Category: CBSE Sample Papers

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

CBSE Sample Papers for Class 12 Maths Term 2 Set 2 with Solutions

Maximum Marks : 40
Time : 2 Hours

General Instructions:

• This question paper contains three sections – A, B and C. Each part is compulsory.
• Section – A has 6 short answer type (SA1) questions of 2 marks each.
• Section – B has 4 short answer type (SA2) questions of 3 marks each.
• Section – C has 4 long answer type questions (LA) of 4 marks each.
• There is an internal choice in some of the questions.
• Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section – A

Question 1.
Find: ∫\(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\) dx
OR
Evaluate: ∫\(\frac{d x}{1-\tan x}\)
Answer:
We have given

OR

Put cos x – sin x = t
Differentiating both sides, we get
⇒ (- sin x – cos x) dx = dt
⇒ (sin x + cos x) dx = – dt
∴ I = \(\frac{-1}{2} \int \frac{d t}{t}+\frac{1}{2} \int d x\)
⇒ I = \(\frac{-1}{2}\)log |t| + \(\frac{x}{2}\) + C
⇒ I = \(\frac{-1}{2}\) log |cos x – sin x| + \(\frac{x}{2}\) + C

Question 2.
Determine the order and degree of the following differential equation \(\frac{d^{2} y}{d x^{2}}\) = cos 3x + sin 3x
Answer:
⇒ \(\frac{d^{2} y}{d x^{2}}\) – cos 3x – sin 3x = 0
The highest order derivative present in the differential equation is \(\frac{d^{2} y}{d x^{2}}\). Therefore, its order is two.
It is a polynomial equation in \(\frac{d^{2} y}{d x^{2}}\) and the power raised to \(\frac{d^{2} y}{d x^{2}}\) is 1.
Hence, its degree is one.

Question 3.
Find a vector r equally inclined to the three axes and whose magnitude is 3√3 units.
Answer:
We have
|\(\vec{r}\)| = 3√3
Since, \(\vec{r}\) is equally inclined to the three axes, direction consines of the unit vector \(\vec{r}\) will be same.
i.e. l = m = n
Now we know that,
l² + m² + n² = 1
⇒ l² + l² + l² = 1 (∴ l = m = n)
⇒ 3l² = 1
⇒ l² = \(\frac{1}{3}\)

Question 4.
Find the value of X such that the line \(\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4}\) is perpendicular to the plane 3x – y – 2z = 7.
Answer:
Given that line \(\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4}\) is perpendicular to plane 3x – y – 2z = 7
∴ \(\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}\)
When a line is perpendicular to a plane, their direction ratios are proportional]
⇒ 2 = – λ = 2
⇒ – λ = 2
⇒ λ = – 2

Question 5.
Three persons A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. Find the probability of two hits.
Answer:
Here,
P(A) = 0.4, P(Ā) = 0.6, P(B) = 0.3,
P(B̄) = 0.7, P(C) = 0.2 and P(C̄) = 0.8
∴ Probability of two hits = P(A).P(B).P(C̄) + P(A).P(B̄) + P(C) + P(Ā).P(B).P(C)
= 0.4 × 0.3 × 0.8 + 0.4 × 0.7 × 0.2 + 0.6 × 0.3 × 0.2
= 0.096 + 0.056 + 0.036 = 0.188

Question 6.
A die is rolled. Consider the events
A = {2, 4, 6}, B = {4, 5}, C = (3, 4, 5, 6}
Find P[A ∪ B/C].
Answer:
Given that, A = {2, 4, 6), B = {4, 5}, C = {3, 4, 5, 6}
Now, A ∪ B = {2, 4, 6} ∪{4, 5} = {2, 4, 5, 6}
So, P(A ∪ B) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
Now, (A ∪ B) ∩ C = {2, 4, 5, 6} ∩ {3, 4, 5, 6} = {4, 5, 6}
So, P[(A ∪ B) ∩ C] = \(\frac{3}{6}=\frac{1}{2}\)
Also P(C) = \(\frac{4}{6}=\frac{2}{3}\)
Required probability = P[A ∪ B)/C] = \(\frac{\mathrm{P}[(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})}\)
= \(\frac{1 / 2}{2 / 3}=\frac{1}{2} \times \frac{3}{2}=\frac{3}{4}\)

Section – B

Question 7.
Evaluate:

Answer:

Question 8.
Solve the following differential equation :
\(\frac{d y}{d x}\) + y = cos x – sin x
OR
Given that \(\frac{d y}{d x}\) = e^-2y and if x = 5, then y = 0. When y = 3, then find the value of x.
Answer:
Given differential equation is dy
\(\frac{d y}{d x}\) + y = cos x – sin x ……..(1)
Here P = 1, Q = cos x – sin x.
I.F. = e^{∫1 dx} = e^x
Solution of given equation will be given by
y.I.F. = ∫Q × I.F.dx + C
⇒ y.e^x = ∫e^x(cos x – sin x) dx + C}
⇒ y.e^x = ∫e^x cos x dx – ∫e^x sin x dx + C
⇒ y.e^x = – e^x sin x – ∫e^x.(- sin x)dx – ∫e^x sin x dx + C
⇒ y.e^x = – e^x sin x + C
⇒ y = – sin x + ce^{– x}

OR

Given:
\(\frac{d y}{d x}\) = e^-2y
⇒ \(\frac{d y}{e^{-2 y}}\) = dx
On integrating both sides
⇒ ∫e^2ydy = ∫dx
⇒ \(\frac{e^{2 y}}{2}\) = x + C
When x = 5 and y = 0, then putting these values in equation (1),
\(\frac{e^{0}}{2}\) = 5 + C
⇒ \(\frac{1}{2}\) = 5 + C
⇒ C = \(\frac{1}{2}\) – 5 = \(\frac{-9}{2}\)

Question 9.
Using vectors, find the area of a triangle with vertices A (1, 1, 2), B (2, 3, 5) and C(1, 5, 5).
Answer:
A(1, 1, 2), B(2, 3, 5) and C (1, 5,5) are the vertices of the given triangle.
\(\overrightarrow{\mathrm{AB}}\) = (2 – 1) î + (3 – 1)ĵ + (5 – 2)k̂
∴ \(\overrightarrow{\mathrm{AB}}\) = î + 2ĵ + 3k̂
and \(\overrightarrow{\mathrm{AC}}\) = (1 – 1)î + (5 – 1)ĵ + (5 – 2)k̂
∴ \(\overrightarrow{\mathrm{AC}}\) = 0î + 4ĵ + 3k̂
Area of ∆ ABC = \(\frac{1}{2}\)

Question 10.
Find the shortest distance between the following pair of lines:
\(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\) and \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\)
OR
Find the cartesian equation of the line passing through the point A (1, 2, -4) and perpendicular to the lines:
Answer:
Given lines are
\(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
and \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\)
Here, (a₁, b₁, c₁) = (1, -2, 1); x₁ = 3, y₁ = 5, z₁ = 7
and (a₂, b₂, c₂) = (7, -6, 1); x₂ = -1, y₂ = -1, z₂ = -1
Now, we know that shortest distance between two lines is given by

Hence, the required shortest distance is 2√29 units.

OR

Equation of line passing through A(1, 2, – 4) and parallel to \(\vec{b}\) = (b₁î + b₂ĵ + b₃k̂)

Since line (1) is perpendicular to the lines (2) and (3).
∴ 2b₁ + 3b₂ + 4b₃ = 0
and b₁ – 3b₂ + 5b₃ = 0
On solving (4) and (5), we get

Section – C

Question 11.
Evaluate: ∫\(\frac{3 x+4}{(x-1)(x+2)(x-3)}\)dx.
Answer:

⇒ (3x + 4) = A(x + 2) (x – 3)+ B (x – 3) (x – 1) + (x + 2) (x – 1) ……… (2)
Put x = 1 in (2)
⇒ 7 = A(3) (- 2) + B (0) + C(0)
⇒ 7 = – 6A
⇒ A = –\(\frac{7}{6}\)
Put x = 3 in (2)
⇒ 13 = 0 + 0 + C (5)(2)
⇒ C = \(\frac{13}{10}\)
Put x = – 2 in (2)
⇒ – 2 = B(- 3) (- 5)
⇒ B = \(\frac{-2}{15}\)
Putting these values in (1)

Question 12.
Find the area of the region enclosed between the parabola 4y = 3x² and the line 3x – 2y + 12 = 0.
OR
Find the area of the region quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4.
Answer:
Given 4y = 3x²
and y = \(\frac{3}{2}\)x + 6
Solving (1) and (2), we get
x = – 2 or x = 4

OR

The area of region bounded by the circle, x² + y² = 4, x = √3y, and the x-axis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is √3, 1
Area ∆ OAB = Area of ∆ OCA + Area of ACB
∴ Area of ∆ OCA = \(\frac{1}{2}\) × OC × AC = \(\frac{1}{2}\) × √3 × 1 = \(\frac{\sqrt{3}}{2}\)

Therefore area enclosed by x-axis, the line x = √3y, and the circle x² + y² = 4 in the first quadrant
= \(\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}\) units.

Question 13.
Find the distance of point – 2î + 3ĵ – 4k̂ from the line \(\vec{r}\) = î + 2ĵ – k̂ + λ( î + 3ĵ – 9k̂) measured parallel to the plane x – y + 2z – 3 = 0.
Answer:
Let the given point be A (- 2î + 3ĵ – 4k̂)
Equation of line is,
\(\vec{r}\) = î + 2ĵ – k̂ + λ( î + 3ĵ – 9k̂) …….. (i)
Let B be the point on the line such that AB is parallel to plane
x-y + 2z – 3 = 0
In cartesian form, equation (1) can be written as,
So, \(\frac{x-1}{1}=\frac{y-2}{3}=\frac{z+1}{-9}\) = λ(say)
x = λ + 1,
y = 3λ + 2
z = – 9λ – 1
Let co-ordinates of B be (λ, + 1, 3λ + 2, – 9λ -1)
Now, equation of line AB can be written as,

The line AB is parallel to the given plane, so it will be perpendicular to plane’s normal.
Direction ratios of AB = (λ + 3, 3λ – 1, – 9λ + 3)
Direction ratio of plane = (1, – 1, 2)
∴ (λ + 3)(1) + (3λ – 1)(-1) + (- 9λ + 3) (2) = 0
⇒ λ + 3 – 3λ + 1 – 18λ + 6 = 0
⇒ – 20λ + 10 = 0
⇒ λ = \(\frac{1}{2}\)
So, point B is

Thus, Distance AB = \(\sqrt{\left(\frac{3}{2}+2\right)^{2}+\left(\frac{7}{2}-3\right)^{2}+\left(\frac{-11}{2}+4\right)^{2}}\)
= \(\frac{\sqrt{59}}{2}\) Units.

Question 14.
In an office three employees Vinay, Sonia and Iqbal process incoming copies of a certain from. Vinay processes 50% of the forms. Sonia processes 20% and Iqbal processes the remaining 30% of the forms. Vinay has an error rate of 0.06%, Sonia has an error rate of 0.04% and Iqbal has an error rate of 0.03%.

Based on the above information answer the following:
(i) The total probability of committing an error in processing the form is?
(ii) The manager of the company wants to do a quality check. During inspection he selects a form at random from the days output of processed forms. If the form selected at random has an error, the probability that the form is NOT processed by Vinay is?
Answer:
p(V) = 50% = \(\frac{50}{100}\)
p(S) = 50% = \(\frac{20}{100}\)
p(I) = 50% = \(\frac{30}{100}\)
Let E be the event that error occurred.
P(E/V) = \(\frac{6}{100}\)
P(E/S) = \(\frac{4}{100}\)
P(E/I) = \(\frac{3}{100}\)

(i) P(E) = P(V) P(E/V) + P(S) P(E/S) + P(I) P(E/I)

(ii)

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 8 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Accountancy Standard Term 2 Set 8 with Solutions

Maximum Marks: 40
Time allowed: 2 Hours

General Instructions:

• This question paper comprises two Parts -A and B. There are 12 questions in the question paper. All questions are compulsory.
• Question nos. 1 to 3 and 10 are Short Answer Type I Questions carrying 2 marks each.
• Question nos. 4 to 6 and 11 are Short Answer Type II Questions carrying 3 marks each.
• Question nos. 7 to 9 and 12 are Long Answer Type Questions carrying 5 marks each.
• 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)

Short Answer (SA) Type I Questions (2 Marks)

Question 1.
Nykaa Ltd, a famous makeup brand purchased furniture from Raj Ltd. and paid to them as follows
(i) By issuing 400, 9% debentures of ₹ 100 each at a discount of 10%.
(ii) By issuing 200, 8% debentures of ₹ 100 each.
(iii) Balance by accepting a bill of exchange of ₹ 1,00,000
Pass the necessary journal entries.
Answer:

Working Note
Purchase Consideration = 400 x 90 + 200 x 100 +1,00,000= ₹ 1,56,000

Question 2.
Medhavi Ltd. took a loan of ₹ 2,00,000 from Axis Bank and issued them 3,000, 9% debentures of ₹ 100 as collateral security. Pass necessary journal entries for the above transactions.
Answer:

Question 3.
Name the head under which, the following will appear in the balance sheet.
(i) Debenture suspense account.
(ii) Securities premium reserve account.
Answer:
(i) Debenture suspense account shall appear under the head ‘non-current assets’ and sub-head ‘other non-current assets in the balance sheet.
(ii) Securities premium reserve shall appear under the head ‘Shareholders’ funds’ and sub-head ‘Reserve and surplus’ in the balance sheet.

Short Answer (SA) Type II Questions (3 Marks)

Question 4.
A, B, and C were partners in a firm sharing profits in the ratio of 2: 2 :1. On 28th February, 2019, their firm was dissolved. From the following information, complete realisation account and partners’ capital accounts.

Answer:

Question 5.
X, Y and Z were partners in a firm sharing profits in the ratio of 5 : 3: 2. Unfortunately, Y met with a tragic car accident in which he died. Goodwill appeared in their books at a value of ₹ 60,000 and general reserve at ₹ 20,000. On the date of his death, the goodwill of the firm was valued at ₹ 2,40,000.

The new profit sharing ratio decided among X and Z was 2 : 3. Record necessary journal entries. Or P, Q, and R were partners in a firm sharing profits in the ratio of 2: 2 :1. P died on 31st March, 2020. The balance sheet of the firm on that date was as under
Balance Sheet
as at 31st March, 2020

On P’s death, a claim of ₹ 12,000 was accepted for workmen’s compensation. Pass necessary journal entries for general reserve, profit and loss account and workmen’s compensation fund, in the books of the firm.
Answer:

Working Notes
1. Calculation of Gaining Ratio = New Ratio – Old Ratio
X =\(\frac{2}{5}-\frac{5}{10}=\frac{1}{10}\) = (Sacrifice), Z=\(\frac{3}{5}-\frac{2}{10}=\frac{4}{10} \) (Gam)

2. Calculation of Share of Goodwill
Y’s Share of Goodwill =2,40000 x \(\frac{3}{10}\) = ₹ 72,000
X’s Share of Goodwill = 2,40,000 x \(\frac{1}{10}\) = ₹ 24,000

Question 6.
Pass the necessary journal entries for the issue of debentures in the following cases
(i) 20,000,12% debentures of ₹ 50 each issued at 10% premium, repayable at 20% premium.
(ii) 15,000,10% debentures of ₹ 100 each issued at 10% premium, repayable at par.
Or Pass the following journal entries for the issue of debentures in the books of Brothers Ltd.
(i) 5,000,10% debentures of ₹ 100 each issued at par repayable at 10% premium.
(ii) 5,000, 10% debentures of ₹ 100 each issued at 10% premium, repayable at 10% premium.
Answer:

Long Answer (LA) Type Questions (5 Marks)

Question 7.
The following is the receipts and payments account of Ananda Hospital for the year ended on 31st March, 2020. You are required to prepare income and expenditure account for the year ended on 31st March 2020 and balance sheet as on that date.
Receipts and Payments Account

Additional Information

Particulars	31st March 2019 (₹)	31st March 2020 (₹)
(i) Subscription-in-arrear	1,400	2,800
(ii) Subscription-in-advance	4,200	3,500
(iii) Stock of Medicines	30,800	35,000
(iv) Amount due to Suppliers of Medicines	19,600	32,200
(v) Value of Equipments	75,600	95,200
(vi) Value of Building	1,90,400	1,79,200

Answer:

Question 8.
Randhir, Rishi, and Rajeev are partners with 2:2:1 ratio.

Additional Information
(i) Rajeev takes retirement.
(ii) New profit sharing ratio between Randhir and Rishi is 11: 9 and the goodwill of the firm is valued at ₹ 40,000.
(iii) Investment is taken over by Rajeev at ₹ 15,000.
(iv) ₹ 6,000 worth unrecorded typewriter is taken by Randhir at ₹ 5,000.
(v) Building increase by 10%.
(vi) ₹ 10,000 paid to Rajeev in cash and the balance transferred to his loan account.
Prepare necessary accounts and balance sheet.
Or A, B, and C are partners with 2:2:1 ratio. The following is their balance sheet

Additional Information
(i) C takes retirement.
(ii) Goodwill of the firm is valued at ₹ 60,000.
(iii) Building undervalued by ₹ 20,000.
(iv) A debtor of ₹ 10,000 became insolvent and nothing is receivable from him.
(v) Provision for outstanding repair bills ₹ 5,000.
(vi) ₹ 10,000 unrecorded creditors brought into account.
(vii) A and B decide to pay off C by taking necessary bank overdrafts.
Prepare revaluation account, partners’ capital accounts, bank account, and balance sheet.
Answer:

Working Notes
GoodwiIofFirm=₹ 40,000, Rajeev’s Share of Goodwill =40,000 x \(\frac{1}{5}\) = ₹ 8,000
Gaining Ratio =New Share – Old Share, Randhir = \(\frac{11}{20}-\frac{2}{5}=\frac{11-8}{20}=\frac{3}{20} \)
Rishi=\(\frac{9}{20}-\frac{2}{5}=\frac{9-8}{20}=\frac{1}{20}\) , Gainrng ratio=3:1

Working Note
Goodwill of firm = ₹ 60,000
C’s share in goodwill 60,000 x \(\frac{1}{5}\) = ₹ 12,000
to be contributed by continuing partners in gaining ratio, i.e. 1:1.

Question 9.
Given below is the information related to Happy Club.

Assets (1st April 2019)	Amt(₹)
Ground and Pavilion	50,000
Sports Equipment	30,000
Furniture	7,000
Stock of Stationery	1,000
Subscription Due	1,200

Prepare the balance sheet as on 1st April 2019 and 31st March 2020.
Answer:

Part B
(Analysis of Financial Statements)

Short Answer (SA) Type I Question (2 Marks)

Question 10.
State whether following would result in inflow, outflow, or no flow of cash and cash equivalents
(i) Purchase of land and building.
(ii) Cash is withdrawn from bank.
Answer:
(i) Purchase of land and building will lead to cash outflow as cash is paid for it.
(ii) Cash withdrawn from a bank will lead to no flow of cash as it is just a movement between cash and cash equivalents.

Short Answer (SA) Type II Question (3 Marks)

Question 11.
From the following balance sheets as at 31st March 2020 and 2019 of Kapoor Ltd. Prepare a comparative balance sheet.
Balance Sheet
as at 31st March 2020 and 2019

Or
Calculate cash flow from operating activities of Batra Ltd. from the following information
(i) Net profit before tax and extraordinary items ₹ 1,86,000
(ii) Depreciation of fixed assets ₹ 30,000.
(iii) Amortisation of goodwill ₹ 20,000.
(iv) Loss on sale of machine ₹ 10,000.
(v) Profit on sale of building ₹ 20,000.
(vi) Dividend received on investment ₹ 10,000.
(vii) Provision for taxation made ₹ 16,000.

The following additional information is also available to you.

Particulars	31st March 2019 (₹)	31st March 2020 (₹)
Accounts Receivable	60,000	50,000
Accounts Payable	30,000	70,000
Prepaid Expenses	5,000	3,000

Long Answer (LA) Type Question (5 Marks)

Question 12.
From the balance sheet and information given below, prepare a cash flow statement.
Balance Sheet
as at 31st March 2020

During the year, machine costing ₹ 8,000 (accumulated depreciation ₹ 2,400) was sold for ₹ 4,000. The provisions for depreciation against machinery as on 31st March 2019 and 31st March 2020 were ₹ 20,000 and ₹ 32,000 respectively.
Net profit for the year amounting to ₹ 36,000.
Answer:

Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Maths Term 2 Set 1 with Solutions

Maximum Marks : 40
Time : 2 Hours

General Instructions:

• This question paper contains three sections – A, B and C. Each part is compulsory.
• Section – A has 6 short answer type (SA1) questions of 2 marks each.
• Section – B has 4 short answer type (SA2) questions of 3 marks each.
• Section – C has 4 long answer type questions (LA) of 4 marks each.
• There is an internal choice in some of the questions.
• Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section – A

Question 1.
Find: ∫\(\frac{\log x}{(1+\log x)^{2}}\) dx [2]
OR
Find: ∫\(\frac{\sin 2 x}{\sqrt{9-\cos ^{4} x}}\) dx
Answer:

Question 2.
Write the sum of the order and the degree of the following differential equation: [2]
\(\frac{d}{d x}\left(\frac{d y}{d x}\right)=5\) = 5
Answer:
Here, \(\frac{d}{d x}\left(\frac{d y}{d x}\right)=5\) = 5
⇒ \(\frac{d^{2} y}{d x^{2}}\) = 5 ⇒ \(\left[\frac{d^{2} y}{d x^{2}}\right]^{1}\) = 5
∴ order = 2 and degree = 1
and sum = 2 + 1 = 3.

Question 3.
If and are unit vectors, then prove that |â + b̂| = 2 cos \(\frac{\theta}{2}\), where θ is the angle between them. [2]
Answer:

Question 4.
Find the direction cosines of the following line:
\(\frac{3-x}{-1}=\frac{2 y-1}{2}=\frac{z}{4}\)
Answer:

Clearly direction ratios are < 1, 1, 4 > and its direction ratios (dr’s) are to be converted to direction cosines.

Question 5.
A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-by-one without replacement. [2]
Answer:
Here, Sample space = 1 red ball + 3 white balls
Here P(not red) = \(\frac{3}{4} \times \frac{2}{3}\) = \(\frac{1}{2}\) → for [x = 0]
P(1 red ball) = \(\frac{1}{4} \times \frac{3}{3}+\frac{3}{4} \times \frac{1}{3}\) = \(\frac{1}{4}+\frac{1}{4}\) = \(\frac{1}{2}\) → for [x = 1]
∴ P(X = 0) = \(\frac{1}{2}\) and P(X = 1) = \(\frac{1}{2}\), so probability distribution table will be

X	0	1
P(X)	\(\frac{1}{2}\)	\(\frac{1}{2}\)

Question 6.
Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack?. [2]
Answer:
Total cards = 52
Total no. of red cards = 26 and red card without jack = 26 – 2 = 24
No. of Jacks = 4
Now drawing of cards one by one without replacement are independent events.
∴ Probability of getting first card or a red card + second card a jack (without replacement)
= [getting a red jack + getting a jack] or [getting a red card (not a jack) + getting a jack]
= \(\left[\frac{2}{52} \times \frac{3}{51}\right]+\left[\frac{24}{52} \times \frac{4}{51}\right]=\frac{1}{26}.\)

Section – B

Question 7.
Find \(\int \frac{x+1}{\left(x^{2}+1\right) x}\) dx. [3]
Answer:
Here, one of the factors in the denominator (x² + 1) is quadratic.

Question 8.
Find the general solution of the following differential equation: [3]
x\(\frac{d y}{d x}\) = y – x sin\(\left(\frac{y}{x}\right)\)
OR
Find the particular solution of the following differential equation, given that y = 0 when x = \(\frac{\pi}{4}\);
\(\frac{d y}{d x}\) + y cot x = \(\frac{2}{1+\sin x}\)
Answer:
Given, x\(\frac{d y}{d x}\) = y – x sin \(\left(\frac{y}{x}\right)\) ……….. (i)
x and y are of same degree (one).
so it is a homogeneous differential equation
put y = vx, on differentiating we get
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Substituting in (i), we get

Integrating on both sides
∫\(\frac{d v}{\sin v}\) = – ∫\(\frac{d x}{x}\)
log |cosec v – cot v| = – log |x| + log k, k > 0, log k is an arbitrary constant
or log |cosec v – cot v| + log |x| = log k
or log |(cosec v – cot v)x| = log k
= |(cosec v – cot v)x| =± k

which is the required general solution.

OR

The differential equation is a linear differential equation
I.F. = e^{∫cot x dx} = e^{log sin x} = sin x
The general solution is given by

Question 9.
If \(\vec{a} \neq \overrightarrow{0}, \vec{a}, \vec{b}=\vec{a}, \vec{c}, \vec{a} \times \vec{b}=\vec{a} \times \vec{c}\), then show that \(\vec{b} \times \vec{c}\).
Answer:

Question 10.
Find the shortest distance between the following lines:
\(\vec{r}\) = (î + ĵ – k̂) + s(2î + ĵ + k̂)
\(\vec{r}\) = (î + ĵ – 2k̂) + t(4î + 2ĵ + 2k̂)
OR
Find the vector and the cartesian equations of the plane containing the point î + 2ĵ – k̂ and parallel to the lines \(\vec{r}\) = (î + 2ĵ + 2k̂) + s(2î – 3ĵ + 2k̂) = 0 and \(\vec{r}\) = (3î + ĵ – 2k̂) + t(î – 3ĵ + k) = 0.
Answer:
Here, the lines are parallel. The shortest distance = \(\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}\right|}{|\vec{b}|}\)
= \(\frac{(3 \hat{k}) \times(2 \hat{i}+\hat{j}+\hat{k}) \mid}{\sqrt{4+1+1}}\) ………. (i)
(3k̂) × (2î + ĵ + k̂) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
0 & 0 & 3 \\
2 & 1 & 1
\end{array}\right|\) = – 3î + 6ĵ
∴ |- 3k̂ + 6ĵ| = \(\sqrt{3^{2}+6^{2}}\) = √45 = 3√5
Hence, the required shortest distance = \(\frac{3 \sqrt{5}}{\sqrt{6}}\) units from (i)

OR

Since, the plane is parallel to the given lines, the cross product of the vectors 2î – 3ĵ + 2 k̂ and î – 3ĵ + k̂ will be a normal to the plane
(2î – 3ĵ + 2 k̂) × (î – 3ĵ + k̂) = \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & 2 \\
1 & -3 & 1
\end{array}\right|\) = 3î – 3k̂
The vector equation of the plane is
\(\vec{r}\) . (3î – 3k̂) = (î + 2ĵ – k̂).(3î – 3k̂)
or \(\vec{r}\) . (î – k̂) = 2
and the cartesian equation of the plane is x – z – 2 = 0.

Section – C

Question 11.
Evaluate:

Answer:
The given definite integral =

Here, x(x – 1) (x – 2) = 0 has three zeroes x = 0, x = 1 and x = 2
Now look at the figure:

For interval – 1 ≤ x ≤ 0
at x = – 1 x(x – 1) (x – 2) = (- 1) (- 1 – 1) (- 1 – 2) = – 6
at x = 0 x(x – 1)(x – 2) = 0 (0 – 1)(0 – 2) = 0

∴ |x³ – 3x² + 2x| = – (x³ – 3x² + 2x) for – 1 ≤ x ≤ 0 ……. (i)
For interval 0 ≤ x ≤ 1
at x = 0
x(x – 1) (x – 2) = 0(0 – 1) (0 – 2) = 0
at x = \(\frac{1}{2}\) ⇒ 0 ≤ \(\frac{1}{2}\) ≤ 1

at x = 1
x(x – 1) (x – 2) = (1) (1 – 1) (1 – 2) = 0
∴ |x³ – 3x² + 2x| = – x³ – 3x² + 2x for 0 < x < 1 ……. (ii)
For interval 1 ≤ x ≤ 2
at x = 1
x(x – 1) (x – 2) = 1(1 – 1) (1 – 2) = 0
at x = \(\frac{3}{2}\) ⇒ 1 ≤ \(\frac{3}{2}\) ≤ 2, we have

at x = 2, we have
x(x – 1) (x – 2) = (2) (2 – 1) (2 – 2) = 0
∴ |x³ – 3x² + 2x| = – (x³ – 3x² + 2x) for 1 ≤ x ≤ 2 ……. (iii)

Question 12.
Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y² = x and the x-axis.
OR
Using integration, find the area of the region {(x, y): 0 ≤ y ≤ √3x, x² + y² ≤ 4} [4]
Answer:
Solving x + y = 2 and y² = x simultaneously, we get the points of intersection as (1, 1) and (4, – 2).
The required area = the shaded area (OABO)


Solving y = √3x and x² + y² = 4, we get the points of intersection as (1, √3) and (- 1, -√3).
The required area
= the shaded area (OABO)

Question 13.
Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane. [4]
Answer:
The equation of the line perpendicular to the plane and passing through the point (1, 2, 0) is
\(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z}{2}\)
The coordinates of the foot of the perpendicular are (μ + 1, – 3μ + 2, 2μ) for some μ.
These coordinates will satisfy the equation of the plane. Hence, we have
μ + 1 – 3(- 3μ + 2) + 2(2μ) = 9
⇒ μ = 1
The foot of the perpendicular is (2, – 1, 2).
Hence, the required distance = \(\sqrt{(1-2)^{2}+(2+1)^{2}+(0-2)^{2}}\)
= √14 units.

Case-Based/Data-Based

Question 14.
An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at sometime within a fixed one-year period with probability 0.6, whereas this probability is 0.2 for a person who is not accident prone. The company knows that 20 percent of the population is accident prone.

Based on the given information, answer the following questions.
(i) What is the probability that a new policyholder will have an accident within a year of purchasing a policy? [2]
(ii) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone? [2]
Answer:
Let E₁ = The policy holder is accident prone.
E₂ = The policy holder is not accident prone.
E = The new policy holder has an accident within a year of purchasing a policy.

(i) P(E) = P(E₁) × P(E/E₁) + P(E₂) × P(E/E₂)
= \(\frac{20}{100} \times \frac{6}{10}+\frac{80}{100} \times \frac{2}{10}=\frac{7}{25}\)

(ii) By Bayes’ Theorem,

Students can access the CBSE Sample Papers for Class 12 Economic 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 Economics Term 2 Set 7 with Solutions

General Instructions :

• This is a subjective question paper containing 13 questions.
• This paper contains 5 questions of 2 marks each, 5 questions of 3 marks each and 3 questions of 5 marks each.
• 2 marks questions are Short Answer Type Questions and are to be answered in 30-50 words.
• 3 marks questions are Short Answer Type Questions and are to be answered in 50-80 words.
• 5 marks questions are Long Answer Type Questions and are to be answered in 80-120 words.
• This question paper contains Case/Source Based Questions.

Maximum Marks: 40
Time: 2 Hours

Short Answer (SA) Type I Questions (2 Marks)

Question 1.
Calculate NNP_FC from the following data

Items	₹ (in crore)
GNPmp	15,000
Depreciation or Consumption of Fixed Capital	1,000
Indirect Taxes	500
Subsidies	150
Net Factor Income from Abroad	650

Or
If the Real GDP is ₹200 and Nominal GDP is ₹350, calculate Price Index (base =100).
Answer:
NNP_FC = GNP_MP – Depredation – Net Indirect Taxes
= 15,000 – 1,000 – (500 – 150) (∵ NIT = Indirect taxes – Subsidies)
= 14,100 – 350
= ₹13,650 crore
Or
Real GDP = ₹200
Nominal GDP = ₹350
Price Index = \(\frac{\text { Nominal GDP }}{\text { Real GDP }} \times 100\)
\(\frac{350}{200}\) x 100 = 175

Question 2.
Explain the role of open market operations in correcting the problem of deflationary gap.
Answer:
Open market operations refers to the sale and purchase of securities by the Central Bank to the commercial bank or general public.

In the situation of deflationary gap, the Central Bank buys securities in the open market and makes payment to the sellers. The money flows out of the Central Bank and ultimately reaches the commerdal banks as deposits. This raises the lending capadty of the banks. People can borrow more. This will raise the level of aggregate demand in the economy.

Question 3.
How is unemployment measured in India?
Or
“Unemployment in India is one of the most faced challenges by the government. Many programmes are initiated for the same. One of the most successful programmes is NREGA.” In the light of the statement, briefly explain the main features of the National Rural Employment Guarantee Act, 2005.
Answer:
Measurement of unemployment (any two)
(i) Usual Status The usual status approach to measuring unemployment uses a reference period of 365 days i.e., one year preceding the date of the survey of NSSO for measuring unemployment.

(ii) Current Weekly Status The current weekly status approach for measuring unemployment uses seven days preceding the date of survey as the reference period.

(iii) Current Daily Status The current daily status approach for measuring unemployment seeks to ascertain the activity status of an individual for each day of the reference week.
Or
The main features of the National Rural Employment Guarantee Act, 2005 (NREGA) is as follows (any two)

• It guarantees the 100 days of wage employment in the financial year to a rural household whose adult members volunteer to do unskilled manual work.
• In this scheme, one-third of the proposed jobs will be reserved for women.
• This scheme initially started in 200 districts. Later on, this scheme will be further extended to 600 districts.
• If the applicant is not employed within 15 days, then he/she will be entitled to a daily unemployment allowance.

Question 4.
Find investment from the following
National Income = ₹600
Autonomous Consumption = ₹150
Marginal Propensity to Consume = 0.70
Or
In an economy, the marginal propensity to consume is 0.75. Investment expenditure in the economy increases by ₹75 crore. Calculate the total increase in national income.
Answer:
Here, Y = ₹600, \(\bar{C}\) = ₹150, MPC or b = 0.70
We know that, Y = C + S or Y – C = S or S = 600 – (\(\bar{C}\) + bY)
= 600 – (150 + 0.7 x 600) = 600 – (150 + 420)
= 600 – 570 = ₹30
As, I = S = ₹30 (At equilibrium level of income)
Hence, Investment = ₹30 Or Here,
MPC = 075, ΔI = ₹75 crore
ΔI = ΔS = ₹75 crore
MPC = 1 – MPC = 1 – 0.75 = 0.25
Now, MPC = \(\frac{\Delta S}{\Delta Y}\) or 0.25 = \(\frac{75}{\Delta Y}\)
Or
ΔY = \(\frac{75}{0.25}\)
Change in National Income (ΔY) = ₹300 crores

Question 5.
Observe the data given below carefully and answer the question
Trends in Employment Pattern (Statuswise) 1972-2012 (in%)

Status	1972-73	2011-2012
Self-employed	61.4	52.0
Regular Salaried Employees	15.4	18.0
Casual Wage Labourers	23.2	30.0
Total	100.0	100.0

Data suggest that the percentage of casual wage workers have increased between 1972-73 to 2011-12. Give reason, why?
Answer:
In recent past, there has been migration that has happened from rural areas to urban areas in search of better jobs. Since, everyone cannot be absorbed as regular salaried employees owing to lack of skills, it has led to increase in percentage of casual workers.

Short Answer (SA) Type II Questions (3 Marks)

Question 6.
When is an economy in equilibrium? Explain, with the help of saving and investment functions.
Or
State whether the following statements are true or false.
(i) APC can never be zero or negative.
(ii) MPC is slope of both consumption and AD curves.
(iii) APS falls with rise in income.
Answer:
The equilibrium level of income or output is that level at which the planned savings and planned investments are equal.

It is derived from aggregate demand and aggregate supply approach.

Aggregate Demand (AD) in a two sector economy is defined as the sum of Consumption Expenditure (C) and Investment Expenditure (I) i.e., AD =C + 1, whereas Aggregate Supply (AS) is defined as the sum of Consumption Expenditure (C) and Saving (S) i.e., AS = C + S.

Mathematically, at equilibrium level of output,
AD = AS or, C + I = C + S
Hence, I = S
or S = I

Determination of Equilibrium Level of Income and Output

Or

1. True, APC is calculated from consumption. Consumption is comprised of autonomous consumption which can never be zero.
2. True, MPC measures the change in consumption demand due to change in income, thus it measures the slope of both consumption and AD curves while investment expenditure is assumed to be constant.
3. False, APS rises with rise in level of income. When income increases, people tend to save a more part of their income.

Question 7.
Elaborate the concept of ‘consumption of fixed capital’ and ‘capital loss’.
Answer:
Consumption of Fixed Capital It refers to the depreciation of fixed assets as they are being used in the production process.

The various causes of consumption of fixed capital includes

• Normal wear and tear
• Accidental damages
• Expected obsolescence

Generally, a provision is made by the producers to meet these expenses.

Capital Loss It refers to the loss in value of fixed assets when these are not being used.

The various causes of capital loss includes

• Natural calamities
• Fall in the market price of assets

No provision can be made for these as they arises unexpectedly.

Question 8.
Observe the data given below and answer the question.

Level of Child Labour (2012-13) Percentage of children aged 5-14 years engaged in work at the state level and the district with the lowest and highest rate in each state.
Source Annual Health Survey Report, A Report On Core And Vital Health Indicators, Part I In which state, highest percentage of children aged between 5 to 14 years are engaged in workforce and why?

Direction Read the following text carefully and answer questions 9 and 10 given below

The 20th century included the inception of modem family planning, which restricted the fertility of hundreds of millions of couples around the world. Due to concerns about the world’s unprecedented rate of population growth in the mid-20th century, some aid agencies and international organisations began to support the establishment of family planning programs.

About 40 years later, in the mid-1990s, large-scale family planning programs were active in 115 countries.

China’s One Child Policy (OCP) is the largest among the World’s family planning programs. In the 1970s, after two decades of explicitly encouraging population growth, policy-makers in China began enacting a series of measures to curb it. The OCP was formally initiated in 1979 and firmly established across the country in 1980.

It was the first time that family planning policy became formal law in China. Differing from birth control policies in many other countries, the OCP assigned a compulsory general ‘one-birth’ quota to each couple, though its implementation has varied considerably across regions for different ethnicities at different times. The policy affected millions of couples and lasted more than 30 years.

According to the World Bank, the fertility rate in China dropped from 2.81 in 1979 to 1.51 in 2000. The reduced fertility rate is likely to have affected the Chinese labour market profoundly. In 1979, the Chinese government formally initiated the OCP to alleviate social, economic, and environmental problems such as the high unemployment rate and scarcity of land resources.

As opposed to many family planning policies in other countries, the OCP was compulsory rather than voluntary. As the name suggests, the policy restricted a couple to having only one child. However, there were some exemptions. The birth quota varied according to residence (urban/rural) and ethnicity (Han/non-Han).

Since Han ethnicity is by far the largest in China, accounting for 93% of the population, the policy mainly restricted the fertility of people with Han ethnicity. In general, Han households in urban regions were only allowed to have one child, while most households in rural areas could have a second child if their first was female (this exception is called the “one-and-a-half-child policy”). Meanwhile, in most regions, households of non-Han ethnicity were allowed to have two or three children, regardless of gender.

In March 1991, to show resoluteness, the Central Government listed family planning among the three basic state policies in China’s Eighth Five Year Plan passed by the National People’s Council. The Eighth Five Year Plan explicitly set a goal of reducing the natural growth rate of the country’s population to less than 1.25% on average during the following decade.

To achieve such a challenging objective, national leaders employed a “responsibility system” to induce subnational or provincial officials to set high fine rates. During the short-period between 1989 and 1992, over half of the country’s provinces (16 out of 30) saw a significant increase in their fine rate, with the average rate increasing from 1.0 to 2.8 times a household’s yearly income.

Indeed, 16 of the 21 significant increases in the policy’s history (i.e., increases of more than one times a household’s income) occurred during this period. Tire amount collected via the policy fine was not made public until recently: the total was about 20 billion RMB yuan (US$ 3.3 billion) among 24 provinces that reported fine rates in 2012.

For example, Guangdong, one of the richest provinces in China, collected 1.5 billion yuan in 2012. Meanwhile, as a comparison, total local government expenditure on compulsory schooling in the province was 10.5 billion.
Answer:
As per the given data, Madhya Pradesh and Odisha are two states where 3.8 and 3.7 percent children belonging to age group of 5 to 14 years are engaged in employment. This is majorly due to lack of education and skills. As the parents are insufficient income earners and doesn’t earn sufficient income for the living, thus, the children have to join the workforce as child labourers to support family income.

Question 9.
Enumerate the effects of the policy of one child norm in China.
Answer:
The important implications of the one child norm are (any three)

• Low population growth.
• Decline in the sex ratio.
• After a few decades, there will be more elderly people in proportion to young people.
• In the long-run, China will have to provide more social security measures with fewer workers.
• It helped China in better management of human capital.

Question 10.
Comment on the growth rate trends witnessed in China and India in the last two decades.
Answer:
China has the second largest GDP (PPP) of $ 22.5 trillion, whereas India’s GDP (PPP) is $ 9.03 trillion.

When many developed countries were finding it difficult to maintain a growth rate of even 5%, China was able to maintain near double-digit growth for more than two decades.

Its growth rate in 1980-90 was 10.3% and for 2005-2013, it was 10.2%.

In the 1980-90, China was having double-digit growth and India was stuck at 5.7% decade growth rate. This rate showed improvement in 2005-2013 at 7.6%, but still it was far behind China’s growth rate.

As far as sector wise contribution to growth is concerned, in 2019, contribution of agriculture to GDP in China was 7% while in India it was 16%.

On the other hand, manufacturing contributes the highest to GDP in China at 41%, whereas in India service sector contributes the highest at around 54%. Thus, China’s growth is mainly contributed by the manufacturing sector and India’s growth by service sector.

In the last two decades, the growth of agriculture sector has declined in both the countries.

In the industrial sector, China has maintained a double-digit growth rate, whereas for India industrial growth rate has declined.

In the case of service sector, China has been able to raise its rate of growth in 2005-13, while service sector growth in India has stagnated.

Long Answer (LA) Type Questions (5 Marks)

Question 11.
(i) Calculate gross domestic product at market price and gross national product at market price from the following data

Items	₹ (in crores)
Compensation of Employees	2,000
Net Exports	(-) 10
Profits	800
Interest	500
Rent	300
Gross National Product at Factor Cost	3,700
Gross Domestic Capital Formation	440
Net Fixed Capital Formation	300
Change-in-stock	40
Factor Income from Abroad	60
Net Indirect Taxes	200

(ii) What precautions are taken, while measuring national income by income method?
Answer:
(i) National Domestic Product at Factor Cost (NDP_FC) = Compensation of Employees + Profits + Interest + Rent
= 2,000 + 800 + 500 + 300
= ₹3,600 crore

Gross Domestic Product at Market Price (GDP_MP) = NDP_FC + Net Indirect Taxes + Consumption of Fixed Capital
= 3,600 + 200 + 100
= ₹3,900 crore

Gross National Product at Market Price (GNP_MP) =GNP_FC + Net Indirect Taxes
= 3,700 + 200
= ₹3,900 crore

*Consumption of Fixed Capital = Gross Domestic Capital Formation – Change-in-stock – Net Fixed Capital Formation
= 440 – 40 – 300
= ₹100 crore

(ii) The following precautions are to be taken while measuring national income by income method (any four)

• Income from illegal activities like smuggling, theft, gambling, etc, should not be included.
• Commission paid on the sale and purchase of second hand goods are to be included.
• Transfer earnings like old age pensions, unemployment allowances, scholarships, pocket expenses, etc. should not be included.
• Imputed rent of owner occupied houses is to be treated along with rent as a component of factor incomes.
• Direct taxes paid like income tax paid by the employees are part of salary, like corporate taxes paid by the producers.

Question 12.
(i) What will happen to an economy, if aggregate demand increases beyond full employment level? Explain using a graph.
(ii) Define full employment and invountary unemployment.
Answer:
(i) If aggregate demand increases beyond full employment level, it will lead to inflationary situations in the economy.

In the above figure,
AD_FE = AD at full employment level
AD_AE = AD above full employment level

The point Q is the equilibrium point where AD AS. But at this point, aggregate demand is more than the aggregate supply in the economy. This difference of actual aggregate demand and aggregate supply i.e., EF is the inflationary gap.

Inflationary Gap = Excess Demand
= AD_AE – AD_FE = EF

(ii) (a) Full Employment It refers to a situation, when all those who are able to work and are willing to work at the existing wage rate are getting work.

In this situation, (at a given wage rate)

Demand for Labour = Supply of Labour and the labour market is cleared.

(b) Involuntary Unemployment It refers to a situation, when some people who are able to work and are willing to work at the existing wage rate are not getting work.

Here, Demand for Labour < Supply of Labour

In this situation, aggregate demand is not enough to induce fuller utilisation of existing resources, so some people are forced to remain unemployed.

Question 13.
There are various schemes launched by the Indian government to eradicate unemployment directly or indirectly. Mention any three anti-unemployment programmes launched by the government to reduce unemployment. How the effectiveness of these schemes can be improved?
Or
“Today development has become a burden on nature/environment.” Comment.
Answer:
(i) Various schemes run by the Indian government to reduce unemployment directly or indirectly are (any three)

• Prime Minister Rozgar Yojana
• Rural Employment Generation Programme
• Swama Jayanti Gram Swarojgar Yojana
• National Rural Employment Guarantee Act

(ii) The effectiveness of these schemes can be improved with the help of following measures (any three)

• Population control
• Create new employment opportunities
• Education
• Proper implementation and right targeting

Or
The present thinking with regard to relationship between nature and development is that there should be maximum exploitation of natural resources for development.

As a result, people are using nature beyond its carrying capacity. Our present technology is creating a number of environmental problems.

A number of non-degradable materials are being produced in present day through the production technology.

Following are the some important reasons responsible for the heavy burden on nature

• Rise in human population in underdeveloped countries.
• Affluent consumption style in developed countries.
• Misuse of production technology in almost all the countries and poor planning of development.
• As a result of above, there is a reckless use of resources creating negative effects on the society.

The negative effects of development on nature are stated below

• Pollution
• Degradation of resources

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Accountancy Standard Term 2 Set 4 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. All questions are compulsory.
• Question nos. 1 to 3 and 10 are Short Answer Type I Questions carrying 2 marks each.
• Question nos. 4 to 6 and 11 are Short Answer Type II Questions carrying 3 marks each.
• Question nos. 7 to 9 and 12 are Long Answer Type Questions earning 5 marks each.
• There is no overall choice. However, an internal choice has been provided in 3 questions of three marks and 1 question of five marks.

Maximum Marks : 40
Time : 2 Hours

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

Short Answer (SA) Type I Questions (2 Marks)
Question 1.
From the following information of a club, show the amounts of match expenses and match fund in the financial statements of the club for the year ended on 31st March, 2021.

Answer:

Note Expenses on account of match exceed the balance in match fund by ? 1,000. This amount will be debited to income and expenditure account.

Question 2.
Why a retiring partner is entitled to a share of goodwill of the firm?
Answer:
When an existing partner of a firm decides to retire from the firm, the remaining partner(s) will gain in future profits. The remaining partner who gains by acquiring an additional right to share future profits must compensate the outgoing partner who sacrifices by foregoing his right to share future profits. The amount of compensation must be equal to the proportionate value of firm’s goodwill.

Question 3.
Kanu, Sunny and Richa are partners in a firm sharing profits and losses in the ratio of 4 : 3 : 1. As per the terms of partnership deed, on the death of any partner, goodwill was to be valued at 3 years’ purchase of average profits of last 5 years which were 2016 – ₹ 40,000, 2017 – ₹ 40,000, 2018 – ₹ 30,000, 2019 – ₹ 40,000 and 2020 – ₹ 50,000. Kanu died on 29th May, 2020 and the new profit sharing ratio decided between Sunny and Richa is 1 : 1. Pass necessary journal entry to adjust Kanu’s share of goodwill and show your workings clearly.
Answer:

Working Notes
1. Calculation of Gaining Ratio
Gaining Ratio = New Share – Old Share
Sunny = \(\frac{1}{2}-\frac{3}{8}=\frac{4-3}{8}=\frac{1}{8}\) ; Richa = \(\frac{1}{2}-\frac{1}{8}=\frac{4-1}{8}=\frac{3}{8}\); Gaining Ratio = 1 : 3

2. Calculation of Goodwill

Goodwill = Average Profit × Number of Year’s Purchase = 40,000 × 3 = ₹ 1,20,000
Kanu’s share of goodwill = 1,20,000 × \(\frac{4}{8}\) = ₹ 60,000
₹ 60,000 will be contributed by Sunny and Richa in their gaining Ratio i.e., 1 : 3.

Short Answer (SA) Type II Questions (3 Marks)
Question 4.
A company purchased an asset of the book value of ₹ 11,000. Purchase consideration was paid by issuing 12% debentures of ? 100 each. Give journal entries assuming that debenture:: have been issued at (i) par (ii) premium of 10%.
Or
Pass the necessary journal entries for the following
(i) Goel Ltd. invited applications for issuing 6,000,12% debentures of ₹ 100 each at a premium of ₹ 50 per debenture. The full amount was payable on application. Application were received for 8,000 debentures.
Applications of 2,000 debentures were rejected and application money was refunded. Debentures were allotted to the remaining applicants.
(ii) Goel Ltd issued 2,000,10% debentures of ₹ 100 each at 10% premium which are redeemable at par.
Answer:


Working Note

Number of Debentures Issued = \(\frac{11,000}{100+10}\) = 100 Debentures
Or

Question 5.
Krishna and Priyansh are partners sharing profits in the ratio of 1 : 1. They decided to dissolve their firm. Pass necessary journal entries for the following transactions
(i) Workmen compensation reserve stood at ₹ 6,000 and liability in respect of it was ascertained at ₹ 7,500.
(ii) Realisation expenses amounted to ₹ 15,000 were paid by the firm on behalf of a partner, Krishna.
(iii) There was an unrecorded furniture of ₹ 6,000 which was taken over by Krishna at ₹ 5,000.
Or
A, B and C are partners. They share profits in capital ratio. Their balance sheet is given below

Additional Information
(i) B takes retirement.
(ii) New ratio of A and C is 1 : 1.
(iii) Goodwill of the firm ₹ 1,20,000.
(iv) Make 5% provision for debtors.
(v) Building increased by 10%.
(vi) Make provision for claim for damages of ₹ 12,000.
(vii) ₹ 20,000 paid to B in cash. Balance transferred to his loan account.
Prepare revaluation account.
Answer:

Or

Question 6.
The following is the receipts and payments account of Queen’s Club for the year ended 31st March, 2020


The following additional information is provided
(i) Subscription include ₹ 22,000 for 2018-19 and ₹ 8,000 for 2020-21.
(ii) Stock of stationery on 31st March, 2019 and 2020 was ₹ 4,200 and ₹ 6,400 respectively.
(iii) Stock of sports material at the beginning and end of the year were ₹ 18,000 and ₹ 31,000 respectively.
(iv) Rent includes ₹ 4,000 paid for March, 2019, rent for March, 2020 is outstanding.
(v) Telephone expenses include ₹ 3,000 as quarterly expenses upto 31st May, 2020.
(vi) The value of building as on 31st March, 2019 was ₹ 8,00,000 and you are required to write-off depreciation @ 5%.
(vii) The value of investments on 31st March, 2019 was ₹ 10,00,000 and the club made similar additional investment during the year on 1st October, 2019.

You are required to prepare income and expenditure account of the club for the year ended 31st March, 2020.
Answer:

Working Note

Long Answer (LA) Type Questions (5 Marks)
Question 7.
X and Y were partners in a firm sharing profits in the ratio of 3 : 2. On 31st March, 2020, the balance sheet of the firm was as follows

The firm was dissolved on 1st April, 2020 and the assets and liabilities were settled as follows
(i) Building was taken over by creditors as their full and final payment.
(ii) Furniture was taken over by Y for cash payment at 5% less than the book value.
(iii) Debtors were collected by a debt collection agency at a cost of ₹ 5,000.
(iv) Stock realised ₹ 70,500.
(v) Y agreed to bear all realisation expenses. For this service, Y is paid ₹ 500. Actual expense of realisation amounted to ₹ 1,000.
Pass necessary journal entries for dissolution of the firm.
Or
A, B and C were partners in a firm sharing profit and loss in the ratio of 4 : 3 : 3. On 31st March, 2021, their balance sheet was as follows

On the above date, A retired and it was agreed that
(i) Debtors of ₹ 4,000 will be written-off as bad debts and a provision of 5% on debtors for bad and doubtful debts will be maintained.
(ii) An unrecorded creditor of ₹ 20,000 will be recorded.
(iii) Patents will be completely written-off and 5% depreciation will be charged on stock, machinery and building.
(iv) B and C will share the future profit in the ratio of 3 : 2.
(v) Goodwill of the firm on A’s retirement was valued at ₹ 5,40,000.
Pass necessary journal entries for the above transactions in the books of the firm on A’s retirement.
Answer:

Working Notes

Or


Working Notes
1. Calculation of Gaining Ratio
Gaining Ratio = New Share – Old Share
B = \(\frac{3}{5}-\frac{3}{10}=\frac{6-3}{10}=\frac{3}{10}\)
C = \(\frac{2}{5}-\frac{3}{10}=\frac{4-3}{10}=\frac{1}{10}\)
Gaining Ratio = 3 : 1

3. Calculation of Sameer’s Share of Goodwill Firm’s goodwill = ₹ 5,40,000; A’s share = 5,40,000 × \(\frac{4}{10}\) 2,16,000
To be contributed by B and C in their gaining ratio i.e., 3 : 1.

4. Loss on Revaluation
It can be ascertained by preparing revaluation account in the following manner

Question 8.
Entertainment Tennis Club presented the following receipts and payments account for the year ended 31 March, 2020.

Additional Information

	1st April, 2019 (₹)	31st March, 2020 (₹)
Tennis Balls on Hand (at cost)	1,600	3,600
Creditors for Refreshments	16,000	12,000
Subscriptions Outstanding	8,000	14,000
Club Courts	2,40,000	–

Prepare an income and expenditure account for the year ended 31st March, 2020 and show the balance sheet as on that date. Also prepare the relevant accounts for computation of various items.
Answer:

Working Notes

Question 9.
Pass the journal entires in following cases
(i) Laxmi Ltd. purchased the assets of Shri Ram Ltd. for ₹ 5,00,000, payable ₹ 1,00,000 in cash and the balance by issue of 7% debentures of ₹ 100.
(ii) PQR Ltd. took over assets of ₹ 5,00,000 and liabilities of ₹ 60,000 of XYZ company for the purchase consideration of ₹ 6,60,000. PQR Company Ltd. paid the purchase consideration by issuing debentures of ₹ 100 each at 10% premium.
(iii) Ashima Ltd purchased furniture from Disha Ltd and paid to Disha Ltd as follows
(a) By issuing 20,000,10% debentures of ₹ 100 each at premium of 10%.
(b) Balance by accepting a bill of exchange of ₹ 40,000 payable after one month.
Answer:

Working Note
Number or Shares Issued = \(\frac{\text { Amount Due }}{\text { Issue Price }}=\frac{4,00,000}{100}\) = 4,000 Debentures


Working Note
Number of Debentures = \(\frac{6,60,000}{110}\) = 6,000 Debentures

Working Note
Purchase Consideration = 20,000 × 110 + 40,000 = ₹ 22,40,000

Part B
(Analysis of Financial Statements)

Short Answer (SA) Type I Question (2 Marks)
Question 10.
State whether the following will result in inflow, outflow or no flow of cash.
(i) Old vehicle written-off.
(ii) Bills receivable endorsed to creditors.
Answer:
(i) No flow of cash. Writing-off old vehicle implies charging depreciation in the profit and loss
account. Here, no cash is involved as depreciation is a non-cash expense.
(ii) Bills receivable endorsed to creditors indicates neither an inflow or an outflow of cash. In this transaction, only current liabilities and current assets balance gets reduced in the balance sheet. Thus, cash balance remain unaffected.

Short Answer (SA) Type II Question (3 Marks)
Question 11.
Prepare comparative statement of profit and loss from the following information

Particulars	31st March, 2021 (₹)	31st March, 2020 (₹)
Revenue from Operations	20,00,000	16,00,000
Purchase of Stock-in-trade	8,80,000	6,20,000
Change in Inventories of Stock-in-trade	50,000	90,000
Other Expenses	10% of cost of Revenue from Operations	8% of Cost of Revenue from Operations
Tax Rate	30%	46%

8% of Cost of Revenue from Operations
46%
Or
Following are the balance sheets of Arihant Tiles Ltd. as at 31st March, 2019 and 31st March, 2020


Notes to Accounts

Additional Information
During the year, a piece of machinery costing ₹ 18,000 on which accumulated depreciation was ₹ 12,000, was sold for ₹ 4,500. Prepare cash flow from investing activity.
Answer:

Working Note

Or

Long Answer (LA) Type Question (5 Marks)
Question 12.
From the following balance sheets of Karunesh Ltd. as on 31st March, 2019 and 2020, . prepare a cash flow statement.


Notes to Accounts

Additional Information
(i) Depreciation charged on building ₹ 20,000 and plant ₹ 10,000.
(ii) Interest paid on debentures ₹ 14,400. New debentures of ₹ 36,000 were issued on 1st October, 2019.
(iii) Interest paid on deposits from public ₹ 19,200 for the year.
Answer:

Working Notes