Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 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 (20 marks)

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

Question 1.
In an examination out of 1000 students, 70% boys and 80% girls are passed, if total pass percentage was 76%, then the number of girls is: [1]
(a) 560
(b) 600
(c) 580
(d) 620
Solution:
(b) 600

Explanation:
Let the number of girls be x and number of boys is 1000 – x.
\(\frac{70}{100}\) (1000 – x)+ \(\frac{80}{100}\) × x = \(\frac{76}{100}\) × 1000
70(1000 – x) + 80x = 76 × 100
70 × 1000 – 70x + 80x = 76 × 1000
10x = 1000 (76 – 70)
10x = 6000
x = 600
Hence the number of girls = 600.

Question 2.
\(\int x^2 e^{x^3}\) dx equals: [1]
(a) \(\frac{1}{3}\) ex3 + C
(b) \(\frac{1}{2}\) ex3 + C
(c) \(\frac{1}{3}\) ex2 + C
(d) \(\frac{1}{2}\) ex2 + C
Solution:
(a) \(\frac{1}{3}\) ex3 + C

Explanation:
Let I = \(\int x^2 e^{x^3} d x\)
Putting x3 = t, we get 3x2 dx = dt
So, I = \(\int x^2 e^{x^3} d x\) = \(\frac{1}{3} \int e^t d t\)
⇒ I = \(\frac{1}{3}\) et + C, or \(\frac{1}{3}\) ex3 + C

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 3.
E(X) of the probability distribution of X is given by [1]

Xi 1 2 3 4 5 6
P(X = xi) 0.05 0.43 0.27 0.12 0.09 0.04

(a) 2.89
(b) 1.6
(c) 5
(d) 7.9
Solution:
(a) 2.89

Explanation:
E(X) = (1 × 0.05) + (2 × 0.43) + (3 × 0.27) + (4 × 0.12) + (5 × 0.09) + (6 × 0.04)
= 0.05 + 0.86 + 0.81 + 0.48 + 0.45 + 0.24
= 2.89

Question 4.
If the function f(x) = x3 – 27x + 5 is strictly increasing when [1]
(a) x < – 3 (b) |x| > 3
(c) x > 3
(d) |x| < 3 Solution: (b) |x| > 3

Explanation:
Given f(x) = x3 – 27x + 5
So, f ‘(x) = 3x2 – 27 = 3 (x2 – 9) = 3 (x + 3)(x – 3)
Here, f ‘(x) > 0 if x > 3 or x < -3 Hence, f(x) is strictly increasing when |x| > 3

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 5.
Let X be a discrete random variable assuming values x1, x2, x3, ……………….. xn, with probabilities p1, p2, p3, ……………….. pn, respectively. Then the variance of X is: [1]
(a) E(X2)
(b) E(X2) + E(X)
(c) E(X2)-[E(X)]2
(d) \(\sqrt{E\left(X^2\right)-[E(X)]^2}\)
Solution:
(c) E(X2)-[E(X)]2

Question 6.
If a pipe fills a tank in 4 hours, what part of the tank will be filled 1.2 hours? [1]
(a) 0.3
(b) 0.35
(c) 0.4
(d) 0.28
Solution:
(a) 0.3

Question 7.
If A and B are square matrices of the same order, then the value of (A + B)(A – B) is equal to : [1]
(a) A2 – B2
(b) A2 – BA – AB – B2
(c) A2 – B2 + BA – AB
(d) A2 – BA + B2 + AB
Solution:
(c) A2 – B2 + BA – AB

Explanation:
(A + B) (A – B) = AA + BA – AB – BB
= A2 + BA – AB – B2

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 8.
The probability of getting exactly a toss of 7 tosses of a coin is: [1]
(a) \(\frac{21}{32}\)
(b) \(\frac{21}{128}\)
(c) \(\frac{21}{64}\)
(d) \(\frac{21}{256}\)
Solution:
(b) \(\frac{21}{128}\)

Explanation:
P(X = 5) = 7C5 (\(\frac{1}{2}\))5 (\(\frac{1}{2}\))2
= 21 (\(\frac{1}{2}\))7, or \(\frac{21}{128}\)

Question 9.
The line y = x + 1 is a tangent to the curve y2 = 4x at the point. [1]
(a) (1, 2)
(b) (2, 1)
(c) (-1, 2)
(d) (1, 2)
Solution:
(a) (1, 2)

Explanation:
For y2 = 4x, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{y}\)
Since the slope of the line y = x + 1, which is tangent to the curve is 1, we equate \(\frac{2}{y}\) to 1 and get y = 2.
Substituting y = 2 in y2 = 4x, we get x = 1
So, the required point is (1, 2)

Question 10.
In a Poisson distribution the sum of mean and variance is 32. Its S.D. is [1]
(a) 4
(b) 8
(c) 128
(d) 256
Solution:
(a) 4

Explanation:
For a P.D, mean = variance
and S.D = \(\sqrt{\text { Variance }}\)
Here, mean = variance = 16
and hence S.D = \(\sqrt{\text { Variance }}=\sqrt{16}\) = 4

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 11.
If the mean and variance of a binomial distribution are 4 and 2 respectively, then the Probability of 2 successes is: [1]
(a) \(\frac{28}{256}\)
(b) \(\frac{219}{256}\)
(c) \(\frac{128}{256}\)
(d) \(\frac{37}{256}\)
Solution:
(a) 256

Explanation :
Here, np = 4 and npq = 2
So, q = \(\frac{1}{2}\) and hence p = \(\frac{1}{2}\) and n = 8
So, P(X = 2) = 8C2 (\(\frac{1}{2}\))2 (\(\frac{1}{2}\))6 = 28(\(\frac{1}{2}\))8
= \(\frac{28}{256}\)

Question 12.
For real values of x, the minimum value of \(\frac{1-x+x^2}{1+x+x^2}\) is [1]
(a) 0
(b) 1
(c) 3
(d) \(\frac{1}{3}\)
Solution:
(d) \(\frac{1}{3}\)

Explanation:
Let y = \(\frac{1-x+x^2}{1+x+x^2}\), x ∈ R
Then, x2 (y – 1) + x(y + 1) + y – 1 = 0 ………………… (1)
Since x is real, discriminant of (1) ≥ 0
⇒ (y + 1)2 – 4(y – 1)(y – 1) ≥ 0
i.e., -3y2 + 10y – 3 ≥ 0
or 3 (y – 3) (y – \(\frac{1}{3}\)) ≤ 0
or (y – 3) (y – \(\frac{1}{3}\)) ≤ 0
or \(\frac{1}{3}\) ≤ y ≤ 3
Hence, minimum value of y is \(\frac{1}{3}\).

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 13.
The mean of the standard normal variable is [1]
(a) zero
(b) 2
(c) np
(d) 1
Solution:
(a) zero

Question 14.
The components of time series attached to long term variation is termed as [1]
(a) seasonal trend
(b) cyclical trend
(c) secular trend
(d) irregular trend
Solution:
(c) secular trend

Question 15.
The order and degree of the differential equation \(\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}+y=\log \left(\frac{d y}{d x}\right)\) are respectively [1]
(a) 2, 1
(b) 1, 2
(c) 2, 0
(d) 2, not defined
Solution:
(d) 2, not defined

Explanation: The order of highest order derivative \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) is 2.
Given differential equation is not a polynomial equation in its derivatives. So, its degree is not defined.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 16.
If f(a + b + x) = f(x), then \(\int_a^b x f(x) d x\) [1]
(a) \(\frac{a+b}{2} \int_a^b f(b+x)\)
(b) \(\frac{a+b}{2} \int_a^b f(b+x)\)
(c) \(\frac{b-a}{2} \int_a^b f(x) d x\)
(d) \(\frac{a+b}{2} \int_a^b f(x) d x\)
Solution:
(a) \(\frac{a+b}{2} \int_a^b f(b+x)\)

Explanation:
Let I = \(\int_a^b\) xf(x) dx
= \(\int_a^b\) (a + b – x) f (a + b – x) dx
= (a + b) \(\int_a^b\) f(a + b – x) dx – \(\int_a^b\) x f(a + b – x) dx
= (a + b) \(\int_a^b\) f(x) dx = \(\int_a^b\) x f(x) dx
= (a + b) \(\int_a^b\) f(x) dx – I
⇒ 2I = (a + b) \(\int_a^b\) f(x) dx
⇒ I = \(\frac{a+b}{2} \int_a^b f(x) d x\)

Question 17.
The formula for simple average of price relatives is: [1]
(a) \(\sum \frac{P_c}{P_b}\) × 100
(b) \(\frac{1}{n} \sum \frac{q_c}{q_b}\) × 100
(c) \(\frac{1}{n} \sum \frac{P_c}{P_b}\) × 100
(d) \(\sum \frac{q_c}{q_b}\) × 100
Solution:
(c) \(\frac{1}{n} \sum \frac{P_c}{P_b}\) × 100

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 18.
While determining trend values by semi¬averages method, the data containing even number of years is divided into : [1]
(a) four parts
(b) three parts
(c) two parts
(d) two equal parts
Solution:
(d) two equal parts

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 (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): Z = 20x1 + 20x2 subject x2 ≥ 0, x1 ≥ 2, + x1 + 2x2 ≥ 8, 3x1 + 2x1 ≥ 15, 5x1 + 2x2 ≥ 20
Out of the comer points of feasible region (8, 0), (\(\frac{5}{2}\), \(\frac{15}{4}\)), (\(\frac{7}{2}\), \(\frac{9}{4}\)) and (0, 10) the minimum value of Z occurs at (\(\frac{7}{2}\), \(\frac{9}{2}\)). [1]
Reason (R):

Corner Points Z = 20x1 + 20x2
(8, 0) 160
(\(\frac{5}{2}\), \(\frac{15}{4}\)) 16/25
(\(\frac{7}{2}\), \(\frac{9}{4}\)) 115 → min.
(0, 10) 200

Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 20.
A and B are two matrices such that both AB and BA are defined. [1]
Assertion (A): (A + B) (A – B) = A2 – B2
Reason (R): (A + B) (A – B) = A2 – AB + BA – B2.
Solution:
(d) (A) is false but (R) is true.

Explanation:
For two matrices A and B, even if both AB and BA are defined, generally AB ≠ BA
(A + B) (A – B) = A2 – AB + BA – B2.
Since, AB ≠ AB, (A + B) (A – B) ≠ A2 – B2
Hence, R is true but A is false.

Section – B (10 Marks)

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

Question 21.
Write any three consecutive odd numbers whose sum lies between 90 and 100. [2]
Solution:
Let three consecutive odd numbers be (2x + 1), (2x + 3) and (2x + 5).
Then, 90 < (2x + 1) + (2x + 3) + (2x + 5) < 100
90 < 6x + 9 < 100
81 < 6x < 91
\(\frac{81}{6}\) < x < \(\frac{91}{6}\)
So, the possible value of x is 14 or 15.
three consecutive odd numbers may be 29, 31, 33 or 31, 33, 35

Question 22.
In hypothesis testing, define‘null hypothesis’ and‘alternative hypothesis’. [2]
OR
(A) Find the critical (or tabulated) value for α = 0.01 with d.f. = 22 for a left – tailed test.
(B) Find the critical (or tabulated) values for α = 0.10 with d.f. = 18 for a two – tailed test.
Solution:
Null and alternative hypotheses are used in statistical hypothesis testing. The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

OR

(A) – 2.51 (B) + 1.73 and -1.73

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 23.
Mr. Ashok has taken a loan of ₹ 5,00,000 at an interest rate of 10%. His EMI is ₹ 10,000 (reducing balance) / Calculate his loan tenure. [Given (1.0075)120 = 2.4514] [2]
Solution:
Using “reducing balance method”
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 4
⇒ (1.0083)n = \(\frac{2.4096}{1.4096}\) = 1.7094
⇒ n log (1.0083) = log (1.7094)
⇒ n = \(\frac{\log (1.7094)}{\log (1.0083)}\)
= \(\frac{0.23284}{0.00359}\) = 65 (approx.)
Thus, the loan’s tenure is 65 months.

Question 24.
Let a, b be distinct positive real numbers. Then, prove that \(\frac{1}{a}+\frac{1}{b}>\frac{2}{a+b}\). [2]
Solution:
a > 0 and b> 0 a2 > 0, b2 > 0
Hence, a2 + b2 > 0
or a2 + b2 + 2ab > 2ab
or (a + b)2 > 2ab
or \(\frac{a+b}{a b}>\frac{2}{a+b}\)
or \(\frac{1}{a}+\frac{1}{b}>\frac{2}{a+b}\)

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 25.
Graph the following system of constraints and shade the feasible region: [2]
3x + 7y – 1 < 1 x > 0, y > 0
OR
Graph the following system of constraints and shade the feasible region:
3x + y ≥ 24 ; x + y ≥ 16; x + 3y ≥ 24
x ≥ 0, y ≥ 0
Solution:
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 5
OR
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 6

Section – C (18 marks )

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

Question 26.
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]
OR
Solve for x, y, z and t, if \(2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{cc}
1 & -1 \\
0 & 2
\end{array}\right]\) = \(3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
Given
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 7
= (qr – ps) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= (qr – ps)I, i.e., R.H.S.

OR

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 8
⇒ 2x + 3 = 9; 2z – 3
= 15; 2y = 12
and 2t + 6 = 18
Solving these equations, we have x = 3, y = 6, z = 9 and t = 6.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 27.
Find the equation of the tangent and the normal to the curve x(2/3) + y(2/3) = 2 at the point (1, 1). [3]
Solution:
x2/3 + y2/3 = 2 gives \(\frac{d y}{d x}=-\frac{y^{1 / 3}}{x^{1 / 3}}\)
⇒ \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = \(\left(-\frac{y^{1 / 3}}{x^{1 / 3}}\right)_{(1,1)}\) = -1
So, the equation of the tangent to the curve at (1, 1) is
y – 1 = -1(x – 1)
i.e., x + y = 2
The equation of the normal to the curve at (1, 1) is
y – 1 = 1 (x – 1)
i.e., x – y = 0

Question 28.
Find: \(\int \frac{x^2}{\sqrt{x^6-1}}\) dx [3]
OR
Draw a rough sketch of y = \(\sqrt{6 x+4}\) and find the area of the region bounded by the curve y = \(\sqrt{6 x+4}\) and the ordinates x = 0 and x = 2.
Solution:
Let I = \(\int \frac{x^2}{\sqrt{x^6-1}}\) dx
= \(\int \frac{x^2}{\sqrt{\left(x^3\right)^2-1}}\) dx
Put x3 = t so that 3x2dx = dt or = x2dx = \(\frac{d t}{3}\)
Thus,
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 9
OR

The rough sketch of the curve is shown on the right.
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 10
= \(\frac{1}{9}\) (64 – 8) = \(\frac{56}{9}\) sq units.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 29.
Ten cartons are taken at random from an automatic packing machine. The mean net weight of the ten cartons is 11.8 kg. and standard deviation is 0.15 kg. Does the sample mean differ significantly from the intended weight of 12 kg? (Given t9 (0.05) = 2.26) [3]
Solution:
Here,
H0 : µ = 12, s = 0.15, n = 10 and \(\bar{X}\) = 11.8
To test H0, the statistic t is
t = \(\frac{\bar{X}-\mu}{s / \sqrt{n-1}}=\frac{11.8-12}{0.15 / \sqrt{9}}\) = -4
The table value of t at α = 0.05 and 9 d.f. is 2.26

Conclusion: Since |t| > tα, the null hypothesis is rejected and we conclude that the sample mean differs significantly from the intended mean of 12 kg.

Question 30.
Construct price index by simple average of price relatives method for the year 2015 by taking year 2012 as base year from the following data: [3]

Commodity A B C D E
Price (in ₹) in 2012 100 80 160 220 40
Price (in ₹) in 2015 140 120 180 240 40

Solution:
We have

Commodity Base year Price (₹) Current year Price (₹) Price relative \(\frac{p_c}{p_b}\) × 100
A 100 140 \(\frac{140}{100}\) × 100 = 140
B 80 120 \(\frac{120}{80}\) × 100 = 150
C 160 180 \(\frac{180}{160}\)× 100 = 112.5
D 220 240 \(\frac{240}{220}\) × 100 = 109.1
E 40 40 \(\frac{40}{40}\) × 100 = 100

Here, Σ(\(\frac{p_c}{p_b}\) × 100) = 611.6 and n = 5
Hence, the required price index by simple average of price relatives method
= \(\frac{\sum\left(\frac{p_c}{p_b} \times 100\right)}{n}=\frac{611.6}{5}\) = 122.32

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 31.
A machine costs a company ₹ 5,75, 000 and its effective life is estimated to be 20 years. A sinking fund is created for replacing the machine at the end of its life time when its scrap realises a sum of ₹ 75,000 only. Calculate what amount should be provided every year out of profits, for the sinking fund if it accumulates an interest of 5% per annum, compounded annually. (Use (1.05)20 = 2.655). [3]
Solution:
Net amount required at the end of 20 years = ₹ 5,75,000 – ₹ 75,000 = ₹ 5,00,000
Now, S = \(\frac{\mathrm{A}}{\mathrm{r}}\) [(1 + r)n – 1] gives
5,00,000 = \(\frac{\mathrm{A}}{0.05}\) [(1.05)20 – 1]
⇒ A = \(\frac{25000}{[2.655-1]}\)
= \(\frac{25000}{1.655}\) = 15105.74
Thus, a sum of ₹ 15105.74 must be put into the sinking fund annually.

Section – D (20 Marks)

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

Question 32.
Suppose that a bond has a face value of ₹ 1000 and will mature in 10 years. The annual coupon rate is 5%, the bond makes semi-annual coupon payments. With a price of ₹ 950, what is the bond’s YTM? [5]
Solution:
Given. F = ₹ 1000 and id = \(\frac{5 \%}{2}\) = 0.025
∴ R = F × id = ₹ (1000 × 0.025) = ₹ 25
ALso, Present value = ₹ 950
Now, Approx. YTM = Approx.
YTM = \(\frac{C+\frac{F-P . V .}{8}}{\frac{F-P . V .}{2}}=\left[\frac{25+\frac{1000-950}{20}}{\frac{1000+950}{2}}\right]\)
= \(\frac{25+2.5}{975}\) = 0.0282
So, Approx. YTM = 0.0282 or 2.82% per half year or 5.64% per annum.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 33.
An amount of ₹ 5000 is put into three different investments at the rate of interest of 6%, 7% and 8% per annum respectively. The total annual income is ₹ 358. If the income from first two investments is ₹ 70 more than the income from the third investment, then find the amount of each investment, using Cramer’s rule. [5]
OR
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60, The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹ 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹ 70 . Find the cost of each item per kg.
Solution:
Let ₹ x, ₹ y and ₹ z be three investments at the rate of interest of 6%, 7% and 8% per annum respectively. According to the problem, we have the following system of linear equations:
x + y + z = 5000, \(\frac{6}{100}\) x + \(\frac{7}{100}\) y + \(\frac{8}{100}\) z
= 358 ; \(\frac{6}{100}\) x + \(\frac{7}{100}\) y = \(\frac{8}{100}\) z + 70
or x + y + z = 5000,
6x + 7y + 8z = 35800 ;
6x + 7y – 8z = 7000
So, D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
6 & 7 & 8 \\
6 & 7 & -8
\end{array}\right|\) Dx = \(\left|\begin{array}{ccc}
5000 & 1 & 1 \\
35800 & 7 & 8 \\
7000 & 7 & -8
\end{array}\right|\),
Dy = \(\left|\begin{array}{ccc}
1 & 5000 & 1 \\
6 & 35800 & 8 \\
6 & 7000 & -8
\end{array}\right|\) ; Dz = \(\left|\begin{array}{ccc}
1 & 1 & 5000 \\
6 & 7 & 35800 \\
6 & 7 & 7000
\end{array}\right|\)
Here, D = 1(-56 – 56) – 1(-48 – 48) + 1(42 – 42)
= -16 ≠ 0;
Dx = 5000(- 56 – 56) – 1 (- 286400 – 56000) + 1 (250600 – 49000)
= – 16000;
Dy = 1(- 286400 -56000) – 5000 (- 48 – 48) + 1 (42000 – 214800)
= – 35200;
Dz = 1(49000 – 250600) – 1 (42000 – 214800) + 5000 (42 – 42)
= – 28800
Using Cromer’s rule, we get
x = \(\frac{D_x}{D}=\frac{-16000}{-16}\) = 1000;
y = \(\frac{D_y}{D}=\frac{-35200}{-16}\) = 2200;
z = \(\frac{D_z}{D}=\frac{-28800}{-16}\) = 1800
Thus, ₹ 1000, ₹ 2200 and ₹ 1800 are the three investments made at the rate of interest of 6%, 7% and 8% per annum respectively.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

OR

Let the cost (per kg) of onion, wheat and rice be ₹ x, ₹ y and ₹ z respectively. Then, according to the problem, we have the following system of equations:
4x + 3y +2z = 60;
2x + 4y + 6z = 90,
6x + 2y + 3z = 70
So, D = \(\left|\begin{array}{lll}
4 & 3 & 2 \\
2 & 4 & 6 \\
6 & 2 & 3
\end{array}\right|\), Dx = \(\left|\begin{array}{lll}
60 & 3 & 2 \\
90 & 4 & 6 \\
70 & 2 & 3
\end{array}\right|\)
Dy = \(\left|\begin{array}{lll}
4 & 60 & 2 \\
2 & 90 & 6 \\
6 & 70 & 3
\end{array}\right|\), Dz = \(\left|\begin{array}{lll}
4 & 3 & 60 \\
2 & 4 & 90 \\
6 & 2 & 70
\end{array}\right|\)
Here, D = 4(12 – 12) -3 (6 – 36) + 2(4 – 24)
= 50 ≠ 0;
Dx = 60(12 – 12) -3 (270 – 420) + 2 (180 – 280) = 250;
Dy = 4(270 – 420) – 60 (6 – 36) + 2 (140 – 540) = 400;
Dz = 4(280 – 180) -3 (140 – 540) + 60 (4 – 24) = 400
Using Cramer’s rule, we get
x = \(\frac{D_x}{D}=\frac{250}{50}\) = 5 ;
y = \(\frac{D_y}{D}=\frac{400}{50}\) = 8 ;
z = \(\frac{D_z}{D}=\frac{400}{50}\) = 8
Hence, the cost (per kg) of onion, wheat and rice are ₹ 5, ₹ 8 and ₹ 8 respectively.

Question 34.
A bond of ₹ 1000 value carries a coupon rate of 10% and a maturity period of 6 years. Interest is payable semi-annually. If the required rate of return is 12%, calculate the value of the bond. [Given (1.06)-12 = 0.497] [5]
Solution:
Here,
F = Face value of the bond = ₹ 1000
n = number of periodic divident payments = 6
i = Annual yield rate = 0.06
R = F × id = ₹ (1000 × 0.05) = ₹ 50
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 = \(\mathrm{R}\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)-n
⇒ V = ₹ {50 \(\left[\frac{1-(1.06)^{-6}}{0.06}\right]\) + 1000(1.06)-5}
= ₹ {833 [1 – (1.06)-6)] + 1000(1.06)-6)}
= ₹ {833 [1 – 0.497] + 1000(0.497)}
= ₹ {419 + 497} ; or ₹ 916
Thus, the value of the bond is ₹ 916.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 35.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/ cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cuttirig machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/ cutting machine for at the most 12 hours. The profit from the sale of a lamp is ₹ 5 and that from a shade is 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? [5]
OR
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:

From/to Cost (in ₹)
A B C
P 160 100 150
Q 100 120 100

How many Units should be transported from each factory to each depot in order that the transportation cost in minimum. What will be the minimum transportation cost?
Solution:
Let x be the number of pedestal lamps and y be the number of wooden shades produced in a day and Z be the total profit of the manufacturer in a day.

Grinding/

Cutting

Sprayer Profit
Pedestal lamps 3 hours 3 hours ₹ 5
Wooden shades 1 hour 2 hours ₹ 3
Time available 12 hours 20 hours

Then, the mathematical formulation of the given LPP is
Maximise Z = 5x + 3y
subject to constraints
2x + y ≤ 12, 3x + 2y ≤ 20, x, y ≥ 0
Let us draw the graph for the system of inequalities representing constraints.
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 11
The feasible region is OABC shown (shaded) in Fig. 7, which is bounded.
The coordinates of the corner points of the feasible region OABC are O(0, 0), A(6, 0), B(4, 4) and C (0, 10).
Let us evaluate the objective function Z = 5x + 3 y at the corner points.

Corner point Corresponding value of Z
O(0, 0) 0
A(6, 0) 32
B(4, 4) 32 (maximum)
C (0, 10) 30

Z is maximum at B(4, ) and the maximum profit is ₹ 32
Thus, the industry should manufacture 4 pedestal lamps and 4 wooden shades to realise maximum profit of ₹ 32.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

OR

Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then, (8 – x – y) units will be transported to the depot at C.
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 12
Hence, we have x, y ≥ 0 and 8 – x – y ≥ 0
i. e., x + y ≤ 8 and x, y ≥ 0
Now, the weekly requirement of the depot at A is 5 units of the commodity. So, (5 – x) units need to be transported to the depot A from the factory at Q. Obviously, 5 – x > 0.
Similarly, (5 – y) units and {6 -(5 – x + 5 – y)}
= (x + y – 4 )units are to be transported from the factory at Q to the depots at B and C respectively. Thus,
5 – y ≥ 0, x + y – 4 ≥ 0
or y < 5, x + y > 4
Let Z be the total transportation cost. Then,
Z = 160x + 100y + 150(8 – x – y) + 100(5 – x) + 120 (5 – y) + 100 (x + y – 4)
= 10(x – 7y + 190)
Thus, the required LPP is
Minimise Z = 10(x -7y + 190)
subject to constraints
x + y ≤ 8, x + y ≥ 4, x ≤ 5, y ≤ 5, x, y ≥ 0
Let us draw the graph for the system of inequalities representing constraints.
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 13
The feasible region is ABCDEF shown (shaded) in Fig. 6, which is bounded.
The coordinates of the corner points of the feasible region ABCDEF are A(0, 4), B(0, 5),C (3, 5), D (5, 3), E (5, 0) and F (4, 0)
Let us evaluate the objective function Z =10(x -7y + 190) at the corner points.

Corner point Corresponding value of Z
A(0, 4) 1620
B (0, 5) 1550 (minimum)
C(3, 5) 1580
D (5, 3) 1740
E (5, 0) 1950
F (4, 0) 1940

Z is minimum at B(0, 5) and minimum cost is ₹ 1500

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

So, the minimum transportation cost is ₹ 1500 when 0,5 and 3 units are transported from factory at P and 5, 0 and 1 units are transported from factory at Q to the depots A, B and C respectively.

Section – E (12 marks)

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

Question 36.
The probability of any ship of a company being destroyed on a certain voyage is 0.02. The company owns 6 ships for the voyage.
Based on the above, answer the following questions:
(A) What is the probability of losing ‘r’ ships? [1]
(B) What is the probability of losing one ship? [1]
(C) What is the probability of losing two ships? [2]
OR
What is the probability of losing at most two ships?
Solution:
(A) 6Cr (0.02)r (0.98)6-r

(B) Probability of losing one ship
= 6C1 (0.02)1 (0.98)5
= 0.10847

(C) Probability of losing two ships
= 6C0 (0.02)0 (0.98)4 = 0.0006

OR

Probability of losing at most two ships
= 6C0 (0.02)0 (0.09)6 + 6C1 (0.02) (0.98)5
= + 6C02 (0.02)2 (0.98)4
= 0.999

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 37.
Two multi storey building (represented by AP and BQ) are on opp site side of a 20 m wide road at point A and B respectively. There is a point R. as shown in figure.
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 1
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 2
(A) What is area of trapezium ABQP? [1]
(B) What is the length of PQ? [1]
(C) Let there be a quantity s, such that s = RP2 + RQ2, then find the minimum value of s. [2]
OR
Find the interval in which f(x) = log (x + 3) is a decreasing function.
Solution:
(A) Area of trapezium
= \(\frac{1}{2}\) (sum of parallel sides) × distance between parallel sides
= \(\frac{1}{2}\) × (16 + 22) × 20 = 380 m2

(B) PQ2 = 62 + 202
= 36 + 400
= 436
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 14
PQ = \(\sqrt{436}\) = 20.8 cm

(C) We have s = RP2 + RQ2
Since, RP2 = 162 + x2
= 256 + x2
and RQ2 = 222 + (20 – x)2
= 484 + 400 + x2 – 40x
∴ s = 2x2 – 40x + 1140
\(\frac{\mathrm{ds}}{\mathrm{dx}}\) = 4x – 40
Foul minimum value,
put \(\frac{\mathrm{ds}}{\mathrm{dx}}\) = 0
4x – 40 = 0
⇒ x = 10
\(\frac{\mathrm{d}^{2} \mathrm{~x}}{\mathrm{dx}^{2}}\) = 4> 0
∴ Minimum value of s
= 2(4)2 – 40 (4) + 1140
= 32 – 160 + 1140
= 1172 – 160
= 1012

OR

We have f(x) = log (x + 3)
there, f(x) is defined for x > – 3
on differentiating w.r.t x,
we get f(x) = \(\frac{1}{x+3}\)
and x > -3
For f(x) to be decreasing function f(x) < 0 and x > – 3.
\(\frac{1}{x+3}\) < 0 and x > – 3
⇒ x < – 5 and x > – 3
Hence, for no value of x, f(x) is decreasing.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

Question 38.
In a Kilometre race, if A gives B, a start of 40 m, than B wins by 19 seconds but if A Kilometre race B, a start of 30 seconds then B wins by 40 meteres.
CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions 3
Let the time taken by A to run 1 km be x sec and time taken by B to run 1 km be y sec:
(A) In first case, find the time taken by B to complete the race? [1]
(B) If A gives a start of 30 seconds to B, then how much time does A run for. [1]
(C) What is the time taken by A to ran a kilometer? [2]
OR
A can run 22.5m while B runs 25m. By what distance B beat A in a kilometre race?
Solution:
(A) If a gives B, a start of 40m, it means that in the same time A runs 1000 metres while B runs (1000 – 40) m = 960 cm
∴B runs 1000 m in y seconds
∵ B runs 960 m in (\(\frac{y}{1000}\) × 960) sec
= \(\frac{24}{25}\) sec.

(B) A had given a start of 30 seconds to B, then A had run for (y – 30) seconds.

(C) From Q(A) B wins by 19 sec
∴\(\frac{24}{25}\) y – x = 19
25x – 24 y = – 475 ………… (i)
In second condition, B wins by 40 m
∴ 1000 – \(\frac{1000}{x}\) (y – 30) = 40
⇒ \(\frac{1000}{x}\) (y – 30) = 960
⇒ \(\frac{25}{x}\) (y – 30) = 24
24x – 25y = – 750 ………………… (ii)
On solving (i) & (ii), we get
x = 125 seconds
So, A complete, race in 125 sec.

CBSE Sample Papers for Class 12 Applied Mathematics Set 4 with Solutions

OR

B beats A by (25 – 22.5) m = 2.5 m in running 25 m Thus, in running 25 m, B beats A by 2.5 m .
∴ In running y km i.e, 1000 m B beats A by (\(\frac{2.5}{25}\) × 1000) m = 100 m.