Raju

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 11 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 11 with Solutions

Maximum Marks : 40
Time : 2 Hours

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. It contains one case study based question. Internal choice has been provided in one question.

Section – A [12 Marks]

Question 1.
Evaluate ∫\(\frac{d x}{x\left(x^{3}+8\right)}\)
Or
The supply function of a producer is given by p = \(\frac{2}{5}\)e^2x, where x denotes thousand units.
Find producer’s surplus when sales are 2000 units.
Answer:
Let I = ∫\(\frac{d x}{x\left(x^{3}+8\right)}\) = ∫\(\frac{x^{2}}{x^{3}\left(x^{3}+8\right)}\)dx
Now, put x³ + 8 = t

[adding and substracting by (4)² in denominator]

Or

Given, the supply function, p = \(\frac{2}{5}\)e^2x
When sales are 2000 units i.e. x₀ = 2, then we get
p₀ = \(\frac{2}{5}\)e⁴ [when x = 2 then p = latex]\frac{2}{5}[/latex]e⁴]

Question 2.
Find a linear trend to the following data by the least squares method,

Or
Find seasonal variations by ratio-to-trend method from the data given below.

Answer:
Here, n = 5 is odd.
Hence, middle year i.e. 1989 taken as origin.

We construct the table as

Let trend line of y (production) on x be
y_t = a + bx (origin 1989) ………(i)

The normal equation for estimating a and b in Eq. (i) are Σy = na + b Σx
⇒ 103 = 5a + b × 0
⇒ 5a = 103 ⇒ a = \(\frac{103}{5}\) = 20.6
and Σxy = aΣx + bΣx²
26 = a x O + b x 40
b = \(\frac{26}{40}\) = 0.65

The linear trend
y_t = a + bx
y_t = 20.6 + 0.65x

Or

For determining seasonal variation by ratio-to-trend method, first we will determine the trend of yearly data and then convert it to quarterly data.
Table for calculating trend values

The equation of the straight line trend is y_t = a + bx.
Since, Σx = 0, a = \(\frac{\Sigma y}{n}=\frac{280}{5}\) = 56
and b = \(\frac{\Sigma x y}{\Sigma x^{2}}=\frac{120}{10}\) = 12
Quarterly increment = \(\frac{12}{4}\) = 3.

Question 3.
What sum of money invested, now would establish a scholarship of ₹ 20000 to be awarded at the starting of each year to a deserving student, if money is worth 8% compounded annually?
Answer:
Since, the payments are made at the starting of each year forever, it is a perpetuity due. Periodic payment (R) = ₹ 20000 Interest is compounded annually, so i = 8% = 0.08
Present value (P) of a perpetuity due is given as, P= R(1 + \(\frac{1}{i}\)) ⇒ P = 20000(1 + \(\frac{1}{0.08}\))
⇒ P = 20000 × 13.5 = 270000

∴ Amount to be invested is ₹ 270000.

Question 4.
A desktop computer costing ₹ 60000 has a useful life of 18 yr. If annual depreciation is ₹ 3000, find its scrap value, using linear method.
Answer:
Given, original cost of desktop computer = ₹ 60000
Useful life = 18 yr
Annual depreciation = ₹ 3000
Let the scrap, value of the desktop computer be ₹ S. Then, using annual depreciation
= \(\frac{\text { Original cost }-\text { Scrap value }}{\text { Useful life in years }}\)
3000 = \(\frac{60000-S}{18}\)
⇒ 54000 = 60000 – S ⇒ S = 6000
Hence, the scrap value of desktop computer is ₹ 6000.

Question 5.
Consider the following hypothesis test.
H₀ : µ ≤ 30
H₁ : µ > 30
A sample of 45 provided a sample mean of 32.4. The population standard deviation is 6.
(i) Compute the value of the test statistic.
(ii) What is the p-value?
Answer:
Given, µ₀ = 30, n = 45, x̄ = 32.4, σ = 6 and α = 0.01
(i) Test statistics,

(ii) ∵ Z = 2.68 > 0
So, p-value of 2.68 = Area under the standard normal curve to the right of Z
= 1 – 0.9963 (using table) = 0.0037
∴ p-value = 0.0037

Question 6.
Find the maximum value of Z = 4x + 2y, subject to constraints are
2x + 3y ≤ 18, x + y ≥ 10 and x, y ≥ 0.
Answer:
The given LPP’s maximize Z = 4x + 2y
Subject to constraints are
2x + 3y ≤ 18, x + y ≥ 10 and x, y ≥ 0
The graph of the above LPP is

From the above graph, we see that for the feasible solutions there is no common area in the first quadrant. Hence, the objective function Z cannot be maximized.

Section – B [12 Marks]

Question 7.
You are given the following data about the life of two brands of bulbs.

Do you think, there is a significant difference in the quality of the two brands of bulbs? (the table value of t for 25 degree of freedom at 5% level of significance is 1.708).
Answer:
Given, x̄₁ = 2100, s₁ = 250 and n₁ = 12,
x̄₂ = 2180, s₂ = 300 and n₂ = 15
Consider, null hypothesis : There is no significant difference in the quality of the two brands of bulbs.

Since, the computed value of t = 0739 is less than the table value of t = 1.708 [at 5% for 25 degree of freedom], we accept the null hypothesis.

Question 8.
The quarterly profits of a small scale industry (? in thousands) is as follows.

Calculate 4-quarterly moving averages, display these and the original figures graphically on the same graph sheet.
Answer:


We draw the following graph from the above data.

A kink is shown near origin on the Y-axis to indicate that the scale on Y-axis starts with 20. The dark curve represents quarterly profits of a small scale industry and dotted curve represents 4-quarterly centered moving averages.

Question 9.
If the interest is compounded continuously at 6% per annum, how much worth ₹ 1000 will be 10 yr? (given, e^0.6 = 1.822)
Answer:
Let P be the principal amount at any time t and the rate of interest be r% per annum compounded continuously.

⇒ P = 1000 × 1.822
= 1822
Hence, in 10 yr, the worth of ₹ 1000 will become ₹ 1822.

Question 10.
The value of a car depreciates by 12.5% every year. By what percent will the value of the car decrease after 3 yr?
Or
A firm anticipates a capital expenditure of ₹ 80000 for a new equipment in 5 yr. How much should be deposited quarterly in sinking fund carrying 12% per annum compounded quarterly to provide for the purchase? [given (1.03)²⁰ = 1.8061]
Answer:
Let the present value of the car be ₹ P, then value of the car after 3 yr
= P(1 – i)ⁱ
= P(1 – \(\frac{12.5}{100}\))³
= P(1 – \(\frac{1}{8}\))³ = P\(\left(\frac{7}{8}\right)^{3}\)

Decrease in the value of the car after 3 yr
= P – P\(\left(\frac{7}{8}\right)^{3}\)
= P(1 – \(\frac{343}{512}\)) = \(\frac{169P}{512}\)

∴ Percentage decrease in the value of the car after 3 yr

Thus, ₹ 2977.29 deposited quarterly in a sinking fund.

Section – C [16 Marks]

Question 11.
The marginal cost function for a product is given by MC = \(\frac{5}{\sqrt{2 x+9}}\) and fixed cost ₹ 20 find the average cost for 10 units of output.
Answer:

It is given that x = 0 and C = 20
From Eq. (i) we get
20 = 5\(\sqrt{2 \times 0+9}\) + K
⇒ 20 = 15 + K
⇒ K = 5

Substituting K = 5 in Eq (i) we get
C = 5\(\sqrt{2 x+9}\) + 5
Now, AC = \(\frac{C}{x}\)

Hence, average cost for 10 units of outputs is ₹ 3.195

Question 12.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Or
A toy company manufactures two types of dolls A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is almost half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of type B by atmost 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively, on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Answer:
Let x passengers travel by executive class and y passengers travel by economy class.

We construct the following table:

So, our problem is to maximise
Z = 1000x + 600y …(i)
Subject to constraints,
x + y ≤ 200 …(ii)
x ≥ 20 …(iii)
y – 4x ≥ 0 ⇒ y ≥ 4x …(iv)
and x, y ≥ 0 ……….(v)

Firstly, draw the graph of the line x + y = 200

Putting (0, 0) in the inequality x + y ≤ 200, we have
0 + 0 ≤ 200 ⇒ 0 ≤ 200 (which is true)

So, the half plane is towards the origin. Secondly, draw the graph of the line y = 4x

Putting (10,0) in the inequality y ≥ 4x, we have
0 ≥ 4 x 10 ⇒ 0 ≥ 40 (which is false)

So, the half plane is towards Y-axis.
Thirdly, draw the graph of the line x = 20

Putting (0, 0) in the inequality x ≥ 20, we have 0 ≥ 20 (which is false)
So, the half plane is away from the origin. Since, x,y ≥ 0
So, the feasible region lies in the first quadrant.

On solving the equations, we get
A (20, 80), B(40, 160) and C (20, 180).
∴ Feasible region is ABCA

The corner points of the feasible region are A(20, 80), B(40,160) and C(20,180). The value of Z at these points are as follows:

Thus, the maximum value of Z is 136000 at B(40, 160).

Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit and the maximum profit is ₹ 136000.
Or
Let the company manufactures x dolls of type A and y dolls of type B, then
x,y ≥ 0 …(i)
x + y ≤ 1200 …(ii)
y ≤ x/2 ⇒ x – 2y ≥ 0 …(iii)
and x ≤ 3y + 600 ⇒ x -3y ≤ 600 …(iv)

Firstly, draw the graph of the line x + y = 1200

Putting (0, 0) in the inequality x + y ≤ 1200, we have
0 + 0 ≤ 1200 ⇒ 0 ≤ 1200 (which is true)

So, the half plane is towards the origin. Secondly, draw the graph of the line x – 2y = 0
Putting (200, 0) in the inequality x – 2y ≥ 0, we have
200 – 2 × 0 ≥ 0 ⇒ 200 ≥ 0 (which is true)

So, the half plane is towards the X-axis.

Thirdly draw the graph of the line x – 3y = 600

Putting (0, 0) in the inequality x – 3y≤ 600, we have
0 – 3 x 0 ≤ 600 ⇒ 0 ≤ 600 (which is true)

So, the half plane is towards the origin.
Since, x,y ≥ 0, so the feasible region lies in the first quadrant.

The point of intersection of lines x – 3y = 600 and x + y = 1200 is B(1050,150), of lines x = 2 y and x + y = 1200 is C (800,400).
Let Z be the total profit, then Z = 12 x +16y

Feasible region is OABCO.
The comer points of the feasible region are A (600,0), B(1050,150) and C(800,400).

The values of Z at these points are as follows.

The maximum value of Z is 16000 at C(800,400).

Thus, 800 and 400 dolls of types A and type B should be produced respectively to get the maximum profit of ₹ 16000.

Question 13.
Face value of bond is 1 1000, coupon rate 4.25% per annum paid semi-annually and matures in 10 yr. If present value of bond is ₹ 918.23. What is the yield to maturity.
Answer:
Given, F = ₹ 1000, r = 4.25%
and i = \(\frac{4.25}{2}\)% = \(\frac{4.25}{200}\) (semi-annually)
So, C = ₹ 1000 × \(\frac{4.25}{200}\) = ₹ 21.25 200
N = 10 × 2 = 20 yr and PV = ₹ 918.23

Approximate Yield to Maturity (YTM)

Approximate YTM = 2.64% (semi-annually)
= 5.28% per annum

Hence, yield to maturity is 5.28 per annum.

Case Based/Data Based

Question 14.
A loan is the leading of money by one or more individual, organisations or other entities to other individuals, organisations etc. The recepient incurs a debt and is usually liable to pay interest on that debt untill it is repaid as well as to repay the principle amount borrowed.

A loan of ₹ 400000 at the interest rate of 6.75% per annum compounded monthly is to be amortized by equal payment at the end of each month for 10 yr, find
(i) the size of each monthly payment
(ii) the principal outstanding at the beginning of 61st month. [given (1.005625)¹²⁰ = 1.9603 and (1.005625)⁶⁰ =1.4001]
Answer:
Given, P = ₹ 400000, n = 12 × 10 = 120 675
and i = \(\frac{6.75}{1200}\) = 0.005625

(i) ∴ Size of monthly payment (EMI)

Hence, monthly payment is ₹ 4593

(ii) We know that principal outstanding at beginning of kth period
= \(\frac{\operatorname{EMI}\left[(1+i)^{n-k+1}-1\right]}{i(1+i)^{n-k+1}}\)
So, principal outstanding at beginning of 61st month

Hence, principal outstanding at beginning of 61st month is ₹ 233336.89

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 9 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 9 with Solutions

Maximum Marks : 40
Time : 2 Hours

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. It contains one case study based question. Internal choice has been provided in one question.

Section – A [12 Marks]

Question 1.
Evaluate ∫\(\frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}}\) dx
Or
The supply function for a commodity is p = 7 + x. Determine producer’s surplus, if 15 units of goods are sold.
Answer:

Or
Given, p = 7 + x and x₀ = 15
When x₀ = 15, then p₀ = 7 + 15 = 22

Question 2.
Find the f-test value for the following two samples from a populations

Or
Out of 250 clusters 10 clusters are selected randomly as given below:

Estimate the population mean.
Answer:
We have, x̄ = \(\frac{5+7+4+9}{4}=\frac{25}{4}\) = 6.25
and ȳ = \(\frac{4+3+2+8}{4}=\frac{17}{4}\) = 4.25
Σ(x – x̄)² = (5 – 6.25)² + (7 – 6.25)² + (4 – 6.25)² + (9 – 6.25)²
= 1.5625 + 0.5625 + 5.0625 + 7.5625 = 14.75
and Σ(y – ȳ)² = (4 – 4.25)² + (3 – 4.25)² + (2 – 4.25)² + (8 – 4.25)²
= 0.0625 + 1.5625 + 5.0625 + 14.0625
= 20.75

We know that

Or

From the above table,
Σx_i = 96 + 121 + 42 + 49 + 53 + 50 + 39 + 47 + 41 + 75 = 613
and Σx_i = 8 + 12 + 4 + 10 + 9 + 3 + 7 + 3 + 8 + 7 = 71
∴ Population (x) = \(\frac{\Sigma x_{i}}{\Sigma n_{i}}=\frac{613}{71}\) = 8.63

Question 3.
Find the present value of perpetuity of ₹ 800 at the end of each quarter, if money is worth 10% compounded quarterly.
Answer:
Given, R = 800 and r = 10% = 0.10
∴ i = \(\frac{0.10}{4}\) = 0.025
∴ Present value of perpetuity,
P = \(\frac{R}{i}=\frac{800}{0.025}\) = ₹ 32000

Question 4.
The demand of a product over 6 weeks is 90, 110, 105, 130, 85 and 102. If 3 months moving average is x and 5 month morving average is y, find x + y.
Answer:
We know that
3 months moving average
= \(=\frac{\text { Sum of latest } 3 \text { trend }}{3}\)
and similarly for 5 months moving average
Now, x = \(\frac{130+85+102}{3}=\frac{317}{3}\) = 10567
and y = \(\frac{110+105+130+85+102}{5}=\frac{532}{5}\) = 1069
∴ x + y = 105.67 + 106.4 = 212.07

Question 5.
Find the effective rate, which is equivalent to a stated rate of 5% compounded semi-annually.
Answer:
Given, r = 5% and p = 2 (half-year)
So, effective rate (per rupee)
= (1 + \(\frac{r}{100 p}\))^p – 1
= (1 + \(\frac{5}{200}\))² – 1
= (1 + 0.025)² – 1
= (1.025)² – 1
= 1.0506 – 1
= 0.0506
∴ The effective rate = (0.0506) × 100% = 5.06%

Question 6.
Maximise Z = x + y, subject to constraints are
x + 4y ≤8, 2x + 3y ≤ 12, 3x + y ≤ 9 and x, y ≥ 0.
Answer:
Here, the given LPP is,

Maximise Z = x + y
Subject to constraints are,
x + 4y ≤8, 2x + 3y ≤ 12, 3x + y ≤ 9 and x, y ≥ 0
On solving x + 4y = 8 and 3x + y = 9, we get
x = \(\frac{28}{11}\) and y = \(\frac{15}{11}\)
From the feasible region, it is clear that coordinates of comer points are (0, 0), (3, 0), \(\left(\frac{28}{11}, \frac{15}{11}\right)\) and (0,2).

Hence, the maximum value is 3 \(\frac{10}{11}\) at \(\left(\frac{28}{11}, \frac{5}{11}\right)\)

Section – B [12 Marks]

Question 7.
The population of a town grows at the rate of 10% per year. Using differential equation, find how long it will take for the population to grow 2 times.
Or
The marginal revenue function for a commodity is given by MR = 8 + 3x – 5x². Find the demand function.
Answer:
Let P0 be the population initially and P be the population after t yr.

According to question,
\(\frac{d}{d t}\)(P) = 10% of P ⇒ \(\frac{d P}{d t}=\frac{1}{10}\) P
⇒ \(\frac{1}{P}\) dP = \(\frac{1}{10}\)dt

On integration both sides, we get
∫\(\frac{1}{P}\) dP = ∫\(\frac{1}{10}\)dt
⇒ log P = \(\frac{1}{10}\)t + C

When t = 0, then P = P₀
∴ logP₀ = 0 + C ⇒ C = log P₀

From Eq.(i)
log P = \(\frac{1}{10}\)t + log P₀
⇒ log\(\frac{P}{P_{0}}=\frac{1}{10}\)t ……….(ii)

When the population grows 2 times, then
P = 2P₀
∴ log\(\left(\frac{2 P_{0}}{P_{0}}\right)=\frac{1}{10}\)t
⇒ t = 10log2
Hence, it takes 10 log 2 yr for the population to grow 2 times.

Or

Given MR = 8 + 3x – 5x²

Question 8.
10 students are selected at random from a college and their heights are found to be 100, 104,108,110,118,120,122,124,126 and 128 cm. In the light of these data, discuss the suggestion that the mean height of the students of the college is 110 cm. [given, t₉ (0.05) = 2.262].
Answer:
Consider, table for the sample mean (x̄) and S. Computation of x̄ and S

We have, \(\sum_{i=1}^{10}\) x_i = 1160 and n= 10
= \(\frac{1}{n} \sum_{i=1}^{10}\)x_i
= \(\frac{1160}{10}\) = 116

From the table, we find that

We have degree of freedom = (10 -1) = 9 It is given that t₉(0.05) = 2.262
∴ Calculated |t| = 1.94 < 2.262 = t₉(005)
i.e. Calculated |t| < tabulated t₉(0.05)
So, we accept the null hypothesis. Hence, the sample mean is same as the population mean. Consequently, the mean height of the students of the college is 110 cm.

Question 9.
From the following data compute 4-yearly moving averages and determine the trend values. Also, find the short-term fluctuations.

Answer:
Table for 4-yearly moving averages.

Question 10.
Mr. Kapoor invested ₹ 25000 in a mutual fund in year 2012. The value of mutual fund increased to ₹ 30000 in year 2016. Calculate the compound annual growth rate (CAGR) of his investment. [given, (1.2)^1/4 = 1.0466]
Answer:
Given, 13V = ₹ 25000, EV = ₹ 30000
and number of year (n) =4 yr
∴ CAGR = \(\left(\frac{\mathrm{EV}}{\mathrm{BV}}\right)^{1 / n}\) – 1
= \(\left(\frac{30000}{25000}\right)^{1 / 4}\) – 1 = \(\left(\frac{6}{5}\right)^{1 / 4}\) – 1
= (1.2)^1/4 – 1 = 1.0466 – 1 = 0.0466
∴ CAGR = (00466 × 100)% = 4.66%

Section – C [16 Marks]

Question 11.
Maximise and minimise Z = 3x – 4y, subject to constraints are
x – 2y ≤ 0, -3x + y ≤ 4, x – y ≤ 6 and x, y ≥ 0.
Answer:
Given, LPP is,
Maximise and minimise Z = 3x – 4y

Subject to constraints are
x – 2y ≤ 0, -3x + y ≤ 4, x – y ≤ 6 and x, y ≥ 0.
[on solving x – y = 6 and x – 2y = 0, we get x = 12 and y = 6]

From the shown graph, for the feasible region, we see that it is unbounded and coordinates of comer points are (0, 0), (12, 6) and (0, 4).

For given unbounded region the minimum value of Z may or may not be -16. So, for deciding this, we graph the inequality.
3x – 4y < -16 and check whether the resulting open half plane has common points with feasible region or not. Thus, from the figure it shows it has common points with feasible region, so it does not have any minimum value. Also, similarly for maximum value, we graph the inequality 3x – 4y > 12 and see that resulting open half plane has no common points with the feasible region and hence maximum value 12 exist for Z = 3x – 4y.

Question 12.
A couple wishes to purchase a house for ₹ 1000000 with a down payment of ₹ 200000. If they can amortize the balance at an interest rate 8% per annum compounded monthly for 10 yr. Find the monthly installment (EMI). [given, (1.006)^-120 = 0.4878]
Answer:
Amortized amount, P = Cost of house – Cash down payment
= 1000000 – 200000 = ₹ 800000
i = \(\frac{0.08}{12}\) = 0.006
n = 10 × 12 = 120 months

Question 13.
Mr. Shiv Chandra wants to know the amount, he should pay for a goldmine expected to yield an annual return of 400000 for the next 10 yr, after which it will be worthless. Find the amount, he should pay for the mine, if he wants to yield 18% annual return on his investment and also set up a sinking fund to replace the purchase price. Assume that the sinking fund earns 10% annually. [given (1.1)¹⁰ = 2.5937]
Or
A bond has a face value of ₹ 1000, matures in 4 yr coupon rate is 4% per annum. The bond makes annual payments. If the yield to maturity is 4%. Find the fair value of bond, [given (1.04)^-4 =0.8551]
Answer:
Given, rate of return on investment = 18% = 0.18
Let x be the purchase price of the mine.
Return on investment of ₹ x per year = 0.18+
Life of mine (n) = 10 yr
Net annual return = ₹ 400000

Sinking fund is created to replace the purchase price.
∴ Amount to be placed into sinking fund each year is ₹ (400000 – 0.18x).
These payments must accumulate to ₹ x at the end of 10 yr.
∴ Amount A = ₹ x,
i = r = 10% = 0.10
Here, n = 10 and R = ₹ (400000 – 0.18.x)

We know that

∴ Purchase price of the goldmine is ₹ 1648125.25

Or

Here, F = ₹ 1000, r = 4%, N = 4 yr and d = 4%

Case Based/Data Based

Question 14.
A long lasting company produces biscuits. Whose marginal revenue functions of a commodity is given by MR = 6x² – 5x + 2.

On the basis of the above information, answer the following questions.
(i) Find the total revenue functions.
Answer:
Given, marginal revenue functions
MR = 6x² – 5x + 2
⇒ \(\frac{d R}{d x}\) = MR = 6x² – 5x + 2
R = ∫(6x² – 5x + 2)dx
R = \(\frac{6 x^{3}}{3}-\frac{5 x^{2}}{2}\) + 2x + K
R = 2x³ – \(\frac{5 x^{2}}{2}\) + 2x + K

(ii) If the productions increases from 10 to 20. Find the increase in their revenue.
Answer:
The increase in revenue, when productions increase from 10 to 20, is

Students can access the CBSE Sample Papers for Class 12 Accountancy 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 Accountancy Standard Term 2 Set 5 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This question paper comprises two Parts – A and B. There are 12 questions in the question paper. Alt1 questions are compulsory.
• Part-A is compulsory for all candidates.
• Part-B has two options i.e., (i) Analysis of Financial Statements and (ii) Computerized Accounting. Students must attempt only one. of the given options.
• There is no overall choice. However, an internal choice has been provided in 3 questions of three marks and 1 question of five marks.

PART-A
(Accounting for Not-for-Profit Organisations, Partnership Firms and Companies)

Question 1.
From the following information provided by Reunion Club, calculate the amount of subscriptions which will be treated as income for the year ended 31^st March, 2021:
(i) Subscriptions collected during the year ended 31^st March, 2021 amounted to ₹ 49,000.
(ii) Subscriptions for the year ended 31^st March, 2021 collected in the year ended 31^stMarch, 2020 amounted to ₹ 3,000.
(iii) Subscriptions unpaid for the year ended 31^st March, 2021 amounted to ₹ 2,000. (2)
Answer:

Question 2.
On dissolution, how will you deal with Partner’s Loan if it appears on:
(A) Assets side of company’s Balance Sheet
(B) Liabilities side of company’s Balance Sheet (2)
Answer:
(A) If Partner’s Loan appears on the Assets side of the Balance Sheet, it indicates that partner has taken loan from the firm. In that case, at the time of dissolution the amount of loan will be transferred to the concerned Partner’s Capital Account. The following Journal Entry will be passed:

(B) If Partner’s Loan appears on the Liabilities side of the Balance Sheet, it indicates that the respective partner or partners have given loan to the firm. In this case, partner’s loan will be paid off after paying all the external liabilities first. The following accounting entry will be passed:

Question 3.
Tanmay, Gautam and Jagat were partners in a partnership firm ‘Jagatam Limited’ sharing profits and losses in the ratio of 6 : 5 : 4. Gautam retired and his capital after making adjustments on account of reserves, revaluation of assets and reassessment of liabilities stood at ₹ 10,01,600. Tanmay and Jagat agreed to pay him ₹ 11,60,000 in full settlement of his claim.

Pass necessary journal entry for the treatment of goodwill at the time of retirement of Gautam. Also show your workings clearly. (2)
Answer:

Working Notes:
(1) Calculation of Profit Sharing Ratio:
Old Profit Sharing Ratio of Tanmay, Gautam and Jagat = 6 : 5 : 4
New Profit Sharing Ratio and Gaining Ratio of Tanmay and Jagat will be 6 : 4 or 3 : 2, as no information provided of future profit sharing ratio between Tanmay and Jagat.

(2) Calculation of Gautam’s Share of Goodwill:
Gautam’s Share of Goodwill = Amount Paid – Gautam’s Capital after Adjustments
= ₹ 11,60,000 – ₹ 10,01,600
= ₹ 1,58,400

Gautam’s share of goodwill will be compensated by gaining partners, Tanmay and Jagat in their Gaining Ratio, i.e., 3 : 2.
Tanmay Will compansate = ₹ 1,58,400 × \(\frac{3}{5}\) = ₹ 95,040
Jagat Will compansate = ₹ 1,58,400 × \(\frac{2}{5}\) = ₹ 63,360

Question 4.
Show how are the following items dealt with while preparing the final accounts of Khadagpur Sports Club for the year ended 31^st March, 2020:
Case I:
(i) Expenditure on construction of Stadium is ₹ 6,00,000.
(ii) The construction work is in progress and has not yet completed.
(iii) Capital Fund as at 31st March, 2019 is ₹ 20,00,000.

Case II:
(i) Expenditure on construction of Stadium is ₹ 6,00,000.
(ii) The construction work is in progress and has not yet completed.
(iii) Stadium Fund as at 31^st March, 2019 is ₹ 10,00,000.
(iv) Capital Fund as at 31^st March, 2019 is ₹ 20,00,000.
OR
How will you treat the following items while preparing Income and Expenditure Account and Balance Sheet of a Not-for-Profit Organisation ?
(A) Donation for Building
(B) Sale of Newspapers
(C) Investment purchased. (3)
Answer:
Case I:

Case II:

OR
(A) Donation for Building: Donation for Building means that the donation received can only be used as a specific donation for building. Specific donation is capitalised and is shown on the Liabilities side of the Balance Sheet of Not-for-Profit Organisation.

(B) Sale of Newspapers: The money paid for newspapers, magazines, periodicals, etc. is a revenue expense and is debited to Income and Expenditure Account. Thus, the amount realised from the sale of newspapers, magazines, periodicals, etc. will be credited to Income and Expenditure Account.

(C) Investment Purchased: Investments purchased are viewed as capital spending and are shown on the Assets side of the Balance Sheet of Not-for-Profit Organisation.

Question 5.
Alpha, Beta and Gamma were partern in a firm sharing profit and losses is the ratio of 2 : 2 : 1. On 31^st March, 2021, their Balance Sheet was as follows:

On 30^th June, 2021, Beta died. The partnership Deed of the firm provided for the folloiwng on the death of a partner:

(i) Goodwill of the firm was to be calculated on the basis of two times the average profit of he
past five years. The profite for the years ended:
31^st March, 2020 – ₹ 1,00,000
31^st March, 2019 – ₹ 1,60,000
31^st March, 2018 – ₹ 22,20,000
31^st March, 2017 – ₹ 2,40,000

(ii) Beta’s share of profit on loss from 1^st April, 2021 till he death was to be calculated on the basis of the profit or loss for the year ended 31^st March, 2021.

You are required to prepare Beta’s Capital Account at the time of her death to be presented to his executors. (3)
Answer:

Working Notes:
(1) Calculation of Beta’s Share of Goodwill:
Average Profit of 5 years = \(\frac{₹ 1,00,000+₹ 1,60,000+₹ 2,20,000+₹ 4,40,000-₹ 3,20,000}{5}\)
= \(\frac{₹ 6,00,000}{5}\) = ₹ 1,20,000
Goodwill of Firm = Average profit × 2
= ₹ 1,20,000 × 2
= ₹ 2,40,000
Beta’s Share of Goodwill = 22,40,000 × \(\frac{2}{5}\)
= ₹ 96,000
which will be contributed by Alpha and Gamma is their gaining ratio, 2 : 1.
Alpha will contribute = ₹ 96,000 × \(\frac{2}{3}\) = ₹ 64,000
Gamma and contribute = ₹ 96,000 × \(\frac{1}{3}\) = ₹ 32,000

(2) Calculation of Beta’s share of profit or loss upto his date of Death:
Beta’s share of loss upto his Date of Death = 3,20,000 × \(\frac{2}{5} \times \frac{3}{12}\)
= ₹ 32,000

Question 6.
On 1^st June, 2019, Rilco Energy Ltd. issued 10,000, 7% Debentures of ₹ 100 each at a discount of 10% redeemable at a premium of 10% at the end of five years. All the debentures were subscribed and allotment was made.
Prepare the Balance Sheet (extract) as at 31^st March, 2020.
OR
Relco India Ltd. purchased machinery costing ₹ 1,35,000. It was agreed that the purchase consideration be paid by issuing 9% Debentures of ₹ 100 each. Give necessary journal entries if the debentures have been issued:
(A) at par
(B) at a discount of 10%. (3)
Answer:

Notes to Accounts:

OR

Working Note:

= \(\frac{1,35,000}{100-10}\)
= 1,350

Question 7.
A and B are partners in a firm sharing profits and losses in the ratio of 3 : 2. On 31st March, 2020, their Balance Sheet was as follows:

The firm was dissolved on 31^st March, 2020 and both the partners agreed to the following:
(i) Investments were taken over by A at an agreed value of ₹ 4,000. He also agreed to settle Mrs. A’s Loan.
(ii) Other assets realised as: Stock – ₹ 2,500; Debtors – ₹ 9,250; Furniture – ₹ 2,250; Plant – ₹ 12,500.
(iii) Expenses of dissolution amounted to ₹ 800.
(iv) Creditors agreed to accept ₹ 18,500 in full settlement of their claim.
You are required to prepare Realisation Account and Partners’ Capital Account of the firm dissolved.
OR
Alka, Binod and Cheenu are partners in a firm ABC Limited, sharing profits and losses as Alka 1/3, Binod 1/2 and Cheenu 1/6 respectively. The Balance Sheet of the firm as on 31^st March, 2019 was as follows:

It was agreed that:
(i) Goodwill will be valued at ₹ 27,000.
(ii) Depreciation of 10% was to be provided on Machinery.
(iii) Patents were to be reduced by 20%.
(iv) Liability on account of Provident Fund was estimated at ₹ 2,400.
(v) Chander took over Investments for ₹ 15,800.
(vi) Amit and Balan decided to adjust their capitals in proportion of their profit sharing ratio by opening Current Accounts.
Prepare Revaluation Account and Partners’ Capital Accounts on Chander’s retirement. (5)
Answer:

OR


Working Notes:
(1) Calcuation of Gaining Ratio:
Old Radio of Alka,Binod anid Cheenu = \(\frac{1}{3}: \frac{1}{2}: \frac{1}{6}\) or 2 : 3 : 1
Cheenu retires from the firm
New Ratio of Alka and Binod = 3 : 2
Gaining Ratio = New Radio – Old Radio
Alka’s Gain = \(\frac{3}{5}-\frac{2}{6}\) = \(\frac{18-10}{30}\) = \(\frac{8}{30}\) (Gain)
Binod’s Gain = \(\frac{2}{5}-\frac{3}{6}\) = \(\frac{12-15}{30}\) = –\(\frac{3}{30}\) (Sacrifice)

(2) Adjustment of Goodwill:
Goodwill of the Firm = ₹ 24,000
Cheenu’s Share of Goodwill = ₹ 24,000 × \(\frac{1}{6}\) = ₹ 4,000
Alka’s Gain in Goodwill = ₹ 24,000 × \(\frac{8}{30}\) = ₹ 6,400
Binod’s Sacrifice in Goodwill = ₹ 24,000 × \(\frac{3}{30}\) = ₹ 2,400

Question 8.
Jagannath Hotels Co. engaged in business of hotels, having chain of hotels in Andhra Pradesh decided to expand their business in other states also. For that the directors of the company decided to start their hotels in nearby states viz. Karnataka, Maharashtra and Goa.
For their underway hotels in Goa, Jagannath Hotels Co. purchased furniture of ₹ 88,00,000 from Deluxxe Furniture Store.
Jagannath Hotels Co. paid 50% of the amount by accepting a bill of exchange in favour of Deluxxe Furniture Store payable after five months. For the balance amount the company issued 10% Debentures of ₹ 100 each at a premium of 10% in favour of Deluxxe Furniture Store.

You are required to answer the following questions:
(A) Pass journal entry which will be passed at the time of purchase of furniture in the books of Jagannath Hotels Co.
(B) Pass journal entry for the payment made through bill of exchange.
(C) Calculate the number of debentures issued to Deluxxe Furniture Store.
(D) Calculate the amount to be transferred to Securities Premium Reserve Account.
(E) Pass journal entry for the allotment of debentures. (5)
Answer:

Working Note:
Purchase Price of Furniture = ₹ 88,00,000
Amount Paid by Bill of Exchange = 50% of ₹ 88,00,000
= \(\frac{50}{100}\) × ₹ 88,00,000 = ₹ 44,00,000

(C) Balance Amount due to Deluxxe Furniture Store = Purchase Price of Furniture – Amount Paid by Bill of Exchange
= ₹ 88,00,000 – (50% of ₹ 88,00,000)
= ₹ 44,00,000
Number of Debentures Issued to Deluxxe Furniture Store

= \(\frac{44,00,000}{₹ 100+(10 \% \text { of } ₹ 100)}\)
= \(\frac{44,00,000}{₹ 110}\) = 40,000
Number of Debentures Issued to Deluxxe Furniture Store = 40,000
Premium per debenture = 10% of ₹ 100 = ₹ 10

(E) Amount transferred to Securities Premium Reserve Account
= Number of Debentures × Premium per debenture = 40,000 × ₹ 10 = ₹ 4,00,000

(E)

Question 9.
Following is the Receipts and Payments of Bengaluru Sports Club for the year ended 31^st March, 2021:

(i) On 31^st March, 2020 subscriptions outstanding was ₹ 6,000 and on 31^st March, 2021 subscriptions outstanding was ₹ 8,000.
(ii) Salary outstanding on 31^st March, 2021 was ₹ 4,000.
(iii) On 1^st April, 2020 the club had Building of ₹ 1,60,000, Furniture of ₹ 40,000,10% Investments of ₹ 90,000 and Sports Equipments of ₹ 50,000. Depreciation charged on these items including purchases was 10%.

Prepare Income and Expenditure Account of Bengaluru Sports Club for the year ended 31^st March, 2021. Also prepare Balance Sheet of the club as at 31^st March, 2021. (5)
Answer:

Workin Note:

PART-B
Option-1
(Analysis of Financial Statements)

Question 10.
State whether the following transactions will result in inflow, outflow or no flow of cash while preparing Cash Flow Statement:
(A) Increase in Marketable Securities by ₹ 5,000.
(B) Decrease in Sundry Debtors by ₹ 15,000. (2)
Answer:
(A) Increase in Marketable Securities by ₹ 5,000 wilL result in no fLow of cash while preparing Cash Flow Statement.
(B) Decrease in Sundry Debtors by ₹ 15,000 will result in inflow of cash while preparing Cash Flow Statement

Question 11.
Prepare Common Size Balance Sheet from the Balance Sheets of Bat Ltd. and Ball Ltd. as at 31^st March, 2021 are:

OR
Prepare Comparative Statement of Profit and Loss from the followng figures:

Answer:

OR

Working Notes:
(1) Computation of Other Income:
Other Income on 31^st March, 2020 = 20% of Revenue from Operations
= 20% of ₹ 40,00,000
= ₹ 8,00,000
Other Income on 31^st March, 2021 = 12% of Revenue from Operations
= 12% of ₹ 60,00,000
= ₹ 7,20,000

(2) Computation of Expenses:
Expenses on 31^st March, 2020 = 60% of Revenue from Operations
= 60% of ₹ 40,00,000
= ₹ 24,00,000
Expenses on 31^st March, 2021 = 70% of Revenue from Operations
= 70% of ₹ 60,00,000
= ₹ 42,00,000

Question 12.
From the following Balance Sheet as at 31^st March, 2021 and Statement of Profit and Loss for the year ended 31^st March, 2021 of Roopakshi Ltd. and additional information, prepare Cash Flow Statement for the year ended 31^st March, 2021:


Notes to Accounts:

Additional Information:
(i) Additional debentures were issued on 1^st October, 2020 of ₹ 5,00,000. On the same date, part of outstanding debentures were redeemed and interest was paid, whereas interest on outstanding debentures was paid on 10^th April, 2021.
(ii) Board of Directors proposed dividend in both the years @ 10%.
(iii) Interim Dividend of ₹ 1,00,000 was paid during the year.
(iv) A fixed asset with original cost of ₹ 1,00,000, on which depreciation till date was provided of
₹ 80,000 was sold at a profit of ₹ 10,000. (5)
Answer:


Working Notes:

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 7 with Solutions

Maximum Marks : 40
Time : 2 Hours

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. It contains one case study based question. Internal choice has been provided in one question.

Section – A [12 Marks]

Question 1.
A population grows at the rate of 5% per year. How long does it take for the population to double?
Answer:
Let P₀ be the initial population and the population after t yr be P.
According to given problem, we get
\(\frac{d P}{d t}=\left(\frac{5}{100}\right)\) × P = \(\frac{d P}{d t}=\frac{P}{20}\)
Now, separating the variables, we get
\(\frac{d P}{P}=\frac{d t}{20}\)
On integrating both sides, we get
l0gP = \(\frac{t}{20}\) + C₁
⇒ P = ^{et/20 + C₁} = e^t/20 – e^C₁ = Ce^t/20, where e^C₁ = C
∵ At t = 0, P = P₀, therefore we have P₀ = C.e⁰

Question 2.
The mean and variance of a random sample of 81 observations were computed as 200 and 144, respectively. Compute the 95% confidence limits for population mean.
Or
The mean weekly sales of mango candy in candy stores was 225.4 mango candy per store. After an advertising campaign the mean weekly sales in 25 stores for a typical week increased to 237.6 and showed a standard deviation of 21.3 was the advertising campaign successful? Level of confidence 5%. [given, t₂₄ (0.05) = 1.711]
Answer:
Given, n = 81, x̄ = 200 and σ² = 144 ⇒ σ = 12
∴ Confidence limits (95%) are
⇒ x̄ ± 1.96\(\left(\frac{\sigma}{\sqrt{n}}\right)\) = x̄ ± 1.96\(\left(\frac{12}{\sqrt{81}}\right)\)
= [200 – 1.96 \(\left(\frac{12}{9}\right)\),200 + 1.96 \(\left(\frac{12}{9}\right)\)]
= [200 – 2.61,200 + 2.61]
= [197.39,202.61]

Or

Given, n = 25, x̄ = 237.6, μ = 225.4 and S = 21.3
Consider, H₀ : μ = 225.4 and H₁ : μ > 225.4

Now, test statistics,
t = \(\frac{\bar{x}-\mu}{S / \sqrt{n}}\)
= \(\frac{237.6-225.4}{21.3 / \sqrt{25}}=\frac{12.2 \times 5}{21.3}\) = 2.86

Since, calculated value of t > tabulated value of t. Hence, H₀ can rejected
⇒ Advertising campaign was successful.

Question 3.
Anand Prasad borrows a sum of ₹ 225000 at an interest rate of 10% (flat) for a tenure of 5 yr. Calculate his EMI.
Or
A sum of ₹ 75000 invested at r% compounded quarterly will provide payments of ₹ 800 each at the end of 3 months. Find the value of r.
Answer:
Given, principal (P) = ₹ 225000
Interest (yearly) = 10% of 225000 = ₹ 22500
Interest for 5 yr = 5 × 22500 =112500
⇒ EMI = \(\frac{225000+112500}{5 \times 12}=\frac{337500}{60}\)
[∵ EMI = \(\frac{\text { Principal + Interest }}{n}\)]
= 5625
∴ Hence, EMI is 15625.

Or

Given, present value (P) = ₹ 75000
Let the rate of interest be r.
Since, interest compounded quarterly
∴ i = \(\frac{r}{4}\)
Periodic payment (R) = ? 800

We know that
P = \(\frac{R}{i}\)
⇒ 75000 = \(\frac{800}{i}\)
⇒ i = \(\frac{800}{75000}\)
⇒ i = 0.010
⇒ \(\frac{r}{4}\) = 0.010 4
⇒ r = 0.040
= 4% (approx.)

Question 4.
Calculate the 3-yearly moving averages of the following data.

Answer:
Calculation of 3-yearly moving averages.

Question 5.
What amount is recieved at the end of every 6 months forever, if ₹ 72000 kept in a bank earns 8% per annum compounded half-yearly.
Answer:
It is given that the present value of perpetuity of ₹ R payable at the end of each 6 months is ₹ 72000. If money is worth 8% compounded semi-annually. Thus,
We have, P = ₹ 72000
and i = \(\frac{8}{200}\) = 0.04
∵ P = \(\frac{R}{i}\)
⇒ R = Pi
⇒ R = 72000 × 0.04 = 2880
Hence, R is ₹ 2880

Question 6.
Find the maximum value of Z for the problem maximise Z = 2x + y subject to constraints are x + y ≤ 2, x ≥ 0, y ≥ 0 is
Answer:
We have, maximize Z = 2 x + y

Subject to constraints are
x + y ≤ 2, x ≥ 0, y ≥ 0

The shaded region shown in fig OAB is bounded and the coordinates of comer points O, A and B are (0, 0), (2,0) and (0, 2), respectively.

Section – B [12 Marks]

Question 7.
A product can be manufactured at a total cost C(x) = \(\frac{x^{2}}{100}\) + 100x + 40, where x is the number of units produced. The price at which each unit can be sold is given by p = (200 – \(\frac{x}{100}\)). Determine the production level x at which the profit is maximum. What is the price per unit and total profit at the level of production?
Answer:
Given, total cost C(x) = \(\frac{x^{2}}{100}\) + 100x + 40
Price of each unit can be sold
p = 200 – \(\frac{x}{100}\)

Price of x unit sold
R(x) = px = 200x – \(\frac{x^{2}}{400}\)

∴ Profit P(x) = R(x) – C(x)
= 200x – \(\frac{x^{2}}{400}-\frac{x^{2}}{100}\) – 100x – 40
⇒ P(x) = \(\frac{x^{2}}{80}\) + 100x – 40 ……..(i)

On differentiating both sides w.r.t. x, we get
P'(x) = \(\frac{2x}{80}\) + 100
= – \(\frac{x}{40}\) + 100

For maxima or minima, put P'(x) = 0
⇒ – \(\frac{x}{40}\) + 100 = 0
x = 4000
Now, P”(x) = – \(\frac{1}{40}\) < 0

Hence, P(x) is maximum when x = 4000.
Total profit at x = 4000, then
P(4000) = \(\frac{-(4000)^{2}}{80}\) + 100(4000) – 40 [from Eq. (i)]
= -200000 + 400000 – 40
= 199960
Hence, the total profit is ₹ 199960.

Question 8.
Ten oil tins are taken at random from an automatic filling machine. The mean weight of the tins is 17.2 kg and standard deviation is 0.62 kg. Does the sample mean differ significantly from the intended weight of 18 kg? [given, t₉(0.05) = 2.26]
Answer:
Given, n = 10, x̄ = 17.2, μ = 18 and S = 0.62
Consider, H₀: μ = 18
H₁: μ ≠ 18
t = \(\frac{\bar{x}-\mu}{S / \sqrt{n}}\)
= \(\frac{17.2-18}{0.62 / \sqrt{10}}\)
= –\(\frac{0.8 \times \sqrt{10}}{0.62}\) = -4.08

Since, calculated value of If |t| is greater than the tabulated value of t. So, H₀ is not accepted. Thus, the difference between sample mean weight and the intended weight is not insignificant.

Question 9.
Below are given the figures of production (in thousand tonnes) of a sugar factory.

Fit a straight line trend by the method of least squares and find trend value for year 2005.
Or
Consider the following data

Calculate 3-days moving average and display these and the original figures on the same graph.
Answer:
Here, n =7 (odd)
So, we shift the origin to the middle of the time period of the 2008.

Let the straight line trend of y on x be
y_t = a + bx …(i)

Now, construct the table as under.

Now, a = \(\frac{\Sigma y}{n}=\frac{785}{7}\) = 112.14
and b = \(\frac{\Sigma x y}{\Sigma x^{2}}=\frac{40}{28}\) = 1.43

So, the required equation of the straight line trend is
y_t = 112.14 + 1.43x

:. Trend value for year 2005,
y_t = 112.14 + 1.43(- 3)
= 112.14 – 4.29
= 107.85

Or

According to the question,
Let the 3-days moving average is m.


On the basis of above data we can draw the following graph:

Question 10.
Mr. Narayan Sankar has invested ₹ 150000 in a financial plan whose compound annual growth rate (CAGR) is 8.5% and he received a final value of ₹ 300000.
Find the period (completed) for which he has invested the amount. [given, log 2 = 0.3010 and log(1.09) = 0.0374]
Answer:
10. Given, EV = ₹ 300000
BV = ₹ 150000 and CAGR = 8.5%

We know that
CAGR = [\(\left(\frac{\mathrm{EV}}{\mathrm{BV}}\right)^{1 / n}\) – 1] × 100
⇒ 8.5 = [\(\left(\frac{300000}{150000}\right)^{1 / n}\) – 1] × 100
⇒ 8.5 = [(2)^1/n – 1] × 100
⇒ 8.5 = (2)^1/n × 100 – 100
⇒ 8.5 + 100 = (2)^1/n × 100
⇒ 108.5 = (2)^1/n × 100
⇒ (2)^1/n = \(\frac{108.5}{100}\) = 1.085
⇒ \(\frac{1}{n}\) log2 = log(1.09)
⇒ n = \(\frac{\log 2}{\log (1.09)}\)
⇒ n = \(\frac{0.3010}{0.0374}\)
⇒ n = 8.0481
⇒ n = 8
∴ The period for which he has invested the amount is 8 yr.

Section – C [16 Marks]

Question 11.
A machine is bought for ₹ 320000. Its effective life is 8 yr, after which its salvage value would be ₹ 25000. It is decided to create a sinking fund to replace this machine at the end of its effective life by making half yearly payments that will earn an interest of 8% per annum compounded half yearly. If it is known that the cost of machine increases by 5% per annum. Calculate the amount of each payment to the sinking fund. [given (1.04)¹⁶ =1.8730 and (1.05)⁸ =1.4774]
Or
A bond has face value of ₹ 10000 and maturity period of 10 yr. The nominal interest rate is 6% per annum. What should be the price of the bond to yield an effective interest of 8%? [given (1.08)^-0 = 0.4631]
Answer:
Let each semi-annually deposit in the sinking fund of ₹ R. Since, the cost of new machine is increases by 5% per annum the cost of present.
Cost of machine at present = ₹ 320000
Cost of machine after increasing 5% per annum after 8 yr
= 32000o\(\left(1+\frac{5}{100}\right)^{8}\)
= 32000 (1.05)⁸
= 320000 × 1.4774
= ₹ 472768
Salvage value of present machine = ₹ 25000
So, net amount required at the end of 8 yr to purchase the new model is ₹ (472768 – 25000) = ₹ 447768

We know that R = \(\frac{i \times S}{(1+i)^{n}-1}\)
Here, S = ₹ 447768, n = 8 × 2 = 16 yr
i = \(\frac{8}{200}\) = 0.04
∴ R = \(\frac{(0.04) \times(447768)}{(1+0.04)^{16}-1}\)
= \(\frac{17910.72}{(1.04)^{16}-1}\)
= \(\frac{17910.72}{18730-1}\)
= \(\frac{17910.72}{0.8730}\) = 20516.28
Thus, ₹ 20516.28 deposited half yearly out of the profit to purchase the new model of the machine.

Or

We have, F = Face value of bond = ₹ 10000
N = Number of period = 10
C = Coupon payment = Annual dividend × 10000
= \(\frac{6}{100}\) × 10000 = 1600
F = Maturity value = Face value = ₹ 10000

Let PV be the price of the bond.

Hence, the price of bond is ₹ 8657.75.

Question 12.
The supply function of a producer is given by p = \(\frac{2}{5}\) e^2x, where x denotes thousand units. Find producer’s surplus when sales are 2000 units.
Answer:
The supply functions is p = \(\frac{2}{5}\) e^2x
When sales are 2000 units i.e., x₀ =2, we get

Question 13.
Anil wants to invest atmost ₹ 12000 in bonds A and B. According to the rules, he has to invest atleast ₹ 2000 in bond A and atleast ₹ 4000 in bond B. If the rate of interest in bond A is 8% per annum and on bond B is 10% per annum, then to maximise the interest, then find the investment in bond A and B.
Answer:
Let the investment in bond A and B are x and y respectively.
Then, our problem is to Maximise Z = 0.08x + 0.10y
Subject to constraints are
x + y ≤ 12000, x ≥ 2000
y ≥ 4000 and x, y ≥ 0

The graph of the above inequality is given by

In the above graph shaded region is the feasible bounded region ABC.
Now, the values of Z at the comer points are given by

From the above table, we see that the maximum value of Z is 1160 which occurs at the point (2000,10000).
Hence, to maximise the interest ₹ 2000 and ₹ 10000 must be invested in bond A and B, respectively.

Case Based/Data Based

Question 14.
EMI is a part of equally divided monthly outgoes to clear off an outstanding loan within a stipulated time frame. For a fixed interest rate loan, the EMI remain fixed for the entire tenure of the loan, provided there is no default or part payment in between. The EMI is used off both the principal and interest components of an outstanding loan, The first EMI has the highest interest component and the lowest principal component.

Rajesh purchased a house from a company for ₹ 2500000 and made a down payment of ₹ 500000. He repays the balance in 25 yr by monthly installments at the rate of 9% per annum compounded monthly, (given (1.0075)_300 = 0.1062)
(i) What are the monthly payment?
(ii) What is the total interest payment?
Answer:
Cost of house = ₹ 2500000
Down payment = ₹ 500000
Principal amount = ₹ (2500000 – 500000)
= ₹ 2000000,
n = 25 × 12 = 300
and i = \(\frac{9}{1200}\) = 0.0075

(i) We know that

Hence, monthly payment is ₹ 16782.27

(ii) We have, EMI = ₹ 16782.27
n = 300 and P = ₹ 2000000
Total interest = n × EMI – P
= 300 × 16782.27 – 2000000
= 5034681 – 2000000
= 3034681
Hence, total interest is ₹ 3034681.

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Accountancy Standard Term 2 Set 7 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This question paper comprises two Parts – A and B. There are 12 questions in the question paper. Alt1 questions are compulsory.
• Part-A is compulsory for all candidates.
• Part-B has two options i.e., (i) Analysis of Financial Statements and (ii) Computerized Accounting. Students must attempt only one. of the given options.
• There is no overall choice. However, an internal choice has been provided in 3 questions of three marks and 1 question of five marks.

PART-A
(Accounting for Not-for-Profit Organisations, Partnership Firms and Companies)

Question 1.
Calculate the amount to be posted to the Income and Expenditure Account of Sangan Sports Academy for the year ended 31^st March, 2021 from the information given:
Stock of Sports Materials on 1^st April, 2020 – ₹ 60,000
Creditors for Sports Materials on 1^st April, 2020 – ₹ 40,000
Amount paid for Sports Materials during the year – ₹ 2,16,000
Stock of Sports Materials on 31^st March, 2021 – ₹ 10,000 (2)
Answer:

Question 2.
Differentiate between ‘Dissolution of Partnership’ and ‘Dissolution of Partnership Firm’ on the basis of:
(A) Economic Relationship
(B) Closure of Books (2)
Answer:
Difference between Dissolution of Partnership and Dissolution of Partnership Firm

Question 3.
Pass journal entries for following transactions on the dissolution of a firm of partners X and Y, after various assets (other than cash) and outside liabilities have been transferred to Realisation Account?
(A) ‘X’ took 50% of the stock at a discount of 20%. Remaining stock was sold at a profit of 30% on csot (Book value of stock given in the Balance Sheet before dissolution was ₹ 8,00,000).
(B) Debtors ₹ 5,28,000. Provision for Doubtful Debts. ₹ 48,000, ₹ 96,000 of the book debts proved
bad. (2)
Answer:

Question 4.
From the following information given by Giani Sports Club, calculate the amount of Subscription received during the year 2020-21.
(i) Subscription credited to Income and Expenditure Account for the year ending 31^st March, 2021 amounted to ₹ 12,00,000. Each member of the club is required to pay an annual subscription of ₹ 12,000.
(ii) During the year 2019-20,12 members paid the subscription amount for the year 2020-21.
(iii) Subscription in arrears as on 1^st April, 2020 amounted to ₹ 64,000.
(iv) During the year 2020-21, 10 members made partial payment of ₹ 1,04,000 towards subscription, 8 members failed to pay the subscription amount and 5 members paid the subscription amount for the year 2021-22.
OR
Following is the Receipts and Payments Account of South Literary Club for the year ended 31^st March, 2021:

Additional Information:
(i) Subscriptions outstanding as on 31^st March, 2020 were ₹ 2,000 and on 31^st March, 2021 were ₹ 2,500.
(ii) On 31^st March, 2021 salary outstanding was ₹ 600 and rent outstanding was ₹ 1,200.
(iii) On 31^st March, 2020 the club owned Furniture of ₹ 15,000 and Books of ₹ 7,000.
You are required to prepare Income and Expenditure Account of South Literary Club for the year ended 31^st March, 2021. (3)
Answer:

Note: Subscription outstanding on 31-03-2021 is ₹ 2,500 which includes the subscription outstanding for the year 2019-20 ₹ 800 (i.e., ₹ 2,000 – ₹ 1,200 received during 2020-21). Hence, the subscription outstanding for the year 2020-21 is ₹ 1,700 (i.e. ₹ 2,500 – ₹ 800). it should be noted that there is a difference between subscription outstanding on 31-03-2021 and subscription outstanding for 2020-21 (which indicates that subscription outstanding only for 2020-21).

Question 5.
Arun, Bimal and Chandan were partners sharing profits in the ratio of 2 : 2 : 1. They decided to dissolve their firm on 31^st March, 2019 when the Balance Sheet was:

Following transactions took place:
(i) Arun took over Stock at ₹ 36,000. He also took over his wife’s loan.
(ii) Bimal took over half of Debtors at ₹ 28,000.
(iii) Chandan took over Investments at ₹ 54,000 and half of Creditors at their book value.
(iv) Remaining Debtors realised 60% of their book value. Furniture sold for ₹ 30,000; Machinery ₹ 82,000 and Land ₹ 1,20,000.
(v) An unrecorded asset was sold for ₹ 22,000.
(vi) Realisation expenses amounted to ₹ 4,000.
Prepare Realisation Account. (3)
Answer:

Question 6.
Pass the necessary journal entries for the issue of debentures for the following transactions:
(A) Renuka Traders Limited took over Plant and Machinery of ₹ 16,00,000 and liabilities of ₹ 6,00,000 from Unnati Limited for a purchase consideration of ₹ 12,00,000. The payment was made by issue of 12% Debentures of ₹ 100 each at 20% premium.

(B) On 1st April, 2021, Surya Limited issued 1,600, 12% Debentures of ₹ 500 each at a premium of 20%, to Chandani Limited for Plant purchased from them costing ₹ 9,60,000.
OR
Akanksha Limited issued 8% Debentures of the face value of ₹ 20,00,000 at a discount of 6% on 1st April, 2016. These debentures are redeemable by annual drawings of ₹ 4,00,000 made on 31^st March each year. The directors decided to write off discount based on the debentures outstanding each year.

Prepare Discount on Issue of Debentures Account of Akanksha Limited for five years. (3)
Answer:

Working Note:

= \(\frac{₹ 12,00,000}{₹(100+20)}\)
= 10,000

OR
Total Discount on the Issue of Debentures = ₹ 20,00,000 × \(\frac{6}{10}\) = ₹ 1,20,000
Since debentures are redeemable by annual drawings of ₹ 4,00,000, the amount of discount written off from Statement of Profit and Loss is determined as follows:

*Ratio has been obtained by dividing Debentures Outstanding by ? 4,00,000.

Question 7.
Aman, Jay and Kant are partners in firm ‘Ajaka Limited’ sharing profits and losses in the ratio of 3 : 2 : 1. They decided to dissolve their firm on 31^st March, 2020, the date on which their Balance Sheet stood as:

The following additional information is given:
(i) The Investments are taken by Aman for ₹ 5,000 in settlement of his loan.
(ii) Assets were realised as follows:
Stock – ₹ 17,500
Debtors – ₹ 14,500
Furniture – ₹ 6,800
Machinery – ₹ 30,300

(iii) Expenses on realisation amounted to ₹ 2,000.
Prepare Realisation Account and Partners’ Capital Account at the time of dissolution of the partnership firm ‘Ajaka Limited’.
OR
On 31^st March, 2022, the Balance Sheet of Pooja, Qureshi and Ross, who were partners in Pukaro Limited was as under:

Qureshi died on 1^st July, 2022. The profit sharing ratio of the partners was 2 : 1 : 1. On the death of a partner, the partnership deed provided for the following:

(i) His share in the profits of the firm till the date of his death will be calculated on the basis of average profits of last three completed years.

(ii) Goodwill of the firm will be calculated on the basis of total profit of last two years.

(iii) Interest on loan given by the firm to a partner will be charged at the rate of 6% p.a. or ₹ 4,000, whichever is more.

(iv) Profits for the last three years 2019-20, 2020-21 and 2021-22 were ₹ 45,000, ₹ 48,000 and ? 33,000 respectively.

You are required to prepare Qureshi’s Capital Account to be rendered to his executors. Also show the working notes clearly. (5)
Answer:

OR

Working Notes:
(1) Calculation of Profit Sharing Ratio:
Old Profit Sharing Ratio of Pooja, Qureshi and Ross = 2 : 1 : 1
New Profit Sharing Ratio of Pooja and Ross = 2 : 1 and
Gaining Ratio of Pooja and Ross = 2 : 1

(2) Calculation of Qureshi’s Share of Goodwill:
Goodwill of Firm = ₹ 48,000 + ₹ 33,000 = ₹ 81,000
Qureshi’s Share of Goodwill = ₹ 81,000 × \(\frac{1}{4}\) = ₹ 20,250
This share of Goodwill will be contributed by Pooja and Ross in their gaining ratio, i.e, 2 : 1.
Pooja will contribute = ₹ 20,250 × \(\frac{2}{3}\) ₹ 13,500
Ross will contribute = ₹ 20,250 × \(\frac{1}{3}\) = ₹ 6,750

(3) Calculation of Qureshi’s Share of Profit till the date of his death:
Average prifit of last three years = \(\frac{45,000+48,000+33,000}{3}\) = ₹ 42,000
Qureshi’s Share of Profit till the date of his death
= Previous year’s Profit × Qureshi’s Share ×
= ₹ 42,000 × \(\frac{1}{4} \times \frac{3}{12}\) = ₹ 2,625

(4) Calculation of Qureshi’s Share in Reserve Fund:
Qureshi’s Share in Reserve Fund = ₹ 2,00,000 × \(\frac{1}{4}\) = ₹ 50,000

(5) Calculation of Amount due on account of Loan given to Qureshi:
Loan given to Qureshi by firm = ₹ 1,00,000
Amount of Interest till 1st July, 2022 = ₹ 1,00,000 × \(\frac{6}{100} \times \frac{3}{12}\) = ₹ 1,500

Total Amount due to firm on 1^st July = Loan given to Qureshi by firm + Amount of Interest
= ₹ 1,00,000 + ₹ 4,000 [As ₹ 4,000 > Amount of Interest]
= ₹ 1,04,000

Question 8.
Pacific Foods Limited, a FMCG company has an equity share capital of ₹ 20,00,000. The company earns a return on investment of 15% on its capital. The company needed funds for diversification. The finance manager had the following two options:
(i) Borrow ₹ 10,00,000 @ 15% p.a. from a bank payable in four equal quarterly instalments starting from the end of the fifth year.
(ii) Issue ₹ 10,00,000, 9% Debentures of ? 100 each to the public at par, redeemable after five years at a premium of 10%.

After all deliberations, on 1st April, 2021, the board of directors of the company opted for option (ii), to increase the return to the shareholders. The Balance Sheet of the company on 1^st April, 2021 shows a balance of ₹ 3,00,000 in Capital Reserve which the company decided to use for writing off the discount on issue of debentures.
You are required to answer the following questions:
(A) Pass journal entry for receipt of application money of debentures.
(B) Pass journal entry to be passed at the time of allotment of debentures.
(C) Pass journal entry to write ofFloss on issue of debentures.
(D) Prepare Loss on Issue of Debentures Account.
(E) Calculate the amount of annual fixed obligation associated with debentures. (5)
Answer:


(E) Interest on 9% Debentures = ₹ 10,00,000 × \(\frac{9}{100}\) = ₹ 90,000

Question 9.
Following is the Receipts and Payments Account of Jalandhar Sports Club for the year ending 31^st March, 2021:

Additional Information:
(i) Subscriptions outstanding was ₹ 1,200 on March 31, 2020 and ₹ 3,200 on March 31, 2021.
(ii) Locker rent outstanding on March 31, 2021 was ₹ 250.
(iii) Salary outstanding on March 31, 2021 was ₹1,000.
(iv) Fixed Deposit was made on 1st January, 2021 @10% p.a.
(v) On April 1, 2020, club has following assets: Building ₹ 36,000, Furniture ₹ 12,000, and Sports Equipments ₹ 17,500. Depreciation on these items is to be charged at 10% (including purchase).
You are required to prepare Income and Expenditure Account for the year ending 31^st March, 2021 and also ascertain the Capital Fund as on 31st March, 2020. (5)
Answer:

PART-B
Option-1
(Ana Lysis of Financial Statements)

Question 10.
Classify the following transactions as Operating Activities for a financial company and a Non- Financial Company:
(A) Payment of Interest
(B) Royalty received (2)
Answer:
(A) Payment of Interest – Financial Company
(B) Royalty received – Non-Financial Company

Question 11.
Prepare Comparative Statement of Profit and Loss from the following information:

OR
You are required to prepare Common Size Balance Sheet of Revati Trading Limited as at 31^st March, 2020 and 2021.

You are required to prepare Common Size Balance Sheet of Revati Trading Limited as at 31^st March, 2020 and 2021. (3)
Answer:

OR

Question 12.
Prepare Cash Flow Statement for the year ended 31st March, 2020 from the following Balance Sheet of Sanskar Limited as at 31^st March, 2020 and additional information provided:

Notes to Accounts:

Additional Information:
(i) An old machinery having book value of? 50,000 was sold for ₹ 60,000.
(ii) Depreciation provided on Plant and Machinery during the year was ₹ 30,000. (5)
Answer:

Working Notes:

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 Applied Mathematics Term 2 Set 5 with Solutions

Maximum Marks : 40
Time : 2 Hours

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. It contains one case study based question. Internal choice has been provided in one question.

Section – A [12 Marks]

Question 1.
Evaluate

Or
The marginal cost function of producing x units of a product is given by
MC = \(\frac{x}{\sqrt{x^{2}+3600}}\). Find the total cost function and the average cost function, if the fixed cost is ₹ 1000.
Answer:

Question 2.
Calculate the 3-yearly moving average of the following data.

Answer:

∴ 3-Yearly moving averages are 3.67,5.33,6.67, 8.33 and 10.33.

Question 3.
Find the effective rate of interest corresponding to 10% nominal rate compounded quarterly, [given (1.025) =1.1038]
Or
Find the present value of a sequence of payments of ₹ 4500 made at the end of every 6 months continuing, if money is worth 8% converted half-yearly.
Answer:
Given, nominal rate of interest,
r = 10% = 0.10
Since, interest is compounded quarterly.
∴ Number of conversion per year (m) = 4
We know that effective rate of interest
r_e = (1 + \(\frac{r}{m}\))^m – 1
= (1 + \(\frac{0.10}{4}\))⁴ – 1
= (1 + 0025)⁴ – 1
= (1025)⁴ – 1
= 1.1038 – 1
= 0.1038
= 10.38%

Or

Given, R = ₹ 4500
and rate of interest (r) = 8% = 0.08
Since, interest is compounded semi-annually.
∴ i = \(\frac{r}{2}=\frac{0.08}{2}\) = 0.04
Present value of an immediate perpetuity is
P = \(\frac{R}{i}\) ⇒ P = \(\frac{4500}{0.04}\)
= ₹ 112500

Question 4.
What sum of money invested now would establish a scholarship of ₹ 12000 to be awarded at the starting of each year to a deserving student, if money is worth 5% compounded annually?
Answer:
Given, R = ₹ 12000
i = \(\frac{5}{100}\) = 0.05

∴ Present value of a perpetuity due,
P = R(1 + i)
= 12000(1 + \(\frac{1}{0.05}\)
= 12000(1 + 20)= 12000 × 21= 252000
∴ Amount to be invested is ₹ 252000.

Question 5.
A random sample of size 25 has 60 as mean. The sum of squares of deviations from mean is 168, can this sample be regard as taken from the population having 64 as mean? Level of significance is 5%.
Answer:
Consider, H₀: µ = 64
H₁: µ > 64
n = 25, x̄ = 60, µ = M
and Σ(x_i – x̄)² = 168

Degree of freedom =25- 1=24
∴ Tabulated value of t = 1.711
Since, calculated value |t| = 7.55 is greater than
the tabulated value, so H₀ is rejected.

Question 6.
Two tailors A and B earn ₹ 300 and ₹ 400 per day, respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day.
To find how many days should each of them work and if it is desired to produced at least 60 shirts and 32 pairs of trousers at a minimum labour cost. Formulate this problem as on LPP.
Answer:
Suppose, tailor A works for x days and tailor B works for y days.

The given data can be written in the tabulor form as follows

Required linear programming problem is Min Z = 300x + 400y
Subject to constraints are 6x + 10y ≥ 60,
4x + 4y ≥ 32 ⇒ x + y ≥ 8 and
x,y ≥ 0

Section – B [12 Marks]

Question 7.
Mr. Amit have set up a sinking fund in order to have ₹ 40000 in 12 yr for his son’s higher education. What amount he has to set aside at the end of every month into the fund paying 6% per annum compounded monthly? [given, (1.005)¹⁴⁴ = 205]
Answer:
Given, A = ₹ 40000
i = \(\frac{r}{12 \times 100}=\frac{6}{12 \times 100}\)
= 0.005
n = 12 × 12 = 144 months

∴ The monthly periodic payments,

Question 8.
Find the f-test value for the following two samples from a population.
Set-I 6, 8, 5, 7
Set-II 4, 6, 5, 8
Answer:
x̄ = \(\frac{6+8+5+7}{4}=\frac{26}{4}\) = 6.5
and ȳ = \(\frac{4+6+5+8}{4}=\frac{23}{4}\) = 5.75

∴ S² = \(\frac{1}{n_{1}+n_{2}-2}\)[Σ(x – x̄)² + Σ(x – ȳ)²]
= \(\frac{1}{6}\)[5 + 8.9325]

Here, Σ(x – x̄)² = (-0.5)² + (1.5)²
+ (-1.5 )² + (0.5 )² = 0.25 + 2.25 + 2.25 + 0.25 = 5
and Σ(x – ȳ)² = (-1.75)² + (0.5)² + (- 0.75)² + (2.25 )²
= 306 + 0.25 + 0.5625 + 5.06 = 8.9325
⇒ S² = \(\frac{1}{6}\) (13.9325) = 2.3220 ⇒ S = 1.52

Question 9.
Consider the following data

Calculate 3-days moving average and display these and the original figures on the same graph.
Answer:
Given, table is shown below:

On the basis of above data, we can draw the following graph.

Question 10.
The demand function of a product is p = 10e^-x. Find the consumer’s surplus, when the market price at p₀ = 1. [given, log₁₀ e = 0.4343]
Or
The demand and supply functions for a commodity are p = x² – 6x +16 and p = \(\frac{x^{2}}{3}+\frac{4 x}{3}\) + 4, respectively. Find the equilibrium point assuming x ≤ 5.
Answer:
The demand function is p = 10e^-x
It is given that p₀ = 1
On putting the value of p₀ = 1 in p = 10e^-x, we get
I = 10e^-x ⇒ e^-x = 10
⇒ x = log_e10
∴ x₀ = log_e10

Then, consumer’s surplus is given by

Or
The demand and supply functions are p= D(x) and p = S(x),
where D(x) = x² – 6x +16 and S(x) = \(\frac{x^{2}}{3}+\frac{4 x}{3}\) + 4

The equilibrium point (x0, p0) is the point at which the demand supply curves intersect
∴D(x) = S(x)
x² – 6x +16 = \(\frac{x^{2}}{3}+\frac{4 x}{3}\) + 4
\(\frac{2}{3}\)x² – \(\frac{22}{3}\)x + 12 + 0
x² – 11x + 18 = 0
(x – 2)(x – 9) = 0 ⇒ x = 2 [∵ x ≤ 5]
On putting x = 2 in D(x), we get
P = (2 )² – 6(2) + 16
p = 4 – 12 + 16 = 8
Hence, (2, 8) is the equilibrium point.

Section – C [16 Marks]

Question 11.
If Mr. Nirav deposits ₹ 250 at the beginning of each month in an account that pays an interest of 6% per annum compounded monthly, how many months will be required for the deposit to amount to at least ₹ 6390?
Answer:
Given, Mr. Nirav deposit amount per month is a = ₹ 250
The total amount of annuity is (M) = ₹ 6390
Rate of interest, r = 6% = \(\frac{6}{100}\) per annum

Let the number of months be n.
We know that

⇒ 0.127164 + 1 = (1.005)ⁿ
⇒ 1.127164 = (1.005)ⁿ
Taking log on both sides of Eq. (i), we get
log 1.127164 = n log 1.005
⇒ n = \(\frac{\log (1.127164)}{\log (1.005)}\) = 24110075
∴Number of months required = 24 months

Question 12.
Mr. Ankit has invested ₹ 6 lakhs in a financial plan and after the completion of 5 yr. He received a sum of ₹ 10 lakhs. Find
(i) the absolute return.
(ii) the simple annual growth rate (SAGR).
(iii) the compound annual growth rate (CAGR). [given, (1.66)^1/5 =1.1067]
Answer:
Given, BV = ₹ 600000
EV = ₹ 1000000
n = 5 yr

Question 13.
In a bank principal increases at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it be worth after 10 yr (given, e^0.5 = 1.648)?
Or
Suppose, it is given that the rate at which some bacteria multiply is proportional to the instantaneous number present. If the original number of bacteria triples in three hours. In how many hours will it be four times.
Answer:
Let P and t be the principal and time, respectively.
It is given that the principal increases continuously at the rate of 5% per year.
So, \(\frac{d P}{d t}\) = 5%of P ⇒ \(\frac{d P}{d t}=\frac{5}{100}\)P
On separating the variable, we get \(\frac{d P}{P}=\frac{1}{20}\) dt

On integrating both sides, we get
∫ \(\frac{d P}{P}\) = ∫ \(\frac{1}{20}\) dt ⇒ log |P| = \(\frac{1}{20}\)t + C …(i)
When t = 0, P = ₹ 1000 (initialy) this
log 1000 = C …(ii)
On putting the value of C in Eq. (i), we get

Hence, after 10 yr, the amount will worth ₹ 1648

Or

Let the original count of bacteria be N₀ and at any time the count of bacteria be N.
Then, \(\frac{d N}{d t}\) ∝ N ⇒ \(\frac{d N}{d t}\) = kN ⇒ \(\frac{d N}{N}\) = kdt

On integrating both sides, we get
∫\(\frac{d N}{N}\) = ∫kdt
⇒ logN = kt + C
⇒ N = e^kt+C
⇒ N = e^Ce^kt
⇒ N = Aêt
When t = O, then N = N₀, we get A = N₀
∴ N = N₀e^kt
When t = 3, N = 3N₀
∴ 3N₀ = N₀e^3k ⇒ e^3k = 3
= e^k = (3)e^1/3
When N = 4N₀
∴ 4N₀ = N₀e^kt
⇒ e^kt = 4
⇒ (3^1/3)^t = 4
⇒ t = \(\frac{3 \log 4}{\log 3}\)h

Case Based/Data Based

Question 14.
A company manufactures three kinds of calculators A, B and C in its two factories I and II. The company has got an order for manufacturing atleast 6400 calculators of kind A, 4000 of kind B and 4800 of kind C. The daily output of factory I is of 50 calculators of kind A, 50 calculators of kind B and 30 calculators of kind C.
The daily output of factory II is 40 calculators of kind A, 20 of kind B and 40 of kind C. The cost per day to run factory I is ₹ 12000 and of factory II is f 15000. Let factory71 run x days and factory II run y days.
On the basis of above information, answer the following question.
(i) Formulate this problem as an LPP.
Answer:
Total cost (in ₹) = 12000x + 15000y
Then, minimise Z = 12000x + 15000y

Subject to constraints are,
50x + 40y ≥ 6400 or 5x + 4y ≥ 640 …(i)
50x +20y ≥ 4000 or 5x +2y ≥ 400 …(ii)
30x + 40y ≥ 4800 or 3x + 4y ≥ 480 …(iii)
and x ≥ 0, y ≥ 0 …………(iv)

(ii) How many days do the factory I has to be in operation to produce the order with the minimum cost?
Answer:
The above Eqs. (i), (ii), (iii) and (iv) can be represented graphically as

Clearly, the feasible region ABCD (shaded region), where comer points are A (160,0), B(80,60), C(32,120)and D(0,200)is unbounded.

The values of Z at comer points are given below:

In the above table, we find that minimum value of Z is 1860000 occur at the point B (80,60).
But we cannot say that it is a minimum value of Z as region is unbounded.

Therefore, we have to draw the graph of the inequality
12000x + 15000y < 1860000
or
12x + 15y < 1860
or
\(\frac{x}{155}+\frac{y}{124}\) < 1

Now, check whether the resulting open half plane has any point common with feasible region. From figure, we see that it has no point in common.

Thus, the minimum value of Z is ₹ 1860000 attained at the point B(80,60).
Hence, factory I should run for 80 days.

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 6 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Accountancy Standard Term 2 Set 6 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This question paper comprises two Parts – A and B. There are 12 questions in the question paper. Alt1 questions are compulsory.
• Part-A is compulsory for all candidates.
• Part-B has two options i.e., (i) Analysis of Financial Statements and (ii) Computerized Accounting. Students must attempt only one. of the given options.
• There is no overall choice. However, an internal choice has been provided in 3 questions of three marks and 1 question of five marks.

PART-A
(Accounting for Not-for-Profit Organisations, Partnership Firms and Companies)

Question 1.
From the following information provided by Reunion Club, calculate the amount of subscriptions which will be treated as income for the year ended 31^st March, 2019:
(i) Subscriptions received during the year ended 31^st March, 2019 amounted to ₹ 25,000.
(ii) Subscriptions outstanding in the beginning of the year ended 31st March, 2019 amounted to ₹ 3,000.
(iii) Subscriptions outstanding for the year ended 31^st March, 2019 amounted to ₹ 5,500.
(iv) Subscriptions not yet collected for the year ended 31^st March, 2019 amounted to ₹ 5,000. (2)
Answer:

Question 2.
Babita, Chetan and Dhanush were partners in a firm sharing profits and losses in the ratio of 1 : 4 : 5. On 31^st March, 2020 the firm was dissolved and on that date the Balance Sheet of the firm showed a loan of ₹ 10,000 given by Chetan’s brother Rohan. Chetan agreed to pay his brother’s loan.

Pass necessary journal entry for the above on the firm’s dissolution. (2)
Answer:

Question 3.
Preeti, Kareena and Rani were partners sharing profits in the ratio of their Capital contribution which were ₹ 6,00,000; ₹ 4,00,000 and ₹ 5,00,000 respectively. Their books are closed on 31^st March every year. Preeti dies on 24th August, 2020.

Under the partnership deed, deceased partner is entitled to her share of profit/loss to the date of death based on the average profits of preceding four years.
Profits were:
2016-17 – ₹ 50,000
2017-18 – ₹ 20,000 (Loss)
2018-19 – ₹ 30,000
2019-20 – ₹ 60,000
Pass necessary journal entry for the treatment of profit and also show your workings clearly. (2)
Answer:

Working Notes:
Average Profit/(Loss) = \(\frac{₹ 50,000-₹ 20,000+₹ 30,000+₹ 60,000}{4}\)
= ₹ 30,000
Number of Days from 01.04.20 to 24.08.20 = 30 + 31 + 30 + 31 + 24 = 146
Preeti’s share of Profit upto the date of her death = ₹ 30,000 × \(\frac{6}{15} \times \frac{146}{365}\)
= ₹ 4,800

Question 4.
Show how are the following items dealt with while preparing the final accounts of Fitness Sports Club for the year ended 31^st March, 2020:
(i) Expenditure on construction of Stadium is ₹ 6,00,000. The construction work is in progress and has not yet completed.
(ii) Stadium Fund as at 31^st March, 2019 is ₹ 10,00,000, and Capital Fund as at 31^stMarch, 2019 is ₹ 20,00,000.
(iii) Donation Received for Stadium on 1st January, 2020 is ₹ 5,00,000.
OR
The book value of Machinery on 1^st April, 2020 is ₹ 1,20,000. Half of this Machinery is sold for ₹ 40,000 on 30th September, 2020. Depreciation is to be charged on Machinery @ 10% p.a.
Calculate loss on sale of Machinery. Show how the loss on sale and depreciation on Machinery will be shown in the Income and Expenditure Account for the year ended 31st March, 2021. (3)
Answer:

OR

Working Note:
Calculation of Profit or Loss on Sale of Machinery

Question 5.
In a partnership firm, Guruprakash and Sadguru were partners sharing profits and losses is the ratio of 3 : 1. On 1st April, 2019 partners decided to dissolve the firm.
After transferring the assets (other than cash) and outsider’s liabilities of the firm to Realisation Account, you are given the following information:
(i) A creditor of ₹ 1,44,000 accepted furniture valued at ₹ 2,00,000 and paid to the firm ₹ 56,000.
(ii) A second creditor for ? 20,000 accepted stock at ₹ 18,000 in full settlement of his claim.
(iii) A third creditor amounting to ₹ 36,000 accepted ₹ 18,000 is cash and investment worth ₹ 17,200 is full settlement of his claim.
(iv) Loss on dissolution of the firm was ₹ 6,000.

Pass the necessary journal entries for the above transaction in the books of the partnership firm assuming that all payments were made by cheque. (3)
Answer:

Question 6.
XYZ Ltd. issued 5,000, 10% Debentures of ₹ 100 each on 1^st April, 2015 at a discount of 10%, redeemable at a premium of 10% after four years.
Pass necessary journal entries in the books of XYZ Ltd. for debenture interest for the year ended 31^st March, 2016, assuming that the interest on debentures was payable half-yearly on 30^th September and 31^st March. The rate of tax is 10%.
OR
Dauji Bros. Limited purchased the business of Mohan Bros. Limited consisting assets of the book value ₹ 20,00,000 and the liabilities of ₹ 2,50,000. It was agreed that the purchase consideration, settled at ? 19,00,000 be paid by issuing 12% Debentures of ₹ 100 each.
Pass journal entries in the books of the firm, if the debentures are issued:
(A) at a discount of 10%; and
(B) at a premium of 10%.
It was agreed that any fraction of debentures be paid in cash. (3)
Answer:


OR

(A) When Debentures are issued at 10% Discount:

Working Note:

= \(\frac{₹ 19,00,000}{₹(100-10)}\)
= \(\frac{₹ 19,00,000}{₹(100-10)}\)
= 21,111.11 or 21,111

(B) When Debentures are issued at 10% Premium:

Working Note:

= \(\frac{₹ 19,00,000}{₹(100+10)}\)
= \(\frac{₹ 19,00,000}{₹ 110}\)
= 17,272.72 or 17,272

Question 7.
Balance Sheet of X, Y and Z who shared profits in the ratio of 5 : 3 : 2, as on 31^st March, 2019 was as follows:

Y retired on 1^st April, 2019 and it was agreed between the partners that:
(i) Goodwill of the firm is valued at ₹ 1,12,500 and Y’s share of it be adjusted into the accounts of X and Z who are going to share future profits in the ratio of 3 : 2.
(ii) Fixed Assets be appreciated by 20%.
(iii) Stock of the firm be reduced to ₹ 75,000.
(iv) Y be paid amount brought in by X and Z so as to make their capitals proportionate to their new profit sharing ratio.
Prepare Capital Account of Partners at the time of Y’s retirement and the Balance Sheet of the New Firm.
OR
Balance Sheet of Leena, Meena and Neena as at 31^st March, 2021, who were sharing profits in the ratio of 5 : 3 :1, was:

The partners decided to dissolve the business on 31^st March, 2021.
The Assets of the business were realised as follows:
(i) Stock realised ₹ 23,400.
(ii) Debtors realised 50% amount.
(iii) Other Fixed Assets were realised at 10% less than their book value.
Bills Payable were settled for ₹ 32,000. There was an Outstanding Bill of Electricity of ₹ 800 which was paid off. Realisation expenses ₹ 1,250 were also paid.
You are required to prepare Realisation Account, Partner’s Capital Accounts and Bank Account of the firm. (5)
Answer:

Working Notes:
(1) Calculation of Gaining Ratio:
Old Profit Sharing Ratio of X, Y and Z = 5 : 3 : 2
New Profit Sharing Ratio of X and Z = 3 : 2
Gaining Ratio = New Profit Sharing Ratio – Old Profit Sharing Ratio
X’s Gain = \(\frac{3}{5}-\frac{5}{10}\) = \(\frac{6-5}{10}-\frac{1}{10}\)
Z’s Gain = \(\frac{2}{5}-\frac{2}{10}=\frac{4-2}{10}=\frac{2}{10}\)
Gaining Ratio of X and Z = 1 : 2

(2) Adjustment of Goodwill:
Goodwill of Firm = ₹ 1,12,500
Y’s share of Goodwill = ₹ 1,12,500 × \(\frac{3}{10}\) = ₹ 33,750
which will be compensated by X and Z in their gaining ratio, i.e., 1 : 2
X will compensate = ₹ 33,750 × \(\frac{1}{3}\) = ₹ 11,250
Z will compensate = ₹ 33,750 × \(\frac{2}{3}\) = ₹ 22,500
(3)

(4) Adjustment of Capital:
Adjusted Capital of X = ₹ 1,65,000 + ₹ 11,250 + ₹ 15,000 – ₹ 11,250 = ₹ 1,80,000
Adjusted Capital of Y = ₹ 84,000 + ₹ 6,750 + ₹ 9,000 + ₹ 11,250 + ₹ 22,500 = ₹ 1,33,500
Adjusted Capital of Z = ₹ 66,000 + ₹ 4,500 + T 6,000 – ₹ 22,500 = ₹ 54,000
New Capital of Firm = ₹ 1,80,000 + ₹ 1,33,500 + ₹ 54,000 = ₹ 3,67,500
X’s New Capital = ₹ 3,67,500 x \(\frac{3}{5}\) = ₹ 2,20,500
Z’s New Capital = X 3,67,500 x \(\frac{2}{5}\) = ₹ 1,47,000
X brings \n = ₹ 2,20,500 – ₹ 1,80,000 = ₹ 40,500
Z brings in = ₹ 1,47,000 – ₹ 54,000 = ₹ 93,000
OR

Question 8.
Suhana Textiles Limited was incorporated on 1st April, 2015 with registered office in Gujarat. The company is growing year by year, and begins to expand its operations throughout India. To expand their business in Rajasthan, directors of the company decided to purchase one of the well- known textile firms of Rajasthan, Jaipuriya Textiles Limited,

On 1^st April, 2021 Suhana Textiles Limited bought the business of Jaipuriya Textiles Limited consisting sundry assets of ₹ 36,00,000 and sundry creditors of ₹ 10,00,000 for a consideration of ₹ 30,72,000.

Suhana Textiles Limited issued 12% Debentures of ₹ 100 each fully paid, at a discount of 4% in satisfaction of purchase consideration to Jaipuriya Textiles Limited. On 10th June, 2021, the company also issued 500,10% Debentures of ₹ 100 each credited as fully paid-up to the promoters for their services to incorporate the company.

You are required to answer the following questions:
(A) Calculate the amount of Goodwill purchased by Suhana Textiles Limited of Jaipuriya Textiles Limited.
(B) Pass journal entry for the purchase of business of Jaipuriya Textiles Limited.
(C) Calculate the number of debentures issued to Jaipuriya Textiles Limited.
(D) Pass journal entry for the allotment of debentures to Jaipuriya Textiles Limited.
(E) Pass journal entry for the allotment of debentures to the underwriters. (5)
Answer:
(A) Amount of Goodwill = Purchase Consideration – (Value of Assets – VaLue of Liabilities)
= ₹ 30,72,000 – (₹ 36,00,000 – ₹ 10,00,000)
= ₹ 4,72,000

Question 9.
From the following Receipts and Payments Account and information given below, prepare Income and Expenditure Account and Balance Sheet of Superior Literacy Organisation as on 31^st March, 2020.

Additional Information:
(i) Subscriptions outstanding as on 31^st March, 2019 was ₹ 8,000 and on 31st March, 2020 was ₹ 6,000.
(ii) Fixed Deposit with Bank was made on 30th September, 2019 @10% p.a.
(iii) On 31st March, 2020 Salary outstanding ? 2,400, and one month Rent paid in advance.
(iv) On 1st April, 2019, Superior Literacy Organisation owned Furniture of ₹ 48,000, Books of ₹
20,000. (5)
Answer:


Working Note:
(1) Calculation of Prepaid Rent:
Prepaid Rent (for Year 2020-21) = ₹ 26,000 x \(\frac{1}{13}\) = ₹ 2,000

PART-B
Option-1
(Analysis of Financial Statements)

Question 10.
Identify the following transactions as belonging to Operating Activities, Investing Activities, Financing Activities or Cash and Cash Equivalents:
(A) Cash paid against Services taken.
(B) Bank Overdraft (2)
Answer:
(A) Cash paid against Services taken – Operating Activities
(B) Bank Overdraft – Financing Activities

Question 11.
Explain some advantages of Common Size Statements.
OR
You are provided with the Comparative Balance Sheet of Picasso Limited with missing information. You are required to fill in the blanks:

(3)
Answer:
The advantages of Common Size Statements are as follows:
(i) Easy to Understand: Common Size Statement helps the users of financial statement to make clear about the ratio or percentage of each individual item to total assets/liabilities of a firm. For example, if an analyst wants to know the working capital position he may ascertain the percentage of each individual component of current assets against total assets of a firm and also the percentage share of each individual component of current liabilities.

(ii) Comparison at a Glance: An analyst can compare the financial performances at a glance since percentage of increase or decrease of each individual component of cost, assets, liabilities etc. are available and he can easily ascertain his required ratio.

(iii) Helpful for Time Series Analysis: A Common Size Statement helps an analyst to find out a trend relating to percentage share of each asset in total assets and percentage share of each liability in total liabilities.

(iv) Helpful in analysing Structural Composition: A Common Size Statement helps the analyst to ascertain the structural relations of various components of cost/expenses/assets/liabilities etc. to the required total of assets/liabilities and capital.
OR

Question 12.
Following is the summarised Balance Sheet of Philips India Ltd. as at 31^st March 2020:


Additional Information:
(i) Investments costing ? 24,000 were sold during the year for ₹ 25,500.
(ii) Provision for Tax made during the year was ₹ 27,000.
(iii) During the year, a part of the Fixed Assets costing ₹ 30,000 was sold for ₹ 36,000. The profits were included in the Statement of Profit and Loss.
(iv) The Interim Dividend paid during the year amounted to ₹ 1,20,000.
You are required to prepare Cash Flow Statement for the year ended 31st March, 2020. (5)
Answer: