CBSE Class 12 Applied Mathematics Question Paper 2025 Solved Code 465

SECTION A

1. \( -41 \bmod 9 \) is [1 Mark]
(A) 5
(B) 4
(C) 3
(D) 0

Answer: (B) 4

Teacher's Note:
a) Write \( -41 = 9 \times (-5) + 4 \), where the remainder must satisfy \( 0 \le r \lt 9 \).
b) A common mistake is to write \( -41 = 9 \times (-4) - 5 \) and answer \( -5 \); the remainder in mod can never be negative.

2. If \( a \gt b \) and \( c \lt 0 \), then which of the following is true ? [1 Mark]
(A) \( a + c \lt b + c \)
(B) \( a - c \lt b - c \)
(C) \( ac \gt bc \)
(D) \( a - c \gt b + c \)

Answer: (D) \( a - c \gt b + c \)

Teacher's Note:
a) Adding or subtracting the same number keeps the sign, so \( a - c \gt b - c \).
b) Since \( c \lt 0 \), \( -c \gt c \), so \( b - c \gt b + c \); hence \( a - c \gt b + c \).
c) Multiplying by a negative number reverses the sign, so (C) is wrong.

3. If A and B are symmetric matrices of the same order, then \( (AB' - BA') \) is a [1 Mark]
(A) symmetric matrix
(B) null matrix
(C) diagonal matrix
(D) skew symmetric matrix

Answer: (D) skew symmetric matrix

Teacher's Note:
a) As \( A' = A \) and \( B' = B \), the matrix is \( P = AB - BA \).
b) \( P' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -P \), so P is skew symmetric.

4. The inverse of matrix \( A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix} \) is [1 Mark]
(A) \( \frac{1}{6} \begin{bmatrix} -4 & 2 \\ -1 & -1 \end{bmatrix} \)
(B) \( \begin{bmatrix} \frac{1}{3} & \frac{1}{6} \\ \frac{2}{3} & -\frac{1}{6} \end{bmatrix} \)
(C) \( \begin{bmatrix} \frac{1}{6} & \frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} \end{bmatrix} \)
(D) \( \begin{bmatrix} -\frac{2}{3} & \frac{1}{6} \\ -\frac{1}{3} & -\frac{1}{6} \end{bmatrix} \)

Answer: (C) \( \begin{bmatrix} \frac{1}{6} & \frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} \end{bmatrix} \)

Teacher's Note:
a) \( |A| = 4(1) - (-1)(2) = 6 \) and \( \text{adj } A = \begin{bmatrix} 1 & 1 \\ -2 & 4 \end{bmatrix} \).
b) For a 2 by 2 matrix, swap the diagonal elements and change the signs of the other two, then divide by \( |A| \).

5. If \( \begin{vmatrix} 2x & 5 \\ 4 & x \end{vmatrix} = \begin{vmatrix} 3 & 5 \\ 4 & 6 \end{vmatrix} \), then the value of \( x \) is [1 Mark]
(A) \( \frac{3}{2} \)
(B) 6
(C) 3
(D) \( \pm 3 \)

Answer: (D) \( \pm 3 \)

Teacher's Note:
a) \( 2x^2 - 20 = 18 - 20 \Rightarrow 2x^2 = 18 \Rightarrow x^2 = 9 \).
b) Do not forget the negative root: \( x = \pm 3 \).

6. The slope of the normal to the curve \( y = \frac{x-3}{x-4} \) at \( x = 6 \) is [1 Mark]
(A) 4
(B) \( -\frac{1}{4} \)
(C) \( -4 \)
(D) \( \frac{1}{4} \)

Answer: (A) 4

Teacher's Note:
a) \( \frac{dy}{dx} = \frac{(x-4) - (x-3)}{(x-4)^2} = \frac{-1}{(x-4)^2} \), which is \( -\frac{1}{4} \) at \( x = 6 \).
b) Slope of normal \( = -\frac{1}{\text{slope of tangent}} = 4 \); option (B) is the tangent slope, a common trap.

7. The rate of change of population P(t) with respect to time (t), where \( \alpha, \beta \) are the constant birth and death rates, respectively, is [1 Mark]
(A) \( \frac{dP}{dt} = (\alpha + \beta)P \)
(B) \( \frac{dP}{dt} = (\alpha - \beta)P \)
(C) \( \frac{dP}{dt} = \frac{\alpha + \beta}{P} \)
(D) \( \frac{dP}{dt} = \frac{\alpha - \beta}{P} \)

Answer: (B) \( \frac{dP}{dt} = (\alpha - \beta)P \)

Teacher's Note:
a) Births add \( \alpha P \) and deaths remove \( \beta P \) per unit time, so the net growth rate is \( (\alpha - \beta)P \).
b) This is the exponential growth model; its solution is \( P = P_0 e^{(\alpha - \beta)t} \).

8. A pair of dice is thrown two times. If X represents the number of doublets obtained, then the expectation of X is [1 Mark]
(A) \( \frac{1}{6} \)
(B) 1
(C) \( \frac{1}{3} \)
(D) \( \frac{11}{36} \)

Answer: (C) \( \frac{1}{3} \)

Teacher's Note:
a) Probability of a doublet in one throw \( = \frac{6}{36} = \frac{1}{6} \).
b) X is binomial with \( n = 2, p = \frac{1}{6} \), so \( E(X) = np = 2 \times \frac{1}{6} = \frac{1}{3} \).

9. The mean of t-distribution is [1 Mark]
(A) 0
(B) 1
(C) 2
(D) not defined

Answer: (A) 0

Teacher's Note:
a) The t-distribution is symmetric about 0, like the standard normal curve.
b) So its mean (and median and mode) is 0.

10. The variations which occur due to change in climate, festivals or weather conditions are known as [1 Mark]
(A) secular variations
(B) cyclic variations
(C) seasonal variations
(D) irregular variations

Answer: (C) seasonal variations

Teacher's Note:
a) Seasonal variations repeat regularly within a year because of weather, festivals and customs.
b) Secular means long-term trend; cyclic means business cycles of more than a year.

11. In a LPP, the maximum value of \( z = 3x + 4y \) subject to the constraints \( x + y \le 40, x + 2y \le 60, x, y \ge 0 \) is [1 Mark]
(A) 120
(B) 140
(C) 150
(D) 130

Answer: (B) 140

Teacher's Note:
a) Corner points are \( (0, 0), (40, 0), (20, 20) \) and \( (0, 30) \).
b) z values are 0, 120, 140 and 120, so the maximum is 140 at \( (20, 20) \).

12. The present value of a sequence of payments of Rs. 100 made at the end of every year and continuing forever, if the money is worth 5% compounded annually, is [1 Mark]
(A) Rs. 2,000
(B) Rs. 20,000
(C) Rs. 5,000
(D) Rs. 12,000

Answer: (A) Rs. 2,000

Teacher's Note:
a) This is a perpetuity: present value \( = \frac{R}{i} \).
b) \( \frac{100}{0.05} = 2000 \), so the present value is Rs. 2,000.

13. The demand function of a monopolist is given by \( p = 30 + 5x - 3x^2 \), where \( x \) is the number of units demanded and p is the price per unit. The marginal revenue when 2 units are sold, is [1 Mark]
(A) Rs. 28
(B) Rs. 23
(C) Rs. 1
(D) Rs. 14

Answer: (D) Rs. 14

Teacher's Note:
a) Revenue \( R = px = 30x + 5x^2 - 3x^3 \), so \( MR = \frac{dR}{dx} = 30 + 10x - 9x^2 \).
b) At \( x = 2 \): \( MR = 30 + 20 - 36 = 14 \). Do not differentiate p itself.

14. If the cost function and revenue function of \( x \) items are respectively given as \( C(x) = 100 + 0.015x^2, R(x) = 3x \), then the value of \( x \) for maximum profit is [1 Mark]
(A) 50
(B) 100
(C) 150
(D) 200

Answer: (B) 100

Teacher's Note:
a) Profit \( P(x) = 3x - 100 - 0.015x^2 \), so \( P'(x) = 3 - 0.03x = 0 \Rightarrow x = 100 \).
b) \( P''(x) = -0.03 \lt 0 \), which confirms a maximum.

15. If a random variable X has the probability distribution
\( P(X = x) = \begin{cases} k, & \text{if } x = 0 \\ 2k, & \text{if } x = 1 \text{ or } 2 \\ 0, & \text{otherwise,} \end{cases} \)
then the value of k is [1 Mark]
(A) \( \frac{1}{3} \)
(B) \( \frac{1}{5} \)
(C) \( \frac{1}{6} \)
(D) \( \frac{1}{4} \)

Answer: (B) \( \frac{1}{5} \)

Teacher's Note:
a) The sum of all probabilities is 1: \( k + 2k + 2k = 1 \).
b) Remember that \( 2k \) is taken twice, once for \( x = 1 \) and once for \( x = 2 \), so \( 5k = 1 \).

16. The test statistic t for testing the significance of differences between the means of two independent samples is given by [1 Mark]
(A) \( t = \frac{\bar{x} - \bar{y}}{\sqrt{s}} \)
(B) \( t = \frac{\bar{x} - \bar{y}}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \)
(C) \( t = \frac{\bar{x} - \bar{y}}{\frac{s}{\sqrt{n-1}}} \)
(D) \( t = \frac{\bar{x} + \bar{y}}{s\sqrt{\frac{1}{n_1} - \frac{1}{n_2}}} \)

Answer: (B) \( t = \frac{\bar{x} - \bar{y}}{s\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \)

Teacher's Note:
a) Here s is the pooled standard deviation of the two samples and the degrees of freedom are \( n_1 + n_2 - 2 \).
b) The numerator is a difference of means and the root has a plus sign; this rules out (D).

17. The effective rate of interest equivalent to a nominal rate of 4% compounded semi-annually, is [1 Mark]
(A) 4.12%
(B) 4.04%
(C) 4.08%
(D) 4.14%

Answer: (B) 4.04%

Teacher's Note:
a) \( r_{\text{eff}} = \left(1 + \frac{i}{m}\right)^m - 1 = (1.02)^2 - 1 = 0.0404 \).
b) The effective rate is always a little more than the nominal rate when compounding is more than once a year.

18. The CAGR of an investment, whose starting value is Rs. 5,000 and it grows to Rs. 25,000 in 4 years, is : [Given \( (5)^{0.25} = 1.4953 \)] [1 Mark]
(A) 49.53%
(B) 14.95%
(C) 495.3%
(D) 1.49%

Answer: (A) 49.53%

Teacher's Note:
a) \( \text{CAGR} = \left(\frac{\text{Final value}}{\text{Initial value}}\right)^{\frac{1}{n}} - 1 = (5)^{0.25} - 1 = 0.4953 \).
b) Convert to percent by multiplying by 100: 49.53%.

19. Assertion (A) : The area of the region bounded by the line \( y - 1 = x \), the \( x \)-axis and the ordinates \( x = -1 \) and \( x = 1 \) is 2 square units.
Reason (R) : The area of the region bounded by the curve \( y = f(x) \), the \( x \)-axis and the ordinates \( x = a \) and \( x = b \) is given by \( \int_{a}^{b} f(x)\, dx \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

Teacher's Note:
a) Area \( = \int_{-1}^{1} (x + 1)\, dx = \left[\frac{x^2}{2} + x\right]_{-1}^{1} = \frac{3}{2} - \left(-\frac{1}{2}\right) = 2 \) sq units.
b) The line \( y = x + 1 \) stays above the \( x \)-axis on \( [-1, 1] \), so the integral in (R) directly gives the area.

20. Assertion (A) : The differential equation representing the family of curves \( y = mx \), m being an arbitrary constant, is \( x\frac{dy}{dx} - y = 0 \).
Reason (R) : For a family of curves, the differential equation is obtained by differentiating the equation of family of curves with respect to \( x \) and then eliminating the arbitrary constant, if any. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).

Teacher's Note:
a) Differentiating \( y = mx \) gives \( \frac{dy}{dx} = m \); putting \( m = \frac{y}{x} \) gives \( x\frac{dy}{dx} - y = 0 \).
b) This is exactly the method stated in (R), so (R) explains (A).

SECTION B

21. (a) The cost of Type I sugar is Rs. 25 per kg and Type II sugar is Rs. 35 per kg. If both Type I sugar and Type II sugar are mixed in the ratio 3:2, find the price per kg of the mixture. [2 Marks]

Answer:
1. Let the price of the mixture be Rs. \( x \) per kg. By the rule of alligation, \( \frac{\text{Quantity of Type I}}{\text{Quantity of Type II}} = \frac{35 - x}{x - 25} \).
2. So \( \frac{35 - x}{x - 25} = \frac{3}{2} \).
3. \( 2(35 - x) = 3(x - 25) \Rightarrow 70 - 2x = 3x - 75 \Rightarrow 5x = 145 \Rightarrow x = 29 \).
4. Price of the mixture is Rs. 29 per kg.

Teacher's Note:
a) Check with weighted average: \( \frac{3 \times 25 + 2 \times 35}{5} = \frac{145}{5} = 29 \).
b) In alligation, the cheaper quantity goes with (dearer price minus mean price); do not swap the differences.

OR

(b) Pipe A can fill a tank in 1 hour and Pipe B can fill it in \( 1\frac{1}{2} \) hours. If both the pipes are opened in the empty tank, how much time will they take to fill the tank ? [2 Marks]

Answer:
1. Pipe A fills the tank in 60 minutes and Pipe B in 90 minutes. Let both together take \( n \) minutes.
2. \( \frac{1}{n} = \frac{1}{60} + \frac{1}{90} = \frac{3 + 2}{180} = \frac{5}{180} = \frac{1}{36} \).
3. So \( n = 36 \) minutes, that is, \( \frac{3}{5} \) hour.

Teacher's Note:
a) Add the work done per unit time by each pipe, never the times themselves.
b) Converting hours to minutes first avoids fractions like \( \frac{2}{3} \) per hour.

22. A boat goes 3.5 km upstream and then returns. Total time taken is 1 hour and 12 minutes. If the speed of the current is 1 km/h, then find the speed of the boat in still water. [2 Marks]

Answer:
1. Let the speed of the boat in still water be \( x \) km/h. Upstream speed \( = (x - 1) \) km/h and downstream speed \( = (x + 1) \) km/h.
2. Total time \( = 1 + \frac{12}{60} = \frac{6}{5} \) h, so \( \frac{3.5}{x - 1} + \frac{3.5}{x + 1} = \frac{6}{5} \).
3. \( 3.5(2x) \times 5 = 6(x^2 - 1) \Rightarrow 35x = 6x^2 - 6 \Rightarrow 6x^2 - 35x - 6 = 0 \).
4. \( (x - 6)(6x + 1) = 0 \Rightarrow x = 6 \) or \( x = -\frac{1}{6} \). Rejecting the negative value, the speed of the boat in still water is 6 km/h.

Teacher's Note:
a) Convert 1 hour 12 minutes into \( \frac{6}{5} \) hour before forming the equation.
b) Check: \( \frac{3.5}{5} + \frac{3.5}{7} = 0.7 + 0.5 = 1.2 \) h.

23. A runs \( \frac{3}{2} \) times as fast as B. If A gives B a start of 40 m, how far must the winning post from the starting point be, so that A and B reach at the same time ? [2 Marks]

Answer:
1. Let the winning post be \( x \) m from the starting point. A runs \( x \) m while B runs \( (x - 40) \) m in the same time.
2. Distances are in the ratio of speeds: \( \frac{x}{\frac{3}{2}} = \frac{x - 40}{1} \Rightarrow x = \frac{3}{2}(x - 40) \).
3. \( 2x = 3x - 120 \Rightarrow x = 120 \).
4. The winning post must be 120 m from the starting point.

Teacher's Note:
a) In the same time, distance covered is proportional to speed.
b) Check: A runs 120 m while B runs 80 m, and \( 120 : 80 = 3 : 2 \).

24. Given \( A = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix} \), find BA. [2 Marks]

Answer:
1. \( BA = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} \)
2. Row 1: \( (2 + 3 + 0,\ 0 + 4 - 10,\ 1 + 5 - 15) = (5, -6, -9) \).
3. Row 2: \( (-10 + 3 + 0,\ 0 + 4 - 10,\ -5 + 5 - 15) = (-7, -6, -15) \).
4. Row 3: \( (2 - 6 + 0,\ 0 - 8 + 8,\ 1 - 10 + 12) = (-4, 0, 3) \).
5. \( BA = \begin{bmatrix} 5 & -6 & -9 \\ -7 & -6 & -15 \\ -4 & 0 & 3 \end{bmatrix} \)

Teacher's Note:
a) The order matters: BA means rows of B multiplied by columns of A.
b) Getting at least 4 entries right earns the first mark, so write each row product clearly.

25. (a) If a fair coin is tossed 6 times, find the probability of getting atleast 4 heads. [2 Marks]

Answer:
1. Here \( n = 6, p = \frac{1}{2}, q = \frac{1}{2} \).
2. \( P(X \ge 4) = {}^{6}C_{4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^2 + {}^{6}C_{5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^1 + {}^{6}C_{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^0 \)
3. \( = (15 + 6 + 1)\left(\frac{1}{2}\right)^6 = \frac{22}{64} = \frac{11}{32} \).

Teacher's Note:
a) "At least 4" means 4, 5 or 6 heads; include all three terms.
b) Write the values of n, p and q first, as this step carries marks.

OR

(b) Given that mean of a normal variate X is 9 and standard deviation is 3, then find :
(i) the z-score of the data point 15
(ii) the data point if its z-score is 4. [2 Marks]

Answer:
1. (i) \( Z = \frac{X - \mu}{\sigma} = \frac{15 - 9}{3} = 2 \).
2. (ii) \( 4 = \frac{X - 9}{3} \Rightarrow X - 9 = 12 \Rightarrow X = 21 \).

Teacher's Note:
a) The z-score tells how many standard deviations a value lies from the mean.
b) For part (ii), rearrange as \( X = \mu + Z\sigma = 9 + 4 \times 3 \).

SECTION C

26. Find the units digit in \( 7^{295} \). [3 Marks]

Answer:
1. \( 7^2 = 49 \equiv -1 \pmod{10} \).
2. \( 7^{295} = (7^2)^{147} \times 7 \).
3. \( (7^2)^{147} \equiv (-1)^{147} \equiv -1 \pmod{10} \).
4. So \( 7^{295} \equiv -1 \times 7 = -7 \equiv 3 \pmod{10} \).
5. Hence the units digit of \( 7^{295} \) is 3.

Teacher's Note:
a) Units digit means the remainder on division by 10, so work in mod 10.
b) Quick check: units digits of powers of 7 repeat as 7, 9, 3, 1; \( 295 = 4 \times 73 + 3 \), so the digit is 3.

27. Two numbers are selected at random (without replacement) from first six positive integers. Let X denotes the smaller of the two numbers obtained. Calculate the mathematical expectation of X. [3 Marks]

Answer:
1. The numbers are 1, 2, 3, 4, 5, 6. Number of ways to select two numbers \( = {}^{6}C_{2} = 15 \). X can take the values 1, 2, 3, 4, 5.
2. Probability distribution:
X: 1 | 2 | 3 | 4 | 5
P(X): \( \frac{5}{15} \) | \( \frac{4}{15} \) | \( \frac{3}{15} \) | \( \frac{2}{15} \) | \( \frac{1}{15} \)
X P(X): \( \frac{5}{15} \) | \( \frac{8}{15} \) | \( \frac{9}{15} \) | \( \frac{8}{15} \) | \( \frac{5}{15} \)
3. \( E(X) = \sum X\,P(X) = \frac{5 + 8 + 9 + 8 + 5}{15} = \frac{35}{15} = \frac{7}{3} \).

Teacher's Note:
a) X = 1 when 1 is paired with any of the 5 larger numbers, X = 2 with any of 4 larger numbers, and so on.
b) Check that the probabilities add up to 1 before finding the expectation.

28. (a) If the mean and variance of a binomial distribution are \( \frac{4}{3} \) and \( \frac{8}{9} \) respectively, then find \( P(x = 1) \). [3 Marks]

Answer:
1. Mean \( = np = \frac{4}{3} \) and variance \( = npq = \frac{8}{9} \).
2. \( q = \frac{npq}{np} = \frac{8}{9} \times \frac{3}{4} = \frac{2}{3} \), so \( p = 1 - \frac{2}{3} = \frac{1}{3} \).
3. \( n \times \frac{1}{3} = \frac{4}{3} \Rightarrow n = 4 \).
4. \( P(x = 1) = {}^{4}C_{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^3 = 4 \times \frac{1}{3} \times \frac{8}{27} = \frac{32}{81} \).

Teacher's Note:
a) Divide variance by mean to get q directly.
b) Each of p, q and n carries half a mark, so show each step.

OR

(b) The mortality rate for a certain disease is 0.007. Using Poisson distribution, calculate the probability for 2 deaths in a group of 400 people. [Use \( e^{-2.8} = 0.0608 \)] [3 Marks]

Answer:
1. Given \( p = 0.007, n = 400 \).
2. \( \lambda = np = 400 \times 0.007 = 2.8 \).
3. \( P(X = 2) = \frac{e^{-\lambda}\lambda^2}{2!} = \frac{(2.8)^2 e^{-2.8}}{2} \).
4. \( = \frac{7.84}{2} \times 0.0608 = 3.92 \times 0.0608 = 0.2383 \) (approx.).

Teacher's Note:
a) Poisson formula: \( P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \) with \( \lambda = np \).
b) Use the given value of \( e^{-2.8} \); do not try to compute it.

29. (a) There are two types of fertilizers \( F_1 \) and \( F_2 \). \( F_1 \) consists of 10% nitrogen and 6% phosphoric acid. \( F_2 \) consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If \( F_1 \) costs Rs. 6 per kg and \( F_2 \) costs Rs. 5 per kg, how much of each type of fertilizer should be used so that the cost is minimum. Formulate a linear programming problem. [3 Marks]

Answer:
1. Let \( x \) kg of fertilizer \( F_1 \) and \( y \) kg of fertilizer \( F_2 \) be used.
2. Objective function: Minimise \( Z = 6x + 5y \).
3. Nitrogen constraint: \( 10\% \text{ of } x + 5\% \text{ of } y \ge 14 \Rightarrow \frac{10x}{100} + \frac{5y}{100} \ge 14 \Rightarrow 2x + y \ge 280 \).
4. Phosphoric acid constraint: \( 6\% \text{ of } x + 10\% \text{ of } y \ge 14 \Rightarrow \frac{6x}{100} + \frac{10y}{100} \ge 14 \Rightarrow 3x + 5y \ge 700 \).
5. Non-negativity: \( x \ge 0, y \ge 0 \).

Teacher's Note:
a) "At least" gives the \( \ge \) sign; a common mistake is to use \( \le \).
b) Clear the percentages by multiplying by 100 and simplify each constraint.

OR

(b) Solve the following linear programming problem graphically :
Maximise \( z = 50x + 30y \)
subject to \( 2x + y \le 18 \)
\( 3x + 2y \le 34 \)
\( x, y \ge 0 \). [3 Marks]

Answer:
1. Draw \( 2x + y = 18 \) through \( (9, 0) \) and \( (0, 18) \), and \( 3x + 2y = 34 \) through \( \left(\frac{34}{3}, 0\right) \) and \( (0, 17) \). The feasible region is the shaded region OABC towards the origin, in the first quadrant.
2. Solving \( 2x + y = 18 \) and \( 3x + 2y = 34 \): \( 4x + 2y = 36 \), so \( x = 2 \) and \( y = 14 \). Corner points are O(0, 0), A(0, 17), B(2, 14) and C(9, 0).
3. Values of z: at O, \( z = 0 \); at A, \( z = 510 \); at B, \( z = 100 + 420 = 520 \); at C, \( z = 450 \).
4. Maximum \( z = 520 \) at \( x = 2, y = 14 \).

Teacher's Note:
a) The graph carries \( 1\frac{1}{2} \) marks and the corner-point table carries 1 mark.
b) Find the intersection point B by solving the two equations; do not read it roughly from the graph.

30. A machinist is making engine parts with axle diameter of 0.7 cm. A random sample of 10 parts shows mean diameter 0.742 cm with a standard deviation of 0.04 cm. On the basis of this sample, find if you would say that the work is inferior. (Given \( t_9(0.05) = 2.262 \)) [3 Marks]

Answer:
1. Given \( \bar{x} = 0.742, \mu = 0.7, n = 10, s = 0.04 \).
2. \( H_0 \): There is no significant difference between \( \bar{x} \) and \( \mu \). \( H_1 \): There is a significant difference between \( \bar{x} \) and \( \mu \).
3. \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n - 1}}} = \frac{0.742 - 0.7}{\frac{0.04}{\sqrt{9}}} = \frac{0.042 \times 3}{0.04} = 3.15 \).
4. Degrees of freedom \( = 9 \) and \( t_9(0.05) = 2.262 \). Since \( |t| = 3.15 \gt 2.262 \), \( H_0 \) is rejected.
5. Hence the work is inferior.

Teacher's Note:
a) The marking scheme also accepts \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \); with it too \( |t| \) is greater than 2.262.
b) Always state \( H_0 \), \( H_1 \), the degrees of freedom and the conclusion in words.

31. Calculate EMI under Flat-Rate System for a loan of Rs. 5,00,000 with 7.5% annual interest rate for 5 years. [3 Marks]

Answer:
1. \( P = \) Rs. 5,00,000.
2. Interest \( I = \frac{PRT}{100} = \frac{500000 \times 7.5 \times 5}{100} = \) Rs. 1,87,500.
3. \( n = 5 \) years \( = 60 \) months.
4. \( \text{EMI} = \frac{P + I}{n} = \frac{500000 + 187500}{60} = \frac{687500}{60} \).
5. EMI \( = \) Rs. 11,458.33.

Teacher's Note:
a) In the flat-rate system, interest is charged on the full principal for the whole period.
b) Convert years into months, since EMI is a monthly payment.

SECTION D

32. (a) If \( A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \), find \( A^{-1} \) and hence solve the following system of linear equations :
\( 2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y - 2z = -3 \) [5 Marks]

Answer:
1. \( |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 \neq 0 \), so \( A^{-1} \) exists.
2. The system is \( \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \), that is, \( AX = B \Rightarrow X = A^{-1}B \).
3. Cofactors: \( A_{11} = 0, A_{12} = 2, A_{13} = 1 \); \( A_{21} = -1, A_{22} = -9, A_{23} = -5 \); \( A_{31} = 2, A_{32} = 23, A_{33} = 13 \). So \( \text{adj}(A) = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \).
4. \( A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{-1}\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \).
5. \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
6. So \( x = 1, y = 2, z = 3 \).

Teacher's Note:
a) The adjoint is the transpose of the cofactor matrix; finding it correctly carries \( 1\frac{1}{2} \) marks.
b) Verify by substitution: \( 2(1) - 3(2) + 5(3) = 11 \).

OR

(b) Using properties of determinants, prove that
\( \Delta = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2abc(a + b + c)^3 \) [5 Marks]

Answer:
1. Applying \( C_1 \to C_1 - C_3 \) and \( C_2 \to C_2 - C_3 \):
\( \Delta = \begin{vmatrix} (b+c)^2 - a^2 & 0 & a^2 \\ 0 & (c+a)^2 - b^2 & b^2 \\ c^2 - (a+b)^2 & c^2 - (a+b)^2 & (a+b)^2 \end{vmatrix} \)
2. Taking \( (a + b + c) \) common from \( C_1 \) and from \( C_2 \):
\( \Delta = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ c-a-b & c-a-b & (a+b)^2 \end{vmatrix} \)
3. Applying \( R_3 \to R_3 - R_1 - R_2 \) and taking 2 common from \( R_3 \):
\( \Delta = 2(a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ -b & -a & ab \end{vmatrix} \)
4. Applying \( C_1 \to aC_1 + C_3 \) and \( C_2 \to bC_2 + C_3 \) (so we divide by \( ab \)):
\( \Delta = \frac{2(a+b+c)^2}{ab} \begin{vmatrix} a(b+c) & a^2 & a^2 \\ b^2 & b(c+a) & b^2 \\ 0 & 0 & ab \end{vmatrix} \)
5. Expanding along \( R_3 \): \( \Delta = \frac{2(a+b+c)^2}{ab} \times ab\left[ab(b+c)(c+a) - a^2b^2\right] = 2(a+b+c)^2 \times ab\left[bc + ab + c^2 + ac - ab\right] \)
\( = 2(a+b+c)^2 \times ab \times c(a + b + c) = 2abc(a + b + c)^3 = \) RHS. Hence proved.

Teacher's Note:
a) Use \( x^2 - y^2 = (x - y)(x + y) \) after the column operations to bring out the factor \( (a + b + c) \).
b) Create two zeros in one row before expanding, so that only one product has to be worked out.
c) Write each row or column operation clearly, since each step carries marks.

33. If the supply function is \( p = 4 - 5x + x^2 \), then find the producer's surplus when price is 18. [5 Marks]

Answer:
1. Given \( p = 4 - 5x + x^2 \) and \( p_0 = 18 \). So \( 18 = 4 - 5x + x^2 \Rightarrow x^2 - 5x - 14 = 0 \).
2. \( (x - 7)(x + 2) = 0 \Rightarrow x = 7 \) (\( x = -2 \) is rejected). So \( x_0 = 7 \).
3. \( p_0 x_0 = 18 \times 7 = 126 \).
4. \( PS = p_0 x_0 - \int_{0}^{7} (x^2 - 5x + 4)\, dx = 126 - \left[\frac{x^3}{3} - \frac{5x^2}{2} + 4x\right]_{0}^{7} \).
5. \( \left[\frac{x^3}{3} - \frac{5x^2}{2} + 4x\right]_{0}^{7} = \frac{343}{3} - \frac{245}{2} + 28 = \frac{686 - 735 + 168}{6} = \frac{119}{6} \).
6. \( PS = 126 - \frac{119}{6} = \frac{756 - 119}{6} = \frac{637}{6} \approx 106.17 \).

Teacher's Note:
a) Producer's surplus \( = p_0x_0 - \int_{0}^{x_0} S(x)\, dx \); for consumer's surplus the order is reversed.
b) Reject the negative quantity \( x = -2 \), as quantity cannot be negative.

34. (a) Compute the seasonal indices by 4-year moving averages from the given data of production of paper (in thousand tons) :
Year : 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010
Index number : 2450 | 1470 | 2150 | 1800 | 1210 | 1950 | 2300 | 2500 | 2480 | 2680 [5 Marks]

Answer:
1. 4-yearly moving totals (placed between the 2nd and 3rd years of each group): 7870, 6630, 7110, 7260, 7960, 9230, 9960.
(For example, \( 2450 + 1470 + 2150 + 1800 = 7870 \) and \( 1470 + 2150 + 1800 + 1210 = 6630 \).)
2. 4-yearly moving averages (total \( \div 4 \)): 1967.5, 1657.5, 1777.5, 1815, 1990, 2307.5, 2490.
3. Centred totals (sum of two consecutive moving averages), placed against 2003 to 2008: \( 1967.5 + 1657.5 = 3625 \), 3435, 3592.5, 3805, 4297.5, 4797.5.
4. Centred moving averages (centred total \( \div 2 \)):
Year: 2003 | 2004 | 2005 | 2006 | 2007 | 2008
Centred moving average: 1812.5 | 1717.5 | 1796.25 | 1902.5 | 2148.75 | 2398.75
5. No centred values exist for the first two years (2001, 2002) and the last two years (2009, 2010).

Teacher's Note:
a) With an even period (4 years), the moving averages fall between years, so they must be centred.
b) Marks are split as: moving totals 1, moving averages \( 1\frac{1}{2} \), centred totals 1, centred moving averages \( 1\frac{1}{2} \).

OR

(b) Fit a straight-line trend by method of least squares for the following data :
Year : 2011 | 2012 | 2013 | 2014 | 2015 | 2016
Production (in tons) : 210 | 225 | 275 | 220 | 240 | 235 [5 Marks]

Answer:
1. Here \( n = 6 \) (even). Take \( X = \frac{x_i - 2013.5}{0.5} \), so X takes the values \( -5, -3, -1, 1, 3, 5 \).
2. Working table:
Year \( (x_i) \): 2011 | 2012 | 2013 | 2014 | 2015 | 2016
Y: 210 | 225 | 275 | 220 | 240 | 235
X: \( -5 \) | \( -3 \) | \( -1 \) | 1 | 3 | 5
\( X^2 \): 25 | 9 | 1 | 1 | 9 | 25
XY: \( -1050 \) | \( -675 \) | \( -275 \) | 220 | 720 | 1175
\( \sum Y = 1405, \sum X = 0, \sum X^2 = 70, \sum XY = 115 \).
3. \( a = \frac{\sum Y}{n} = \frac{1405}{6} = 234.17 \) (approx.).
4. \( b = \frac{\sum XY}{\sum X^2} = \frac{115}{70} = 1.64 \) (approx.).
5. Required trend line: \( Y = 234.17 + 1.64X \), where \( X = \frac{x - 2013.5}{0.5} \).
Trend values: 225.97 | 229.25 | 232.53 | 235.81 | 239.09 | 242.37 (for 2011 to 2016).

Teacher's Note:
a) For an even number of years, take the origin midway (2013.5) and the unit as half a year, so that \( \sum X = 0 \).
b) The correct table carries \( 2\frac{1}{2} \) marks; always state the origin and unit of X with the final equation.

35. A machine costs Rs. 1,00,000 and its effective life is estimated to be 12 years. A sinking fund is created for replacing the machine by a new model at the end of its life time when its scrap realizes a sum of Rs. 5,000 only. Find what amount should be set aside at the end of each year, out of the profits for the sinking fund if it accumulates at 5% effective. [Use \( (1.05)^{12} = 1.7958 \)] [5 Marks]

Answer:
1. Amount needed after 12 years \( = 100000 - 5000 = \) Rs. 95,000.
2. The sinking fund is an annuity of 12 annual payments R at 5% per year, so \( A = R\left[\frac{(1 + i)^n - 1}{i}\right] \).
3. \( 95000 = R\left[\frac{(1.05)^{12} - 1}{0.05}\right] = R\left[\frac{1.7958 - 1}{0.05}\right] = R \times \frac{0.7958}{0.05} \).
4. \( R = \frac{95000 \times 0.05}{0.7958} = \frac{4750}{0.7958} \).
5. \( R = \) Rs. 5,968.84 (approx.). (Using the approximation \( \frac{4750}{0.8} \), R \( = \) Rs. 5,937.50 is also accepted.)

Teacher's Note:
a) Subtract the scrap value first; the fund needs to cover only Rs. 95,000.
b) Use the amount (future value) formula of an annuity, not the present value formula.

SECTION E

36. A man has an expensive square-shaped piece of golden board of side 36 cm. He wants to turn it into a box without top by cutting a square from each corner and folding the flaps. Let \( x \) cm be the side of square, which is cut from each corner.

[Figure: Left - a square board of side 36 cm with a small square of side \( x \) cm marked at each corner and dashed fold lines joining them. Right - the open box formed by folding up the flaps, with its height marked as \( x \).]

Based on the above information, answer the following questions :

(i) Find the expression for the volume (V) of open box in terms of \( x \). [1 Mark]

Answer: The base of the box is a square of side \( (36 - 2x) \) cm and its height is \( x \) cm, so \( V = x(36 - 2x)^2 \) cubic cm.

Teacher's Note:
a) A square of side \( x \) is cut from both ends of each side, so the side reduces by \( 2x \).
b) The side of the cut square becomes the height of the box.

(ii) Find \( \frac{dV}{dx} \). [1 Mark]

Answer:
1. \( \frac{dV}{dx} = (36 - 2x)^2 + x \times 2(36 - 2x)(-2) = (36 - 2x)(36 - 2x - 4x) \).
2. \( = (36 - 2x)(36 - 6x) = 12(18 - x)(6 - x) \).

Teacher's Note:
a) Use the product rule and the chain rule; do not forget the factor \( -2 \) from differentiating \( 36 - 2x \).
b) Factorising the result makes part (iii) very quick.

(iii) Find the value of \( x \) for which the volume (V) is maximum. [2 Marks]

Answer:
1. \( \frac{dV}{dx} = 0 \Rightarrow 12(18 - x)(6 - x) = 0 \Rightarrow x = 18 \) or \( x = 6 \).
2. \( x = 18 \) is rejected, as then \( 36 - 2x = 0 \) and there is no box. So \( x = 6 \).
3. \( \frac{d^2V}{dx^2} = 12(18 - x)(-1) + 12(-1)(6 - x) = -12(24 - 2x) \); at \( x = 6 \), \( \frac{d^2V}{dx^2} = -144 \lt 0 \).
4. Hence the volume is maximum for \( x = 6 \) cm.

Teacher's Note:
a) Always give a reason for rejecting \( x = 18 \).
b) The second derivative test (negative value) confirms a maximum.

OR

(iii) Find the maximum volume of the open box. [2 Marks]

Answer:
1. \( \frac{dV}{dx} = 0 \Rightarrow x = 18 \) or \( x = 6 \); \( x = 18 \) is rejected, so \( x = 6 \).
2. \( \frac{d^2V}{dx^2} = 12(18 - x)(-1) + 12(-1)(6 - x) = -144 \lt 0 \) at \( x = 6 \), so V is maximum there.
3. Maximum \( V = 6(36 - 12)^2 = 6 \times (24)^2 = 6 \times 576 = 3456 \) cubic cm.

Teacher's Note:
a) First find the value of \( x \), then substitute it in V.
b) Write the unit as cubic cm, since V is a volume.

37. There are two factories located one at P and the other at 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 4, 4 and 6 units of the commodity while the production capacity of the factories at P and Q are 9 and 5 units respectively. The cost of transportation per unit is given as :
Cost (in Rs.)
From / To | A | B | C
P | 160 | 100 | 150
Q | 100 | 120 | 100
Based on the above information, answer the following questions :
Let \( x \) units and y units of the commodity be transported from factory P to the depots at A and B respectively, then

[Figure: Network diagram - Factory P (9U) at the top with arrows to depots A (4U), B (4U) and C (6U); the arrow P to A is labelled \( x \) and the arrow P to B is labelled y. Factory Q (5U) at the bottom has arrows to A, B and C.]

(i) Find (in terms of \( x \) and y) how many units of commodity be transported from factory P to depot C. [1 Mark]

Answer: Factory P has 9 units, out of which \( x \) go to A and \( y \) go to B. So units from P to C \( = 9 - (x + y) = 9 - x - y \).

Teacher's Note:
a) Total supply \( 9 + 5 = 14 \) equals total demand \( 4 + 4 + 6 = 14 \), so P sends all its 9 units.
b) Write the answer in terms of both \( x \) and \( y \).

(ii) Find how many units of commodity be transported from factory Q to A, B and C respectively. [1 Mark]

Answer:
1. Q to A \( = 4 - x \) and Q to B \( = 4 - y \).
2. Q to C \( = 6 - (9 - x - y) = x + y - 3 \).

Teacher's Note:
a) Each depot's requirement minus what it gets from P is supplied by Q.
b) Check: \( (4 - x) + (4 - y) + (x + y - 3) = 5 \), which is Q's capacity.

(iii) Using (i) and (ii), find the total transportation cost z. [2 Marks]

Answer:
1. \( z = 160x + 100y + 150(9 - x - y) + 100(4 - x) + 120(4 - y) + 100(x + y - 3) \).
2. \( = 160x + 100y + 1350 - 150x - 150y + 400 - 100x + 480 - 120y + 100x + 100y - 300 \).
3. \( z = 10x - 70y + 1930 \).

Teacher's Note:
a) Multiply each route's quantity by its unit cost from the table, then add all six terms.
b) Collect the \( x \) terms, \( y \) terms and constants separately to avoid errors.

OR

(iii) Using (i) and (ii), find the constraint inequalities for minimum cost z. [2 Marks]

Answer:
1. P to C is non-negative: \( 9 - x - y \ge 0 \Rightarrow x + y \le 9 \).
2. Q to C is non-negative: \( x + y - 3 \ge 0 \Rightarrow x + y \ge 3 \).
3. Q to A and Q to B are non-negative: \( 4 - x \ge 0, 4 - y \ge 0 \Rightarrow x \le 4, y \le 4 \).
4. Also \( x \ge 0, y \ge 0 \).

Teacher's Note:
a) Every quantity sent along a route must be non-negative; each such condition gives one constraint.
b) Do not forget the non-negativity constraints \( x, y \ge 0 \).

38. Ramesh borrowed a home loan amount of Rs. 7,00,000 from a bank at an interest of 12% per annum for 30 years, to be paid in monthly installments.
Based on the above information, answer the following questions :

(i) Write the formula for calculating EMI by reducing balance method. [1 Mark]

Answer: \( E = \frac{P\,i}{1 - (1 + i)^{-n}} \) or \( E = \frac{P\,i\,(1 + i)^n}{(1 + i)^n - 1} \), where P is the principal, \( i \) is the interest rate per month and \( n \) is the number of monthly installments.

Teacher's Note:
a) Both forms of the formula are correct; they differ only by multiplying by \( (1 + i)^n \).
b) Define each symbol when you write the formula.

(ii) Write the values of P, i and n respectively. [1 Mark]

Answer: \( P = \) Rs. 7,00,000, \( i = \frac{12}{1200} = 0.01 \) per month and \( n = 12 \times 30 = 360 \) months.

Teacher's Note:
a) For monthly installments, divide the annual rate by 12 and multiply the years by 12.
b) A common mistake is to use \( i = 0.12 \) and \( n = 30 \).

(iii) Find the EMI. [Use \( (1.01)^{-360} = 0.02781668 \)] [2 Marks]

Answer:
1. \( E = \frac{700000 \times 0.01}{1 - (1.01)^{-360}} = \frac{7000}{1 - 0.02781668} \).
2. \( E = \frac{7000}{0.97218332} = \) Rs. 7,200.29 (approx.). (Using the approximation \( \frac{7000}{0.97} \), E \( = \) Rs. 7,216.49 is also accepted.)

Teacher's Note:
a) Substitute P, i and n from part (ii) into the formula of part (i).
b) Subtract the given value from 1 carefully before dividing.

OR

(iii) If the loan is to be returned in 20 years, find EMI. [Use \( (1.01)^{-240} = 0.09180584 \)] [2 Marks]

Answer:
1. Now \( n = 12 \times 20 = 240 \) months, so \( E = \frac{700000 \times 0.01}{1 - (1.01)^{-240}} = \frac{7000}{1 - 0.09180584} \).
2. \( E = \frac{7000}{0.90819416} = \) Rs. 7,707.60 (approx.). (Using the approximation \( \frac{7000}{0.91} \), E \( = \) Rs. 7,692.31 is also accepted.)

Teacher's Note:
a) Only \( n \) changes; P and \( i \) stay the same.
b) A shorter loan period gives a higher EMI but less total interest.

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