Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Informatics Practices Term 2 Set 1 with Solutions
Time: 2 Hours
Maximum Marks: 40
General Instructions:
- The question paper is divided into 3 sections -A, B and C
- Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
- Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
- Section C comprises of 4 questions of 4 marks each. It contains one case study based question. Internal choice has been provided in one question.
Section – A (2 Marks)
Question 1.
The marginal revenue function for a commodity is given by MR = 9 + 2x – 6x2. Find the demand function.
OR
The marginal cost of producing x pairs of tennis shoes is given by the following function:
MC = 50 + \(\frac{300}{x+1}\)
If the fixed cost is ₹ 2000, then find the total cost function.
Answer:
Given, MR = 9 + 2x – 6x2
Let the demand function be p,
We know that,
TR =f(x + 2x – 6x2)dx
TR = 9x + x2 – 2x3+ C
where C is Integration Constant
When x = 0, TR = 0, so C = 0
TR = 9x + x2 – 2x3 = px
⇒ px = 9x + x2 – 2x3
⇒ p = 9 + x – 2x2
which is the required demand function.
OR
Let the total cost function be TC
TC = ∫(50 + \(\frac{300}{x+1}\))dx
TC = 50x + 300log|x + 1| + C
If x = 0, TC = ₹ 2000
So 2000 = 300(log 1) + C
⇒ C = 2000
So TC = 50x + 300log(x + 1) + 2000
Hence, the required total cost function
TC = 50x + 300log(x + 1) + 2000
Question 2.
Find the present value of a perpetuity of ₹ 600 at end of each quarter, if money is worth 8% compounded quarterly.
Answer:
Let the present value of perpetuity be P
Given, R = ₹ 600
i = \(\frac{0.08}{4}\) = 0.02
Present value of perpetuity = P = \(\frac{R}{i}\)
⇒ P = \(\frac{600}{0.02}\) = ₹ 30.000
Question 3.
What effective rate is equivalent to a nominal rate of 8% per annum compounded quarterly?
OR
Find the present value of an annuity of ₹ 1000, payable at the end of each year for 5 years, if money is worth 6%compounded annually. [Given (1.06)-5 = 0.7473]
Answer:
Let the effective rate be r(subscript){eff}
Since, reff = (1 + \(\frac{r}{m}\))m – 1
(Given r = 0.08)
= (1 + \(\frac{0.08}{4}\))4 – 1
= (1.02)4 – 1
= 0.0824 or 8.24%
So effective rate is 8.24% compounded quarterly.
OR
Present value of ordinary annuity
Question 4.
A sampling distribution of the sample means X̄ is formed from a population with mean weight μ = 60 kg and standard deviation σ = 9kg. What is the expected value and standard deviation of X̄ t if sample size is 36?
Answer:
Let the expected value be E, standard deviation be SD and sample size be n
√n = E(X̄) = 6kg
Standard deviation of
X̄ = SD(X̄)
= \(\frac{\sigma}{\sqrt{n}}=\frac{9}{6}\)
= 1.5kg
Question 5.
Find the trend values using 3 yearly moving average for the loans sanctioned to farmers by a particular branch of a bank in a village.
The data is given as follows:
Answer:
Question 6.
The feasible region of the L.PP Min Z = 3x + 2y subject to constraints
2x + y ≥ 6,x – y ≥ 0, x ≥ 0, y ≥ 0 is given below:
Determine the optimal solution Justify your answer.
Answer:
The feasible region is given as follows:
The values at the corner points P and Q can be shown as:
| Corner points | Z = 3x + 2y |
| P (2, 2) | 10 |
| Q (3, 0) | 9 |
The smallest value of Z is 9. Since, the feasible region is unbounded, we draw the graph of 3x + 2y < 9. The resulting open half plane has points common with feasible region, therefore Z = 9 is not the minimum value of Z. Hence, the optimal solution does not exist.
Section – B (3 marks each)
Question 7.
The supply function for a commodity is given by p = x2 + 4x + 3, where x is the quantity supplied at the price p. Find the producer’s surplus, when the price of the commodity is ₹ 48.
Answer:
Substituting, p0 = ₹ 48 in p = x2 + 4x + 3 We get x0 = 5
Question 8.
The following table shows the quarterly sales (in X crore) of a real estate company. Compute the trend by quarterly moving averages.
OR
Fit a straight line trend by the method of least squares and also estimate the trend for the year 2023.
Answer:
The trend value are given by 4 quarterly centered moving average.
OR
a = \(\frac{\Sigma Y}{n}=\frac{532}{7}\) = 76
b = \(\frac{\Sigma X Y}{\Sigma X^{2}}=\frac{672}{28}\) = 24
YC = a + bX, YC = 76 + 24X
Estimated sales = YC for 2023 = 76 + 24 × 6 = ₹ 220 lacs
Question 9.
A machine produces washers of thickness 0.50 mm. To determine whether the machine is in proper working order, a sample of 10 washers is chosen for which the mean thickness is 0.53 mm and the standard deviation is 0.03 mm. Test the hypothesis at 5% level of significance so that the machine is working in proper order. [Given critical value, t0.025 = 2.262 at v(d.f) = 9]
Answer:
Define Null hypothesis H0 and alternate hypothesis Hx as follows:
H0: μ = 0.50 mm
H1 : μ ≠ 0.50 mm
Thus, a two-tailed test is applied under hypothesis H0,
We have,
t = \(\frac{\bar{X}-\mu}{S} \sqrt{n-1}\)
= \(\frac{0.53-0.50}{0.03}\) × 3 = 3
Since, the calculated value of t = 3 does not lie in the interval – t0.025 to t0.025 i.e., – 2.262 to 2.262 for 10 – 1 = 9 degree of freedom. So we reject H0 at 0.05 level. Hence, we conclude that machine is not working properly.
Question 10.
A person invested ₹ 15000 in a mutual fund and the value of investment at the time of redemption was ₹ 25000. If CAGR for this investment is 8.88%, then calculate the time period for which the given amount was invested? [Given log(1.667) = 0.2219 & log(1.089) = 0.037]
Answer:
We Know
CAGR = \(\left[\left(\frac{F V}{I V}\right)^{\frac{1}{n}}-1\right]\) × 100
where, IV = Initial value of investment
FV = Final value of investment
⇒ 8.88 = \(\left[\left(\frac{25000}{15000}\right)^{\frac{1}{n}}-1\right]\) × 100
⇒ 0.0888 = (\(\frac{5}{3}\))1/n – 1
⇒ 1.089 = (1.667)1/n
⇒ \(\frac{1}{n}\)log(1 .667) = log(1 .089)
⇒ n(0.037) = 0.2219
⇒ n = 5.99 ≈ 6 years
Section – C (4 marks each)
Question 11.
S & D chemicals produces two products, an alkaline solution and a base oil that are sold as raw material to the companies manufacturing soaps and detergents. On the basis of current inventory levels and estimated demand for the coming month, S & D’s management has decided that combined production of alkaline solution and the base oil must be at least 3500 gallons. S & D chemicals are also committed to supply 1250 gallons of alkaline solution to one of its major customer. The alkaline solution and base oil requires respectively 2 hours and 1 hour of processing time per gallon. The total processing time available for the coming month is 6000 hours. The production cost is ₹ 200 per gallon for the alkaline solution and ₹ 300 per gallon for the base oil.
Formulate the above as an L.EP and solve it by using graphical method, to help S & D chemicals for determining the minimum production cost.
Answer:
Let the company produces x and ij gallons of alkaline solution and base oil respectively, also
let C be the production cost.
Min C = 200x + 300y
Subject to constraints:
x + y ≥ 3500 …(1)
x ≥ 1250 …(2)
2x + y ≤ 6000 …(3)
x,y ≥ 0
| Corner points | C = 200x + 300y |
| P (1250,2250) | ₹ 9,25,000 |
| Q (1250, 3500) | ₹ 13,00,000 |
| R (2500,1000) | ₹ 8,00,000 |
Minimum cost is 8,00,000 when 2500 gallons of alkaline solutions & 1000 gallons of base oil are manufactured.
Question 12.
A machine costing ₹ 50,000 is to be replaced at the end of 10 years, when it will have a salvage value of ₹ 5000. In order to provide money at that time for a machine costing the same amount, a sinking fund is set up. If equal payments are placed in the fund at the end of each quarter and the fund earns at the rate of 8% compounded quarterly, then what should each payment be? [Given (1.02)40 = 2.208]
Answer:
The amount of sinking fund S at any time is given by
S = R\(\left[\frac{(1+i)^{n}-1}{i}\right]\)
where R = periodic payment,
i = Interest per period,
n = number of payments
S = Cost of machine – Salvage value
= ₹ 50,000 – ₹ 5000 = ₹ 45,000
i = \(\frac{8 \%}{4}\) = 0.02
⇒ 45000 = R\(\left[\frac{(1+0.02)^{40}-1}{0.02}\right]\)
⇒ 45000 = R\(\left[\frac{2.208-1}{0.02}\right]\)
⇒ R = \(\frac{900}{1.208}\)
⇒ R = ₹ 745.03
Question 13.
A couple wishes to purchase a house for ₹ 15,00,000 with a down payment of ₹ 4,00,000. If they can amortize the balance at an interest rate of 9% per annum compounded monthly for 10 years, then find the monthly installment (EMI). Also find the total interest paid. [Given (1.0075)-120 = 0.4079 ]
OR
A X 2000,8% bond is redeemable at the end of 10 years at ₹ 105. Find the purchase price to yield 10% effective rate. [Given (1.1)-10 = 0.3855]
Answer:
P = Cost of house – Cash down payments
P = ₹ 15,00,000 – ₹ 4,00,000 = ₹ 11,00,000
Total interest paid = nR – R
= 13933.5 × 120 – 11,00,000 = ₹ 5,72,020
OR
Face value of bond, F = ₹ 2000
Redemption value C = 1.05 × 2000 = ₹ 2100
Nominal rate = 8%
R = C × id = 2000 × 0.08 = ₹ 160
Number of periods before redemption i.e., n = 10
Annual yield rate, i =10% or 0.1
Purchase price,
V = R\(\left[\frac{1-(1+i)^{-n}}{i}\right]\) + C(1 + i)-n
= 160\(\left[\frac{1-(1+0.1)^{-10}}{0.1}\right]\) + 2100(1 + 0.1)-10
= 160 × 6.14 + 2100 × 0.3855
= ₹ 1792
Thus, present value of the bond is ₹ 1792
Case Study
Question 14.
General anaesthesia is used for major operations to cure the patients and conduct pain free surgeries. Propofol is a commonly used anaesthetic injected for major operations such as knee replacement or open heart surgery. It also acts as a sedative and an analgesic.
A patient is rushed to operation theatre for a 2-hour cardiac surgery. A person is anaesthetized when its blood stream contains at least 3 mg of propofol per kg of body weight. The rate of change of propofol (x), in the body is proportional to the quantity of propofol present at that time. Based on the above information. Answer the following questions:
(i) Show that the propofol given intravenously is eliminated exponentially from the patients’ blood stream. (2)
(ii) What dose of propofol should be injected, to induce unconsciousness in a 50 kg adult for a two hours operation? (2)
(Given (2)1/5 = 1.1487 & assume half-life of propofol = 5 hours )
Answer:
∵ \(\frac{d x}{d t}\) ∝ x,
\(\frac{d x}{d t}\) = -kx
⇒ ∫\(\frac{d x}{d t}\) = ∫-kdt
⇒ log x = -kt + C
⇒ x = e-kt+c
⇒ x = λ-kt where e0 = λ
(i) Let x = x0 at t = 0
∵ x0 = λ
x = x0 e-kt
where x0 = original quantity
(ii) Let the half life be t and amount of propofol be x0
x = x0 e-kt …(i)
Now, \(\) = x0 e-5k ( half life = 5 hours)
⇒ e-5k = \(\frac{1}{2}\)
⇒ ek = 21/5
The quantity of propofol needed in a 50 kg adult at the end of
2 hours = 50 × 3 = 150 mg
⇒ 150 = x0e-2k [using ..(i)]
⇒ x0 = 150 (ek)2
⇒ x0 = 150(21/5)2 = 150 × 1.3195
x0 = 197.93mg
So, 197.53 mg of propofol is needed.