ii), we get y = 4x

2
, which is the required solution.
3(1+ x )

27. Face value of the bond C = ₹600

Nominal rate of interest i = 8% or 0.08
As dividends are paid semi-annually
Therefore, Rate of interest per period id = 0.08

2
= 0.04
Therefore, periodic dividend payment R = C × id = 600 × 0.04 = 24
So, semi-annual dividend R is ₹24
Yield rate is 8% = 0.08, compounded semi annually
0.08
Therefore i = 2
= 0.04
No. of years n = 5
Therefore, no. of dividend periods (n) = 5 × 2 = 10
Purchase price (V) of the bond is given by
−n
∣ 1−(1+i) ∣
V = R∣ i
∣ + C (1 + i )
−n

∣ ∣
−10
∣ 1−(1+0.04) ∣ −10
= 24 ∣ ∣ + 600(1 + 0.04)
∣ 0.04 ∣
−10
∣ 1−(1.04) ∣ −10
= 24 ∣ ∣ + 600(1.04)
∣ 0.04 ∣

Page 12 of 21
 1−0.6755
= 24 [ ∣ +600(0.6755)
0.04

= 194.7 + 405.3 = 600

Therefore, purchase price of bond is ₹600.
28. Let P denote the profit function. Then,
dP

dx
= MR - MC
⇒
dP
= (50 - 4x) - (-10 + x)
dx
2

⇒
dP

dx
= 60 - 5x and d P

2
= -5
dx

For maximum value of P, we must have

dP

dx
= 0 ⇒ 60 - 5x = 0 ⇒ x = 12
2

Clearly, d P

2
= -5 < 0 for all x.
dx

So, profit P is maximum when 12 units are produced. Thus, the profit maximization level of output is 12 units.
dP
Now, dx
= 60 - 5x
⇒ P = ∫ (60 − 5x)dx + k ... [On intergrating]
⇒ P = 60x - x + k ... (i)
5

2
2

where k is the constant of integration

It is given that the company suffers a loss of ₹ 1000, if its product does not sell at all i.e. P = -1000 at x = 0. Substituting these
values in (i), we obtain k = -1000.
Putting k = -1000 in (i), we obtain:
P = 60x - x + 1000 5

2
2

This is the total profit function. For break-even points

P = 0 ⇒ 60x - 5

2
x
2
+ 1000 = 0 ⇒ 5x2 - 120x + 2000 = 0
⇒ x2 - 24x + 400 = 0
This equation does not give real values of x. So, there is no break-even point.
29. Let X denote the income. Then X is normally distributed with mean μ = ₹ 750 and standard deviation σ = ₹ 50. Let Z be the
standard normal variate. Then,
X−μ X−750
Z= σ
or, Z = 50
668−750
When X = 668, we obtain: Z = 50
= −
82

50
= -1.64
Now, P(X > 668)
= P(Z > -1.64)
= P(-1.64 < Z ≤ 0) + P(Z ≥ 0) = P(0 ≤ Z < 1.64) + 0.5 = 0.4495 + 0.5 = 0.9495
Thus, 94.95% persons had income exceeding ₹ 668
832−750
When X = 832, we obtain: Z = = 1.64 50

∴ P(X > 832)

= P(Z > 1.64)
= P(Z ≥ 0) - P(0 ≤ Z < 1.64) = 0.5 - 0.4495 = 0.0505
Thus, 5.05 % persons had income exceeding ₹ 832
Now, probability of selecting a person out of richest 100 persons = 100

10000
= 0.01
In order to find the lowest income among the richest 100, we have to find the value k of X such that P (X ≥ k) = 0.01
k−750
When X = k, we obtain Z = 50
= Z1(Say)
Now, P(X > k) = 0.01
= P( Z ≥ Z1) = 0.01
= 0.5 - P(0 ≤ Z ≤ Z1) = 0.01
= P(0 ≤ Z ≤ Z1) = 0.49
= Z1 = 2.33
k−750
= 50
= 2.33 ⇒ k = 750 + 50 × 2.33 ⇒ k = 866.5
Hence, the lowest income among the richest 100 was ₹ 866.50
OR
We have,
Sum of the mean and variance = 25

Page 13 of 21
 ⇒ np + npq = 25

⇒ np(1 + q) = 25

3
...(i)
Product of the mean and variance = 50

3
50
⇒ np(npq) = 3
...(ii)
Dividing eq. (ii) by eq. (i), we have,
np(npq)
50 3
= ×
np(1+q) 3 25

npq
⇒ = 2
1+q

⇒ npq = 2(1 + q)
⇒ np(1 - p) = 2(2 - p)
2(2−p)
⇒ np =
(1−p)

25
Substituting this value in np + npq = 3
, we have,
2(2−p) 25
(2 − p) =
(1−p) 3

⇒ 6(4 - 4p + p2) = 25 - 25p

⇒ 6p2 + p - 1 = 0
⇒ (3p - 1) (2p + 1) = 0
1 −1
⇒ p =
3
or 2
1
As p cannot be negative, therefore possible value of p is 3
2
q=1-p= 3

⇒ np + npq = 25

1 2 25
⇒ n( ) (1 + ) =
3 3 3

⇒ n = 15
r 15−r

∴ P(X = r) = 15
Cr (
1

3
) (
2

3
) , r = 0, 1, 2 .... 15
30. Note that the number of years is Odd
⇒ n = odd

Procedure:
i. Take middle year values as As i.e. A = 2017
ii. Find X = xi - A

iii. Find X2 and XY

Year Index number (Y) X = xi - A = xi - 2017 X2 XY Trend value Yt = a + bX

2014 145 -3 9 -435 182.1 + (-3) × 16.6 = 132.3

2015 140 -2 4 -280 182.1 + (-2) × 16.6 = 148.9

2016 150 -1 1 -150 182.1 + (-1) × 16.6 = 165.5

2017 190 0 0 0 182.1 + (0) × 16.6 = 182.1

2018 200 1 1 200 182.1 + (1) × 16.6 = 198.7

2019 220 2 4 440 182.1 + (2) × 16.6 = 215.3

2020 230 3 9 690 182.1 + (3) × 16.6 = 231.9

n=7 ∑Y = 1275 ∑X =0 ∑X
2
= 28 ∑ XY = 465 ∑ Yt = 1274.7
∑Y
a=
n
=
1275

7
= 182.14
∑ XY
and b = 2
=
465

28
= 16.6
∑X

Therefore, the required equation of the straight-line trend is given by

y = a + bX
⇒ y = 182.1 + 16.6 X

31. We have,
μ = Population mean = 110, X = Sample mean = 105
¯

Page 14 of 21
 17
2
n = Sample size = 17 and, ∑ (x i
¯
− X) = 1225
i=1
n
1 2
2 ¯
∴ s = ∑ (xi − X )
n
i=1
−−−−−−
⇒ s
2
=
1225

17
= 72.0588 ⇒ s = √72.0588 = 8.4887
We define, Null Hypothesis H0: There is no significant difference between the sample mean and population means i.e. assumption
that mean of the population is 110 cm is valid.
Alternate hypothesis H1: Assumption that mean of the population is 110 cm is not valid. Let t be the test statistic given by
¯
X −μ 105−110 −−−−− −5×4
t= s
⇒ t= 8.4887
× √17 − 1 =
8.4887
= -2.3561
√n−1

⇒ |t| = 2.3561
The sample statistic follows Student's t -distribution with v = (17 - 1) = 16 degrees of freedom.
We shall now compare this calculated value with the tabulated value of t for 16 degrees of freedom at 5% and 1% levels of
significance.
At 5% level of significance: It is given that t16(0.05) = 2.12
We find that Calculated |t| = 2.3561 > 2.12 = t16(0.05)
i.e. Calculated |t| > Tabulated t16(0.05)
So, we reject the null hypothesis at 5% level of significance. Hence, the assumption that the population has a mean of 110 cm is
not correct.
The confidence limits at 5% level of significance are
s s
¯
X − t16(0.05) and X
¯
+ t16(0.05)
√n−1 √n−1

8.4887 8.4887
or 105 - 4
× 2.12 and 105 + 4
× 2.12
or, 105 - 4.499 = 100.501 and 105 + 4.499 = 109.499
The confidence interval is [100.501,109.499]
At 1% level of significance: It is given that t16(0.01) = 2.921
Clearly, calculated |t| < tabulated t16(0.01)
So, we accept the null hypothesis at 1% level of significance. Hence, the assumption that the mean of the population is 110 cm is
valid.
The confidence limits at 1% level of significance are
¯
X − t16(0.01) and X
s ¯
+ t16(0.01) s

√n−1 √n−1

or, 105 - 8.4887

4
× 2.921 and 105 + 8.4887

4
× 2.921
or, 105 - 6.199 = 98.801 and 105 + 6.199 = 111.199
The confidence interval at 1% level of significance or at 99% confidence level is [98.801, 111.199]
Section D
32. Let x units of item A and y units of item B be manufactured. Therefore, x, y ≥ 0
As we are given,
Item Number of hours required by the machine

I II III

A 1 2 1
5
B 2 1 4

Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day.
According to the question, the constraints are
x + 2y ≤ 12
2x + y ≤ 12
x+ y≥55

He makes a profit of ₹6.00 on item A and ₹4.00 on item B. Profit made by him in producing x items of A and y items of B is 6x +
4y
Total profit Z = 6x + 4y which is to be maximized
Thus, the mathematical formulation of the given linear programming problem is

Page 15 of 21
Max Z = 6x + 4y, subject to
x + 2y ≤ 12
2x + y ≤ 12
5
x+ 4
y≥5
x, y ≥ 0
First, we will convert the inequations into equations as follows:
5
x+ 2y = 12, 2x + y = 12, x + y = 5, x = 0 and y = 0
4

The region represented by x + 2y ≤ 12

The line x + 2y = 12 meets the coordinate axes at A(12, 0) and B(0, 6) respectively. By joining these points, we obtain the line x +
y = 12. Clearly (0, 0) satisfies the x + 2y = 12. So, the region which contains the origin represents the solution set of the
inequation x + 2y ≤ 12
The region represented by 2x + y ≤ 12
The line 2x + y = 12 meets the coordinate axes at C(6, 0) and D(0, 12) respectively. By joining these points, we obtain the line 2x
+ y = 12. Clearly (0, 0) satisfies the 2x + y = 12. So, the region which contains the origin represents the solution set of the
inequation 2x + y ≤ 12
The region represented by x + y ≥ 5 5

The line x + 5

4
y ≥ 5 meets the coordinate axes at E(5, 0) and F(0, 4) respectively. By joining these points, we obtain the line x +
5 5

4
y = 5. Clearly (0, 0) satisfies the x + 4
y ≥ 5. So, the region which does not contain the origin represents the solution set of the
5
inequation x + 4
y≥5
The region represented by x ≥ 0, y ≥ 0:
Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations
x ≥ 0 and y ≥ 0
The feasible region determined by the system of constraints
x + 2y ≤ 12, 2x + y ≤ 12, x + y ≥ 5, x, y ≥ 0 are as follows:
5

Thus the maximum profit is of ₹40 obtained when 4 units each of items A and B are manufactured
The corner points are D(0, 6), I(4, 4), C(6, 0), G(5, 0), and H(0, 4). The values of Z at these corner points are as follows:
Corner points Z = 6x + 4y

D 24

I 40

C 36

G 30

H 16

Page 16 of 21
 The maximum value of Z is 40 which is attained at I(4, 4).
OR
The data given in the problem can be summarized in the following tabular form:
Nutrient constituent
Product Const in ₹
X Y Z

A 36 3 20 20

B 6 12 10 40

Minimum Required 108 36 100

Let x units of product A and y units of product B are bought to fulfill the minimum requirement of X, Y and Z and to minimize
the cost.
The mathematical formulation of the above problem is as follows:
Minimize Z = 20x + 40y
Subject to 36x + 6y ≥ 108
3x + 12y ≥ 36
20x + 10y ≥ 100
and, x, y, z ≥ 0
The set of all feasible solutions of the above LPP is represented by the feasible region shaded darkly in Figure. The coordinates of
the corner points of the feasible region are A2 (12, 0), P1 (4, 2), P2 (2, 6) and B1 (0,18).

Now, we have to find a point or points in the feasible region which give the minimum value of the objective function. For this, let
us give some value to Z, say 20, and draw a dotted line 20 = 20x + 40y. Now, draw lines parallel to this line which have at least
one point common to the feasible region and locate a line that is nearest to the origin and has at least one point common to the
feasible region. Clearly, such a line is Z1 = 20x + 40y and it has a point P1(4, 2) common with the feasible region. Thus, Z1 = 20x
+ 40y is the minimum value of Z, and the feasible solution which gives this value of Z is the comer P1(4, 2) of the shaded region.
The values of the variables for the optimal solution are x = 4, y = 2. Substituting these values in Z = 20x + 40y, we get Z = 160 as
the optimal value of Z.
Hence, 2 units of product A and 4 units of product B are sufficient to fulfill the minimum requirement at a minimum cost of ₹160
33. First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the
equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality.
You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y-intercepts always.
x+y≥9
x 0 5 9

y 9 4 0

Page 17 of 21
 3x + y ≥ 12
x 0 2 4

y 12 6 0
x ≥ 0, y ≥ 0

34. Let X be a random variable denoting the total number of green balls drawn in three draws without replacement. Clearly, there may
be all green, 2 green, 1 green or no green at all. Therefore, X can take values 0, 1, 2, and 3. Let G denote the event of getting a i

green ball in ith draw.

Now, we have,
P (X = 0) = Probability of getting no green ball in three draws
¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ 5 4 3 5
⇒ P (X = 0) =P (G 1 ∩ G2 ∩ G3 ) = P (G1 )P (G2 /G1 )P (G3 /G1 ∩ G2 ) =
8
×
7
×
6
=
28

F (X = 1) = Probability of getting one green ball in three draws

¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯
⇒ P (X = 1) = P ((G 1 ∩ G2 ∩ G3 ) ∪ (G1 ∩ G2 ∩ G3 ) ∪ (G1 ∩ G2 ∩ G3 ))

¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯
⇒ P (X = 1) = P (G 1
∩ G2 ∩ G3 ) + P (G1 ∩ G2 ∩ G3 ) + P (G1 ∩ G2 ∩ G3 ) +P (G1 )P (G2 /G1 )P (G3 /G1 ∩ G2 )

⇒ P(X=1)= 3

8
×
5

7
×
4

6
+
5

8
×
3

7
×
4

6
+
5

8
×
4

7
×
3

6
=
15

28

¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯
P (X = 2) = P((G 1 ∩ G2 ∩ G3 ) ∩ (G1 ∩ G2 ∩ G3 ) ∪ (G1 ∩ G2 ∩ G3 ))

¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯
⇒ P (X = 2) = P(G 1) P (G2 /G1 ) P (G3 /G1 ∩ G2 ) + P (G1 )P (G2 /G1 ) P (G3 /G1 ∩ G2 )

¯
¯¯¯¯
¯ ¯
¯¯¯¯
¯
+P (G1 ) P (G2 /G1 ) P (G3 /G1 ∩ G2 )

3 5 5 3 3 5 15
⇒ P (X = 2) = 8
×
2

7
×
6
+
8
×
7
×
2

6
+
8
×
7
×
2

6
=
56

and,
G2 G3
P (X = 3) = P(G 1 ∩ G2 ∩ G3 ) = P (G1 ) P (
G1
)P (
G1 ∩ G2
) =
3

8
×
2

7
×
1

6
=
1

56

Therefore, the probability distribution of the number of green balls is given by

X 0 1 2 3
5 15 15 1
P(X) 28 28 56 56

OR
i. We know that the sum of all the probabilities in a probability distribution is always
unity. Therefore,we have,
P (X = 0) + P (X = 1) + ....+ P { X = 7 ) = 1
⇒ 0 + k + 2k + 2k + 3 k + k2 + 2k2 + 7 k2 + k = 1
⇒ 10k2 + 9k - 1 = 0
⇒ (10k -1 ) (k + 1) = 0

⇒ 10k - 1 = 0
1
⇒ k =
10

ii. P (X < 6) = P(X = 0) + P(X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5)

⇒ P (X < 6) = 0 + k + 2k + 2k + 3k+ k2
⇒ P (X < 6) = k2 + 8k

Page 18 of 21
 2
1 8 1
⇒ P (X < 6) = (
10
) +
10
...[∵ k = 10
]
81
⇒ P (X < 6) =
100

iii. P (X ≥ 6) = P (X = 6) + P (X = 7)
⇒ P (X ≥ 6) = 2k 2
+ 7k
2
+ k

⇒ P (X ≥ 6) = 9k2 + k
⇒ P (X​​≥ 6) =
9

100
+
1

10
...[∵ k = 1

10
]
19
⇒ P (X ≥ 6) =
100

iv. P (0 <X <5) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

⇒ P (0 < X < 5 ) = k + 2k + 2k + 3k

⇒ P (0 <X <5) = 8k
8 4 1
⇒ P (0 < X < 5) =
10
=
5
...[∵ k = 10
]
35. i. Given P = ₹ 400000, n = 120, i = 6.75

1200
= 0.005625
120
400000×0.005625×(1.005625)
∴ EMI = 120
(1.005625) −1

= 400000×0.005625×1.9603

0.9603
= ₹4593.
ii. Principal outstanding at the beginning of 61 rnonths
n−k+1 120−61+1
EMI[(1+i) −1] 4593[(1.005625) −1]

=
n−k+1
= 120−61+1
i(1+i) 0.005625(1.005625)

4593(1.4001−1)
= 0.005625×1.4001
= ₹ 233336.89
n−k+1
EMI [(1+i) −1]

iii. Interest paid in 61st payment = n−k+1

(1+i)

4593×0.4001
=
1.4001
= ₹1312.52
iv. Principal paid in 61st payment = EMI - Interest paid in 61st period
= ₹ 4593 - ₹ 1312.52 = ₹ 3280.48
v. Total interest paid = n × EMI - P
= 120 × 4593 - 400000 = ₹ 151160.
Section E
36. Read the text carefully and answer the questions:
A factory owner wants to construct a tank with rectangular base and rectangular sides, open at the top, so that its depth is 2 m and
capacity is 8 m3. The building of the tank costs ₹280 per square metre for the base and ₹180 per square metre for the sides.

(i) Given volume = 8 m3 and height = 2 m

So, x × y × 2 = 8 ⇒ xy = 4
(ii) Area of sides of the tank = 2(x + y) × h = 4(x + y) m2
∴ The cost of construction of the sides of the tank = ₹ 180 × 4(x + y) = ₹720 (x + y)
The cost of construction of the base of the tank = ₹ x × y × 280 = ₹ 4 × 280 = ₹ 1120 (using part (x × y × 2 = 8 ⇒ xy
= 4))
So, C = ₹ [1120 + 720 (x + y)]
⇒ C = 1120 + 720 (x + 4

x
)

(iii)
2
dC d C 5760

dx
= = 0 + 720 (1 − 4

2
) and 2
=
3
.
x dx x

= 0 ⇒ x2 = 4 ⇒ x = 2
dC
Now, dx
= 0 ⇒ (1 − 4

2
)
x

2
d C 5760
[
2
] = 8
>0
dx x=2

⇒ C is minimum when x = 2
OR

Page 19 of 21
 The least cost of the tank = ₹ [1120 + 720 (2 + 4

2
)]

= ₹(1120 + 2880) = ₹ 4000

37. Read the text carefully and answer the questions:
An equated monthly installment (EMI) is a set monthly payment provided by a borrower to a creditor on a set day, each month.
EMIs apply to both interest and principal each month, and the loan is paid off in full over some years.
How is EMI calculated?
There are two ways in which EMI can be calculated. These methods are:
The flat rate method: When the loan amount is progressively being repaid, each interest charge is computed using the
original principal amount in the flat rate method.
The reducing balance method: The reducing balance technique, compared to the flat rate method, determines the interest
payment according to the outstanding principal.
Example:
A loan of ₹250000 at the interest rate of 6% p.a. compounded monthly is to be amortized by equal payments at the end of each
month for 5 years.
(Given (1.005)60 = 1.3489, (1.005)21 = 1.1104)
(i) Given, P = ₹ 250000, i = = 0.005 and n = 5 × 12 = 60 12×100
6

60
250000×0.005×(1.005)
EMI = 60
(1.005) −1

250000×0.005×1.3489
=
0.3489
= ₹ 4832.69
6
(ii) Given, P = ₹ 250000, i = 12×100
= 0.005 and n = 5 × 12 = 60
Principal outstanding at beginning of 40th month
60−40+1 21
EMI[(1+i) −1] 4832.69×[(1.005) −1]

= =
60−40+1 21
i(1+i) 0.005×(1.005)

4832.69×[1.1104−1] 4832.69×0.1104
=
0.005×1.1104
= 0.005×1.1104
= ₹ 96096.72
6
(iii)Given, P = ₹ 250000, i = 12×100
= 0.005 and n = 5 × 12 = 60
60−40+1
EMI[(1+i) −1]

Interest paid in 40th payment = 60−40+1

(1+i)

21
4832.69×[(1.005) −1]

= 21
= 4832.69×0.1104

1.1104
= ₹ 480.48
(1.005)

OR
Given, P = ₹ 250000, i = 12×100
6
= 0.005 and n = 5 × 12 = 60
Principal paid in 40th payment = EMI - Interest paid in 40th payment
= 4832.69 - 480.48 = ₹ 4352.21
38. Here, |A| = -(-4 - 3) - (12 + 1) + 2(9 - 1)
= 7 - 13 + 16 = 10 ≠ 0
T
−7 −13 8 −7 2 3
⎡ ⎤ ⎡ ⎤

⇒ adj(A) = ⎢ 2 −2 2⎥ = ⎢ −13 −2 7⎥
⎣ ⎦ ⎣ ⎦
3 7 −2 8 2 −2

−7 2 3
⎡ ⎤
Hence A −1
=
1

10
⎢ −13 −2 7⎥
⎣ ⎦
8 2 −2

−1 1 2 −7 2 3 1 0 0
⎡ ⎤⎡ ⎤ ⎡ ⎤
−1 1
AA = ⎢ 3 −1 1 ⎥ ⎢ −13 −2 7 ⎥ = ⎢0 1 0⎥
10
⎣ ⎦⎣ ⎦ ⎣ ⎦
−1 3 4 8 2 −2 0 0 1

OR
The matrix equation AX = B is
1 −1 1 x 4
⎡ ⎤⎡ ⎤ ⎡ ⎤

⎢2 1 −3 ⎥ ⎢ y ⎥ = ⎢ 0 ⎥
⎣ ⎦⎣ ⎦ ⎣ ⎦
1 1 1 z 2

|A| = 10
′
4 −5 1 4 2 2
⎡ ⎤ ⎡ ⎤

adj A = ⎢ 2 0 −2 ⎥ = ⎢ −5 0 5⎥

⎣ ⎦ ⎣ ⎦
2 5 3 1 −2 3

Page 20 of 21
 4 2 2
⎡ ⎤

A-1 =
1
Here 10
⎢ −5 0 5⎥
⎣ ⎦
1 −2 3

x 4 2 2 4 2
⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤

So, ⎢ y ⎥ = 1

10
⎢ −5 0 5 ⎥ ⎢ 0 ⎥ = ⎢ −1 ⎥
⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦
z 1 −2 3 2 1

Thus, x = 2, y = -1, z = 1

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