CBSE Class 12 Applied Mathematics Sample Paper 2026 27 with Solutions PDF Download

SECTION - A (20 x 1 = 20)

This section comprises of 18 multiple choice questions and two assertion and reason type questions of 1 mark each.

1. Given two integers \( a, b \) such that \( a \lt b \) and \( c \) be any non-zero integer then which of the following always holds true: [1 Mark]
(A) \( a - c \gt b - c \)
(B) \( ac \lt bc \)
(C) \( ac \gt bc \)
(D) \( a + c \lt b + c \)

Answer: (D) \( a + c \lt b + c \)

Teacher's Note:
a) Adding the same number to both sides of an inequality keeps the inequality, whether the number is positive or negative.
b) Multiplying by \( c \) is not safe: if \( c \) is negative, the inequality sign reverses.

2. Vessel A contains milk and water in the ratio 4 : 5. Vessel B contains milk and water in the ratio 5 : 1. In what ratio should quantities be taken from both vessels to form a mixture in which milk and water are in the ratio 5 : 4? [1 Mark]
(A) 5 : 2
(B) 5 : 3
(C) 4 : 1
(D) 1 : 4

Answer: (A) 5 : 2

Teacher's Note:
a) Milk fractions: vessel A \( = \frac{4}{9} \), vessel B \( = \frac{5}{6} \), required mixture \( = \frac{5}{9} \).
b) By alligation, required ratio \( = \left( \frac{5}{6} - \frac{5}{9} \right) : \left( \frac{5}{9} - \frac{4}{9} \right) = \frac{5}{18} : \frac{1}{9} = 5 : 2 \).

3. Number of symmetric matrices of order \( 3 \times 3 \) with each entry 1 or \( -1 \) is [1 Mark]
(A) 512
(B) 64
(C) 8
(D) 4

Answer: (B) 64

Teacher's Note:
a) A \( 3 \times 3 \) symmetric matrix \( \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & f \end{bmatrix} \) has only 6 independent entries.
b) Each entry has 2 choices, so the number of matrices is \( 2^{6} = 64 \).

4. In a car race, car A beats car B by 45 km, car B beats car C by 50 km, and car A beats car C by 90 km then the length of the race course is [1 Mark]
(A) 400 km
(B) 450 km
(C) 500 km
(D) 550 km

Answer: (B) 450 km

Teacher's Note:
a) Let the course be \( d \) km. Then \( \frac{S_A}{S_C} = \frac{S_A}{S_B} \times \frac{S_B}{S_C} \) gives \( \frac{d}{d - 90} = \frac{d}{d - 45} \times \frac{d}{d - 50} \).
b) \( (d - 45)(d - 50) = d(d - 90) \Rightarrow d^2 - 95d + 2250 = d^2 - 90d \Rightarrow 5d = 2250 \Rightarrow d = 450 \) km.

5. If \( x = \log t \) and \( y = t^2 \), then value of \( \frac{d^2y}{dx^2} \) at \( t = \frac{1}{\sqrt{2}} \) is [1 Mark]
(A) 4
(B) 2
(C) \( 2\sqrt{2} \)
(D) 1

Answer: (B) 2

Teacher's Note:
a) \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{1/t} = 2t^2 \).
b) \( \frac{d^2y}{dx^2} = \frac{d}{dt}(2t^2) \times \frac{dt}{dx} = 4t \times t = 4t^2 \), which equals \( 4 \times \frac{1}{2} = 2 \) at \( t = \frac{1}{\sqrt{2}} \).

6. Which of the following statements about the future value (FV) of a perpetuity is true? [1 Mark]
(A) The future value can be calculated by compounding the present value to a specific future date.
(B) The future value is equal to twice the present value multiplied by number of payments.
(C) The future value is always equal to the total cash flows received up to the future date.
(D) The future value of perpetuity is undefined.

Answer: (D) The future value of perpetuity is undefined.

Teacher's Note:
a) A perpetuity pays forever, so there is no final date at which its future value can be found.
b) Only the present value of a perpetuity is finite: \( P = \frac{R}{i} \).

7. The order of the differential equation whose general solution is given by \( y = (a + b)\log(x + c) - de^{x + k} \) where a, b, c, d and k are arbitrary constants, is [1 Mark]
(A) 4
(B) 5
(C) 3
(D) 2

Answer: (C) 3

Teacher's Note:
a) Write \( y = A\log(x + c) - Be^{x} \), where \( A = a + b \) and \( B = d \cdot e^{k} \).
b) Only three essential arbitrary constants \( (A, B, c) \) remain, so the order is 3. Do not simply count the letters given.

8. The Marginal Revenue (MR) of a company is given by the function \( MR(x) = 20e^{0.5x} \) where \( x \) is the number of units sold. The additional revenue generated when sales increase from 2 units to 4 units is: [1 Mark]
(A) \( 20e^{2} \)
(B) \( 20(e^{2} - e^{1}) \)
(C) \( 40e^{2} \)
(D) \( 40(e^{2} - e^{1}) \)

Answer: (D) \( 40(e^{2} - e^{1}) \)

Teacher's Note:
a) Increase in revenue \( = \int_{2}^{4} 20e^{0.5x}\,dx = \left[ \frac{20e^{0.5x}}{0.5} \right]_{2}^{4} \).
b) \( = 40(e^{2} - e^{1}) \). Remember to divide by 0.5 while integrating.

9. In the context of standard reducing-balance EMI, over the loan period, which of the following is correct? [1 Mark]
(A) Principal and Interest components both remain constant.
(B) Principal component decreases, and Interest component increases.
(C) Principal component increases, and Interest component decreases.
(D) Principal component remains constant, and Interest component decreases.

Answer: (C) Principal component increases, and Interest component decreases.

Teacher's Note:
a) Interest is charged on the outstanding balance, which falls after every EMI.
b) As the EMI stays fixed, the smaller interest part leaves a bigger principal part.

10. If you are willing to pay Rs. 40,000 today for an investment that promises to pay Rs. 3,200 at the end of every year indefinitely, then the implied rate of return is [1 Mark]
(A) 6 %
(B) 8 %
(C) 10 %
(D) 12.5 %

Answer: (B) 8 %

Teacher's Note:
a) For a perpetuity, \( P = \frac{R}{i} \Rightarrow 40000 = \frac{3200}{i} \).
b) \( i = 0.08 \), so the rate of return is 8 %.

11. A foreign research lab observes that the number of microscopic defects on a solar-panel surface follows a Poisson distribution with mean 2 defects per 100 \( cm^2 \). If a scientist inspects an area of 250 \( cm^2 \), what is the probability that exactly 1 defect is found? [1 Mark]
(A) \( 6e^{-5} \)
(B) \( 5e^{-5} \)
(C) \( 4e^{-5} \)
(D) \( 3e^{-2} \)

Answer: (B) \( 5e^{-5} \)

Teacher's Note:
a) First scale the mean to the new area: \( \lambda = \frac{2}{100} \times 250 = 5 \).
b) \( P(X = 1) = e^{-5} \times \frac{5^{1}}{1!} = 5e^{-5} \).

12. Which of the following statement(s) is/are true:
I: The mean of a population is denoted by \( \bar{x} \).
II: The population mean is a statistic. [1 Mark]
(A) I only
(B) II only
(C) Both I and II
(D) Neither I nor II

Answer: (D) Neither I nor II

Teacher's Note:
a) The population mean is denoted by \( \mu \); \( \bar{x} \) denotes the sample mean.
b) A population measure is a parameter; a measure calculated from a sample is a statistic.

13. Two random variables X and Y have the same mean \( \mu = 50 \). If \( Var(X) = 100 \) and \( Var(Y) = 25 \), which statement is TRUE? [1 Mark]
(A) X is more consistent than Y
(B) Y is more consistent than X
(C) Both have equal consistency
(D) Cannot be determined with the given data

Answer: (B) Y is more consistent than X

Teacher's Note:
a) Variance measures the spread of values around the mean.
b) Lower variance means values are closer to the mean, so Y (variance 25) is more consistent.

14. A trend line is considered the best fit when the sum of squares of residuals is [1 Mark]
(A) Maximum
(B) Minimum
(C) Positive
(D) Negative

Answer: (B) Minimum

Teacher's Note:
a) This is the principle of the method of least squares.
b) A sum of squares can never be negative, so options (C) and (D) do not describe a best fit.

15. A sample of 100 cereal boxes has mean = 505g and SD = 15g. The 99% confidence interval for the true mean \( \mu \) is (501.125, 508.875) grams.
Which of the following is the correct interpretation of this interval? [1 Mark]
(A) 99% of repeated samples will have means in the interval (501.125,508.875) grams.
(B) The probability that the mean weight, \( \mu \), is between 501.125 and 508.875 grams is 0.99.
(C) We are 99% confident that the calculated interval of (501.125, 508.875) grams contains the true mean weight \( \mu \).
(D) Approximately 99% of all boxes have a weight between 501.125 and 508.875 grams.

Answer: (C) We are 99% confident that the calculated interval of (501.125, 508.875) grams contains the true mean weight \( \mu \).

Teacher's Note:
a) A confidence interval is a statement of confidence about the true population mean.
b) It does not describe the weights of individual boxes, so option (D) is wrong.

16. Let X be a random variable following the binomial distribution \( B(6, p) \). If \( 2P(X = 4) = 3P(X = 3) \), then the value of p is: [1 Mark]
(A) \( \frac{1}{3} \)
(B) \( \frac{1}{2} \)
(C) \( \frac{2}{3} \)
(D) \( \frac{3}{4} \)

Answer: (C) \( \frac{2}{3} \)

Teacher's Note:
a) \( 2 \times {}^{6}C_{4}\, p^4 q^2 = 3 \times {}^{6}C_{3}\, p^3 q^3 \Rightarrow 30p^4q^2 = 60p^3q^3 \Rightarrow p = 2q \).
b) \( p = 2(1 - p) \Rightarrow p = \frac{2}{3} \).

17. Suppose a random variable \( X \) follows the binomial distribution with parameters \( n \) and \( p \), where \( 0 \lt p \lt 1 \). If \( \frac{P(X = r)}{P(X = n - r)} \) is independent of \( n \) and \( r \), then \( p \) equals [1 Mark]
(A) \( \frac{1}{2} \)
(B) \( \frac{1}{3} \)
(C) \( \frac{1}{5} \)
(D) \( \frac{1}{7} \)

Answer: (A) \( \frac{1}{2} \)

Teacher's Note:
a) \( \frac{P(X = r)}{P(X = n - r)} = \frac{{}^{n}C_{r}\, p^{r} q^{n - r}}{{}^{n}C_{n - r}\, p^{n - r} q^{r}} = \left( \frac{p}{q} \right)^{2r - n} \), since \( {}^{n}C_{r} = {}^{n}C_{n - r} \).
b) This is independent of \( n \) and \( r \) only when \( \frac{p}{q} = 1 \), i.e. \( p = q = \frac{1}{2} \).

18. A firm manufacturing two types of lubricants whose non-negative production quantities, represented by decision variables \( x \) and \( y \) satisfy the following constraints:
\( x \ge 2, y \le 3, x + y = 5 \)
Then the feasible region of the above LPP is: [1 Mark]
(A) A triangle
(B) A quadrilateral
(C) A line segment
(D) unbounded

Answer: (C) A line segment

Teacher's Note:
a) The equality \( x + y = 5 \) forces every feasible point to lie on this line.
b) With \( x \ge 2 \) and \( y \ge 0 \), the feasible region is the segment of \( x + y = 5 \) from \( (2, 3) \) to \( (5, 0) \).

Assertion-Reason Based Questions

Directions: Question numbers 19 and 20 are Assertion and Reason based questions carrying 1 mark each. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer from the options (A), (B), (C) and (D) as given below:

19. Assertion (A): Let \( A \) be a non-singular square matrix of order 3. If \( |A| = 4 \), then the value of \( |2 . adj(A)| \) is 128.
Reason (R): For a matrix \( A \), \( |adj(A)| = |A|^{n - 1} \) and \( |kA| = k|A| \) [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false but Reason (R) is true.

Answer: (C) Assertion (A) is true but Reason (R) is false.

Teacher's Note:
a) \( |2 \cdot adj(A)| = 2^{3}|adj(A)| = 8 \times 4^{2} = 128 \), so the Assertion is true.
b) The Reason is false because \( |kA| = k^{n}|A| \), not \( k|A| \).

20. Assertion (A): If two probability distributions have the same mean, they must have the same variance.
Reason (R): Variance measures the spread of data around the mean and is calculated as \( Var(X) = E(X^2) - [E(X)]^2 \) [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true but (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false but Reason (R) is true.

Answer: (D) Assertion (A) is false but Reason (R) is true.

Teacher's Note:
a) Two distributions can have the same mean but different variances, so the Assertion is false.
b) The Reason correctly states the meaning and formula of variance.

SECTION - B (5 x 2 = 10)

This section comprises of 5 very short answer (VSA) type questions of 2 marks each.

21. (A) If \( x \) is an odd integer, show that \( x^2 \equiv 1 (mod\, 8) \) [2 Marks]

Answer:
1. As \( x \) is an odd integer, \( x = 2n + 1 \), where \( n \in Z \).
2. \( x^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1 \).
3. \( n(n + 1) \) is a product of two consecutive integers, so it is even. Let \( n(n + 1) = 2m \).
4. Then \( x^2 = 4(2m) + 1 = 8m + 1 \), which leaves remainder 1 when divided by 8.
Hence \( x^2 \equiv 1 (mod\, 8) \).

Teacher's Note:
a) The key step is stating that the product of two consecutive integers is always even.
b) Quick check: \( 3^2 = 9 = 8 + 1 \) and \( 5^2 = 25 = 24 + 1 \).

OR

(B) If \( 3^{333} \equiv x\ mod\ 11 \), find \( x \). [2 Marks]

Answer:
1. \( 3^5 = 243 = 22 \times 11 + 1 \), so \( 3^5 \equiv 1 (mod\, 11) \).
2. \( (3^5)^{66} \equiv 1^{66} (mod\, 11) \Rightarrow 3^{330} \equiv 1 (mod\, 11) \).
3. \( 3^{330} \times 3^3 \equiv 27 (mod\, 11) \Rightarrow 3^{333} \equiv 27 (mod\, 11) \).
4. \( 27 = 2 \times 11 + 5 \), so \( 3^{333} \equiv 5 (mod\, 11) \).
Therefore, \( x = 5 \).

Teacher's Note:
a) First find the smallest power of 3 that leaves remainder 1 on division by 11.
b) Write \( 333 = 5 \times 66 + 3 \) and reduce the final remainder so that it lies between 0 and 10.

22. An asset was purchased for Rs. 120,000 with an expected salvage value of Rs. 10,000 and 10-year useful life. What is the asset's book value at the end of sixth year? [2 Marks]

Answer:
1. Under linear depreciation, \( D = \frac{C - S}{n} = \frac{120000 - 10000}{10} = \frac{110000}{10} = 11000 \).
2. Annual depreciation is Rs. 11,000.
3. Book value at the end of 6 years \( = C - 6D = 120000 - 6(11000) = 54000 \).
The book value at the end of the sixth year is Rs. 54,000.

Teacher's Note:
a) Subtract the salvage value from the cost before dividing by the useful life.
b) Book value falls by the same amount each year under the linear (straight line) method.

23. A man rows 12 km downstream and 12 km upstream, taking 4 hours in total. If the speed of the stream is 4 km/h, find the speed of the man in still water. [2 Marks]

Answer:
1. Let the speed of the man in still water be \( x \) km/h. Speed downstream \( = (x + 4) \) km/h and speed upstream \( = (x - 4) \) km/h.
2. According to the question, \( \frac{12}{x + 4} + \frac{12}{x - 4} = 4 \Rightarrow \frac{1}{x + 4} + \frac{1}{x - 4} = \frac{1}{3} \).
3. \( \frac{x - 4 + x + 4}{(x + 4)(x - 4)} = \frac{1}{3} \Rightarrow 6x = x^2 - 16 \Rightarrow x^2 - 6x - 16 = 0 \).
4. \( (x - 8)(x + 2) = 0 \Rightarrow x = 8 \) or \( x = -2 \). But \( x \ne -2 \).
Hence the speed of the man in still water is 8 km/h.

Teacher's Note:
a) Time = distance \( \div \) speed; add the two times and equate to 4 hours.
b) Reject the negative root, as speed cannot be negative.

24. (A) The number of customer complaints received by an e-commerce company follows a Poisson distribution. It is observed that the probability of receiving exactly 2 complaints in a day is equal to the probability of receiving exactly 3 complaints in a day. Find the mean and the variance of the distribution. [2 Marks]

Answer:
1. Given \( P(X = 2) = P(X = 3) \Rightarrow e^{-m}\frac{m^2}{2!} = e^{-m}\frac{m^3}{3!} \).
2. \( 3m^2 = m^3 \Rightarrow m = 3 \) (as \( m \ne 0 \)).
3. In a Poisson distribution, mean = variance = \( m \).
So, mean = variance = 3.

Teacher's Note:
a) Cancel \( e^{-m} \) and \( m^2 \) from both sides; \( m \) cannot be 0 for a Poisson distribution.
b) Remember that the mean and the variance of a Poisson distribution are equal.

OR

(B) A metro station observes that trains arrive at a particular platform at an average rate of 20 trains per hour during peak hours.
(i) What is the expected number of trains arriving in a 6-minute interval?
(ii) Find the probability that exactly 2 trains arrive in a 6-minute interval.
[Use \( e^{-2} = 0.1353 \)] [2 Marks]

Answer:
(i) Rate per 6 minutes \( = 20 \times \frac{6}{60} = 2 \) trains. So \( m = 2 \).
(ii) \( P(X = 2) = e^{-2}\frac{2^2}{2!} = 2 \times 0.1353 = 0.2706 \).

Teacher's Note:
a) Convert the hourly rate to the 6-minute interval before using the Poisson formula.
b) \( \frac{2^2}{2!} = 2 \), so the answer is simply \( 2e^{-2} \).

25. The feasible region for an LPP is shown as in the graph below:
Frame all the constraints of the LPP represented by the above graph. [2 Marks]

[Figure: Graph with x-axis from -1 to 8 and y-axis from 0 to 8. Line AB joins A(0, 6) and B(6, 0); line CD joins D(0, 4) and C(8, 0); line EF joins F(0, 8) and E(4, 0). The lines meet at G(2, 4), H and I(4, 2). The shaded feasible region is the small triangle GHI lying between the three lines.]

Answer:
1. Equation of AB: \( \frac{x}{6} + \frac{y}{6} = 1 \), i.e. \( x + y = 6 \).
2. Equation of CD: \( \frac{x}{8} + \frac{y}{4} = 1 \), i.e. \( x + 2y = 8 \).
3. Equation of EF: \( \frac{x}{4} + \frac{y}{8} = 1 \), i.e. \( 2x + y = 8 \).
4. The shaded region lies on the origin side of AB and away from the origin for CD and EF.
Constraints are: \( x + y \le 6,\ x + 2y \ge 8,\ 2x + y \ge 8 \).

Teacher's Note:
a) Use the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \) with the intercepts read from the graph.
b) Test a point inside the shaded region, such as \( (3, 3) \), to fix the direction of each inequality.

SECTION - C (6 x 3 = 18)

This section comprises of 6 short answer (SA) type questions of 3 marks each.

26. (A) The price of a stock S (in dollars) t days after its IPO is modeled by \( S(t) = \frac{4t^2 + 1}{t} \), \( t \gt 0 \). Show that the stock price decreased for the first half of the opening day and then started increasing thereon. [3 Marks]

Answer:
1. \( S(t) = \frac{4t^2 + 1}{t} = 4t + \frac{1}{t} \Rightarrow S'(t) = 4 - \frac{1}{t^2} = \frac{4t^2 - 1}{t^2} = \frac{(2t - 1)(2t + 1)}{t^2} \).
2. For decreasing: \( S'(t) \le 0 \Rightarrow (2t - 1)(2t + 1) \le 0 \Rightarrow -\frac{1}{2} \le t \le \frac{1}{2} \). As \( t \gt 0 \), \( S(t) \) is decreasing for \( 0 \lt t \le \frac{1}{2} \), i.e. for the first half of the opening day.
3. For increasing: \( S'(t) \ge 0 \Rightarrow t \le -\frac{1}{2} \) or \( t \ge \frac{1}{2} \). As \( t \gt 0 \), \( S(t) \) is increasing for \( t \ge \frac{1}{2} \), i.e. after the first half of the opening day.
Hence, the stock price decreased for the first half of the opening day and then started increasing thereon.

Teacher's Note:
a) \( t \) is in days, so the first half of the opening day means \( 0 \lt t \le \frac{1}{2} \).
b) Since \( t^2 \gt 0 \), the sign of \( S'(t) \) depends only on \( (2t - 1)(2t + 1) \).

OR

(B) Consider a simple market where quantity demanded is \( Q_d = 15 - 2P \) and quantity supplied is \( Q_S = 3P - 5 \). The rate of change of price i.e., \( \frac{dP}{dt} \) is half of the excess demand i.e. \( (Q_d - Q_S) \).
(i) Show that \( \frac{dP}{dt} = \frac{1}{2}(20 - 5P) \).
(ii) If \( P = 2 \) when \( t = 0 \), then show that \( P = 2(2 - e^{-2.5t}) \) [3 Marks]

Answer:
(i) \( \frac{dP}{dt} = \frac{1}{2}(Q_d - Q_S) = \frac{1}{2}[(15 - 2P) - (3P - 5)] = \frac{1}{2}(20 - 5P) \).
(ii) Separating the variables: \( \int \frac{1}{20 - 5P}\,dP = \frac{1}{2}\int dt + C \)
\( \Rightarrow \frac{\log|20 - 5P|}{-5} = \frac{1}{2}t + C \Rightarrow 2\log|20 - 5P| = -5t - 10C \) ...(1)
When \( t = 0,\ P = 2 \): \( 2\log|20 - 10| = -10C \Rightarrow C = -\frac{1}{5}\log 10 \).
Using this in (1): \( 2\log|20 - 5P| = -5t + 2\log 10 \Rightarrow \log\left( \frac{20 - 5P}{10} \right) = -2.5t \)
\( \Rightarrow \frac{20 - 5P}{10} = e^{-2.5t} \Rightarrow 20 - 10e^{-2.5t} = 5P \)
\( \Rightarrow P = 4 - 2e^{-2.5t} = 2(2 - e^{-2.5t}) \).

Teacher's Note:
a) Do not forget the factor \( -\frac{1}{5} \) when integrating \( \frac{1}{20 - 5P} \).
b) Find the constant C from the initial condition before removing the logarithm.

27. A Rs. 1000 bond with a 4% coupon rate is 5 years until maturity and pays quarterly interest. If the annual market yield to maturity is 6%, what is the purchase price of the bond? [Given: \( 1.015^{-20} = 0.742 \)] [3 Marks]

Answer:
1. Quarterly values: \( n = 4 \times 5 = 20 \), \( i = \frac{6}{400} = 0.015 \), face value F = Rs. 1000.
2. Annual coupon = \( 1000 \times 4\% \) = Rs. 40, so quarterly coupon C = Rs. 10. Redemption price R = face value = Rs. 1000.
3. Purchase price \( P = C\left[ \frac{1 - (1 + i)^{-n}}{i} \right] + R(1 + i)^{-n} = 10\left[ \frac{1 - (1.015)^{-20}}{0.015} \right] + 1000(1.015)^{-20} \).
4. \( = 10\left[ \frac{1 - 0.742}{0.015} \right] + 1000 \times 0.742 = 172 + 742 = 914 \).
The purchase price of the bond is Rs. 914.

Teacher's Note:
a) Convert the coupon, the yield and the number of periods to quarterly values first.
b) The price is the present value of the coupons plus the present value of the redemption amount.
c) The price is below face value because the yield (6%) is more than the coupon rate (4%).

28. (A) The overall increase in a company's market share was 33.1%. Management reports the growth rate as a CAGR of 8.5%. Assuming the CAGR is accurate, how many years did it take for the market share to grow this much?
[Use \( \log(1.331) = 0.124 \) and \( \log(1.085) = 0.035 \)] [3 Marks]

Answer:
1. Let PV of the share be Rs. 100. Then final value \( = 100 + 0.331 \times 100 = 133.10 \).
2. \( CAGR = \left[ \left( \frac{FV}{PV} \right)^{\frac{1}{n}} - 1 \right] \times 100 \Rightarrow 8.5 = \left[ (1.331)^{\frac{1}{n}} - 1 \right] \times 100 \Rightarrow 1.085 = (1.331)^{\frac{1}{n}} \).
3. \( \log(1.085) = \frac{1}{n}\log(1.331) \Rightarrow n = \frac{\log(1.331)}{\log(1.085)} = \frac{0.124}{0.035} = 3.5 \) (approx.)
Required number of years = 3.5 years (approx.)

Teacher's Note:
a) A 33.1% increase means the final value is 1.331 times the starting value.
b) Take logarithms to bring \( n \) down from the power.

OR

(B) A company needs to accumulate Rs. 45,000,000 in \( n \) years to replace a fleet of vehicles. They decide to make annual deposits of Rs. 5,00,000 at the end of each year into a sinking fund. If the interest rate is 15% compounded annually, then in how many years will it take to reach the target?
[Use \( \log(14.5) = 1.161 \) and \( \log(1.15) = 0.061 \)] [3 Marks]

Answer:
1. S = Rs. 45,000,000, P = Rs. 5,00,000, \( r = 15\% \Rightarrow i = 0.15 \).
2. Sinking fund: \( S = P \times \frac{(1 + i)^n - 1}{i} \Rightarrow 45000000 = 500000 \times \frac{(1.15)^n - 1}{0.15} \).
3. \( 90 \times 0.15 = (1.15)^n - 1 \Rightarrow 13.50 = (1.15)^n - 1 \Rightarrow (1.15)^n = 14.5 \).
4. \( n\log(1.15) = \log(14.5) \Rightarrow n = \frac{1.161}{0.061} = 19.03 \) years (approx.)
So, the required number of years = 19.03 years (approx.)

Teacher's Note:
a) Deposits at the end of each year make this the amount of an ordinary annuity.
b) Divide 45000000 by 500000 first to get 90; this keeps the numbers small.

29. For the matrix \( A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \), show that \( A^3 - 7A - 6I = O \), where I is an identity matrix of order 2 and O is a zero matrix of order \( 2 \times 2 \). Hence find \( A^{-1} \). [3 Marks]

Answer:
1. \( A^2 = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} \)
2. \( A^3 = A^2 A = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix} \)
3. \( A^3 - 7A - 6I = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix} - \begin{bmatrix} 7 & 14 \\ 14 & 7 \end{bmatrix} - \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \)
4. Multiplying \( A^3 - 7A = 6I \) by \( A^{-1} \): \( A^{-1} = \frac{1}{6}(A^2 - 7I) = \frac{1}{6}\left( \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \right) = \frac{1}{6}\begin{bmatrix} -2 & 4 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \)

Teacher's Note:
a) Find \( A^{-1} \) from the given identity; the question says "Hence", so do not use the adjoint method.
b) Check: \( A A^{-1} = I \) confirms the answer.

30. In an examination, 2000 students appeared and the mean of the normal distribution of marks is 30 with standard deviation as 6.25. Find how many students are expected to score:
(i) between 20 and 40 marks?
(ii) less than 25 marks?
[Use \( P(0 \le Z \lt 1.6) = 0.4452 \) and \( P(0 \le Z \le 0.8) = 0.2881 \)] [3 Marks]

Answer:
Mean \( \mu = 30 \), standard deviation \( \sigma = 6.25 \).
(i) For \( X = 20,\ Z = \frac{20 - 30}{6.25} = -1.6 \); for \( X = 40,\ Z = \frac{40 - 30}{6.25} = 1.6 \).
\( P(20 \lt X \lt 40) = P(-1.6 \lt Z \lt 1.6) = 2P(0 \le Z \lt 1.6) = 2 \times 0.4452 = 0.8904 \)
Number of students \( = 2000 \times 0.8904 = 1780.8 \approx 1781 \) students.
(ii) For \( X = 25,\ Z = \frac{25 - 30}{6.25} = -0.8 \).
\( P(X \lt 25) = P(Z \lt -0.8) = 0.5 - P(0 \le Z \le 0.8) = 0.5 - 0.2881 = 0.2119 \)
Number of students \( = 2000 \times 0.2119 = 423.8 \approx 424 \) students.

Teacher's Note:
a) Use \( Z = \frac{X - \mu}{\sigma} \) and the symmetry of the normal curve about 0.
b) Number of students must be a whole number, so round the final answer.

31. A nutritionist claims that the average daily calorie intake of Indian teenagers is 2200 calories. To verify this claim, a random sample of 25 teenagers was selected, and their daily calorie intake was recorded. The sample mean was found to be 2080 calories with a sample standard deviation of 280 calories.
Test at the 5% level of significance whether the nutritionist's claim is valid. Assume that the calorie intake follows a normal distribution.
[Use \( t_{24}(0.025) = 2.064 \)] [3 Marks]

Answer:
1. \( \bar{x} = 2080,\ \mu = 2200,\ n = 25,\ s = 280 \).
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. \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{2080 - 2200}{280/\sqrt{25}} = \frac{-120}{56} = -2.143 \).
4. df = 24 and \( t_{24}(0.025) = 2.064 \). Since \( |t| = 2.143 \gt 2.064 \), the null hypothesis is rejected.
So, the nutritionist's claim is not valid.

Teacher's Note:
a) State both hypotheses clearly; each carries marks.
b) Compare the absolute value of t with the table value at \( n - 1 = 24 \) degrees of freedom.

SECTION - D (4 x 5= 20)

This section comprises of 4 long answer (LA) type questions of 5 marks each.

32. Under monopoly, the quantity sold and market price are determined by demand function \( p = 274 - x^2 \). If the marginal cost for a profit maximising monopolist is equal to \( 4 + 3x \), find the consumer's surplus. [5 Marks]

Answer:
1. Revenue \( R(x) = xp = x(274 - x^2) = 274x - x^3 \), so \( MR = R'(x) = 274 - 3x^2 \). Given \( MC = 4 + 3x \).
2. Let P be the profit. \( \frac{dP}{dx} = MR - MC = 274 - 3x^2 - 4 - 3x = 270 - 3x - 3x^2 \).
3. For maximum profit, \( \frac{dP}{dx} = 0 \Rightarrow 3(90 - x - x^2) = 0 \Rightarrow x = 9 \) or \( x = -10 \). But \( x \ne -10 \), so \( x = 9 \).
4. \( \frac{d^2P}{dx^2} = -3 - 6x \lt 0 \) at \( x = 9 \), so P is maximum at \( x = 9 \). When \( x_0 = 9 \), \( p_0 = 274 - 9^2 = 193 \).
5. C.S. \( = \int_{0}^{x_0} p\,dx - x_0 p_0 = \int_{0}^{9} (274 - x^2)\,dx - 9 \times 193 = \left[ 274x - \frac{x^3}{3} \right]_{0}^{9} - 1737 = 2223 - 1737 = 486 \).
The consumer's surplus is Rs. 486.

Teacher's Note:
a) A monopolist maximises profit where MR = MC; this gives the equilibrium quantity.
b) Check the second derivative to confirm a maximum.
c) \( \int_{0}^{9} (274 - x^2)\,dx = 2466 - 243 = 2223 \).

33. (A) A textile factory manufactures three types of fabrics: Cotton (C), Silk (S), and Wool (W). These fabrics must pass through three processing units: Weaving \( (U_1) \), Dyeing \( (U_2) \), and Packaging \( (U_3) \).
The per hour processing speeds of the units are as follows:
Unit \( U_1 \) (Weaving): 20 units of Cotton, 40 units of Silk, 20 units of Wool.
Unit \( U_2 \) (Dyeing): 25 units of Cotton, 50 units of Silk, 50 units of Wool.
Unit \( U_3 \) (Packaging): 20 units of Cotton, 20 units of Silk, 40 units of Wool.
The total available processing time for the units are 20 hours for Weaving \( (U_1) \), 12 hours for Dyeing \( (U_2) \), and 19 hours for Packaging \( (U_3) \).
Using Cramer's Rule, find the number of units of Cotton, Silk, and Wool that can be produced if the available time is fully utilized. [5 Marks]

Answer:
1. Let \( x, y, z \) be the units of Cotton, Silk and Wool produced.
For \( U_1 \): \( \frac{x}{20} + \frac{y}{40} + \frac{z}{20} = 20 \), i.e. \( 2x + y + 2z = 800 \)
For \( U_2 \): \( \frac{x}{25} + \frac{y}{50} + \frac{z}{50} = 12 \), i.e. \( 2x + y + z = 600 \)
For \( U_3 \): \( \frac{x}{20} + \frac{y}{20} + \frac{z}{40} = 19 \), i.e. \( 2x + 2y + z = 760 \)
2. \( D = \begin{vmatrix} 2 & 1 & 2 \\ 2 & 1 & 1 \\ 2 & 2 & 1 \end{vmatrix} = 2(1 - 2) - 1(2 - 2) + 2(4 - 2) = 2 \ne 0 \)
3. \( D_x = \begin{vmatrix} 800 & 1 & 2 \\ 600 & 1 & 1 \\ 760 & 2 & 1 \end{vmatrix} = 800(1 - 2) - 1(600 - 760) + 2(1200 - 760) = 240 \)
4. \( D_y = \begin{vmatrix} 2 & 800 & 2 \\ 2 & 600 & 1 \\ 2 & 760 & 1 \end{vmatrix} = 2(600 - 760) - 800(2 - 2) + 2(1520 - 1200) = 320 \)
5. \( D_z = \begin{vmatrix} 2 & 1 & 800 \\ 2 & 1 & 600 \\ 2 & 2 & 760 \end{vmatrix} = 2(760 - 1200) - 1(1520 - 1200) + 800(4 - 2) = 400 \)
6. \( x = \frac{D_x}{D} = 120,\ y = \frac{D_y}{D} = 160,\ z = \frac{D_z}{D} = 200 \).
Hence, the firm can produce 120 units of Cotton, 160 units of Silk and 200 units of Wool with the available time fully used.

Teacher's Note:
a) Time taken = units \( \div \) speed per hour; add the times for each unit and equate to the hours available.
b) Check in the first equation: \( 2(120) + 160 + 2(200) = 800 \).

OR

(B) "Cyber-Safe" has upgraded their encryption protocol. They now use a more complex Encoding Matrix (E) to encrypt the 3-letter keyword that unlocks their central server. The encryption follows the formula: C = E . P, where P is the column matrix of the letter positions (A = 1, B = 2, C = 3, ... , Z = 26).
You are an interceptor who has received an encrypted message matrix C and you have managed to crack the Encoding Matrix E.
The Encoding matrix E is given as \( E = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{bmatrix} \)
The intercepted encrypted matrix C is given as \( C = \begin{bmatrix} 52 \\ 36 \\ 35 \end{bmatrix} \)
Using matrix algebra, determine the original numerical matrix P. Also decode the numbers into letters to reveal the 3-letter secret code word. [5 Marks]

Answer:
1. \( P = E^{-1}C \), where \( P = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \).
2. \( |E| = 2(0 - 2) - 1(1 - 0) + 1(2 - 0) = -3 \ne 0 \), so \( E^{-1} \) exists.
3. \( E^{-1} = \frac{1}{|E|} adj\,E = \frac{1}{-3}\begin{bmatrix} -2 & 1 & 1 \\ -1 & 2 & -1 \\ 2 & -4 & -1 \end{bmatrix} \)
4. \( P = \frac{1}{-3}\begin{bmatrix} -2 & 1 & 1 \\ -1 & 2 & -1 \\ 2 & -4 & -1 \end{bmatrix}\begin{bmatrix} 52 \\ 36 \\ 35 \end{bmatrix} = \frac{1}{-3}\begin{bmatrix} -33 \\ -15 \\ -75 \end{bmatrix} \Rightarrow x = 11,\ y = 5,\ z = 25 \)
5. \( 11 \to K,\ 5 \to E,\ 25 \to Y \).
So, the secret code word is 'KEY'.

Teacher's Note:
a) Remember to transpose the cofactor matrix to get adj E.
b) Check: \( EP = \begin{bmatrix} 22 + 5 + 25 \\ 11 + 25 \\ 10 + 25 \end{bmatrix} = \begin{bmatrix} 52 \\ 36 \\ 35 \end{bmatrix} = C \).

34. A farmer can buy local goats at Rs. 400 each and hybrid goats at Rs. 600 each. A local goat produces milk worth Rs. 190 per week, while a hybrid goat produces milk worth Rs. 250 per week. The weekly maintenance cost of each goat is Rs. 60. The farmer has at most Rs. 12,000 to spend on buying goats and can maintain not more than 22 goats.
Formulate the Linear Programming Problem and find how many local goats and hybrid goats the farmer should buy to maximize the weekly profit? Also, find the maximum profit. [5 Marks]

Answer:
1. Let the number of local and hybrid goats be \( x \) and \( y \). Profit on a local goat = Rs. (190 - 60) = Rs. 130; profit on a hybrid goat = Rs. (250 - 60) = Rs. 190.
2. LPP: Maximize \( Z = 130x + 190y \), subject to \( 400x + 600y \le 12000 \) or \( 2x + 3y \le 60 \); \( x + y \le 22 \); \( x \ge 0,\ y \ge 0 \).
3. Graph: draw the lines \( 2x + 3y = 60 \) (through \( (30, 0) \) and \( (0, 20) \)) and \( x + y = 22 \) (through \( (22, 0) \) and \( (0, 22) \)). The feasible region is the bounded region on the origin side of both lines in the first quadrant, with corner points \( (0, 0), (22, 0), (6, 16) \) and \( (0, 20) \).
4. Corner points and values of Z:
O \( (0, 0) \): 0
\( (22, 0) \): 2860
\( (0, 20) \): 3800
\( (6, 16) \): 3820 (maximum)
5. Hence the maximum weekly profit of Rs. 3,820 is obtained when the farmer buys 6 local goats and 16 hybrid goats.

Teacher's Note:
a) Profit per goat = milk value - maintenance cost; the buying price enters only the budget constraint.
b) The corner \( (6, 16) \) comes from solving \( x + y = 22 \) and \( 2x + 3y = 60 \) together.
c) The marking scheme gives 2 marks for the correct graph and 1 mark for the corner point table.

35. (A) The following data represents the production of wheat in India (in Million Tonnes) over a period of 9 years. Due to climatic variations, the production fluctuates annually:
Year: 2015 | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 | 2023
Production (in Million Tonnes): 86 | 92 | 98 | 100 | 103 | 108 | 109 | 107 | 110
Calculate the trend values using the 4-Year Centred Moving Average method for the given data. [5 Marks]

Answer:
1. 4-yearly moving totals (placed between the 2nd and 3rd years of each group): 376, 393, 409, 420, 427, 434.
2. 4-yearly moving averages (totals \( \div \) 4): 94, 98.25, 102.25, 105, 106.75, 108.5.
3. Centred totals (sum of two adjacent moving averages): 192.25, 200.5, 207.25, 211.75, 215.25.
4. Centred moving averages (centred totals \( \div \) 2), which are the trend values:
2017: 96.125
2018: 100.25
2019: 103.625
2020: 105.875
2021: 107.625
5. No trend values can be found for 2015, 2016, 2022 and 2023.

Teacher's Note:
a) For an even period, the moving averages fall between two years, so they must be centred.
b) Marks: 4-yearly moving total 1, moving average 1½, centred total 1, centred moving average 1½.
c) Check the first total: \( 86 + 92 + 98 + 100 = 376 \).

OR

(B) The "Digital India" initiative has led to a massive surge in digital payment transactions across the country. The following data represents the volume of digital transactions (UPI, NEFT, IMPS, debit card etc.) in India (in Crores) for 6 consecutive years. The growth has been significant due to the push for a digital economy.
Year: 2018 | 2019 | 2020 | 2021 | 2022 | 2023
Volume (in crores): 2071 | 3134 | 4572 | 5554 | 8839 | 13462
(i) Fit a straight-line trend to the above data using the method of Least Squares.
(ii) Estimate the likely volume of transactions for the year 2025, assuming the current trend continues. [5 Marks]

Answer:
(i) Take \( X = \frac{x_i - 2020.5}{0.5} \).
Year \( (x_i) \): 2018 | 2019 | 2020 | 2021 | 2022 | 2023
Y: 2071 | 3134 | 4572 | 5554 | 8839 | 13462
X: -5 | -3 | -1 | 1 | 3 | 5
\( X^2 \): 25 | 9 | 1 | 1 | 9 | 25
XY: -10355 | -9402 | -4572 | 5554 | 26517 | 67310
\( n = 6,\ \sum Y = 37632,\ \sum X = 0,\ \sum X^2 = 70,\ \sum XY = 75052 \)
\( a = \frac{\sum Y}{n} = \frac{37632}{6} = 6272 \)
\( b = \frac{\sum XY}{\sum X^2} = \frac{75052}{70} = 1072.17 \) (approx.)
Required line: \( Y = a + bX = 6272 + 1072.17X \)
(ii) For 2025, \( X = \frac{2025 - 2020.5}{0.5} = 9 \).
Estimated trend value \( = 6272 + 1072.17 \times 9 = 15921.53 \) crores.

Teacher's Note:
a) With an even number of years, take the origin midway (2020.5) and use half-year units so that \( \sum X = 0 \).
b) The table carries 2½ marks, so show every column clearly.
c) Remember the origin (2020.5) and the unit (half year) when stating the trend line.

SECTION - E (3 x 4 = 12)

This section comprises of 3 case-study/passage-based questions of 4 marks each with sub parts. The first two case study questions have three sub parts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two sub parts of 2 marks each

36. Three pipes A, B and C can fill a tank in 12 hours, 15 hours and 20 hours respectively. Pipe A is kept open continuously. Pipes B and C in that order are open alternately for one hour each.
Based upon the above information, answer the following questions:

[Figure: Photograph of a concrete water tank with three pipes labelled A, B and C pouring water into it; pipe A is the widest with the largest flow, B is medium and C is the thinnest with a small flow.]

(i) Find the part of the tank filled in one hour when pipes A and C are open together. [1 Mark]

Answer: Part filled by A and C in 1 hour \( = \frac{1}{12} + \frac{1}{20} = \frac{5 + 3}{60} = \frac{2}{15} \).

Teacher's Note:
a) A pipe that fills a tank in n hours fills \( \frac{1}{n} \) of it in one hour.
b) Use the LCM 60 to add the fractions quickly.

(ii) Find the part of the tank filled in one hour when pipes A and B are open together. [1 Mark]

Answer: Part filled by A and B in 1 hour \( = \frac{1}{12} + \frac{1}{15} = \frac{5 + 4}{60} = \frac{3}{20} \).

Teacher's Note:
a) Add the one-hour work of the two pipes.
b) Simplify \( \frac{9}{60} \) to \( \frac{3}{20} \).

(iii) (A) Find the part of tank filled by the three pipes in the first 6 hours. [2 Marks]

Answer:
1. Pipe A is always open, and B and C are open alternately for 1 hour each.
2. Part filled in 2 hours \( = \frac{1}{12} + \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = \frac{2}{12} + \frac{1}{15} + \frac{1}{20} = \frac{17}{60} \).
3. Part filled after three full cycles (6 hours) \( = 3 \times \frac{17}{60} = \frac{17}{20} \).

Teacher's Note:
a) Treat every 2 hours as one cycle: A with B, then A with C.
b) 6 hours = 3 complete cycles.

OR

(iii) (B) Find the part of the unfilled tank at the end of 7 hours. [2 Marks]

Answer:
1. Part filled in 2 hours \( = \frac{1}{12} + \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = \frac{17}{60} \).
2. Part filled after three full cycles (6 hours) \( = 3 \times \frac{17}{60} = \frac{17}{20} \).
3. Part left to be filled \( = 1 - \frac{17}{20} = \frac{3}{20} \).
4. In the 7th hour, pipes A and B fill \( \frac{1}{12} + \frac{1}{15} = \frac{3}{20} \), which is exactly the unfilled part.
So, the tank is completely filled at the end of 7 hours and the unfilled part is 0.

Teacher's Note:
a) The 7th hour starts a new cycle, so pipe B (not C) is open with A.
b) Compare the part filled in the 7th hour with the part left after 6 hours.

37. The cost of building an office block, \( x \) floors high, consists of three components:
Rs. 80 crores for the land
Rs. 2 crores per floor
Extra structural costs Rs. \( (0.08x) \) crores for each floor, where \( x \) is the total number of floors.
Based on the above information, answer the following questions:

(i) Find the total cost of building the office block \( x \) floors high. [1 Mark]

Answer: Total cost \( TC(x) = 80 + 2x + (0.08x)x = 80 + 2x + 0.08x^2 \) crores.

Teacher's Note:
a) The extra structural cost is \( 0.08x \) for each of the \( x \) floors, so it is \( 0.08x^2 \) in total.
b) The land cost of 80 crores is a fixed cost.

(ii) Find the average cost per floor in terms of \( x \). [1 Mark]

Answer: Average cost \( AC(x) = \frac{\text{Total cost}}{x} = \frac{80}{x} + 2 + 0.08x \) crores.

Teacher's Note:
a) Average cost per floor = total cost \( \div \) number of floors.
b) Divide each term of \( TC(x) \) by \( x \).

(iii) (A) Find, how many floors the office block should have if the average cost per floor is to be minimized?
[Use \( \sqrt{10} = 3.2 \)] [2 Marks]

Answer:
1. \( AC(x) = \frac{80}{x} + 2 + 0.08x \Rightarrow AC'(x) = -\frac{80}{x^2} + 0 + 0.08 = \frac{0.08x^2 - 80}{x^2} \).
2. For minima, \( AC'(x) = 0 \Rightarrow x^2 - 1000 = 0 \Rightarrow x = \pm 10\sqrt{10} \). Since \( x \ne -10\sqrt{10} \), \( x = 10\sqrt{10} = 10(3.2) = 32 \).
3. \( AC''(x) = \frac{160}{x^3} \gt 0 \) for all \( x \gt 0 \).
So, for minimum average cost per floor, 32 floors should be built.

Teacher's Note:
a) Minimise the average cost, not the total cost.
b) Show the second derivative test to confirm a minimum.

OR

(iii) (B) Find, for how many floors the average cost keeps decreasing?
[Use \( \sqrt{10} = 3.2 \)] [2 Marks]

Answer:
1. \( AC(x) = \frac{80}{x} + 2 + 0.08x \Rightarrow AC'(x) = \frac{0.08x^2 - 80}{x^2} \).
2. For decreasing, \( AC'(x) \le 0 \Rightarrow x^2 - 1000 \le 0 \) (as \( x^2 \gt 0 \) for all \( x \)).
3. \( (x - 10\sqrt{10})(x + 10\sqrt{10}) \le 0 \Rightarrow -10\sqrt{10} \le x \le 10\sqrt{10} \).
4. Since \( x \gt 0 \) (number of floors cannot be negative), \( 0 \lt x \le 10\sqrt{10} \Rightarrow 0 \lt x \le 32 \).
So, for integral values of floors, the average cost decreases till the building of 32 floors.

Teacher's Note:
a) A function decreases where its derivative is negative (or zero at the end point).
b) Use the given value \( \sqrt{10} = 3.2 \) to get a whole number of floors.

38. A couple wishes to purchase a house for Rs. 15,00,000 with a down payment of Rs. 4,00,000. If they amortize the balance at an interest rate of 9% per annum compounded monthly for 10 years.

(i) Find the monthly installment using the flat rate method. [2 Marks]

Answer:
1. Loan amount P = Rs. (15,00,000 - 4,00,000) = Rs. 11,00,000; \( i = \frac{9}{12 \times 100} = 0.0075 \); \( n = 10 \times 12 = 120 \).
2. Under flat rate, EMI \( = P\left( i + \frac{1}{n} \right) = 1100000\left( 0.0075 + \frac{1}{120} \right) \).
3. \( = 1100000 \times 0.0075 + \frac{1100000}{120} = 8250 + 9166.67 = 17416.67 \).
The monthly installment is Rs. 17,416.67 (the marking scheme rounds \( \frac{1}{120} \) to 0.0083 and gets Rs. 17,380).

Teacher's Note:
a) Flat rate: interest is on the full loan for the whole period, \( 1100000 \times 0.09 \times 10 = 990000 \), and EMI \( = \frac{1100000 + 990000}{120} = 17416.67 \).
b) The marking scheme writes \( 1100000 \times 0.0158 = 17380 \); the exact value is Rs. 17,416.67, as \( \frac{1}{120} = 0.008333... \)

(ii) Find the monthly installment using the reducing balance method.
[Use \( (1.0075)^{120} = 2.451 \) and \( \frac{2.451}{1.451} = 1.69 \)] [2 Marks]

Answer:
1. EMI \( = \frac{P \times i \times (1 + i)^n}{(1 + i)^n - 1} = \frac{1100000 \times 0.0075 \times (1.0075)^{120}}{(1.0075)^{120} - 1} \).
2. \( = \frac{1100000 \times 0.0075 \times 2.451}{2.451 - 1} = 1100000 \times 0.0075 \times 1.69 = 13942.50 \).
The monthly installment is Rs. 13,942.50.

Teacher's Note:
a) Use the given value \( \frac{2.451}{1.451} = 1.69 \) directly to avoid long division.
b) The reducing balance EMI is lower than the flat rate EMI because interest is charged only on the outstanding balance.