CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 5 with Solutions

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Informatics Practices Term 2 Set 5 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 cost of producing x units of a book set of class XII is given by MC = 150 + \(\frac{1}{x+3}\). If the cost is ₹ 2200, find the total cost function, [log 3 = 0.4771]
OR
Evaluate: ∫\(\frac{d x}{5-8 x-x^{2}}\)
Answer:
We know that,
Cost Function, C(x) = ∫MC(c)dx where MC is
Marginal Cost
∴ C(x) = ∫(15o+ \(\frac{1}{x+3}\))dx
= ∫150 dx + ∫\(\frac{1}{x+3}\)dx
= 150x + log |x +3| + C

If x = 0, C(x) = 2200
∴ 2200 = 0 + log |0 + 3|+ C
⇒ 2200 = 0.4771 +C
⇒ C = 2199.5
Thus, C(x) = 150x + log |x + 3| + 2199.5

OR

Common Made Error:
Some candidates make errors in simplifying the expression before integration.

Answering Tip:
When denominator is given in quadratic form, then make it perfect squaie then apply the property of integral.

Question 2.
The value of a machine purchased two years ago, depreciates at the annual rate of 10%. If its present value is ₹ 50,000, then find its value after 2 years.
OR
A company buys a machine at a cost of ₹ 5000. The company decides on a salvage value of ₹ 1000 and a useful life of 5 years, find depreciation.
Answer:
Given, P = ₹ 50,000, i = 10%p.a.
⇒ i = \(\frac{10}{100}\) = 0.1
So, value after 2 years = 50,000 × (1 – 0.1)2
= 50,000 × 0.81
= ₹ 40500

OR

Since,
Annual Depreciation expense
= \(\frac{(\text { Asset cost }-\text { Residual Value })}{\text { Useful life of the asset }}\)
= \(\left(\frac{₹ 5000-₹ 1000}{5}\right)\)
= ₹ 800

Question 3.
Mr. Roy purchased a home for ₹ 200,000 in 2012. In the next few years, homes in neighbourhood have been selling well due to the new shopping plaza a couple of miles away, which increased the market value of his home. So in 2017, he decided to down size and sell his home. Based on the current market value during this time, he was able to sell his home for ₹ 250,000. Find his rate of return.
Answer:
Given, Current value = 250,000
Original value = 200,000
Rate of return
= \(\frac{\text { Current Value-Original Value }}{\text { Original Value }}\) × 100
= \(\frac{250,000-200,000}{200,000}\) × 100

Question 4.
A sample of 100 students is chosen from a large group of students. The average height of these students is 162 cm and standard deviation (S.D.) is 8 cm. Obtain the standard error for the average height of large group of students of 160 cm?
Answer:
Given n = 100, x̄ = 162 cm, s = 8 cm is known in this problem
Since, σ is unknown, so we consider \(\widehat{\mathrm{σ}}\) = s and Φ = 160 cm

Therefore the standard error for the average height of large group of students of 160 cm is 0.8.

Question 5.
The wages of the certain factory workers are given as below. Using 3 yearly moving average indicate the trend in wages.

Answer:
Calculation of Trend Values by method of 3 yearly Moving Average

Question 6.
The feasible region for an L.EE is shown in the following figure. The CB is parallel to OA.

Write down the constraints for L.EE
Answer:
The required constrains for L.EF are
y ≥ 2x
y – 2x ≤ 2
x ≤ 5
x ≥ 0, y ≥ 0

Section – B (3 marks each)

Question 7.
Find the particular solution of the differential equation log\(\left(\frac{d y}{d x}\right)\) = 3x + 4y, given that y = 0, when x = 0.
Answer:
Given differential equation can be written as
\(\frac{d y}{d x}\) = e^{(3x + 4y)} = e^3x.e^4y
∴ ∫e^-4y = ∫e^3x dx
or
\(\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}\) + C

∴ 4e^3x + 3e^-4y + 12C = 0
Taking x = 0, y = 0, we get
C = \(\frac{-7}{12}\)
∴The Solution is 4e^3x + 3e^-4y – 7 = 0

Question 8.
The following table shows the quarterly sales (in ₹ crore) of a real estate company. Compute the trend by quarterly moving averages.

Or
Fit the straight line trend to the following time series data:

Also, tabulate the trend values.
Answer:

Or

Now, a = \(\frac{\Sigma y}{n}=\frac{439}{5}\) = 87.8
b = \(\frac{\Sigma x y}{\Sigma x^{2}}=\frac{21}{10}\) = 2.1

Hence, trend equation is y_t = 87.8 + 2.1x
Y₂₀₁₇= 87.8 + 2.1 (-2) = 83.6
Y₂₀₁₈ = 87.8 + 2.1 (- 1) = 85.7
Y₂₀₁₉ = 87.8 + 2.1 (0) = 87.8
Y₂₀₂₀ = 87.8 + 2.1 (1) = 89.9
Y₂₀₂₁ = 87.8 + 2.1 (2) = 92.0

Question 9.
The heights of six randomly chosen red roses are in cm 63,65,68,69, 71 and 72. Those of ten randomly chosen yellow roses are 61, 62, 65, 66, 69, 70, 71, 72 and 73. Find f-statistics for the given data. Given that mean for first data set is 68 and mean for second data set is 67.8.
Answer:
Construct the following table for standard errors:

where, x̄ = 68 and ȳ = 67.8
Now, compute the standard error, s using formula:

Now using t-test formula:

= 0.0512 × 1.936
= 0.099
Hence, t-test value for the two data sets is 0.099.

Commonly Made Error: Calculation errors are found in the standard error table. Sometimes students apply wrong formula for standard error.

Answering Tip: Learn the all formula thoroughly.

Question 10.
Aritra invested ₹ 1,000 in a fund for three years. While the total Net Asset value (N.A.V) remained ₹ 1,000 for the first year, it increased to ₹ 1,100 in the second year. Upon maturity of this fund, final N.A.V. stood at ₹ 1,300. Find compounded annual growth rate. Given that (1.3)^1/3 = 1.0913.
Answer:

Year	NAV (in ₹)
I	1000
II	1100
III	1300

C.A.G.R = \(\left(\frac{\text { Final value }}{\text { Initial value }}\right)^{1 / n}\) – 1
= \(\left(\frac{13,00}{1,000}\right)^{\frac{1}{3}}\) – 1 = 1.0913 – 1
= 0.0913 = 9.13%

Section – C (4 marks each)

Question 11.
A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

One kg of food X costs ₹ 24 and one kg of food Y costs ₹ 36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.
Answer:
Let x be the number of units of food X and y be the number of units of food Y be mixed to obtain the desired diet. Then L.P.P. of the given problem is:
Minimise, Z = 24x + 36y
Subject to the constraints,
x + 2y ≥ 10, 2x + 2y ≥ 12
i.e., x + y ≥ 6 and 3x + y ≥ 8
x ≥ 0, y ≥ 0
We draw the lines x + 2y = 10, x + y = 6, 3x + y = 8 and obtain the feasible region (unbounded and convex) as shown in the figure.

Thus, corner points are A(0,8), B(1, 5), C(2,4) and D(10,0).
The values of Z(in ₹) at these points are given in the following table:

Corner Point	Objective Function Z = 2A× + 36y
A(0, 8)	Z = 24 × 0 + 36 × 8 = 288
B(1, 5)	Z = 24 × 1 + 36 × 5 = 204
C(2, 4)	Z = 24 × 2 + 36 × 4 = 192(Min.)
D(10, 0)	Z = 24 × 10 + 36 × 0 = 240

As the feasible region is unbounded, we draw half plane 24x + 36y < 192 i.e., 2x + 3y < 16 and note that there is no point common with the feasible region. The minimum value of Z is ₹ 192. It occurs at C(2,4).
i.e., when 2 kg of food X and 4 kg of food Y are mixed to get the desired diet.

Commonly Made Error: Many candidates took incorrect inequality sign, hence they got incorrect feasible region and their corner points were also incorrect. Some candidates did not show any graphical representation of the inequalities. In some cases, the representation of the problem was not up to the mark and the work was not systematic resulting in candidates missing the point of minimum cost.

Answering Tip: Students should practice the sketching of lines. They should practice to express the line in intercept form i.e., \(\frac{x}{a}+\frac{y}{b}\) = 1, so that sketching will be easy. Correct feasible region and its plotting is important.

Question 12.
Ram took a home loan of ₹ 50 lakhs at an 8.50 % interest for a 20 years loan tenure. What would be his E.M.I. ? Given (1.0070)²⁴⁰ = 5.3342
Answer:

Question 13.
Find the purchase price of a ₹ 800, 7% bond, dividends payable semi-annually redeemable at par in 5 years, if the yield rate is to be 7% compounded semi-annually.
Given (1.035)^-10 = 0.7089
OR
(i) What sum of money invested now could establish a scholarship of ₹ 5000 which is to be awarded at the end of every year forever, if money is worth 8% per annum.
(ii) A bond that matures in 3 years has coupon rate of 10% per annum and has a face value of ₹ 1000.
Find the fair value of bond if the yield to maturity is 5%.
[Given, (1.05)^-3 = 0.8638]
Answer:
Face value of the bond, C = ₹ 800 the graph of the
Nominai rate Gf interest, i = 7% or 0.07
As dividends are paid semi-annually
Therefore, Rate of interest per period
i_d = \(\frac{0.07}{2}\) = 0.035
Therefore, periodic dividend payment,
R = C × i_d
= 800 × 0.035 = 28
So, semi-annual dividend R is ₹ 28.
Yield rate is 7% = 0.07, compounded semi-annually
∴ i = \(\frac{0.07}{2}\) = 0.035
No. of years n = 5
∴ no. of dividend periods n = 5 × 2 = 10
Purchase price (V) of the bond is given by

Thus, purchase price of bond is ₹ 567.35

OR

(i) It is a flat perpetuity, so
Present value of perpetuity
= \(\frac{\text { Cash flow }}{\text { Interest rate }}\)
= \(\frac{5000}{\frac{8}{100}}\)
= \(\frac{5000 \times 100}{8}\)
= ₹ 62500

(ii) Given P = ₹ 1000, r = 10% per annum, N = 3 years, i = \(\frac{5}{100}\) = 0.05
So, coupon rate = ₹ 1000 × \(\frac{10}{100}\) = ₹ 100

Case Study

Question 14.
The second new species named Puntius euspilurus is an edible freshwater fish found in the Mananthavady river in Wayanad. The epithet euspilurus is a Greek word referring to the distinct black spot on the caudal fin. The slender bodied fish prefers fast flowing, shallow and clear waters and occurs only in unpolluted areas. It appears in great numbers in paddy fields during the onset of the Southwest monsoon.

Suppose that the supply schedule of this Fish is given in the table below which follows a linear relationship between price and quantity supplied.

Price P per kg (in ₹)	Quantity (X) of Fish supplied (in kg)
.25	800
20	700
15	600
10	500
5	400

Suppose that this Fish can be sold only in the Kerala. The Kerala demand schedule for this Fish is as follows and there is a linear relationship between price and quantity demanded.

Price P per kg (in ₹)	Quantity (X) of Fish Demanded (in kg)
25	200
20	400
15	600
10	800
5	1000

(i) Find the equations of supply and demand curves. (2)
Answer:
From the given table of price and supply, we have the following price and supply coordinates i.e., (25,800), (20,700), (15,600), (10,500), (5,400)
All these coordinates, satisfy the equation
p = – 15 + \(\frac{x}{20}\)
From the given table of price and demand, we have the following price and demand coordinates, i.e., (25, 200), (20, 400), (15, 600), (10,800), (5,1000)
All of these coordinates, lie on the equation
p = 30 – \(\frac{x}{40}\)

(ii) What is the value of x at equilibrium ? Also, find the equilibrium. (2)
Answer:
At equilibrium, Demand = Supply
Therefore, 30 – \(\frac{x}{40}\) = – 15 + \(\frac{x}{20}\)
⇒ 30 + 15 = \(\frac{x}{20}+\frac{x}{40}\)
⇒ 45 = \(\frac{2 x+x}{40}\)
⇒ 3x = 45 × 40
⇒ x = 600
At x = 600
P= 30 – \(\frac{600}{40}\)
= 30 – 15
p = 15