Students must start practicing the questions from CBSE Sample Papers for Class 12 Applied Mathematics with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Applied Mathematics Set 6 with Solutions
Maximum Marks: 80 Marks
Time Allowed : 3 Hours
General Instructions:
- This question paper contains five sections A, B, C, D and E. Each section is compulsory.
- Section – A carries 20 marks weightage, Section – B carries 10 marks weightage, Section – C carries 18 marks weightage, Section – D carries 20 marks weightage and Section – E carries 3 case-based with total weightage of 12 marks.
- Section A: It comprises of 20 MCQs of 1 mark each.
- Section B: It comprises of 5 VSA type questions of 2 marks each.
- Section C: It comprises of 6 SA type of questions of 3 marks each.
- Section D: It comprises of 4 LA type of questions of 5 marks each.
- Section E: It has 3 case studies. Each case study comprises of 3 case-based questions, where 2 VSA type questions are of 1 mark each and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks question in each case-study.
- Internal choice is provided in 2 questions in Section – B, 2 questions in Section – C, 2 questions in Section – D. You have to attempt only one of the alternatives in all such questions.
Section – A (20 Marks)
(All questions are compulsory. No internal choice is provided in this section)
Question 1.
In a race, A gives B a start of 25 m making length of race for B a distance of 175 m. What is the total length of the race? [1]
(a) 130 m
(b) 150
(c) 180 m
(d) 200 m
Solution:
(d) 200 m
Question 2.
A die is tossed twice and getting a number greater than 4 is considered as success. Then, the mean of the probability distribution of the number of successes is [1]
(a) \(\frac{2}{3}\)
(b) \(\frac{8}{9}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{2}{9}\)
Solution:
(a) \(\frac{2}{3}\)
Explanation:
Let the random variable X denotes the number of successes(S) obtained on tossing a die.
Then, X takes values 0,1,2. Also,
P(S) = \(\frac{2}{6}\) i.e, \(\frac{1}{3}\)
and \(P(\bar{S})\) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
P(X = 0) = \(P(\bar{SS})\) = \(\frac{2}{3}\) × \(\frac{2}{3}\) = \(\frac{4}{9}\)
P(X = 2) = P(SS) = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)
P(X = 1) = \(P(S \bar{S})+P(\bar{S} S)\)
= \(\frac{1}{3}\) × \(\frac{2}{3}\) + \(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{4}{9}\)
Thus,
E(X) = (0 × \(\frac{4}{9}\)) + (1 × \(\frac{4}{9}\)) + (2 × \(\frac{1}{9}\))
= \(\frac{6}{9}\) or \(\frac{2}{3}\)
Question 3.
The surface area of a ball is increasing at the rate of 2 sq.cm/sec. The rate at which the radius is increasing when the surface area is 1671 sq.cm/sec, is: [1]
(a) 0.125 cm/sec
(b) 0.25 cm/sec
(c) 0.5 cm/sec
(d) 1 cm/sec
Solution:
(a) 0.125 cm/sec
Explanation:
A Here, S = 4πr2 …………… (1)
⇒ \(\frac{\mathrm{dS}}{\mathrm{dt}}\) = 8πr \(\frac{\mathrm{dr}}{\mathrm{dt}}\) ……………… (2)
For S = 16π, r is given by 4πr2 = 16π
⇒ r = 2
From (2) we have
\(\frac{d r}{d t}=\frac{1}{8} \pi r \frac{d S}{d t}\)
⇒ \(\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)_{r=2}=\left(\frac{1}{8 {\mathrm{Ar}}} \frac{\mathrm{dS}}{\mathrm{dt}}\right)_{r=2}\)
= \(\frac{1}{16 \pi}\)
Question 4.
Present value of perpetuity of ₹R payable at the beginning of every period forever at a rate of i per period is given by [1]
(a) R + \(\frac{\mathrm{R}}{\mathrm{i}}\)
(b) R(1 + i)
(c) R+ Ri
(d) Ri (1 + i)
Solution:
(a) R + \(\frac{\mathrm{R}}{\mathrm{i}}\)
Question 5.
A tap can fill a tank in 12 hours and a leakage can empty the tank in 20 hours. If tap and leakage both work together, then how long will it take to fill the tank? [1]
(a) 30 hours
(b) 15 hours
(c) 20 hours
(d) 28 hours
Solution:
(a) 30 hours
Question 6.
A The optimal value of the objective function is attained at the points: [1]
(a) given by the intersection of inequalities with x – axis
(b) given by comer points of the feasible region
(c) given by the intersection of inequalities with axes only
(d) none of these.
Solution:
(b) given by comer points of the feasible region
Question 7.
A die is tossed twice and getting a number greater than 4 is considered as success. Then, themean of the probability distribution of the number of successes is [1]
(a) \(\frac{2}{3}\)
(b) \(\frac{8}{9}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{2}{9}\)
Solution:
(a) \(\frac{2}{3}\)
Explanation:
Let the random variable X denotes the number of successes(S) obtained on tossing a die.
Then, X takes values 0, 1, 2. Also,
P(S) = \(\frac{2}{6}\) i.e, \(\frac{1}{3}\)
and \(P(\bar{S})\) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
P(X = 0) = \(P(\bar{SS})\) = \(\frac{2}{3}\) × \(\frac{2}{3}\) = \(\frac{4}{9}\)
P(X = 2) = P(SS) = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)
P(X = 1) = \(P(S \bar{S})+P(\bar{S} S)\)
= \(\frac{1}{3}\) × \(\frac{2}{3}\) + \(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{4}{9}\)
Thus,
E(X) = (0 × \(\frac{4}{9}\)) + (1 × \(\frac{4}{9}\)) + (2 × \(\frac{1}{9}\))
= \(\frac{6}{9}\) or \(\frac{2}{3}\)
Question 8.
Let C be the redemption value of a bond. Then, its present value is given by [1]
(a) C + (1 + i)-n
(b) C + (1 + i)n
(c) C (1 + i)-n
(d) C (1 + i)n
Solution:
(c) C (1 + i)-n
Question 9.
The area under the standard normal curve which lies to the right of z = -0.56 is: [1]
(a) 0.7123
(b) 0.2877
(c) – 0.2123
(d) 0.2123
Solution:
(a) 0.7123
Explanation:
Here,
P(Z > – 0.56 ) = 0.5 + P(0 < Z < 0.56)
= 0.5 + 0.2123 = 0 .7123
Question 10.
The sum of money invested now would establish a scholarship of ₹ 10,000 to be awarded at the starting of each year to a deserving student, if money is worth 5% compounded annually is: [1]
(a) 2,10,000
(b) 2,50,000
(c) 2,40,000
(d) 2,60,000
Solution:
(a) 2,10,000
Explanation:
We know that the present value of perpetuity of ₹ R payable in the beginning of every period forever at a rate of i per period is given by
R + \(\frac{\mathrm{R}}{\mathrm{i}}\), i.e., P∞ = R + \(\frac{\mathrm{R}}{\mathrm{i}}\).
Here, R = 10,000, i per period = \(\frac{5}{100}\) = 0.05
So, P∞ = R + \(\frac{\mathrm{R}}{\mathrm{i}}\) = 10,000 + \(\frac{10,000}{0.05}\) = ₹ 2,10,000
So, an amount of ₹ 2,10,000 must be invested.
Question 11.
\(\int \frac{10 x^9+10^x \log _e \log _e 10}{x^{10}+10^x}\) dx is equal to [1]
(a) 10x – x10 + C
(b) 10x + x10 + C
(c) (10x – x10)-1 + C
(d) log(10x + x10 + C)
Solution:
(d) log(10x + x10 + C)
Explanation:
Let I = \(\int \frac{10 x^9+10^x \log _e \log _e 10}{x^{10}+10^x}\) dx
Put x10 + 10x = t.
Then, (10x9 + 10x × loge 10) dx = dt
⇒ I = \(\int \frac{10 x^9+10^x \log _e \log _e 10}{x^{10}+10^x}\)
dx = \(\int \frac{1}{t}\)
dt = log |t| + C
or log (x10 + 10x + C)
Question 12.
A random variable X has the Poisson distribution with mean 2. Then, P(X > 1.5) is: [1]
(a) 2e-2
(b) 3e-2
(c) 1 – 2e-2
(d) 1 – 3e-2
Solution:
(d) 1 – 3e-2
ExpLanation: A Here, P(X = r) = \(\frac{2^r}{r !}\) . e-2
⇒ P(X > 1.5) = 1 – {P(X = 0) +P(X = 1)}
= 1 – {\(\frac{2^0}{0 !}\) . e-2 + \(\frac{2^1}{1 !}\) . e-2}
= 1 – {e-2 + 2.e-2} = 1 – 3e-2
Question 13.
If |x – 2| > 7, then [1]
(a) x ∈ (-5, 9)
(b) x ∈ (-5, 9)
(c) x ∈ (-∞, 5) ∪ (9, ∞)
(d) x ∈ (-∞, -5) ∪ (9, ∞)
Solution:
(c) x ∈ (-∞, 5) ∪ (9, ∞)
Explanation:
By definition, |x – 2| > 7 means x – 2 > 7 and x – 2 < – 7
This further gives, x > 9 and x < – 5, or x, or x ∈ (-∞, 5) ∪ (9, ∞)
Question 14.
One hundred identical coins each with probability p showing up heads are tossed once. If 0 < p < 1 and the probability of heads on 50 coins is equal to that of heads showing on 51 coins, then the value of p is: [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{49}{101}\)
(c) \(\frac{50}{101}\)
(d) \(\frac{51}{101}\)
Solution:
(d) \(\frac{51}{101}\)
Explanation:
Here, it is given that
100C50 (p)50(1 – p)50 = 100C51 (p)51(1 – p)49
⇒ 100C50 (p)50(1 – P) = 100C50 (p)51 (p)
⇒ \(\frac{1-p}{50}\) = \(\frac{p}{51}\)
⇒ p = \(\frac{51}{101}\)
Question 15.
If p and q are the order and degree of the differential equation y2 (\(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\))2 + 3x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + x2y2 = x2 then [1]
(a) p < q (b) p = q (c) 2p = q (d) p > q
Solution:
(b) p = q
Explanation:
The order of highest order derivative \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) is 2. So, p = 2.
The exponent of the highest order derivative \(\frac{\mathrm{d}^{2} \mathrm{~y}}{\mathrm{dx}^{2}}\) in the equation is 2. So, q = 2.
Question 16.
Let X be a normal variable with mean 11 and S.D. 1.5. If P( X > a) = 0.09, then the value of a is: [1]
(a) 0.73
(b) 0.83
(c) 13.01
(d) 0.30
Solution:
(c) 13.01
Explanation:
Z = \(\frac{X-11}{1.5}\)
Let for X = a, Z = z
It is given that
P(X > a) = 0.09
P(Z > z) = 0.09
P(Z > 0) – P(0 < Z < z)
= 0.09
P(0 < Z < z) = 0. 41
z = 1.34
Hence, a = 13.01
Question 17.
The total revenue if (₹) received from the sale of x units of a product is given by [1]
R(x) = 3x2 + 36x + 5
The marginal revenue, when x = 15 is:
(a) 116
(b) 96
(c) 90
(d) 126
Solution:
(d) 126
Explanation:
Here, R(x) = 3x2 + 36x + 5 gives
\(\frac{\mathrm{dR}}{\mathrm{dx}}\) = 6x + 36
We need to determine \(\frac{\mathrm{dR}}{\mathrm{dx}}\) at x = 15
(\(\frac{\mathrm{dR}}{\mathrm{dx}}\))x = 15 = (6x + 36)r=15 = 126
Question 18.
The comer points of the feasible region of a LPP are (0, 3), (3, 2) and (0, 5). Then, the minimum value of Z = 11 x + 7y is: [1]
(a) 14
(b) 21
(c) 33
(d) 35
Solution:
(b) 21
Explanation:
Here,
| Comer point | Corresponding value of Z |
| (0, 3) | 11(0)+ 7(3) = 21 |
| (3,2) | 11(3)+ 7(2) = 47 |
| (0, 5) | 11(0)+ 7(5) = 35 |
The minimum value of z is 21.
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 and (R) is true.
Question 19.
Assertion (A): if an LPP attains its maximum value at two corner points of the feasible region then it attains maximum value at infinitely many points. [1]
Reason (R): The value of the objective function of a LPP is same at two corners then it is came at every point on the line joining two corner points.
Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Question 20.
Assertion (A): It A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-1 & 4 \\
0 & 5
\end{array}\right]\) (A + B)2 = A2 + 2AB + B2. [1]
Reason (R): AB ≠ BA.
Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
∴ (A + B)2 = A2 + 2AB + B2
Section – B (10 Marks)
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 21.
Let a, b be distinct positive real numbers. Then, prove that \(\frac{a+b}{2}\) > \(\sqrt{a b}\). [2]
Solution:
Since a and b are distinct positive real numbers,
(\(\sqrt{a}\) – \(\sqrt{b}\))2 > 0
a + b – 2 \(\sqrt{a b}\) > 0
a + b > 2 \(\sqrt{a b}\)
or \(\frac{a+b}{2}\) > \(\sqrt{a b}\)
Question 22.
Find the present value of a perpetuity of ₹ 750 payable at the beginning of each year, if money is worth 5% p.a. [2]
Solution:
We know that the present value of perpetuity of ₹ R payable in the beginning of every period forever at a rate of i per period is given by
R + \(\frac{\mathrm{R}}{\mathrm{i}}\), i.e., P∞ = R + \(\frac{\mathrm{R}}{\mathrm{i}}\)
Here, R = 750, i per period = \(\frac{5}{100}\)= 0.05
So, P∞ = R + \(\frac{\mathrm{R}}{\mathrm{i}}\) = 750 + \(\frac{750}{0.05}\) = ₹ 15750
So, present value of the perpetuity is ₹ 15750.
Question 23.
Radhika has taken a personal loan of ₹ 2,00,000 for 2 years at an interest rate of 20% p.a. which is to be paid in equal monthly instalments. How much monthly instalments Radhika will pay? [2]
[Given \(\left(\frac{61}{60}\right)^{-24}\) = 0.6725335725]
Solution:
Here, P = 2,00,000, i per period = \(\frac{20}{1200}\) = \(\frac{1}{60}\)
and n = 2 × 12 = 24 months
EMI = P × r × \(\frac{1}{1-(1+r)^{-n}1}\)
= 2,00,000 × \(\frac{1}{60}\) × \(\frac{1}{1-\left(\frac{61}{60}\right)^{-24}}\)
= 2,00,000 × \(\frac{1}{60}\) × \(\frac{1}{1-0.6725335725}\)
= 2,00,000 × \(\frac{1}{60}\) × \(\frac{1}{0.327466428}\)
= 10179
So, Radhika’s monthly instalment is of amount ₹ 10179.
Question 24.
Construct a matrix A= [aij]2 × 3 whose elements aij are given by [2]
(A) aij = i – 3 j
(B) aij = |2i – 3j|
OR
Find the values of x, y, z and w for which the following two matrices are equal:
(A) \(\left[\begin{array}{ccc}
2 x-3 y & z-w & 3 \\
1 & x+4 y & 3 z+4 w
\end{array}\right]\) = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
1 & 6 & 29
\end{array}\right]\)
(B) \(\left[\begin{array}{cc}
3 x+4 y & 2 \\
z+w & 2 z-w \\
x-2 y & -1
\end{array}\right]\) = \(\left[\begin{array}{cc}
2 & 2 \\
5 & -5 \\
4 & -1
\end{array}\right]\)
Solution:
Here, matrix A is \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\)
(A) It is given that aij = i – 3j
So, a11 = -3 = -2; a12 = 1 – 6 = -5;
a13 = 1 – 9 = -8: a21 = 2 – 3 = -1;
a22 = 2 – 6 = -4; = -4;
a23 = 2 – 9 = – 7;
Thus, A = \(\left[\begin{array}{lll}
-2 & -5 & -8 \\
-1 & -4 & -7
\end{array}\right]\)
(B) It is given that aij = |2i – 3j|
So, a11 = |2 – 3| = 1; a12 = |2 – 6| = 4;
a13 = |2 – 9| = 7;
a21 = |4 – 3| = 1:
a22 = |4 – 6| = 2;
a23 = |4 – 9| = 5
Thus, A = \(\left[\begin{array}{lll}
1 & 4 & 7 \\
1 & 2 & 5
\end{array}\right]\)
OR
(A) As two matrices are equal,,we have their corresponding elements equal. So, 2x -3y = 1; z – w = -2; x + 4y = 6 and 3z + 4w = 29 Solving these four equations, we get x = 2, y = 1, z = 3 and w = 5.
(B) As two matrices are equal, we have their corresponding elements equal. So, 3x + 4y = 2;z+w=5;x-2y = 4 and 2z – w = -5 Solving these four equations, we get x = 2, y = – 1, z = 0 and w = 5.
Question 25.
Construct index number for price for the year 2019 with 2016 as the base year from the following data by taking quantities in the base year as weights. [2]
| Commodity | 2016 | 2019 | ||
| Price | Quantity | Price | Quantity | |
| A | 2 | 8 | 4 | 6 |
| B | 5 | 10 | 6 | 5 |
| C | 4 | 14 | 5 | 10 |
| D | 2 | 19 | 2 | 3 |
OR
The daily attendance for a week, is given below for a class of 75 students
53, 48, 64, 68, 54, 70, 72
Calculate the 3-day moving averages and plot these on a graph paper.
Solution:
We have
Required index number
= \(\frac{\sum p_c q_b}{\sum p_b q_b}\) × 100 = \(\frac{200}{160}\) × 100 = 125
OR
To calculate 3 – day moving average, we construct the following table:
| Day (t) | Attendance (t) | 3-day total | 3-day moving average |
| l | 53 | ||
| 2 | 48 | 165 | 55 |
| 3 | 64 | 180 | 60 |
| 4 | 68 | 186 | 62 |
| 5 | 54 | 192 | 64 |
| 6 | 70 | 196 | 65.33 |
| 7 | 72 |
Thus, 3- day moving averages are 55, 60, 62, 64 and 65.33
Section – C (18 Marks)
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 26.
Using Cramer’s rule, solve the system of equations: [3]
5x – 7y + z = 11
6x – 8y – z = 15
3x + 2y – 6z = 7
OR
The sum of three number is 6. If we multiply the third number by 3 and add second number to it , we get 11. By adding first and third numbers, we get twice the second number. Find the numbers.
Solution:
= 5(-55 – 30) + 7(42 – 45) + 11(12 + 24)
= -55
Since D = 55 ≠ 0, the system of equations has a unique solution given by
x = \(\frac{\mathrm{D}_x}{\mathrm{D}}\) = \(\frac{55}{55}\) = 1
y = \(\frac{\mathrm{D}_y}{\mathrm{D}}\) = \(\frac{-55}{55}\) = 1
z = \(\frac{\mathrm{D}_z}{\mathrm{D}}\) = \(\frac{-55}{55}\) = 1
Hence, x = 1, y = -1, z = -1 is the solution of the given system of equations.
OR
Let x, y and z be three numbers. According to the problem, we have the following system of linear equations:
x + y + z = 6, y + 3z = 11, x + z = 2y or x +y + z = 6, y + 3z = 11, x-2y + z = 0
So, D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right|\), Dx = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
11 & 1 & 3 \\
0 & -2 & 1
\end{array}\right|\)
Dy = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
0 & 11 & 3 \\
1 & 0 & 1
\end{array}\right|\), Dz = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
0 & 1 & 11 \\
1 & -2 & 0
\end{array}\right|\)
Here, D = 1(1 + 6) – 1(0 – 3) + 1(0 – 1)
= 9 ≠ 0;
Dx = 6(1 + 6) – 1(11 – 0) + 1(- 22 – 0)
= 9;
Dy = 1(11 – 0) – 6(0 – 3) + 1(0 – 11)
= 18;
Dz = 1(0 + 22) – 1(0 – 11) + 6(0 – 1)
= 27
Using Cramer’s rule, we get
x = \(\frac{\mathrm{D}_x}{\mathrm{D}}\) = \(\frac{9}{9}\) = 1;
y = \(\frac{\mathrm{D}_y}{\mathrm{D}}\) = \(\frac{18}{9}\) = 2;
z = \(\frac{\mathrm{D}_z}{\mathrm{D}}\) = \(\frac{27}{9}\) = 3
Hence, the three numbers are 1, 2 and 3.
Question 27.
Find the interval in which f(x) = x/log x is strictly increasing or strictly decreasing. [3]
OR
The total cost of manufacturing x pocket radios per day is ₹ (x2 + 35x + 25) and rate at which may be sold to a distributor is ₹ 1/2( 100 – x) each. What should be the daily output to attach maximum total prof it?
Solution:
For f(x) = \(\frac{x}{\log x}\)
{Here, f(x) is defined for all x > 0, x ≠ 1}
f'(x) = \(\frac{\log x-1}{(\log x)^2}\)
Now, f'(x) = 0 gives \(\frac{\log x-1}{(\log x)^2}\) = 0, i.e., x = e
| Sub interval | Test Point | Test Value | Sign of f'(x) | Conclusion | |
| (A) | (0, e) | 2 | f'(2) = –\(\frac{31}{48}\) | (-) | f’(x) < 0 |
| (B) | (e, ∞) | 3 | f'(3) = \(\frac{10}{48}\) | (+) | f'(x) > 0 |
Hence, f(x) is strictly increasing on (e, ∞) and strictly decreasing on (0, e) – {1}.
OR
Here, the profit function P(x) is given by
P(x) = SP. – CP. = \(\frac{1}{2}\) x(100 – x) – (\(\frac{x^2}{4}\) + 35x + 25)
= \(\frac{3x^2}{4}\) + 15x – 25
P (x) = –\(\frac{3x}{2}\) + 15 and P”(x) = –\(\frac{3x}{2}\) < 0, for all x.
Equating P'(x) to zero, we have –\(\frac{3x}{2}\) + 15 = 0,
i.e., x = 10
P(x) is maximum hence x = 10
Hence, profit is maximum when daily output is 10 units.
Question 28.
Find the particular solution of the differential equation (x2 – 1) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2xy = \(\frac{2}{x^2-1}\), given that y = 0 when x = 0. [3]
Solution:
Given (x2 – 1) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2x y = \(\frac{2}{x^2-1}\) we have
\(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2 x}{x^2-1} y=\frac{2}{\left(x^2-1\right)^2}\)
It is a linear differential equation of the form \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + Py = Q, where P = \(\frac{2x}{x^2-1}\) and
Q = \(\frac{2}{\left(x^2-1\right)^2}\)
We are given that y = 0 when x = 0. So, by (1), C = 0
Hence, the required particular solution is y . (x2 – 1) = log \(\left|\frac{x-1}{x+1}\right|\)
Question 29.
A boatman rows to a place 45 km distant and back in 20 hours. He finds that he can row 12 km with the stream in the same time as 4 km against the stream. Find the speed of the boatman in still water and the speed of the stream. [3]
Solution:
Let the speed of the boat in still water be x km/h; and the speed of the stream be y km/h. Then,
45/(x + y) + 45/(x – y)
= 20 and 12/(x + y)
= 4/(x – y)
Solving these equations, we get
x = 6 and y = 3
So, the speed of the boat in still water be 6 km/h; and the speed of the stream be 3 km/h.
Question 30.
A sum of ₹ 130 is deposited into a provident fund account at the end of every month. Calculate the total amount of this fund at the end of 15 years, if it earns interest @ 4% p.a. [Given (1.0033)180 = 1.81] [3]
Solution:
Here, A = 130, r = \(\frac{4}{1200}\) = 0.0033
and n = 15 × 12 = 180
So, S = \(\frac{\mathrm{A}}{\mathrm{r}}\)[(1 + r)n -1] gives
S = \(\frac{130}{0.0033}\) [(1.00 3 3)180 -1]
⇒ S = 39394 [1.81 – 1] = 31909
Thus, the total amount of the fund is ₹ 31909.
Question 31.
A cistern has three pipes A, B and C . Pipes A and B are inlet pipes whereas C is an outlet pipe. Pipes A and B can fill the cistern separately in 3 hours and 4 hours respectively while pipe C can empty the completely filled cistern in 1 hour, If the pipes A,B and C are opened in order at 5, 6 and 7 a.m. respectively, at what time will the cistern be empty? [3]
Solution:
Let the cistern be emptied ‘ri hours after 5 a.m. Clearly, pipes A and B fill the tank for ‘n’ hours and ‘n – 1′ hours respectively, while pipe C empties the tank for ‘n – 2’ hours.
\(\frac{n}{3}\) + \(\frac{n-1}{4}\) – \(\frac{n-2}{1}\) = 0
n = \(\frac{21}{5}\), or 4\(\frac{1}{5}\)
Thus, the cistern will be emptied after 4 hours 12 minutes, i.e., at 9 : 12 a.m.
Section – D (20 Marks)
(All questions are compulsory. In case of internal choice, attempt any one question only)
Question 32.
Test scores in mathematics, in two samples of students of the same class from two schools were as given below: [5]
| School A | School B |
| 65 | 57 |
| 45 | 39 |
| 63 | 44 |
| 47 | 76 |
| 84 | 49 |
| 37 | 83 |
| 48 | 91 |
| 52 | 68 |
| 95 | 69 |
| 26 | |
| 68 | |
| 30 |
Can it be concluded that, on the average, the performance of students in the two schools is at par. Also, state the assumption made about the test of hypothesis. (Given t19(0.05) = 2.09)
Solution:
The null hypothesis and alternative hypothesis can be written as :
H0 : There is no significant difference in the performance of students of the two schools,
i.e., µ1 = µ2
There is a significant difference in the performance of students of the two schools,
i.e., µ1 ≠ µ2
Here, n1 = 12 and n2 = 9
\(\overline{X}\) = \(\frac{65+45+63+47+84+37+48+52+95+26+68+30}{12}\)
= \(\frac{660}{12}\) = 55;
\(\overline{Y}\) = \(\frac{57+39+44+76+49+83+91+68+69}{9}\)
= \(\frac{576}{9}\) = 64;
For the calculation of S2, we have the following table:
Thus,
The table value of t at a = 0.05 and (12 + 9 – 2) = 19 d.f. is 2.09
Conclusion: Since |t| < ta, the null hypothesis is accepted, i.e., the performance of the students of two schools is at par at 5% level of significance.
Question 33.
Using integration, find the area of the region bounded by the line y = 3x + 2 and the Ordinates x = -1 and x = 1 [5]
OR
Find: \(\int \frac{x^2}{(x-1)^3(x+1)}\) dx
Solution:
⇒ x2 A(x – 1)2(x + 1) + B(x – 1)(x + 1) + C (x + 1) + D (x – 1)3 ………………….. (2)
Putting (x – 1) = 0, i.e x = 1 in (2). we have
1 = 2C ⇒ C = \(\frac{1}{2}\)
Putting (x + 1) = 0, i.e x = -1 in (2). we have
1 = -8D ⇒ D = –\(\frac{1}{8}\)
Putting x = 0 in (2), we have 0 = A – B + C – D
⇒ A – B = –\(\frac{5}{8}\)
Putting x = 2 in (2). we have 4 = 3A + 3 B + 3C +D
⇒ A + B = \(\frac{7}{8}\)
Now, A – B = –\(\frac{5}{8}\) and A + B = \(\frac{7}{8}\) give
A = \(\frac{1}{8}\), B = \(\frac{3}{4}\)
Putting these values of A, B, C and D in (1).
we have
\(\frac{x^2}{(x-1)^3(x+1)}=\frac{1 / 8}{x-1}+\frac{3 / 4}{(x-1)^2}\) \(+\frac{1 / 2}{(x-1)^3}+\frac{-1 / 8}{x+1}\)
Question 34.
A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of ₹ 8000 on each piece of model A and ₹ 12000 on each piece of Model B. How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week? [5]
OR
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum amount of vitamin A?
Solution:
Let x be the number of pieces of Model A and y be the number of pieces of Model B.
Then, the mathematical formulation of the given LPP is
Maximise Z = 8000x + 12000y
subject to constraints
3x + 4y ≤ 60, x + 3y ≤ 30, x, y ≥ 0
Let us draw the graph for the system of inequalities representing constraints.
The feasible region is OABC shown (shaded) in Fig. 3, which is bounded.
The coordinates of the corner points of the feasible region OABC are 0(0, 0), A(20, 0), B(12, 6) and C (0, 10).
Let us evaluate the objective function Z = 8000 x + 12000 y at the corner points.
| Corner point | Corresponding value of Z |
| 0 (0, 0) | 0 |
| A (20, 0) | 160000 |
| B (12, 6) | 168000 (maximum) |
| C (0, 10) | 120000 |
Z is maximum at B(12, 6) and the maximum profit is ₹ 168000.
Thus, the company should produce 12 pieces of Model A and 6 pieces of Model B to realise maximum profit of ₹ 1,68,000
OR
Let x packages of Food P, y packages of Food Q be mixed and Z be the cost of the mixture.
| Food P | Food Q | Requirement | |
| Calcium (units) | 12 | 3 | At least 240 |
| Iron (units) | 4 | 20 | At least 460 |
| Cholesterol (units) | 6 | 4 | At most 300 |
| Vitamin (units) | 6 | 3 | ? |
Then, the mathematical formulation of the given LPP is
Minimise Z = 6x + 3y (Vitamin A)
subject to constraints
12x + 3y ≥ 240 or 4x + y ≥ 60
4x + 20y ≥ 460 or x + 5y ≥ 115
6x + 4y ≤ 300 or 3x + 2y ≤ 150
x, y ≥ 0
Let us draw the graph for the system of inequalities representing constraints.
The feasible region is LMN shown (shaded) in Fig. 4, which is bounded.
The coordinates of the corner points of the feasible region LMN are L(2, 72), M(15, 20) and N (40, 15).
Let us evaluate the objective function Z = 6x + 3 y at the corner points.
| Corner point | Corresponding value of Z |
| L(2, 72) | 228 |
| M(15, 20) | 150 (minimum) |
| N (40, 15) | 285 |
Z is minimum at M(15, 20) and minimum quantity of vitamin A in the mixture is 150 units
Thus, the optimal mixing strategy for the dietician would be to mix 15 units of vitamin in Food P and 20 units of vitamin A in Food Q.
Question 35.
Kanika owns ₹ 1000 face value bond, bearing a coupon rate of ₹ 75 and matures after 5 years. The bond is currently priced at ₹ 970. Given an appropriate yield rate of 10%, should Kanika hold or sell the bond? [Given (1.1)-5 = 0.6209) [5]
Solution:
Here,
F = Face value of the bond = ₹ 1000
n = number of periodic dividend payments = 5
i = Annual yield rate = 0.1
R = ₹ 75
Since the bond is to be redeemed at par,
C = Redemption price or Maturity value = Face value = ₹ 1000
Let V be the purchase value of the bond. Then,
V = R\(\left[\frac{1-(1+i)^{-n}}{i}\right]\) + F(1 + i)-n
V = ₹ {75\(\left[\frac{1-(1.1)^{-5}}{0.1}\right]\) + 1000(1.1)-5}
= ₹ {750[1 – (1.1)-5)] + 1000(1.1)-5)]
= ₹ {750[1 – 0.6209] + 1000(0.6209)}
= ₹ {284.30 + 620.90}, or ₹ 905.20
Since the market price is higher than the actual value of the bond, it is better for Kanika to sell the bond.
Section – E (12 marks)
(All questions are compulsory. In,case of internal choice, attempt any one question only)
Question 36.
For a two-sector economy with production sectors I and II, the intersectoral demand and final demand are as follows:
| Receiving sector → ↓ Producing sector | I | II | Final Denand |
| I | 264 | 410 | 206 |
| II | 528 | 204 | 292 |
(A) What is the total output of sectors I and II? [1]
(B) What is the technology matrix A? [2]
OR
What is leontief matrix (I – A)? Then find (I – A)-1
(C) What is the Hawkins – Simon condition? [1]
Solution:
(A) Here, total output of sector I = 264 + 410 + 206 = 880
total output of secyor II
= 528 + 204 + 292 = 1024
∴ The total output is 1904.
(B) Here, the technology coefficients are:
a11 = \(\frac{x_{11}}{x_1}\) = \(\frac{264}{880}\) = 0.3,
a21 = \(\frac{x_{21}}{x_1}\) = \(\frac{528}{880}\) = 0.6
a12 = \(\frac{x_{12}}{x_21}\) = \(\frac{410}{1024}\) = 0.4,
a22 = \(\frac{x_{22}}{x_2}\) = \(\frac{204}{1024}\) = 0.2
∴ The technology matrix A is : \(\left[\begin{array}{ll}
0.3 & 0.4 \\
0.6 & 0.2
\end{array}\right]\)
OR
Leontief Matrix = I – A
(C) Here, |I – A| = 0.7 × 0.8 – 0.6 × 0.4
= 0.56 – 0.24 = 0.32 > 0
Also, the elements of the diagonal of (I – A) are positive, is the required condition.
Question 37.
Let X denote the number of hours Mr.Nutan study during a randomly selected school day. The probability that X can take the values x has the following form, where k is some unknown constant.
Based on the above, answer the following questions:
(A) What is the value of k? [1]
(B) Find P( X = 2) [1]
(C) Find P( X ≥ 2) and P(n ≤ 2) [2]
OR
Evaluate E(x)
Solution:
The given probability distribution of X is:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | k | 2k | 2k | k |
(A) We know that Σ P(X) = 1. So, 6k + 0.1 = 1
k = 0.15
(B) P(X = 2) = 2k= 2(0.15) = 0.3
(C) P( X ≥ 2) = P(x = 2) + P(X = 3) + P(X = 4)
= 2k + 2k + k = 5k = 5(0.15) = 0.75
P (x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2)
= 0.1 + k + 2k
= 3k + 0.1
= 3(0.15) + 0.1
= 0.55
OR
E(X) = (0 × 0.1) + (1 × k) +(2 × 2k) +(3 × 2k) + (4 × k)
= 15k
= 15(0.15)
= 2.25
Question 38.
In the figure given below, O is the origin, A is the point (2, 1), B is the point (2, 7) and C is the point (6, 3). The shaded region R. is defined by three inequalities. One of these three inequalities is x + y < 9.
(A) Find the other two inequalities. [2]
OR
Area of the region R.
(B) Given that the point (x, y) is in the region R, find the maximum value of Z = 3x + y. [1]
(C) Given that the point (x, y) is in the region R, find the minimum value of P = 5x – 2y. [1]
Solution:
(A) In the figure, the region is bounded by the line x + y ≤ 9 and the lines
y – 1 = \(\frac{7-1}{2-2}\) (x – 2), i.e., x = 2; and
y – 1 = \(\frac{3-1}{6-2}\) (x – 2) i.e., 2y = x.
Now, the feasible region of the line x = 2 does not contain origin, so its inequality is x ≥ 2.
For the line 2y = x, take any point from the feasible region, say (3, 3).
Now, at (3, 3), 2y – x
= 2(3) – 3
= 6 – 3
= 3 ≥ 0.
So, inequality of the line 2y = x is 2y – x ≥ 0, or 2y ≥ x.
OR
Area of region R = Area of ∆ABC
= \(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\)
= \(\frac{1}{2}\left|\begin{array}{lll}
2 & 1 & 1 \\
6 & 3 & 1 \\
2 & 7 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) |2(3 – 7) – 1(6 – 2) + 1 (42 – 6)|
= 12 sq. units
(B) The corner points of the feasible region R are: A(2, 1), B(2, 7) and C(6, 3).
So, ZA = 7, ZB = 13, ZC = 21
∴ Zmax = 21
(C) The corner points of the feasible region R are: A(2, 1), B(2, 7) and C(6, 3).
So, PA = 8, PB = – 4, PC = 24
∴ Pmin = -4