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Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 10 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Applied Mathematics Term 2 Set 10 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.
The marginal revenue function of a commodity is given by MR = 6 – 5x + x².
Find the demand function.
Or

Answer:
We have, MR = 6 – 5x + x²
Now, R = ∫(MR) dx + K
= ∫(6 – 5x + x²)dx + K
= 6x – \(\frac{5 x^{2}}{2}\) + \(\frac{x^{3}}{3}\) + k
When x = 0, then R = 0.
∴ K = 0
Thus, R = 6x – \(\frac{5}{2}\)x² + \(\frac{x^{3}}{3}\) and the
corresponding demand function is given by

where p is the price, when the number of units sold x.

OR

Here, redefined the given integrand in given interval (0, 4).
For, 0 < x < 4, |x| = x
0 < x ≤ 2 , |x – 2| = – (x – 2)
2 ≤ x < 4 , |x – 2| = (x – 2)
0 < x < 4, |x – 4| = – (x – 4)

= 8 + 2 + 2 + (16 – 8)
= 8 + 2 + 2 + 8 = 20

Question 2.
A group of 5 patients treated with medicines P weigh 10, 8, 12, 6, 4 (in kg). Second group of 7 patients treated with medicine Q weigh 14, 12, 8,10, 6, 2,11 (in kg). Comment on rejection of hypothesis with significance level 5%. [given, t_(10,0.05) = 1.812]
Answer:
Consider,
H₀: µ₁ = µ₂ and H₁: µ₁ > µ₂
[where, µ₁ and µ₂ denotes population means for given two groups]


Given, t_(10,0.05) = 1812
Since, t_cal value < t_tab value
Hence, null hypothesis (H₀) may be accepted with 5% significance.

Question 3.
A machine, the useful life of which is estimated to be 12 yr, cost ₹ 15000. Calculate the scrap value at the end of its life, depreciation being charged at 8% per annum [given, (0.92)¹² = 0.3676].
Or
What sum of money is needed to invest now, so as to get ₹ 8000 at the beginning of every month forever, if the money is worth 6% per annum compounded monthly?
Answer:
We have,
C = ₹ 15000, n = 12 yr and r = 8% = 0.08.
We know that S = C (1 – r)ⁿ
S = 15000(1 – 0.08)¹²
= 15000 (0.92 )¹²
= 15000 × 0.3676
= ₹ 5514
∴ The scrap value is ₹ 5514.

Or

Given, R = ₹ 8000 and r = \(\frac{6}{12}\) % = 0.5% per month.
So, i = \(\frac{0.5}{100}\) = 0.005
The given annuity is a perpetuity of second type.
So, P = R + \(\frac{R}{i}\) = 8000 + \(\frac{8000}{0.005}\)
= 8000 + 1600000
=1608000
Hence, ₹ 1608000 are needed to invest now to get ₹ 8000 at the beginning of every month forever

Question 4.
A firm has to transport 1200 packages using large vans, which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ₹ 400 and each small van is ₹ 200. Not more than ₹ 3000 is to be spent on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.
Answer:
Let the firm has x number of large vans and y number of small vans. From the given information, we have following corresponding constraint table.

Thus, we see that objective function for minimum cost is Z = 400 x + 200y.
Subject to constraints are,
200x + 80y ≥ 1200 [package constraint]
⇒ 5x + 2y ≥ 30 …(i)
and 400x + 200y ≤ 3000 [cost constraint]
⇒ 2x + y ≤ 15 …(ii)
and x ≤ y [van constraint] …(iii)
and x ≥ 0, y ≥ 0
[non-negative constraints] …(iv)
Thus, required LPP to minimise cost is minimise Z = 400x + 200y
Subject to constraints are,
5x + 2y ≥ 30, 2x + y ≤ 15, x ≤ y, x ≥ 0 and y ≥ 0

Question 5.
Construct 3-yearly moving averages from the following data.

Answer:
Calculate of 3-yearly moving averages

Question 6.
Find the effective rate that is equivalent to a nomial rate of 8% compounded semi-annually.
Answer:
We have,

⇒ r_e = (1 + 0.04)² – 1
⇒ r_e = (1.04)² – 1
= 1.0816 – 1
= 0.0816
Hence, the effective rate is 8.16%.

Section – B [12 marks]

Question 7.
The time series given below shows the number of TV sold by a company since 2003.

Find the linear equation that describes the trend in the number of TV sold.
Also, estimate the sale of TV in 2014.
Answer:
Here, n = 10 (even)
So, origin is mean of two middle years
i.e., \(\frac{2007+2008}{2}\) = 2007.5

Table for fitting linear equation

The linear equation would be of the form
y_t = a + bx
The two normal equations shall be
Σy = m + bΣx
Σxy = aΣx + bΣx²
Since, Σx = 0, a = \(\frac{\Sigma y}{n}=\frac{956}{10}\) = 95.6
and b = \(\frac{\Sigma x y}{\Sigma x^{2}}=\frac{1978}{330}\) = 5.994
Thus, the linear equation is y_t = 95.6 + 5.994 x
For 2014, x would be + 13
y₂₀₁₄ = 95.6+ 5.994(13)
= 95.6 + 77.92 = 173.52 thousand
Hence, the expected number of TV that would be sold in 2014 shall be ₹ 173520.

Question 8.
Consider the following hypothesis
H₀ : µ = 275
H₀ : µ ≠ 275
A sample of 60 provided a sample mean of 287.6. The population standard deviation is 15 and level of significance α = 0.05. Check the hypothesis test, given above using interval estimation.
Answer:
Given, µ₀ = 275, n = 60, x̄ = 287.6, σ = 15
and α = 005. So, Z_α/2 = Z_0.025 = 1.96
Confidence interval = x̄ ± Z_α/2. \(\frac{\sigma}{\sqrt{n}}\)
= 287.6 ± 1.96 × \(\frac{15}{\sqrt{60}}\)
= 287.6 ± 3.8
So, confidence interval is
(287.6 – 38,287.6 + 38) i.e. (2838,291.4).
Since, the hypothesized value of population mean µ₀ = 275 lies in the confidence interval (2838, 291.4), so H₀ cannot be rejected.

Question 9.
The present value of a perpetual income of ₹ R at the end of every month, the first being received at the end of 3 yr is ₹ 75000. Find the value of R, if money is worth 15% per annum compounded monthly [given, (1.0125)^-35 = 0.6474]
OR
A bond of face value of ₹ 1200 has a coupon rate 10% per annum paid semi-annually and matures in 6 yr. If the present value of the bond is ₹ 1560, find the yield to maturity.
Answer:
Clearly, it is the case of deferred perpetuity.
Present value (P) = ₹ 75000
Periodic payment = ₹ R
Rate of interest (r) = 15% = 0.15
Interest is compounded monthly, so we have
i = \(\frac{r}{12}=\frac{0.15}{12}\) = 0.0125
Since, the first payment is receivable at the end of 3 yr, the period of deferment (n) = 35.

∴ Approximate yield to maturity (YTM) = 0.0217
or 2% per half year or 4% per annum.

Question 10.
The demand functions for a commodity is given by p = 20 – 2x – x². Find the consumed surplus, when market demand is 3.
Answer:
It is given that x₀ = 3
On substituting x = 3 and p = p₀ in demand function
p = 20 – 2x – x², we get
p₀ = 20 – 6 – 9 = 5

⇒ CS = (60 – 9 – 9) -15
⇒ CS = 27
Hence, consumer’s surplus is 27.

Section – C [16 Marks]

Question 11.
Mr. Taneja purchase a house of ₹ 4500000 with a down payment of ₹ 500000 and balance in EMI for 25 yr. If bank charges 6% per annum compounded monthly. Calculate the EMI. [given, (1.005)³⁰⁰ = 4.4650].
OR
Find the purchase price of a ₹ 20000 bond, redeemable at the end of 10 yr at 110 and paying annual dividends at 4%, if the yield rate is to be 5% effective.
[given, (1.05)^-10 = 0.6139]
Answer:
Cost of house = ₹ 4500000
Down payment = ₹ 500000
∴ Balance = ₹(4500000 – 500000) = ₹ 4000000
So, p = ₹ 4000000, i = \(\frac{6}{1200}\) = 0.005 and n = 25 × 12 = 300 months

Hence, EMI of Mr. Taneja is ₹ 25772.
Total interest paid = x(EMI) – P
= 300(25772) – 4000000
= 7731600 – 4000000
= 3731600
Hence, total interest paid is ₹ 3731600

Or

We have, n = Number of period = 10
and i yield rate = \(\frac{5}{100}\) = 0.05
R = Annual dividend
= 4% of face value
= ₹ \(\left(\frac{4}{100} \times 20000\right)\) = ₹ 800
The bond is redeemed at 110.
Therefore, the redemption price of the bond is 110% of its face value.
Thus, C = Redemption price
= ₹\(\left(20000 \times \frac{110}{100}\right)\) = ₹ 22000
Let V be the purchase price of the bond. Then,

= [16000(1- (1.05)^-10] + 22000(1.05)^-10]
= [16000(1 – 0.6139) + 22000 (0.6139)]
= [16000(0.3861) +1350500]
= (617700 +1350500)
= ₹ 19683.40
∴ The purchase price of bond is 119683.40

Question 12.
A population grows at the rate of 10% per year. How long does it take for the population to double? Use differential equation for it.
Answer:
Let P₀ be the initial population and P be the population after t yr.

Question 13.
Diwakar invested ₹ 20000 in a mutual fund in year 2010. The value of mutual fund increased to ₹ 32000 in year 2015. Calculate the compound annual growth rate (CAGR) of his investment, [given, (1.6)^1/5 = 1.098]
Answer:
Given, the begining value of the investment = ₹20000
Fund value of the investment = ₹ 32000
Number of years = 5 yr

= 0.098 = (0.098 × 100)% = 9.8%
Hence, CAGR is 9.8%.

Case Based/Data Based

Question 14.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 h of machine time and 3 h of craftman’s time in its making, while a cricket bat takes 3 h of machine’s time and 1 h of craftman’s time. In a day, the factory has the availability of not more than 42 h of machine’s time and 24 h of craftman’s time.

(i) What number of rackets and bats must be made, if the factory is to work at full capacity?
(ii) If the profits on rackets and on bats is ₹ 20 and ₹ 10, respectively. Find the maximum profit of the factory, when it works at full capacity.
Answer:
Let the number of rackets and the number of cricket bats to be made in a day be x and y, respectively. Construct the following table.

Construct the following table.

The machine time is not available for more than 42 h.
∴ 1.5x + 3y ≤ 42
The craftman’s time is not available for more than 24 h.
∴ 3x + y ≤ 24
The profit on rackets is ₹ 20 and on bats is ₹ 10.
∴ Maximum Z = 20x + 10y …(i)
Subject to constraints are,
15x + 3y ≤ 42 ……(ii)
3x + y ≤ 24
and x, y ≥ 0 ……. (iv)
Firstly draw the graph of the line
1.5x + 3y = 42

On putting (0, 0) in the inequality 1.5x + 3y ≤ 42, we have
1.5 × 0 + 3 × 0 ≤ 42
⇒ 0 ≤ 42 (which is true)
So, the half plane is towards the origin.
Since, x, y ≥ 0
So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line
3x + y = 24

On putting (0, 0) in the inequality 3x + y ≤ 24, we have
3 × 0 + 0 ≤ 24 ⇒ 0 ≤ 24 (which is true)
So, the half plane is towards the origin.
On solving equations 1.5x + 3y = 42 and 3x + y = 24, we get B (4,12).
∴ Feasible region is OABCO.
The comer points of the feasible region are
0 (0, 0), A (8, 0), B(4,12) and C (0,14).
The values of Z at these points are as follows.

Thus, the maximum profit of the factory, when it works to its full capacity is ₹ 200.
Hence, 4 tennis rackets and 12 cricket bats must be made, if the factory is to work at full capacity.

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics 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 Applied Mathematics Term 2 Set 8 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.
The feasible region (shaded) for a LPP is shown in the following figure

Find the maximum value of Z = 3x + 4y.
Answer:
As clear from the graph comer points are O, A, E, D with coordinates (0, 0), (52, 0), (44, 16) and (0, 38).

Hence, the maximum value of Z is 196 at E(44, 16)

Question 2.
Find the force of interest corresponding to the effective rate 11%.
[given, log(1.11) = 0.0453 and log e = 0.4342]
Answer:
Given, r_e = 11% = 0.11
We know that, r_e = e^r – 1
∴ 0.11 = e^r – 1
⇒ e^r = 1.11
Taking log on both sides,
r log e = log(1.11)
⇒ r(0.4342) = 0.0453
⇒ r = \(\frac{0.0453}{0.4342}\)
= 0.1043 = 10.43%
∴ The force of interest is 10.43%.

Question 3.
An asset costing ₹ 95000 is expected to have a useful life of 8 yr, has final annual depreciation of ₹ 8000. Find the final scrap value of asset.
Answer:
Given, D = ₹ 8000, C = ₹ 95000 and n = 8 yr
We know that, D = \(\frac{C-S}{n}\)
∴ 8000 = \(\frac{95000-S}{8}\)
⇒ 95000 – S = 8000 × 8
⇒ 95000 – S = 64000
⇒ S = 95000 – 64000
⇒ S = ₹ 31000
∴ Final scrap value of the asset is ₹ 31000.

Question 4.
A manufacturer’s marginal cost is \(\frac{600}{\sqrt{5 x+36}}\). Find the cost involved to increase production from 150 units to 450 units.
Or
The marginal revenue function of a commodity is given MR = 12 – 3x² + 4x.
Find the total revenue from the sale of 4 units.
Answer:
Given, marginal cost (MC) = \(\frac{600}{\sqrt{5 x+36}}\)
⇒ \(\frac{d C}{d x}=\frac{600}{\sqrt{5 x+36}}\)
⇒ \(\frac{600}{\sqrt{5 x+36}}\)
Cost involved to increases production from 150 units to 450 units is,

= 240 [√2286 – √786]
= 240 (47.81 – 28D3)
= 240 × 1978
= ₹ 4747.2
Or
We have, MR = 12 – 3x² + 4x
∵ ∫dr = ∫MR dx [∵ MR = \(\frac{d R}{d x}\)
We know that when x = 0, then R = 0
∵ K = 0
R = 12x – x³ + 2x² [from Eq. (i)]
Now, the total revenue from the sale of 4 units is given by put x = 4
R = 48 – 64 + 32 = 16

Question 5.
Consider the following hypothesis
H_o: μ = 315 and H_a: μ ≠ 315
A sample of 60 provided a sample mean of 324.6. The population standard is 14 and level of significance α = 0.05. Check the hypothesis test given above, using interval estimation.
Answer:
Given, μ_o = 315, n = 60, x̄ = 324b
σ = 14 and α = 0.05
∴ Confidence interval = x̄ ± Z_α/2\(\frac{\sigma}{\sqrt{n}}\)
= 324.6 ± 1.96 × \(\frac{14}{\sqrt{60}}\) [∵ Z_α/2 = Z_0.025 = 1.96]
∴ Confidence interval
= 324.6 ± 1.96 × 131
= 324.6 ± 3.54
= (324.6 – 3.54,324b + 3.54)
= (32136, 328.14)
Since, the hypothesis value of population mean p0 = 315 does not lies in confidence interval, so H₀ rejected.

Question 6.
For the following series of observations

Calculate 5- yearly moving averages.
Or
The data below shows the figures of income of a municipal corporation over the years.

Calculate 4-yearly centred moving averages.
Answer:

∴ 5-yearly moving averages are 2.4, 2.8, 3.6 and 3.8
OR

∴ 4-yearly centred moving averages are 5.375, 5.125 and 4.875.

Section – B [12 Marks]

Question 7.
The average nicotine content of a particular brand of cigarette is supposed to be 2.2 mg per cigarette. A smoker of this brand wants to see, if it is really 2.2 mg or more. With this purpose the nicotine contents of 50 cigarettes are measured. The sample mean based on 50 cigarettes is 2.95 mg per cigarette. The population standard deviation is 0.8 mg. Do the data provide enough evidence that the average nicotine content exceeds 2.2 mg? Test at 5% level using p-value approach and critical value approach, [given, Z_0.05 = 1.645]
Answer:
Consider H₀ : μ = 2.2
H_a :μ > 2.2
x̄ = 2.95, μ₀ = 2.2,
n = 50 and σ = 0.8
∴ Test statistics,

Since, |t| > Z_0.05
So, reject H₀ at 5% level.
Means provide sufficient evidence to conclude that average nicotine content exceeds 1.5 mg per cigarette.

Question 8.
Fit a straight line trend for the following data and find the trend values. Estimate the sales for 2018.

Answer:
Table for fitting straight line trend

The equation of the straight line trend is
y_t = a + bx
∴ a = \(\frac{\Sigma y}{n}=\frac{435}{7}\) = 62.143
∴ b = \(\frac{\Sigma x y}{x^{2}}=\frac{273}{28}\) = 9.75
y = 62.143 + 9.75x
Hence, y₂₀₀₆ = 62.143 + 9.75(-3)
= 62.143 – 29.25 = 32893
y₂₀₀₇ = 62.143 + 9.75 (- 2)
= 62.143 – 19.5 = 42.643, etc.
For 2018, x shall be + 9,
y₂₀₁₈ = 62.143 + 9.75(9)
= 62.143 + 87.75 = 149.893
The estimated sales for the year 2018 is ₹ 149.893 lakh.

Question 9.
Rohit has taken a loan of ₹ 600000 with interest rate 10% for the period of 12 yr. Find (reducing) EMI for the above data. [given (1.0083)¹² = 1.1043]
Or
What nominal rate compounded quarterly will be equivalent to 12% compounded continuously? [given e^0.12 = 1.1274]
Answer:
Given, P = ₹ 600000, n = 12 yr
Interest rate = 10% (per annum)
Monthly interest rate (r) = \(\)
= \(\frac{0.10}{12}\) = 0.0083

OR

⇒ \(\left(1+\frac{r}{4}\right)\) = antilog (0.0134)
= 1.031
⇒ \(\frac{r}{4}\) = 0.031
⇒ r = 0.124
⇒ r = 12.4%

Question 10.
A manufacturer’s marginal revenue function is given by MR = 275 – x – 0.3x². Find the increase in the manufacturer’s total increase in revenue, if the production is increased from 10 to 20 units.
Answer:
We have
⇒ MR = 275 – x – 0.3x²
⇒ \(\frac{d R}{d x}\) = MR
⇒ \(\frac{d R}{d x}\) = 275 – x – 0.3x²
We have to find the total increase in revenue, if the production is increased from 10 to 20 units i.e. we have to
R(20) – R(10)

Hence, increase in revenue is ₹ 1900

Section – C [16 Marks]

Question 11.
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 h and that of second machine is 9 h per day.
Each unit of product A requires 3 h on both machines and each unit of product B requires 2 h on first machine and 1 h on second machine. Each unit of product A is sold at a profit of ₹ 7 and B at a profit of ₹ 4. Find the production level per day for maximum profit graphically.
Answer:
Let the manufacturer produces the products A and B be x and y units, respectively.
We construct the following table

Here, total profit Z = 7x + 4y
i.e. maximise Z = 7x + 4y …(i)
Subject to the constraints are
3x + 2y ≤ 12
3x + y ≤ 9 and x ≥ 0, y ≥ 0
Now, consider the given inequations as equations
3x + 2y = 12 …(ii)
3x + y = 9 …….. (iii)
Table for line 3x + 2y = 12 or y = \(\frac{12-3 x}{2}\) is

It passes through the points (0, 6) and (4, 0).
On putting (0, 0) in the inequality 3x + 2 y ≤ 12,
we get
0 + 0 ≤ 12 ⇒ 0 < 12 [true]
So, the half plane is towards the origin.
Table for line 3x + y = 9 or y = 9 – 3x is

It passes through the points (0, 9) and (3, 0).
On putting (0, 0) in the inequality 3x + y ≤ 9, we get
0 + 0 ≤ 9 ⇒ 0 ≤ 9 [true]
So, the half plane is towards the origin.
Also, x ≥ 0 and y ≥ 0, so the region lies in 1st quadrant.
Now, the intersection point of lines (i) and (ii) is
(3x + 2y) – (3x + y) = 12 – 9
⇒ y = 3
and 3x = 12 – 2 × 3
⇒ 3x = 12 – 6
⇒ x = 2
Thus, the point of intersection is B (2, 3).
The graph of inequations is shown below

Here, we see that OABC is a required feasible region, whose comer points are O(0,0), A(3, 0), B(2, 3) and C(0, 6).
The values of Z at these comer points are as follows

Hence, for maximum profit, manufacturer produce 2 units of product A and 3 units of product B.

Question 12.
A machine costs a company ₹ 52000 and its effective life is estimated to be 25 yr. A sinking fund is created for replacing the machine by a new model at the end of its life time, when its scrap realizes a sum of ₹ 2500 only. The price of the new model is estimated to be 25% more than the price of present one. Find what amount should be set aside at the end of each year out of the profit for the sinking fund, if it accumulates at 3.5% compound annually? [given (1.035)²⁵ = 23632]
Answer:
Let ₹ R be the amount set aside each year. Since, the cost of new machine is 25% more than the cost of present.
Cost of machine = ₹ 52000
Cost of new machine after increasing the 25% of the cost of machine,
i.e. 52000 + 25% of 52000
= 52000 \(\left(1+\frac{25}{100}\right)\)
= 52000 × \(\frac{5}{4}\) = ₹ 65000
Scrap value of the present machine = ₹ 2500
So, net amount required at the end of 25 yr to purchase the new model = ₹ (65000 – 2500)
= ₹ 62500
We know that, R = \(\frac{i \times S}{(1+i)^{n}-1}\)
Here, S = ₹ 62500, n = 25 yr and i = \(\frac{3.5}{100}\) = 0.035

Thus, ₹ 1604.68 are set aside each year out of the profit to purchase the new model of the machine.

Question 13.
A bond of face value of ₹ 1000 matures in 5 yr. Interest is paid semi-annually and bond is priced to yield 8% per annum. If the present value of bond is ₹ 1100, find the annual coupon rate, [given (1.04)^-10 = 0.6761]
Or
A person buys a house for which be agrees to pay ₹ 25000 at the end of each month for 8 yr. If money is worth 12% converted monthly, what is the cash price of house? [given (1.01)^-96 = 0.3847]
Answer:
Let the annual coupon rate is 6%.
Given, F = ₹ 1000, then C = ₹ 1000 × \(\frac{r}{200}\) = ₹ 5 r
d = 8% per annum or 4% per half-year
i = \(\frac{4}{100}\) = 0.04
N = 5 × 2 = 10 yr and PV = ₹ 1100
∴ PV = \(\frac{C\left[1-(1+i)^{-N}\right]}{i}\) + F(1 + i)^-N

Hence, coupon rate is 10.47%.
Or
Given, EMI = ₹ 25000, n = 12 × 8 = 96 and
i = \(\frac{12}{1200}\) = 0.01
Let P be the cash price of house.

Hence, the cash price of house is ₹ 1538250.

Case Based/Data Based

Question 14.
Suppose the height and radius of cylindrical pot be 10 cm and ‘r’ cm. The volume of a cylindrical pot changes at a constant rate. At initial time, the radius of the pot is 3 cm. On the basis of above information, answer the following questions.
(i) Write the solution of differential equation formed by the given statement.
Answer:
According to the given statement,
\(\frac{d}{d t}\) (volume of cylinder) = k,
where k is some positive constant.
\(\frac{d}{d t}\) (πr²h) = k
⇒ \(\frac{d}{d t}\) [πr²(10)] = k [∵ h = 10 cm given]
⇒ 10π dr² = kdt
On integrating both sides, we get
10π ∫d(r²) = k∫dt
⇒ 10π(r²) = k(0) + C …(i)
Initially at t = 0, r = 3, then
10π(3)² = k(0) + C
⇒ C = 90π
Put C = 907t in Eq. (i), we get
10πr² = kt + 90π

(ii) After 5 s, the radius of pot becomes 6 cm. Then, find the value of constant.
Answer:
When t = 5 s, r = 6 cm, then
10π(6)² = k( 5) + 90π [∵ 10πr² = kt + 90π]
⇒ 5k= 360π – 90π
⇒ 5 k = 270π
⇒ k = 54π

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics 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 Applied Mathematics Term 2 Set 6 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.
If the marginal cost function at x units of output is given by MC = \(\frac{p}{\sqrt{p x+q}}\), where p, q are constants and the fixed cost of production is zero, find the total cost function.
Or
Find the order and degree of differential equation \(\frac{d^{3} y}{d x^{3}}+\left(\frac{d y}{d x}\right)^{1 / 3}\) = \(x^{\frac{1}{3}}\).
Answer:
We have,
MC = \(\frac{p}{\sqrt{p x+q}}\)
⇒ \(\frac{d C}{d x}=\frac{p}{\sqrt{p x+q}}\) ⇒ dc = \(\frac{p}{\sqrt{p x+q}}\) dx
On integrating both sides, we get
⇒ C = ∫\(\frac{p}{\sqrt{p x+q}}\) dx + K
⇒ C = 2\(\sqrt{p x+q}\) + K …….(i)
where K is the constant of integration.
When x = 0, then fixed cost of production is zero,
i.e. C = 0
Putting x = 0 and C = 0 in Eq. (i), we get
0 = 2√q + K
⇒ K = – 2√q
Putting K = – 2√q in Eq. (i), we find that the total cost function is given by

Question 2.
Samples of two different types of bulbs were tested for length of life and the following data were obtained

Is the difference in mean life of two types of bulbs significant?
Answer:
Given, n₁ = 8, n₂ = 7, s₁ = 36, s₂ = 40, x̄₁ = 1234 and x̄₂ = 1136
Let us take the null hypothesis that there is no significant difference in the mean life of the two types of bulbs.
Test statistics,


From table with v = 8 + 7 – 2 =13, test statistics t_o.05 = 2.16. Since, the calculated value of t is greater than the tabulated value. Therefore, the null hypothesis is rejected.
Hence, there is a significant difference in the mean life of two types of bulbs.

Question 3.
A man borrowed some money and paid back in 3 equal installments of ₹ 2160 each. What sum did he borrow, if the rate of interest charged was 20% per annum compounded annually? Find also the total interest charged, [given, (1.20)^-3 = 0.5786]
Or
Find the present value of a sequence of payments of ₹ 2000 made at the end of every 6 months and continuing forever, if money is worth 10% per annum compounded semi-annually.
Answer:

Hence, the man borrowed ₹ 4551.12
In 3 yr the amount paid = ₹ (3 × 2160) = ₹ 6480
∴ Total interest charged = ₹ (6480 – 455L12)
= ₹ 1928.88
Or
Given, R = ₹ 2000, r = \(\frac{10}{2}\)% = 5% (per half)
So, i = 5 /100 = 0.05
∴ Present value,
P = \(\frac{R}{i}=\frac{2000}{0.05}\) = ₹ 40000

Question 4.
If Σ y = 91, n = 7, Σ x² – 28 and Σ xy = 33, then find the equation of the trend line.
Answer:
We know, a = \(\frac{\Sigma y}{n}\)
∴ a = \(\frac{91}{7}\) = 13 and b = \(\frac{\Sigma x y}{\Sigma x^{2}}=\frac{33}{28}\) = 1.179
∴ Equation of trend line,
y_t = a + bx = 13 + 1.179x.

Question 5.
Neera borrows a sum of ₹ 150000 at an interest rate of 10% (flat) for a tenure of 3 yr. Find her EMI.
Answer:
Given, principal = ₹ 150000
Interest (yearly) = 10% of ₹ 150000 = ₹ 15000
∴ Interest of 3 yr = ₹ 45000
∴ EMI = \(\frac{\text { Principal + Interest }}{\text { Period of months }}\)
= \(\frac{150000+45000}{3 \times 12}\) = \(\frac{195000}{36}\)
= ₹ 5416.67

Question 6.
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is atmost 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300. Formulate on LPP, for finding how many of each should be produced daily to maximise the profit? It is being given that atleast one of each must be produced.
Answer:
Let number of necklaces and bracelets produced by firm per day be x and y, respectively.
Clearly, x ≥ 0 and y ≥ 0
∵ Total number of necklaces and bracelets that the firm can handle per day is atmost 24.
∴ x + y ≤ 24
Since, it takes one hour to make bracelet and half an hour to make a necklace and maximum number of hour available per day is 16.
∴ \(\frac{1}{2}\)x + y ≤ 16 ⇒ x + 2y ≤ 32
Let Z be the profit function.
Then, Z = 100x + 300y
∴ The given LPP reduces to
Maximise Z = 100x + 300y
Subject to constraints,
x + y ≤ 24
x + 2y ≤ 32
and x, y ≥ 0

Section – B [12 Marks]

Question 7.
Evaluate \(\frac{x^{2}-3 x+1}{\sqrt{1-x^{2}}}\).
Or

Answer:

Question 8.
Pramod invested ₹ 21000 in a mutual fund and the value of investment at the time of redemption was ₹ 35000. If CAGR for this investment is 7.55%. Calculate the time period for which the amount was invested?
[given log(1.667) = 0.2219 and log(1.0755) = 0.0316]
Answer:
Given, BV = ₹ 21000
EV = ₹ 35000 and CAGR = 7.55%

⇒ n = 7.022 = 7
Hence, the time period is 7 yr.

Question 9.
A random sample of size 16 has 53 as mean. The sum of squares of deviations from mean is 150. Can this sample be regard as taken from the population having 56 as mean? Level of significance is 5% (given t₁₅ (0.05) = 1.753)
Answer:
Consider: H₀: µ = 56
H_a: µ > 56
Here, n = 16, x̄ = 53 and Σ(x_i – x̄)² = 150
∴ S² = \(\frac{1}{n-1}\) Σ(x_i – x̄)² = \(\frac{150}{15}\) = 10 ⇒ S = √10 = 3.1622
Test statistics, t = \(\frac{\bar{x}-\mu}{(S / \sqrt{n})}=\frac{53-56}{3.1622 / \sqrt{16}}=\frac{(-3) \times 4}{3.1622}\) = – 3.795
Since, calculated value of |t| = 3.795 is greater than tabulated value, so H₀ can be rejected.

Question 10.
The number of letters (in hundred) received in a town on each day of a fortnight is given below

Calculate the 7-days moving average and display it on a graph.
Answer:
Given, table is shown below:

Let the moving average for 7 days is m, then

On the basis of above data we can draw the following graph for moving average

Section – C [16 Marks]

Question 11.
The demand and supply function are p = 25 – x² and p = 2x + 1, respectively. Find the
(i) the equilibrium point
(ii) the consumer’s surplus at the equilibrium points.
Answer:
Given, the demand function and supply are
p = D(x) = 25 – x² and p = S(x) = 2x +1
(i) At equilibrium point D(x) = S(x)
⇒ 25 – x²₀ = 2x₀ + 1
⇒ x²₀ +2x₀ – 24 = 0
⇒ (x₀ – 4) (x₀ + 6) = 0
⇒ x₀ = 4 [∵ x₀ = -6]
Now, putting x₀ = 4 in S(x), we get p₀ = 9
Hence, the equilibrium point (x₀, p₀) is (4,9)

(ii) The consumer’s surplus at the equilibrium point (4, 9) is given by

Question 12.
A firm anticipates on expenditure of ₹ 300000 for plant modernization at the end of 15 yr from now. How much should the company deposite at the end of each year into a sinking fund earning interest 6% per annum? [given, (1.06)¹⁵ = 23966 ]
Answer:
Given, A = ₹ 300000

⇒ R [(1.06)15 – 1] = 300000 × 0.06
⇒ R (2.3966 – 1) = 18000
⇒ R(1.3966) = 18000
⇒ R = \(\frac{18000}{1.3966}\)
= 12888.44
Hence, the company deposite at the end of each year into a sinking fund ₹ 12888.44.

Question 13.
A dietician wishes to mix two types of foods in such a way that the vitamin contents of mixture contains atleast 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C, while food II contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs ₹ 5 per kg to purchase food I and ₹ 7 per kg to purchase food II. Find the minimum cost of such a mixture. Formulate above as an LPP and solve it graphically.
Or
Solve the following LPP graphically.
Maximise Z = 3x + 2y, subject to constraints are x + 2y ≤ 10, 3x + y ≤ 15 and x, y ≥ 0.
Also, determine the area of the feasible region.
Answer:
The given data can be put in the tabular form as follows

Suppose the diet contains x kg of food I and y kg of food II.
Then, the required LPP is minimise
Z = 5x + 7y
Subject to constraints 2x + y ≥ 8, x + 2y ≥ 10 and x ≥ 0, y ≥ 0
On considering the inequalities as equations,
we get
2x + y = 8 …… (i)
and x + 2y = 10 …… (ii)
Table for line 2x + y = 8 is

So, it passes through the points (0, 8) and (4, 0).
On putting (0,0) in the inequality 2x + y ≥ 8,
we get
0 ≥ 8 [which is false]
So, the half plane is away from the origin.
Table for line x + 2y = 10 is

So, it passes through the points (10,0) and (0, 5).
On putting (0,0) in the inequality x + 2y ≥ 10,
we get
0 ≥ 10 [which is false]
So, the half plane is away from the origin.
Also, x, y ≥ 0, so the feasible region lies in the first quadrant.
On solving Eq. (i) and Eq. (ii), we get
x = 2 and y = 4
So, these lines intersect at P(2, 4).
The graphical representation of the system of inequations given below

From the graph, the feasible region is BPC which is unbounded.
The values of Z at comer points are as follows

From table, the minimum value of Z is 38 at P(2,4).
As the feasible region is unbounded, therefore 38 may or may not be the minimum value of Z. For this, we draw a dotted graph of the inequality 5 x + 7y < 38 and check whether the resulting half plane has point in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x + 7y < 38.
Hence, the minimum cost is ₹ 38, when x = 2 and y = 4.
Or
Our problem is to maximise Z = 3x + 2 y …(i) Subject to constraints,
x + 2y ≤ 10 …(ii)
3x + y ≤ 15 …(iii)
and x, y ≥ 0 …(iv)
Table for line x + 2y = 10 is

So, the line passes through the points (0, 5) and (4, 3).
On putting (0,0) in the inequality x + 2y ≤ 10, we get
0 + 2 × 0 ≤ 10
⇒ 0 ≤ 10, which is true.
So, the half plane is towards the origin.
Table for line 3x + y = 15 is

So, the line passes through the points (4, 3) and (5,0).
On putting (0,0) in the inequality 3x + y ≤ 15, we get
3 × 0 + 0 ≤ 15
⇒ 0 ≤ 15, which is true.
So, the half plane is towards the origin.
Also, x, y ≥ 0, so the region lies in the first quadrant.
On solving equations x + 2y = 10 and 3x + y = 15, we get x = 4 and y = 3
So, the intersection point is 6(4,3).
∴ Feasible region is OABCO

The corner points of the feasible region are 0(0,0), A(5,0), 6(4,3) and C(0,5).
The values of Z at the corner points are given below

Therefore, the maximum value of Z is 18 at the point B(4, 3).
∴ Area of feasible region
= Area of ∆BOC + Area of ∆OAB
= \(\frac{1}{2}\) × OC × BD + \(\frac{1}{2}\) × OA × BE
= \(\frac{1}{2}\) × 5 × 4 + \(\frac{1}{2}\) × 5 × 3
= 10 + \(\frac{15}{2}\) = 10 + 7.5
= 17.5 sq units

Case Based/Data Based

Question 14.
An entertainment company has gained much popularity all over the countary. They buy a car to travel to their destination. The purchase price is ₹ 1000000. The car will be depreciated linearly over 10 yr and will have a scrap value of ₹ 500000.

On the basis of above information, answer the following questions.
(i) Find the rate of depreciation.
Answer:
∴ Rate of depreciation
= \(\frac{500000-1000000}{10-0}\)
= – \(\frac{500000}{10}\)
= – 50000
∴ Rate of depreciation is ₹ 50000

(ii) When will the car be worth ₹ 800000?
Answer:
According to the question,
800000 = – 500001 +1000000
⇒ – 200000 = -50000 t
⇒ t = 4 yr

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 3 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. 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.
How are the following dealt with while preparing Income and Expenditure Account for the year ended 31st March 2021? (2)
Receipts and Payments Account (An Extract)

Answer:
Income and Expenditure Account (An Extract) For the year ended 31st March, 2021

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

Question 3.
Star and Moon are two partners sharing profits in the ratio of 2 :1. Give the journal entry at the time of dissolution in the following cases.
(A) Deferred revenue advertising expenditure appered at ₹ 90,000
(B) Profit and Loss A/c was appearing on the assets side of balance sheet at ₹ 1,80,000.
(C) An unrecorded investment realized at ₹ 18,000.
(D) Partner moon paid to a creditor ₹ 60,000 (2)
Answer:

Question 4.
How will you deal with the Entrance Fees while preparing the final accounts for the year ended 31st March, 2021, in each of the following cases:
(A) During the year 2020-21, Entrance Fees received ₹ 2,50,000. It is the policy of the club to treat the Entrance Fees as capital receipts.
(B) During the year 2020-21, Entrance Fees received ₹ 2,50,000. According to the policy of the club, 40% of the Entrance Fees is to be capitalized.
OR
From the following information provided by Moonlight Club, calculate the amount of subscriptions which will be treated as income for the year ended 31st March 2019:
(i) Subscriptions received during the year ended 31st March, 2019 amounted to ₹ 80,000.
(ii) Subscriptions outstanding in the beginning of the year ended 31st March 2019 amounted to ₹ 5,000.
(iii) Subscriptions not yet collected for the year ended 31st March 2019 amounted to ₹ 8,000.
(iv) Subscriptions received in advance for the year ended 31st March, 2020 amounted to ₹ 2,000. (3)
Answer:

Question 5.
In a partnership firm, Abhi, Shiva, and Vishnu were partners sharing profits and losses is the ratio of 5 :3: 2. The balance sheet of the firm as at 31st March 2020 is as under:

On 1st April 2020, Shiva retires from the firm and the new profit sharing ratio decided between Abhi and Vishnu is 3 :1. Following adjustments are agreed:
(i) Land and Building be written up by ₹ 88,000.
(ii) Plant and Machinery be reduced by ₹ 88,000.
(iii) Stock be written down by ₹ 8,440.
(iv) An amount of ₹ 8,800 included is Sundry debtors be written off as it is no longer receivable.
(v) A provision for doubtful debts be maintained at 5%.
(vi) There was an outstanding amount of Repairs of ₹ 4,800.
(vii) An amount of ₹ 5,600 included in sundry creditors be written back as no longer payable,
(viii) An old furniture written-afF previously was sold for ₹ 16,000 as scrap.
You are required to prepare the Revaluation Account of the firm, to give effect to the above adjustment. (3)
Answer:

Question 6.
On 9th April, 2021, Sunshine Ltd. issued 500, 10% Debentures of ₹ 1,000 each credited as fully paid-up to the promoters for their services to incorporate the company. On 18th May, 2021, the company issued 100, 10% Debentures of ₹ 1,000 each credited as fully paid-up to the underwriters towards their commission.
Pass necessary journal entries in the books of the company.
OR
Suhana Limited purchased machinery from Vikrant Manufacturers Limited. The company paid the vendors by issue of some equity shares and debentures and the balance through Bill payable on acceptance in their favour payable after three months. The accountant of the company, while journalizing the above-mentioned transactions left some items blank. You are required to fill in the blanks. (3)

Answer:

Question 7.
Following is the Balance Sheet of A and B as at 31st March, 2021:

The firm was dissolved on the above date under the following arrangement:
(i) A promised to pay off Mrs. A’s Loan and took Stock at ₹ 12,000.
(ii) B took half the Investments @ 10% discount.
(iii) Book Debts realised ₹ 57,000.
(iv) Trade Creditors and Bills Payable were due on average basis of one month after 31st March, but were paid immediately on 31st March @ 2% discount per annum.
(v) Plant realised ₹ 75,000; Building ₹ 1,20,000; Goodwill ₹ 18,000 and remaining Investments ₹ 13,500.
(vi) An old typewriter, written off completely from the firm’s books, now estimated to realise f 900. It was taken by B at this estimated price.
(vii) Realisation expenses were ₹ 3,000.
Prepare Realisation Account and Capital Accounts of Partners.
OR
Amit, Balan and Chander were partners in a firm sharing profits and losses in the proportion of \(\frac{1}{2}, \frac{1}{3} \text { and } \frac{1}{6}\) respectively. Chander retired on 31st March 2022. The Balance Sheet of the firm on the date of Chander’s retirement was as follows:
Balance Sheet
as on 31st March 2022

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:


Working Notes:
(1) Value of Investments taken over by B:
Book Value of Investments taken over by B = 50% of Total Investments
= \(\frac{50}{100} \times ₹ 30,000\)
= ₹ 15,000

Value of Investments taken over by B = Book Value of Investments taken over x \(\frac{90}{100}\)
=15,000x \(\frac{90}{100}\)
= ₹ 13,500
[Since Investment is taken over by B at a discount of 10%]

(2) Calculation of Amount paid to Creditors:
Amount paid to Creditors = Amount to be paid to Credftors – 2% Discount for 1 month
= ₹ 90,000- \(\left(₹ 90,000 \times \frac{2}{100} \times \frac{1}{12}\right)\)
= ₹ 90.000 – ₹ 150
= ₹ 89,850

(3) Calculation of Amount paid to BillS Payable:
Amount paid to Bills Pagable = Amount to be paid to Bills Payable – 2% Discount for 1 month
= ₹ 24,000 – \(\left(₹ 24,000 \times \frac{2}{100} \times \frac{1}{12}\right)\)
= ₹ 24,000 – ₹ 4O
= ₹ 23,960

Working Notes:
(1) Adjustment of Goodwill:
Goodwill of Firm = ₹ 27,000
Chander’s share of Goodwill = ₹ 27,000 x \(\frac{1}{6}\) = ₹ 4,500
which will be compensated by Amit and Balan in their Gaining Ratio. i.e, 3 : 2
Amit will compensate = ₹ 4,500 x \(\frac{3}{5}\) = ₹ 2,700
Balan wilL compensate = ₹ 4,500 x \(\frac{2}{5} \) = ₹ 1,800

(2) Adjustment of CapitoL:
Adjusted Old Capitol of Amit = ₹ 40,000 + ₹ 300 + ₹ 4, 500 – ₹ 2,700
= ₹ 42,100
Adjusted Old Capital of Batan = ₹ 36,500 + ₹ 200 + ₹ 3,000 — ₹ 1,800
= ₹ 37,900
Total Adjusted Capital of Amit and Batan = ₹ 42,100 + ₹ 37,900 = ₹ 80,000
New Profit Shoring Ratio of Amit and Batan = 3: 2
Amit’s New Capital = ₹ 80,000 x \(\frac{3}{5}\) = ₹ 48,000
BaLan’s New Capital = ₹ 80,000 x \(\frac{2}{5}\) = ₹ 32,000
Note: Since, here no information is given regarding the share acquired by Amit and Batan, therefore, their gaining ratio is same as their new profit sharing ratio, i.e 3: 2

Question 8.
Goel Iron Casts Industries engaged in production of Iron rods having registered office in Chandigarh was incorporated on 1st April 2016. To expand their business, company purchased a Plant from Faujdar Metal Limited to increase the production of iron rods. Goel Iron Casts Industries paid the amount to Faujdar Metal Limited as follows:
(i) By issuing 16,000,12% Debentures of ₹ 100 each at a discount of 10%.
(ii) By issuing 80,000, Equity Shares of ₹ 100 each at a premium of 10%.
(iii) Balance by accepting a bill of exchange of ₹ 4,00,000 payable after two months.

You are required to answer the following questions:
(A) Calculate the amount Goel Iron Casts Industries paid to Faujdar Metal Limited by issuing 12% Debentures.
(B) Calculate the purchase price of Plant.
(C) Calculate the amount of annual fixed obligation associated with debentures.
(D) Pass journal entry which will be passed at the time of purchase of Plant in the books of Goel Iron Casts Industries.
(E) Pass journal entry for the allotment of debentures. (5)
Answer:
(A) Amount Paid to Faujdar Metal limited by Issue of 12% Debentures
= 16,000 x [₹ 100 – (10% of ₹ 100)]
= 16,000 x ₹ 90 = ₹ 14,40,000

(B) Purchase Price of Plant = Amount Paid by Issue of Debentures + Amount Paid by Issue of Equity
Shares + Amount Paid by Bill of Exchange
= {16.000 x [₹ 100 – (10% of ₹ 100)J} + { 80,000 x [₹ 100 + (10% of ₹ 100)1} + ₹ 4,00,000
= (16.000 x ₹ 90) + (80,000 x ₹ 110) +₹ 4,00,000
= ₹ 14,40,000 +₹ 88,00.000 + ₹ 4,00,000
= ₹ 1,06,40,000

(C) Interest on 12% Debentures = ₹ 16,00,000 x \(\frac{12}{100}\) =₹ 1,92,000

(D) In the Books of Goel Iron Casts Industries Journal

Question 9.
Receipts and Payments Account of Sunderban Society Club for the year ended 31st March 2021 is as follows:
Receipts and Payments Account

Additional Information:

Depreciate furniture by 20% and Sports Equipment by 30%.
You are required to prepare an Income and Expenditure Account for the year ended 31st March, 2021 and ascertain the Capital Fund on 31st March, 2020. (5)
Answer:
In the Books of Sunderban Society Club
Balance Sheet
as on 31st March. 2020

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

Question 10.
State which of the following transactions would result in inflow, outflow or no flow of Cash and Cash Equivalents:
(A) Decrease in Cash Credit.
(B) Sale of Current Investments. (2)
Answer:
(A) Decrease in Cash Credit wilt results in outflow of Cash and Cash Equivalents.
(B) Sale of Current Investments wilt result in no fLow of Cash and Cash Equivalents.

Question 11.
From the following Statement of Profit and Loss of Gokulnath Traders Limited as at 31st March 2020 and 2021, you are required to prepare Comparative Statement of Profit and Loss as at 31st March 2020 and 2021:

OR
You are provided with the Common Size Balance Sheet of Sushiksha Limited as at 31st March, 2022 with missing information. You are required to fill in the blanks:
In the Books of Sushiksha Limited
Common Size Balance Sheet
as at 31st March 2022 (3)

Answer:
Gotuknath Traders limited
Comparative Statement of Profit and Loss
for the years ended 31 March 2020 and 2021

OR
In the Books of Sushikshci Limited
Common Size BaLance Sheet
as at 31 March 2022

Question 12.
From the following Balance Sheet of Tamalika Limited as at 31st March 2019 and additional information, prepare Cash Flow Statement for the year ended 31st March 2019:
Statement of Profit and Loss
for the year ended 31st March 2019

Notes to Accounts:

Additional Information:
(i) During the year, Machinery costing ₹ 1,40,000 (accumulated depreciation provided thereon ₹ 1,10,000) was sold for ₹ 20,000.
(ii) During the year, Non-Current Investments costing ₹ 80,000 were sold at a profit of ₹ 16,000.
(iii) Additional debentures were issued on 31st March, 2019. (5)

Answer:
Tamatikci Limited Cash Flow Statement
for the year ended 31s March, 2021

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics 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 Applied Mathematics Term 2 Set 4 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.
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 grows 3 times.
Or
The marginal cost of producing x pairs of tennis shoes is given by MC = 60 + \(\frac{400}{x+1}\).
If the fixed cost is ₹ 3000, find the total cost function.
Answer:
Let P₀ be the population initially and P be the population after t yr,

On integration both sides, we get
log P = \(\frac{1}{10}\) t + C
When t = 0, then P = P₀
∴ log P₀ = C
From Eq. (i), log P = \(\frac{t}{10}\) + log P₀
⇒ \(\log \left(\frac{P}{P_{0}}\right)=\frac{t}{10}\)
When the population grows 3 times i.e. P = 3 P₀, then log\(\left(\frac{3 P_{0}}{P_{0}}\right)\) = \(\frac{t}{10}\) ⇒ t = 10 log 3
Thus, it takes 10 log 3 years for the population to grow 3 times.

Or

C = 60x + 400 log |(x + 1)| + K ………. (i)
If x = 0, C = ₹ 3000
∴ 3000 = 0 + 400 log(1) + K
⇒ K = 3000
∴ C = 60x + 400 log |(x + 1)| + 3000 [from Eq. (i)]

Question 2.
A random sample of 36 workers is drawn from a factory to test, if their average monthly wages is ₹ 1500 or not. The population wages distribution is assumed to be normal with standard deviation of ₹ 300. The average monthly wages based on the 36 observations comes out to be ₹ 1800. Perform a test at 5% level of significance. Calculate test statistics.
Answer:
Consider, H₀ : µ = 1500
H₁ : µ ≠ 1500
x̄ = 1800, µ₀ = 1500,
σ = 300 and n = 36
∴ Test statistics,

Question 3.
Find the present value of perpetuity of ₹ 900 at end of each quarter, if money is worth 6% compounded quarterly.
Or
To what sum will ₹ 16000 accumulate in 8 yr, if invested at an effective rate of 10%. [given (1.08)⁸ = 1.85]
Answer:
Given, R = ₹ 900
i = \(\frac{0.06}{4}\) = 0.015
Present value of perpetuity,
P = \(\frac{R}{i}=\frac{900}{0.015}\) = ₹ 60000
Or
We have, P = ₹ 16000,
i = \(\frac{8}{100}\) = 0.08
∴ S = P(1 + i)ⁿ
S = 16000(1 + 0.08)⁸
= 16000(1.08)⁸
= 16000(1.85) = 29600
Hence, the required sum is ₹ 29600

Question 4.
At what rate of interest will be present value of a perpetuity of ₹ 500 payable at the end of every 6 months be ₹ 10000?
Answer:
Let the rate of interest be r% per annum.
Then, i = \(\frac{r}{200}\)
Given, R = ₹ 500 and P = ₹ 10000
Using P = \(\frac{R}{i}\)
⇒ i = \(\frac{R}{P}\)
⇒ \(\frac{r}{200}=\frac{500}{10000}\) ⇒ r = 10
Hence, rate of interest is 10% per annum.

Question 5.
The corner points of the feasible region determined by the following systems of linear inequalities
2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0 and y ≥ 0 are (0, 0), (5, 0), (3,4) and (0, 5).
Let Z = px + qy, when p, q ≥ 0, then find the relation between p and q so that the maximum of Z occurs at both points (3, 4) and (0, 5).
Answer:
The values of Z = px + qy at the points (3,4) and (0,5) are 3p + 4q and 5q, respectively.
As, Z has maximum value of both points (3, 4) and (0, 5), we get
3p + 4q = 5q ⇒ 3p = q ⇒ 3p – q = 0
which is the required relation between p and q.

Question 6.
The production of a cement by a firm in 1 to 9 is given below.

Calculate the 3-yearly moving average.
Answer:
Calculation of 3-yearly moving average

Section – B [12 Marks]

Question 7.
Consider a hypothetical population comprising only four values 4, 6, 8 and 10. Find point estimation of population variance (σ²). Also, find standard error of sample variance (S²).
Answer:

Question 8.
Calculate the quarterly trend values by method of least square for quarterly data for last 5 yr given below.

Or
Apply the method of least squares to obtain the trend values from the following data.

Also, predict the sales for the year 2013.
Answer:
Here, n = 5 (odd)
So, middle year i.e. 1966 is taken as origin, we will fit linear trend equation between average quarterly values (y) and time variable x (in year).

The normal equations for fitting linear trend equations.

∴ Linear trend equation is y_t = 112 + 23 x
From Eqs. (i) and (ii), yearly increment in trend value = b = 23
Quarterly increment = \(\frac{23}{4}\) = 5.75
Here, n = 5 (odd)
So, middle year i.e. 2008 is taken as origin.

The equation of the straight line trend is
y_t = a + bx
Since, Σx = 0, a = \(\frac{\Sigma y}{n}\) = \(\frac{550}{5}\) = 110
and b = \(\frac{\Sigma x y}{\Sigma x^{2}}\) = \(\frac{-20}{10}\) = – 2
The required equation is,
y_t = 110 – 2x
For x = – 2,
y_t = 110 – 2(-2) = 114
Thus, for 2007 the trend value will be
114 – 2 = 112.

Question 9.
Surbhi borrowed a home loan amount of ₹ 150000 from a bank at an interest rate of 12% per annum for 3 yr. Find the monthly installment amount. She has to pay to the bank, [given, (1.01)^-36 = 0.698]
Answer:
If principle, P = ₹ 150000
i = \(\frac{12}{1200}\) = 0.01 and n = 12 × 3 =36
Let E be the monthly installment paid by Surbhi,

Question 10.
Supply function of a producer is given by 40 p = (x + 15)². Find the producer’s surplus, when the market price is ₹ 40.
Answer:
Given, the supply function is
40p = (x + 15)²
At p₀ = 40, we get
40 × 40 = (x₀ +15)²
⇒ x₀ + 15 = 40 ⇒ x₀ = 25
∴ Producer’s surplus,

Section-C [16 Marks]

Question 11.
Consider a bond with a coupon rate of 10% and annual coupons. The par value is ₹ 1000 and the bond has 5 yr to maturity. The yield to maturity is 11%. What is the value of the bond? [given, (1.11)^{– 5} = 0.593451]
Answer:
Given, face value of the bond (F) = ₹ 1000
Coupon rate (id) (annual) = 10% = 0.1
∴ R = F × id = 1000 × 0.1 = ₹ 100
Number of periods before redemption (n) = 5
and yield rate (i) = 11% =0.11.
We know that purchase value

= 100(3.6959) + 593.451
= 369.59 + 593.40
= 962.99 ≅ 963
∴ The value of the bond is ₹ 963.

Question 12.
A propeller costs ₹ 180000 and its effective life is estimated to be 10 yr. A sinking fund is created for replacing the propeller by a new model at the end of its lifetime, when its scrap realises a sum of ₹ 34000 only. The price of the new model is estimated to be 30% more than the price of the present one. What amount should be put into the sinking fund at the end of each year, if it accumulates at 4% per annum compound interest? [given, (1.04)¹⁰ = 1.480]
Answer:
According to the question,
The present value of the machine is ₹ 180000.
Since, the price of new model is 30% more than the price of present machine.
∴ Price of new model 130
= \(\frac{130}{100}\) × 180000 = ₹ 234000
Scrap value of the present machine is ₹ 34000.
The net amount (A) to be paid after 10 yr for new model = 234000 – 34000 = ₹ 200000
Let the value of each sinking fund is a.
The rate of interest for 10 yr is
i = \(\frac{4}{100}\)
We know that
A = \(\frac{R}{i}\)[(1 + i)ⁿ – 1] …….. (i)
Now, we put the value of A, i and n in Eq. (i) to get the value of each instalment

Hence, the value of each sinking fund is ₹ 16666.67 to be paid at the end of each year.

Question 13.
There are two types of fertilisers F₁ and F₂, F₁ consists of 10% nitrogen and
6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁ costs ₹ 6 per kg and F₂ costs ₹ 5 per kg. Determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost? What is the minimum cost?
Or
Two godowns A and B have a grain storage capacity of 100 quintals and 50 quintals, respectively. They supply it to 3 ration shops D, E and F, whose requirements are 60, 50 and 40 quintals, respectively. The cost of transportation per quintal from godowns to the shops are given in the following table

How should the supplies be transported in order that the transportation cost is minimum?
Answer:
Suppose, the farmer uses x kg of F₁ and y kg of F₂. We have to construct the following table

Total cost of fertilisers,
Z = 6x + 5y
So, our problem is to minimise
Z = 6x + 5 y
Subject to constraints are,
\(\frac{x}{10}+\frac{y}{20}\) ≥ 14
⇒ 2x + y ≥ 280
\(\frac{6 x}{100}+\frac{10 y}{100}\) ≥ 14
⇒ 3x + 5y ≥ 700
and x, y ≥ 0
Table for line 2x + y = 280 is

So, line passes through the points (0, 280) and (140, 0)

On putting (0, 0) in the inequality
2 x + y ≥ 280, we get
2 × 0 + 0 ≥ 280
⇒ 0 ≥ 280, which is not true.
So, the half plane is away from the origin.
Table for line 3x + 5y = 700 is

So, line passes through the points (0, 140) and \(\left(\frac{700}{3}, 0\right)\)
On putting (0, 0) in the inequality 3x + 5y ≥ 700,
we get 3 × 0 + 5 × 0 ≥ 700
⇒ 0 ≥ 700, which is not true.
So, the half plane is away from the origin.
Also, x, y ≥ 0, so the region lies in the 1 quadrant.
On solving the equations
2x + y = 280 and 3x + 5y = 700, we get B(100,80).
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are
A\(\left(\frac{700}{3}, 0\right)\), B(100,80)and C(0,280).
The values of Z at the comer points are given below

As, the feasible region is unbounded, therefore 1000 may or may not be the minimum value of Z. For this, we draw a graph of the inequality 6x + 5y < 1000 and check whether the resulting half plane has points in common with the : feasible region or not.

It can be seen that the feasible region has no common point with 6x + 5y < 1000. Therefore, 100 kg of fertiliser F1 and 80 kg of fertiliser F₂ should be used to minimise the cost and the minimum cost is ₹ 1000.

Or

Let godown A supplies x and y quintals of grain to the shops D and E, respectively.
Then, (100 – x – y) will be supplied to shop F.
The requirement at shop D is 60 quintals, since x quintals are transported from godown A.
Therefore, the remaining (60 – x) quintals are transported from godown B. Similarly, (50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shops E and F, respectively. The given problem can be represented diagrammatically as follows

Let Z be the total cost of transportation, then
Z = 6x + 3y + 2.50 (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 [(x + y) – 60]
= 6x + 3y + 250 – 2.50x – 2.50y + 240 – 4x + 100 – 2 y + 3x + 3 y – 180
= 2.50 x + 1.50y + 410 …(i)
Subject to constraints,
60 – x ≥ 0
or x ≤ 60 …(ii)
50 – y ≥ 0
or y ≤ 50 …(iii)
100 – (x + y) ≥ 0
or x + y ≤ 100 …(iv)
x + y – 6 ≥ 0
or x + y ≥ 60 …(v)
and x, y ≥ 0 …(Vi)
Table for line x + y = 100 is

So, line passes through the points (100, 0) and (0, 100).
On putting (0, 0) in the inequality x + y ≤ 100, we get
0 + 0 ≤ 100 ⇒ 0 ≤ 100, which is hue.
So, the half plane is towards the origin.
Table for line x + y = 60 is

So, line passes through the points (0, 60) and (60, 0).
On putting (0, 0) in the inequality x + y ≥ 60, we get
0 + 0 ≥ 60 ⇒ 0 ≥ 60, which is not hue.
So, the half plane is away from the origin.
Now, draw the graph of lines x = 60 and y = 50, which is perpendicular to X and Y-axes, respectively.
Clearly, the half planes x ≤ 60 and y ≤ 50 is towards Y and X-axes, respectively.
Also, x, y ≥ 0, so the region lies in the I quadrant.
The points of intersection of lines corresponding to Eqs. (ii), (iii), (iv) and (v) are
A(60, 0), B (60, 40), C (50, 50) and D (10, 50).
The graphical representation of these lines is given below

The shaded region in the graph represents the feasible region and its comer points are
A (60,0), B (60,40), C (50,50) and D (10,50).
The values of Z at the comer points are given below

The minimum value of Z is 510 at D (10,50).
Thus, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals and 40 quintals, respectively and from B to D, E and F is 50 quintals, 0 quintal and 0 quintal, respectively.
The minimum cost is ₹ 510.

Case Based/Data Based

Question 14.
Following paragraph given to student by the teacher.
The given integral ∫ f(x)dx can be transformed into another form by changing the independent variable x to t by substituting x = g(t).
Consider, I = ∫ f(x) dx
On putting x = g(t), so that \(\frac{d x}{d t}\) = g'(t)
We write, dx = g'(t)dt
Thus, I = ∫ f(x)dx = ∫ f(g(t))g'(t)dt
This change of variable formula is one of the important tools available to us in the name of integration by substitution.
On the basis of above information, answer the following questions.
(i) Evaluate \(\int-\frac{d x}{x^{2}-16}\).
Answer:

(ii) Evaluate \(\int \frac{x}{\sqrt{32-x^{2}}}\) dx.
Answer:
Let I = \(\int \frac{x}{\sqrt{32-x^{2}}}\) dx.
On putting
32 – x² = t
⇒ – 2xdx = dt
⇒ x dx = \(\frac{-1}{2}\) dt
Now,

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Accountancy Standard Term 2 Set 3 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 for the year ended 31st March 2021 from the information given:
Stock of Stationery on 1st April 2020 – ₹ 3,000
Creditors for Stationery on 1st April 2020 – ₹ 2,000
Amount paid for Stationery during the year – ₹ 10,800
Stock of Stationery on 31st March 2021 – ₹ 500. (2)
Answer:

Question 2.
Pass the necessary journal entries for the following transactions on the dissolution of a partnership firm:
(A) Saran, a partner is paid remuneration of ₹ 40,000 for the dissolution of the firm. Realization expenses of ₹ 64,000 are borne by the firm.
(B) Sanam one of the partners was to receive 2% of the value of assets realized as remuneration for completing the dissolution work and was to bear realization expenses. Realization expenses were ₹ 8,000 paid by Sanam. The assets including cash at Bank ₹ 24,000, realized ₹ 12, 24,000. (2)
Answer:

Question 3.
A, B and C are partners in a firm sharing profit in 3: 2 :1. The firm closes its book on 31st March every year. B died on 12th June 2020. On B’s death, the goodwill of the firm was valued at ₹ 2,40,000. B’s share in the profits of the firm till the date of death from the last Balance Sheet was to be calculated on the basis of previous year’s profit which was ₹ 6,00,000.
Fill in the missing figures in the following journal entries.
(2)
Answer:

Working Notes:
B’s share of profit will his date of death
= ₹ 6,00,000 x \(\frac{73}{365} \times \frac{2}{6}\)
= 40,000

Question 4.
Distinguish between Receipts and Payments Account and Income and Expenditure Account on the basis of:
(A) Nature
(B) Period
(C) Nature of Items
OR
From the information given below, prepare Receipts and Payments Account of Harnaam Club for the year ending on 31st March 2021:

(3)
Answer:
Difference between Receipts and Payments Account and Income and Expenditure Account:

Question 5.
Alia, Kiara, and Sonam are partners is Sunrise Limited sharing profit and losses is the ratio of \(\frac{4}{9}: \frac{1}{3}: \frac{2}{9} \) On 1st April 2021 Kiara gave a notice to retire from the firm. Alia and Sonam decided to share the future profits equally.

The capital accounts of Alia and Sonam after all adjustment showed a balance of ₹ 86,000 and ₹ 1,61,000 respectively. The total amount to be paid to Kiara was ₹ 19,10,000. This amount was to be paid by Alia and Sonam is such a way that their capitals become proportionate to their new profit sharing ratio. Pass the necessary journal entries in the books of Sunrise Limited for the above transactions. Show your working clearly. (3)
Answer:


Working Note:
(1) CalcuLation of total capitaL of new firm:
₹
Adjusted Capital of Alia 8,60,000
Act Adjusted Capital of Sonam 16,10,000
Amount to be paid to Kiaro 19,10,000
Total Capital of New Firm 43,80,000

(2) Calculation of Cash to be brought in or Withdrawn by Alio and Sonam:

Question 6.
On 1st April 2021, Jasmine Limited issued 10,000, 8% Debentures of ₹ 100 each at a discount of 5% redeemable at a premium of 15% at the end of five years. All the debentures were subscribed and allotment was made. The company had balance in Securities Premium Reserve of ₹ 80,000.
OR
Malvika Limited purchased furniture worth ₹ 6,60,000 from Sunaina Limited and paid to Suniana Limited as follows:
(i) 50% of the amount by accepting a bill of exchange payable after one month.
(ii) Balance by issuing 8% debentures of ₹ 100 each at a premium of 10%.
Pass necessary journal entries in the books of Malvika Limited for the purchase of furniture and making payment to Sunaina Limited. (3)
Answer:

Working Note:
Purchase consideration = ₹ 6,60,000
Amount paid by Bilis Payable = 50% of ₹ 6,60,000
= ₹ 3,30,000
Amount paid by issuing debentures = ₹ 6,60,000 – ₹ 3,30,000
= ₹ 3,30,000
Number of debenture issued = \(\frac{\text { Amount paid by Issuing Debentures }}{\text { Issue Price per Debenture }} \)
= \(\frac{₹ 3,30000}{(₹ 100+₹ 10)} \)
= \(\frac{₹ 3,30000}{₹ 110}\)
= 3,000

Question 7.
X, Y, and Z decided to dissolve their partnership firm on 31st March 2021. Their profit sharing ratio was 3:2:1 and their Balance Sheet was as under:

It is agreed as follows:
(i) The stock of value of ₹ 41,660 are taken over by X for ₹ 35,000 and he agreed to discharge bank loan.
(ii) The remaining stock was sold at ₹ 14,000.
(iii) Debtors amounting to ₹ 10,000 realised ₹ 8,000. The remaining debtors realized 50% at their book value.
(iv) Land is sold for ₹ 1,10,000.
(v) The cost of realization amounted to ₹ 1,200.
(vi) There was a typewriter not recorded in the book’s worth of ₹ 6,000 which was taken over by one of the Creditors at this value.
Prepare Realisation Account, Partners’ Capital Accounts, and Cash Account to close the books of the firm.
OR
Kanika, Disha, and Kabir were partners in a partnership firm sharing profits and losses in the ratio
of 2 :1: 1. On 31st March 2019, their Balance Sheet was as under:
Balance Sheet
as at 31st March 2019

Kanika retired on 1st April 2019. For this purpose, the following adjustments were agreed upon:
(i) Goodwill of the firm was valued at 2 years’ purchase of average profits of three completed years preceding the date of retirement. The profits for the year were:
2016- 17 ₹ 1,00,000
2017- 18 ₹ 1,30,000
(ii) Fixed Assets were to be increased to ₹ 3,00,000.
(iii) Stock was to be valued at 120%.
(iv) The amount payable to Kanika was transferred to her Loan Account.
You are required to prepare a Revaluation Account, Capital Accounts of the partners, and the Balance Sheet of the reconstituted firm. (5)
Answer:





Working Note:
Calculation of Goodwill:
Goodwill = Average Profits x Number of Years’ Purchase
Average Profits = \(\frac{\text { Total Profits }}{\text { Number of Years }}\)
= \(\frac{₹ 1,00,000+₹ 1,30,000- ₹ 20,000}{3}\)
= \(\frac{₹ 2,10,000}{3}\)
= ₹ 70,000

Goodwill of the Firm = 70,000 x 2 = 1,40,000
Kanika’s share of Goodwill = 1,40,000 x \(\frac{2}{4}\) = ₹ 70,000
which will be borne by gaining partners in their gaining ratio
Disho will compensate = 70,000 x \(\frac{1}{2}\) = ₹ 35,000
Kabir wilt compensate = 70,000 x \(\frac{1}{2}\) = ₹ 35,000
Note: Since no information is given about the share of the gain, it is assumed that the old partners are gaining in their old profit sharing ratio.

Question 8.
Swadeshi Bites Limited, a FMCG company appointed Mr. Abhimanyu Roy as the Marketing IHead of the company, with a target to enter the international market. Mr. Abhimanyu Roy discussed the ways and means to achieve target of the company with his marketing team.

After reviewing the suggestions given by all the marketing team members and consultation with Finance Head an additional fund of ₹ 2,62,50,000 is required to penetrate their roots in the international market.
After all discussions, on 1st April 2021, the board of directors of Swadeshi Bites Limited decided to issue 9% Debentures of ₹ 1,000 each to the public at a premium of 5%, redeemable after 5 years at a premium of 10% to raise the additional fund.

You are required to answer the following question:
(A) Calculate the number of debentures to be issued to raise additional funds.
(B) Pass journal entry to be passed at the time of allotment of debentures.
(C) Pass journal entry to write off loss 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:
(A) Additional. Funds Raised = 2,62,50,000
Number of Debentures Issued to raise Additional Funds = \(\frac{\text { Additional Funds Raised }}{\text { Issue Price }} \)
= \(\frac{₹ 2,62,50,000}{₹(1,000+50)}\)
= 25,000.


(E) Interest on 9% Debentures = ₹ 2,50,00,000 x\(\frac{9}{100}\) =₹ 22,50,000

Question 9.
From the following Receipts and Payments Account of Harvinder Social Club and the information supplied, prepare Income and Expenditure Account for the year ended 31st March 2020 and Balance Sheet as at that date:
Receipts and Payments Account

Additional Information:
(i) The club has 500 members each paying an annual subscription of ₹ 1,000. Subscriptions Outstanding on 31st March 2019 were ₹ 1,20,000.
(ii) On 31st March 2020, Salaries Outstanding amounted to ₹ 40,000. Salaries paid in the year ended 31st March 2020 included ₹ 1,20,000 for the year ended 31st March 2019.
(iii) On 1st April, 2019, the club owned Land & Building valued at ₹ 40,00,000; Furniture ₹ 4,00,000 and Books ₹ 4,00,000.
(iv) Provide depreciation on Furniture at 10%. (5)
Answer:

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

Question 10.
State which of the following transactions will result in inflow, outflow or no flow of Cash and Cash Equivalents:
(A) Sale of Marketable Securities for cash at par.
(B) Declaration of Final Dividend ₹ 50,000. (3)
Answer:
(A) Sale of Marketable Securities for cash at par will result in no flow of Cash and Cash Equivalents.
(B) Declaration of Final Dividend ₹ 50,000 wilt result in no flow of Cash and Cash Equivalents.

Question 11.
From the following information provided of Star Ltd., for the year ended 31st March 2020 and
Following is the Balance Sheet of Sumona Trading Limited as at 31st March 2019 and 2020:

OR
Following is the Balance Sheet of Surnona Trading Limited os oct 31 March 2019 and 2020:

You are required to prepare Comparative Balance Sheet of Sumona Trading Limited as at 31st March 2019 and 2020. (3)
Answer:

Question 12.
From the following Balance Sheet of Samta Ltd., as at 31st March 2021, prepare Cash Flow Statement for the year ended 31st March 2021:


Additional Information:
(i) During the year a piece of machinery costing ₹ 60,000 on which depreciation charged was ₹ 20,000 was sold at 50% of its book value.
Depreciation provided on tangible assets ₹ 60,000.
(ii) Income tax of ₹ 45,000 was provided.
(iii) Additional Debentures were issued at par on 1st October 2020 and Bank Loan was repaid on the same date.
(iv) At the end of the year, Preference Shares were redeemed at a premium of 5%. (5)
Answer:

(3) CaLcuLation of Interest on Debentures
Interest on Debentures = On ₹ 1,50000 @ 8% for 12 months + On ₹ (2.60,000 – 1,50,000) @ 8% for 6 months
= \(\left(₹ 1,50,000 \times \frac{8}{100}\right)+\left(₹ 1,10,000 \times \frac{8}{100} \times \frac{6}{12}\right)\)
= ₹ 12.000 + ₹ 4,400
= ₹ 16,400.

(4) Calculation of Interest on Bank Loan:
Interest on Bank Loan = On ₹ 50,000 @ 8% for 6 months + On ₹ 40,000 @ 8% for 6 months
= \(\left(₹ 50,000 \times \frac{8}{100} \times \frac{6}{12}\right)+\left(₹ 40,000 \times \frac{8}{100} \times \frac{6}{12}\right)\)
= ₹ 2,000 + ₹ 1,600
= ₹ 3,600

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics 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 Applied Mathematics Term 2 Set 2 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 ∫ log x dx.
Or
The marginal cost function of manufacturing x units of a commodity is 6 + 10x – 6x². The total cost of producing one unit of commodity is ₹ 12. Find the total cost function.
Answer:
Let I = ∫ log dx
I = ∫log x ∙ 1 dx
On applying integration by parts, we get
I = log x ∫1 dx – ∫[\(\frac{d}{d x}\) log x ∫1 dx]dx
I = log x ∙ x – ∫\(\frac{1}{x}\) ∙ x dx + C
I = x log x – ∫1 dx + C
I = x log x – x + C
Or
We have, MC = 6 + 10x – 6x²
∴ \(\frac{d C}{d x}\) = MC
On integrating both sides, we get
∫dC = ∫(6 + 10x – 6x²)dx
C = 6x + 5x² – 2×3³ + K ……. (i)
It is given that the cost of producing one unit of the commodity is ₹ 12, i.e. x = 1 and C = 12
Putting x = 1 and C = 12 in Eq. (i), we get
12 = 6 + 5 – 2 + K
K = 3
On substituting the value of K in Eq. (i), we get
C = 6x + 5x² – 2x³ + 3

Question 2.
Sanchit borrowed a home loan amount of ₹ 5000000 from a bank at an interest rate of 12% per annum for 30 yr. Find the equated monthly installment amount, Sanchit has to pay to the bank, [given (1.01) ^-360 = 0.02781].
Or
A bike costing ₹ 55000 has a useful life of 15 yr. If annual depreciation is ₹ 3000.
Find its scrap value, using linear method.
Answer:
Given, P = ₹ 5000000, i = \(\frac{12}{1200}\) = 0.01
and n = 12 × 30 = 360
Let E be the equated monthly installment Sanchit has to pay to the bank.
∴ E = \(\frac{P \times i}{1-(1+i)^{-n}}\)
Here, E = \(\frac{5000000 \times 0.01}{1-(1+0.01)^{-360}}\)
= \(\frac{50000}{1-(1.01)^{-360}}\) = \(\frac{50000}{1-0.02781}\)
= \(\frac{50000}{0.97219}\) = 51430.27
Hence, Sanchit EMI is ₹ 51430.27

Or

Given, cost of bike = ₹ 55000,
Useful life = 15 yr
and Annual depreciation = ₹ 3000
Let scrap value of bike is ₹ S.
Then, annual depreciation = \(\frac{\text { Original cost }-\text { Scrap value }}{\text { Useful life in years }}\)
⇒ 3000 = \(\frac{55000-S}{15}\)
⇒ 45000 = 55000 – S
⇒ S = 55000 – 45000 = 10000
Hence, scrap value is ₹ 10000

Question 3.
Find the present value of a perpetuity of ₹ 5280 payable at the beginning of each year, if money is worth 8% effective.
Answer:
Given, R = ₹ 5280, i = \(\frac{8}{100}\) = 0.08
and let P be the present value of the given perpetuity.
∴ P = R + \(\frac{R}{i}\)
= 5280 + \(\frac{5280}{0.08}\)
= 5280 + 66000
= 71280
Hence, present value of the given perpetuity is ₹ 71280.

Question 4.
A sampling distribution of the sample means X̄ is formed from a population with mean weight µ = 55 kg and standard deviation σ = 8 kg. What is expected value and standard deviation of X̄, if sample size is 49?
Answer:
We have,
µ = 55 kg
∴ E(X̄) = 55 kg
Standard deviation of X̄ = D(X̄) = \(\frac{\sigma}{\sqrt{n}}\)
= \(\frac{8}{\sqrt{49}}\) = \(\frac{8}{7}\) = 1.142 kg

Question 5.
Using 3-yearly moving averages, compute the trend values and short term fluctuations for the following data.

Answer:
Computation of trend value

Question 6.
Determine the maximum value of Z = 11x + 7y, subject to the constraints are 2x + y ≤ 6, x ≤ 2 and x, y ≥ 0.
Answer:
We have,
Maximise Z = 11 x + 7y …….. (i)
Subject to the constraints,
2x + y ≤ 6 ……. (ii)
x ≤ 2 ……. (iii)
and x, y ≥ 0 ……. (iv)
We see that, the feasible region as shaded determined by the system of constraints (ii) to (iv) is OABC and is bounded. So, now we shall use comer point method to determine the maximum value of Z.

Hence, the maximum value of Z is 42 at C(0, 6).

Section -B [12 Marks]

Question 7.
Two batches of the same product are tested for their mean life. Assuming that, the lives of the product follow a normal distribution with an unknown variance, test the hypothesis that the mean life is the same for both the branches, given the following information.

[given, t₁₆ (0.05) = 2.120]
Or
The Educational Testing Service conducted a study to investigate difference between the scores of male and female students on the Scholastic Aptitude Test. The study identified a random sample of 562 female and 852 male students, who had achieved the same high score on the mathematics portion of the test. That is, female and male students were viewed as having similarly high abilities in mathematics. The verbal scores for the two samples are as given
Female students, x̄₁ = 557 and s₁ = 83
Male students, x̄₂ = 545 and s₂ = 78
Do the data support the conclusion that given a population of female students and a population of male students with similarly high mathematics abilities, the female students will have a significantly higher verbal ability? Test at a 5% level of significance. What is your conclusion?
Answer:
Given, n₁ = 10, n₂ = 8, x̄₁ = 750, x̄₂ = 820,
s₁ = ₹ 12 and s₂ = ₹ 14
Consider, Null hypothesis H₀ : Mean life is same for both the batches.
Alternate hypothesis H₁ : Two batches have different mean lives.
Test statistics,

Since, calculated value |t| = 10.763 > tabulated value
t₁₆(0.05) = 2.120
So, rejected the null hypothesis at 5% level of significance.
Hence, the mean life for both the batches is not the same.

Or

We have, x̄₁ = 557, x̄₂ = 545, s₁ = 83, s₂ = 78,
n₁ = 562 and n₂ = 852
Consider, the null hypothesis be that there is no significant difference between male and female verbal ability,
H₀: µ₁ – µ₂ > 0 and H₁: µ₁ – µ₂ < 0
Test statistics,

Here, the computed value of Z is greater than the table value of Z i.e. 1.96. So, reject the null hypothesis. Hence, there is a significant difference between the male and female verbal ability or female students have higher verbal ability.

Question 8.
The production of a soft drink company in thousand of litres during each month of a year is as follows

Calculate the 5-monthly moving average and show these moving averages on a graph.
Answer:
According to the question,

Let the 5-monthly moving average of the Jan-Dec production is m.
So,

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

Question 9.
Mr. Amit invested ₹ 10000 in a mutual fund and the value of investment at the time of redemption was ₹ 25000. If compound annual growth rate (CAGR) for this investment is 9.15%. Calculate the time period for which the amount was invested? [given, log(2.5) = 0.3979 and log(1.0915) = 0.0380]
Answer:
Given, BV = ₹ 10000, EV = ₹ 25000 and CAGR = 9.15%
We know that

Taking log on both sides, we get
\(\frac{1}{n}\) log(2.5) = log(1.0915)
⇒ n = \(\frac{\log (2.5)}{\log (1.0915)}=\frac{0.3979}{0.0380}\)
⇒ n = 10.47 = 10.5 yr

Question 10.
The demand function for a commodity is p = 22 – x – x². Find the consumer’s surplus at equilibrium price p₀ = 10.
Answer:
Given, the demand function is p = 22 – x – x²
at the equilibrium price p₀ = 10
We have,
10 = 22 – x₀ – x²₀
⇒ x²₀ + x₀ – 12 = 0
⇒ x²₀ + 4x₀ – 3x₀ – 12 = 0
⇒ x₀(x₀ + 4) – 3(x₀ + 4) = 0
⇒ (x₀ – 3) (x₀ + 4) = 0
⇒ x₀ = 3 or x₀ = – 4
Since, x₀ ≠ – 4
⇒ x₀ = 3
∴ Consumer’s surplus (CS) at p₀ = 10

Hence, consumer’s surplus is 22.5.

Section – C [16 Marks]

Question 11.
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 maximize the profit for the airline. What is the maximum profit?
Or
Determine graphically the minimum value of the objective function Z = – 50x + 20y, subject to constraints are 2x – y ≥ – 5, 3x + y ≥ 3, 2x – 3y ≤ 12 and x, y ≥ 0.
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 x ≥ 20
y – 4x ≥ 0 or y ≥ 4x
and x, y ≥ 0
Table for line x + y = 200 is

So, line passes through,the points (0, 200) and (200,0).
On putting (0, 0) in the inequality x + y ≤ 200, we get
0 + 0 ≤ 200 ⇒ 0 ≤ 200, which is true.
So, the half plane is towards the origin.
Table for line y = 4x is

So, line passes through the points (0, 0) and (20, 80).
On putting (10, 0) in the inequality y ≥ 4x, we get
0 ≥ 4 x 10 ⇒ 0 ≥ 40, which is not true.
So, the half plane is towards Y-axis.
Now, draw the graph of the line x = 20.
On putting (0,0) in the inequality x ≥ 20,
we get
0 ≥ 20, which is not true.
So, the half plane is away from the origin.
Also, x, y ≥ 0, so the region lies in the I quadrant.
On solving Eqs. (ii), (iii) and (iv), we get
A(20,80), B(40,160) and C (20,180).
∴ Feasible region is ABCA
The comer points of the feasible region are
A(20,80), 6(40,160) and C(20,180).

The value of Z at the corner points are given below

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 if 136000.
Or
Given, objective function is
Minimise Z = – 50 x + 20y
Subject to constraints,
2x – y ≥ – 5 … (i)
3x + y ≥ 3 ……(ii)
2x – 3y ≤ 12 ……. (iii)
and x, y ≥ 0 …(iv)
Table for line 2 x – y = – 5 is

So, the line passes through the points (\(\frac{-5}{2}\), 0) and (0, 5).
On putting (0, 0) in the inequality 2x – y ≥ – 5, we get
0 – 0 ≥ – 5
⇒ 0 ≥ – 5, which is true.
So, the half plane is towards the origin.
Table for line 3x + y = 3 is

So, the line passes through the points (0, 3) and (1, 0).
On putting (0, 0) in the inequality 3x + y ≥ 3, we get
0 + 0 ≥ 3
⇒ 0 ≥ 3, which is not true.
So, the half plane is away from the origin.
Table for line 2x – 3y = 12 is

On putting (0, 0) in the inequality 2 x – 3y ≤ 12,
we get Z
0 – 0 ≤ 12
⇒ 0 ≤ 12, which is time.
So, the half plane is towards the origin.
Also, x ≥ 0 and y ≥ 0, so the region lies in the I quadrant.

On drawing the graph of each linear equation, we get the following graph. In I quadrant, these equations has no intersection point.

Thus, we get the common shaded region ABCD, which gives the feasible region and it is unbounded. The comer points of feasible region are A(0, 5), B(0, 3), C(1, 0) and D(6, 0).
The value of Z at comer points are given below

Here, feasible region is unbounded, so the minimum and maximum value may or may not be exist.
Now, we draw a dotted line of inequation
– 50x + 20y < – 300
or – 5x + 2y < – 30
Here, we see that half plane determined by – 5x + 2y < – 30 has a point in common with the feasible region.
Hence, no minimum value exists.

Question 12.
A person borrows ₹ 68962 on the condition that the will repay the money with compound interest at 5% per annum in 4 equal annual installments, the first one being payable at the end of the first year. Find the value of each installment.
[given (1.05)^-4 = 0.8227]
Answer:
According to the question, the borrow amount (principal amount = P) is ₹ 68962 .
Rate of interest, i = 5% per annum or \(\frac{5}{100}\)
Total number of installments (n) = 4
Let the value of each installment is a.
We know that
P = \(\frac{a}{i}\) [1 – (1 + i)^-n] …….. (i)

0.1773
a = 19447.82
So, the value of each installment is ₹ 1944732.

Question 13.
A company has issued a bond having face value of ₹ 120000, carrying an annual dividend rate of 8% and maturing in 15 yr. If the prevailing market rate of interest is 10% and the bond is redeemed at par, find the bond value, [given (1.1)^-15 = 0.2394)]
Answer:
Given, face value (F) = ₹ 120000
n = 15
and i = \(\frac{10}{100}\) = 0.1
Here, R = Annual dividend
= 8% of face value
= \(\frac{8}{100}\) × 120000 = ₹ 9600
If a bond is redeemed at par, then C = F

where C is redemption price or maturity value.

= 96000[1 – (1.1)^-15] + 120000(L1)^-15
= 96000(1 – 0.2394) + 120000 × 0.2394
= 96000 × 0.7606 + 28728
= 73017.6 + 28728
= ₹ 101745.6

Case Based/Data Based

Question 14.
In Angrish Academy School, class teacher of Xllth class teaches the topic of definite integration.

If f(x) is the continuous function, integral of f(x) over the interval [a, b] is denoted by


On the basis of above information, answer the following questions.
(i)

Answer:

(ii)

Answer:

Students can access the CBSE Sample Papers for Class 12 Computer Science 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 Computer Science Term 2 Set 8 with Solutions

Maximum Marks: 35
Time: 2 hours

Instructions:

• The question paper is divided into 3 sections – A, B and C.
• Section A, consists of 7 questions (1-7). Each question carries 2 marks.
• Section B, consists of 3 questions (8-10). Each question carries 3 marks.
• Section C, consists of 3 questions (11-13). Each question carries 4 marks.
• Internal choices have been given for question numbers 7, 8 and 12.

Section – A
(Each question carries 2 Marks)

Question 1.
Give any two operations that are performed on data structure.
Answer:
Insertion It means addition of a new data element in a data structure.
Deletion It means removal of a data element from a data structure. The data element is searched before its removal.

Question 2.
(i) Expand the following
(a) SMTP
(b) HTML
Answer:
(a) SMTP – Simple Mail Transfer Protocol
(b) HTML – HyperText Markup Language

(ii) Which topology contains a backbone cable running through the whole length of the network?
Answer:
Bus topology

Question 3.
Explain any four constraints used in DBMS.
Answer:
There are some types of constraints are

1. NOT NULL Constraint It ensures that a column cannot store NULL value.
2. UNIQUE Constraint It is used to uniquely identify each record in a database.
3. PRIMARY KEY Constraint It ensures that a column have a unique identity, which helps to find a particular record in a table and no column that is part of the primary key constraint can contain a NULL value.
4. FOREIGN KEY Constraint It designates a column or combination of columns as a foreign key and establishes its relationship with a primary key in different tables.

Question 4.
A result is extracted from the database using the cursor object by giving the following statement.

d = cursor . fetcha 11 ( )

(i) How many records will be returned by fetchall() method?
Answer:
All records

(ii) What will be the datatype of object d after the given command is executed?
Answer:
List of tuples

Question 5.
Observe a table ITEM and give the output of following queries.

(i) SELECT SUM (Qty) FROM ITEM WHERE Name LIKE “S%;
Answer:

(ii) SELECT MAX (Qty) FROM ITEM;
Answer:

(iii) SELECT Itemno, Name FROM ITEM WHERE Qty < 50 AND Name LIKE “%u%”;
Answer:

(iv) SELECT Name, Date FROM ITEM WHERE Date LIKE “%12%”;
Answer:

Question 6.
(i) Which component of database system consists of various secondary storage devices on which data is stored?
Answer:
Hardware

(ii) Which type of language of SQL provides statements for creation and deletion of the database tables, views?
Answer:
DDL (Data Definition Language)

Question 7.
Write answer the following questions, which are based on table PERSON given below.

(i) What is cardinality and degree of a relation?
(ii) Display all the person record of age 18 to 22.
Or
(i) Alter table add CITY column in it as a type character.
(ii) Identify the candidate keys of table PERSON.
Answer:
(i) Cardinality – 5, Degree – 4
(ii) SELECT *FR0M PERSON WHERE AGE BETWEEN 18 AND 22;
Or
(i) ALTER TABLE PERSON ADD CITY CHAR(20);
(ii) PID, FNAME, LNAME

Section – B
(Each question carries 3 Marks)

Question 8.
Suppose STACK is allocated 6 memory locations and initially STACK is empty (Top = 0).
Give the output of the program segment.

AAA = 4 BBB = 6
Push (STACK, AAA)
Push (STACK, 4)
Push (STACK, BBB + 2)
Push (STACK, AAA + BBB)
Push (Top > 0):
Element = STACK, pop ( ) print (Element)

Or

Write the Push operation of stack containing person names. Notice that the name should only accept characters, spaces and period (.) except digits. Assume that Pname is a class instance attribute.
Answer:
Output
10
10
8
4
4

Or

def insert():
name_pattern = re.compile (r “[A - Za - zs.]")
while True : n = input ("Enter name:”)
while name_pattern. search (n): :
print (“Invalid name”)
print (“Enter name correctly”)
n = input( )
Sname. append (n)
c = input (“Enter more name <y/n>”).upper ( )
if (c! = ‘y’):
break

Question 9.
(i) Write the syntax to add a column into a table using SQL command.
Answer:
ALTER TABLE table_name ADD
column_name datatype;

(ii) Explain the following SQL commands in brief.
(a) UPDATE
Answer:
UPDATE It is used to modify existing record in a table.
e.g. Change city value of a person to DELHI for PID 101.
UPDATE TABLE SET CITY=‘DELHI’
WHERE PID = 101;

(b) INSERT
Answer:
INSERT It is-used to insert new values in the existing table.
e.g. Add one row with values in a PERSON table.
INSERT INTO PERS0N(PID, FNAME, LNAME, AGE) VALUES(108, ‘Geeta’, ‘Sharma’, 23);

Question 10.
Create a database HUMAN in MySQL. Also, create a table PERSON with given constraints and data types for each column.
PID – Numeric and primary key, FNAME – Character max length 20, cannot be empty,
LNAME – Character max 20 length, AGE – Numeric and must be at least 18.
Answer:
CREATE DATABASE HUMAN;
CREATE TABLE PERSON (
PID INT,
FNAME CHAR(20) NOT NULL,
LNAME CHAR(20),
AGE INT, CHECK (AGE > = 18),
PRIMARY KEY(PID));

Section C
(Each question carries 4 Marks)

Question 11.
Consider the following tables STUDENT and STREAM. Write SQL commands for the statements (i) to (iv).

(i) To display the name of streams in alphabetical order from table STREAM.
Answer:
SELECT STRNAME FROM STREAM ORDER BY STRNAME;

(ii) To display the number of students whose POINTS are more than 5.
Answer:
SELECT COUN(*) FROM STUDENT WHERE POINTS > 5;

(iii) To update GRADE to A’ for all those students, who are getting more than 8 as POINTS.
Answer:
UPDATE STUDENT SET GRADE = ‘A’ WHERE POINTS > 8;

(iv) ARTS+MATHS stream is no more available. Make necessary change in table STREAM.
Answer:
DELETE FROM STREAM WHERE STRNAME = ‘ARTS + MATHS’;

Question 12.
(i) Explain FTP and telnet network protocols in brief.
Or
Define terms – internet and intranet.
Answer:
FTP The File Transfer Protocol is a standard communication protocol used for the transfer of computer files from a server to a client on a computer network. FTP is built on a client-server model architecture using separate control and data connections between the client and the server.

Telnet This is also called as Remote login. It is used to make connection between remote computers. After establishment of connection with remote host, the telnet client becomes a virtual terminal. It provides an error free connection.

Or

Internet Internet is a called as network of networks, where many networks are connected together to make a big worldwide network, which can be accessed by the general public, anywhere anytime. The network connected can be related with business, finance, game, education, etc. These network can share their data, information, documents on the networks using WWW.

Intranet This type of network belongs to only one organisation. Means if the devices of one organisation are connected with each other with an objective to share data with each other then it is called as Intranet.

(ii) Answer the following
(a) Write any two benefits of using computer network.
(b) What do you mean by data transfer rate?
Answer:
(a) Two benefits of using computer network are

1. File Sharing Computer network helps connected user is sharing of data and files with each other.
2. User Communication Connected user can send and receive messages with each other, using email, chat or other applications of messaging.

(b) Data Transfer Rate Data travels in the form of signals over a channel. One signal carries one or more bits over the channel. Data transfer rate is the number of bits transmitted between source and destination in one second. It is also known as bit rate. It is measured in terms of bits per second (bps).

Question 13.
A School in Dehradun has three blocks A, B and C. It is required to setup a network in the school. The head office of the school is in Delhi.

Distance between blocks are as follows
A to C – 100m
A to B – 40m
B to C – 55m
Dehradun Campus to Delhi Head Office – 350km

Number of computers installed at each building

(i) Suggest most suitable place to install a server.
Answer:
Most suitable building is B block to install a server because it has maximum number of computers.

(ii) Suggest a layout to connect these blocks for wired channel in Dehradun campus.
Answer:

(iii) Suggest placement of switch device in network.
Answer:
Switch needs in every building as they help in sharing of bandwidth.

(iv) The School is planning to make an admission office in Chandigarh far from approximately 300 km, which type of network will be formed (LAN, MAN, WAN)?
Answer:
WAN will be the network as the distance is beyond LAN or MAN.

Students can access the CBSE Sample Papers for Class 12 Accountancy 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 Accountancy Standard Term 2 Set 1 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.
Following information has been provided by M/s Achyut Health Care.
You are required to calculate the amount of medicines consumed during the year 2020-21:

(2)
Answer:
Amount of medicines consumed during the year 2020 – 21:

Question 2.
Distinguish between ‘Dissolution of Partnership’ and ‘Dissolution of Partnership Firm’ based on:
(A) Settlement of assets and LiabiLities
(B) Economic relationship (2)
Answer:

Explanation: Difference between dissolution of Partnership and Dissolution of Partnership Firm:

Question 3.
Suresh, Ramesh and Tushar were partners of a firm sharing profit in the ratio of 6:5:4. Ramesh retired and his capital after making adjustments on account of reserves, revaluation of assets and reassessment of liabilities stood at ₹ 2,50,400.
Suresh and Tushar agreed to pay him ₹ 2,90,000 in full settlement of his claim. Pass necessary journal entry for the treatment of goodwill. Show workings clearly. (2)
Answer:

Working Note:
Ramesh’s share of Goodwill = ₹ 2,90,000 – ₹ 2,50,400 = ₹ 39,600

Working Notes:
(1) Old Profit Sharing Ratio of Suresh, Ramesh and Tushar = 6:5:4.
New Profit Sharing Ratio and Gaining Ratio of Suresh and Tushar will be 6 : 4 or 3 : 2, as no information is provided for future profit-sharing ratio between Suresh and Tushar.

2. Calculation of Ramesh’s Share of Goodwill:
Ramesh’s Share of Goodwill = Amount Paid – Ramesh’s Capital after Adjustments
= ₹ 2,90,000 – ₹ 2,50,400 = ₹ 39,600
which will be compensated by Suresh and Tushar in their Gaining Ratio, i.e., 3 : 2.
Suresh will compensate = ₹ 39,600 x \(\frac{3}{5}\)
= ₹ 23,760
Tushar will compensate = ₹ 39,600 x \(\frac{2}{5}\)
= ₹ 15,840

Question 4.
From the following information given by Modern Dance Academy, calculate the amount of Subscription received during the year 2020-21.
(i) Subscription credited to Income & Expenditure A/c for the year ending 31^st March 2021 amounted to ₹ 3,00,000 and each member is required to pay an annual subscription of ₹ 3,000.
(ii) Subscription in arrears as on 1^st April 2020 amounted to ₹ 16,000.
(iii) During the year 2020-21,10 members made partial payment of ₹ 26,000 towards subscription, 8 members failed to pay the subscription amount and 5 members paid the subscription amount for the year 2021-22.
(iv) During the year 2019-20,12 members paid the subscription amount for the year 2020-21.
OR
Following information is given by Alchemy Medical College, Library department for the year 2020-21.

Show the accounting treatment of the above-mentioned items in the Balance Sheet of the Alchemy Medical College as at 31st March 2021.(3)
Answer:
Calculation of amount of Subscription received during the year 2020-21


Working Note:
Interest on Books and JournaLs Investments = 4,00.000 x \(\frac{7}{100}\) = 28,000
Accrued Interest 28,000 – 13,000 = 15.000
Explanation: Calculation of Amount of Subscription received during the year 2020-21

OR
In the books of Alchemy Medical College Balance Sheet (An Extract) as at 31st March, 2021

Working Note:
Interest on Books and JournaLs Fund Investment = ₹ 4,00,000 x \(\frac{7}{100}\) = ₹ 28,000
Accrued Interest on Books and Journals Fund Investrrients = ₹ 28,000 – ₹ 13,000 = ₹ 15,000.

Question 5.
Harihar, Hemang and Harit were partners with fixed capitals of ₹ 3,00,000, ₹ 2,00,000 & ₹ 1,00,000 respectively.
They shared profits in the ratio of their fixed capitals. Harit died on 31st May 2020, whereas the firm closes its books of accounts on 31st March every year.
According to their partnership deed, Harit’s representatives would be entitled to get share in the interim profits of the firm on the basis of sales.

Sales and profit for the year 2019-20 amounted to ₹ 8,00,000 and ₹ 2,40,000 respectively and sales from 1st April 2020 to 31st May 2020 amounted to ₹ 1,50,000.
The rate of profit to sales remained constant during these two years. You are required to:
(A) Calculate Harit’s share in profit.
(B) Pass journal entry to record Harit’s share in profit. (3)
Answer:
(A) Ratio of Profit to sales = \(\frac{2,40,000}{8,00,000}\) x 100 = 30%
Profit upto the date of death = 1,50,000 x 30% = ₹ 45,000
Profit sharing Ratio = 3:2:1
Harit’s Share of Profit = 45 000 x \(\frac{1}{6}\) = ₹ 7,500
Alternative: Harit’s Share of Profit = \(\frac{2,40,000}{8,00,000}\) x 1,50,000 x \(\frac{1}{6}\) =₹ 7,500
OR
Journal

Explanation:
(A) Sales for the year 2019-20 = ₹ 8,00,000
Profit for the year 2019-20 = ₹2,40,000
Sales from 1st April, 2020 to 31st May 2020 = ₹ 1,50,000
Ratio of Profit to Sales \(\frac{2,40,000}{8,00,000}\) x 100 = 30%
Profit upto Harit’s date of death = ₹ 1,50,000 x 30% = ₹ 45,000
Harit’s Share of Profit upto his date of death = ₹ 45,000 x \(\frac{1}{6}\) = ₹ 7,500
(B) Journal

Question 6.
Videsh Ltd. purchased a running business of Vibhu Enterprises for a sum of ₹ 12,00,000. Videsh Ltd. paid ₹ 60,000 by drawing a promissory note in favour of Vibhu Enterprises., ₹ 1,90,000 through bank draft, and balance by issue of 8% debentures off 100 each at a discount of 5%.
The assets and liabilities of Vibhu Enterprises consisted of Fixed Assets valued at ₹ 17,30,000 and Trade Payables at ₹ 3,20,000.
You are required to pass necessary journal entries in the books of Vedesh Ltd.
OR
Youth Ltd. took a loan of ₹ 15,00,000 from State Bank of India against the security of tangible assets. In addition to principal security, it issued 10,000 11% debentures of ₹ 100 each as collateral security. Pass necessary journal entries for the above transactions, if the company decided to record the issue of 11% debentures as collateral security and show the presentation in the Balance Sheet of Youth Ltd. (3)
Answer:



Working Note:
Number of Debentures Issued = \(\frac{\text { Amount Due }}{\text { Issue Price }} \)
= \(\frac{₹ 9,50,000}{₹(100-5)}\)
= 10,000
OR
In the Books of Youth Ltd. Journal

Question 7.
Madhav, Madhusudan, and Mukund were partners In Jaganath Associates. They decided to dissolve the firm on 31 March 2021. Pass necessary journaL entries for the following transactions after various assets (other than cash) and third-party liabilities have been transferred to realization account:
(i) Old machine fully written off was sold for ₹ 42,000 while a payment of ₹ 6,000 is made to bank for a bill discounted being dishonored.
(ii) Madhusudan accepted an unrecorded asset of ₹ 80,000 at ₹ 75,000 and the balance through cheque, against the payment of his Loan to the firm of ₹ 1,00,000.
(iii) Stock of book value of ₹ 30,000 was taken by Madhav, Madhusudan, and Mukund in their profit sharing ratio.
(iv) The firm had paid realization expenses amounting to ₹ 5,000 on behalf of Mukund.
(v) There was a vehicle loan of ₹ 2,00,000 which was paid by surrender of asset to the bank at an agreed value of ₹ 1,40,000 and the shortfall was met from firm’s bank account
OR
Gini, Bini, and Mini were in partnership sharing profits and Losses In the ratio of 5: 2: 2. Their
Balance Sheet as at 31 March 2021 was as follows:

On 31^st March, 2021, Gini retired from the firm. All the partners agreed to revalue the assets and liabilities on the following basis:
(i) Bad debts amounted to ₹ 5,000. A provision for doubtful debts was to be maintained at 10% on debtors.
(ii) Partners have decided to write off existing goodwill.
(iii) Goodwill of the firm was valued at ₹ 54,000 and be adjusted into the Capital Accounts of Bini and Mini, who will share profits in future in the ratio of 5: 4.
(iv) The assets and liabilities valued as: Inventories ₹ 1,30,000; Machinery ₹ 82,000; Furniture ₹ 1,9 5,000 and Building ₹ 6,00,000.
(v) Liability of ₹ 23,000 is to be created on account of Claim for Workmen Compensation.
(vi) There was an unrecorded investment in shares of ₹ 25,000. It was decided to pay off Gini by giving her unrecorded investment in full settlement of her part payment of ₹ 28,000 and remaining amount after two months.
Prepare Revaluation Account and Partners’ Capital Accounts as on 31st March 2021. (5)
Answer:





OR
Working Notes:
(1) Calculation of Gaining Ratio:
Old Profit Sharing Ratio of Gini, Bini and Mini = 5:2:2
New Profit Sharing Ratio of Bini and Mini = 5:4
Gaining Ratio = New Profit Sharing Ratio – Old Profit Sharing Ratio
Bini’s Gain = \(\frac{5}{9}-\frac{2}{9}=\frac{3}{9}\)
mini’s Gain = \(\frac{4}{9}-\frac{2}{9}=\frac{2}{9}\)
Gaining Ratio of Bini and Mini = 3:2

(2) Calculation of Gini’s Share of Goodwill:
Goodwill of Firm = ₹ 54,000
Gini’s Share of Goodwill = ₹ 54,000 x \(\frac{5}{9}\) = ₹ 30,000
which will be compensated by Bini and Mini in their gaining ratio, i.e., 3 : 2
Bini will compensate = ₹ 30,000 x \(\frac{3}{5}\) = ₹ 18,000
Mini will compensate = ₹ 30,000 x \(\frac{2}{5}\) = ₹ 12,000

(3) Adjustment of Workmen’s Compensation Reserve:
Amount of Workmen’s Compensation Reserve = ₹ 32,000
Claim for Workmen’s Compensation = ₹ 23,000
Workmen’s Compensation Reserve to be distributed among partners = ₹ 32,000 – ₹ 23,000
= ₹ 9,000.

Question 8.
Yogadatra Ltd. (pharmaceutical company) appointed marketing expert, Mr. Kartikay as the CEO of the company, with a target to penetrate their roots in the rural regions. Mr. Kartikay discussed the ways and means to achieve target of the company with financial, production and marketing departmental heads and asked the finance manager to prepare the budget.

After reviewing the suggestions given by all the departmental heads, the finance manager proposed requirement of an additional fund of ₹ 52,50,000. Yogadatra Ltd. is a zero-debt company. To avail the benefits of financial leverage, the finance manager proposed to include debt in the capital structure.

After deliberations, on April 1, 2020, the board of directors had decided to issue 6% Debentures of ₹ 100 each to the public at a premium of 5%, redeemable after 5 years at ₹110 per share.
You are required to answer the following questions:
(A) Calculate the number of debentures to be issued to raise additional funds.
(B) Pass Journal entry for the allotment of debentures.
(C) Pass Journal entry to write off loss on issue of debentures.
(D) Calculate the amount of annual fixed obligation associated with debentures.
(E) Prepare Loss on Issue of Debentures Account. (5)
Answer:
(A) Number of Debentures to be issued = \(\frac{52,50,000}{105}\) = 50,000
(B) In the Books of Yogadatra Ltd. Journal

(D) Interest on 6% Debentures = ₹ 50,00,000 x \(\frac{6}{100}\) = ₹ 3,00,000

Explanation: (A) Additional Funds Raised = ₹ 52,50,000
Number of Debentures Issued to raise Additional Funds = \(\frac{\text { Additional Funds Raised }}{\text { Issue Price }} \)
= \(\frac{₹ 52,50,000}{₹(100+5)}\) = 50,000

(B) In the Books of Yogadatra Ltd.

(C) In the Books of Yogadatra Ltd.
Journal

(D) Interest on 6% Debentures = ₹ 50,00,000 x \(\frac{6}{100}\) = ₹ 3,00,000

Question 9.
From the following Receipts and Payments Account and additional information provided by Ramanath Club, Prepare Income and Expenditure Account for the year ending on 31st March 2021.
Receipts and Payments Account for the year ending 31st March 2021.

Additional Information:
(i) Subscription received during the year includes ₹ 25,000 as donation for Building.
(ii) Telephone bill unpaid as on March 31, 2020 was ₹ 4,000 and on March 31, 2021, ₹ 2,600.
(iii) Value of 8% Government Securities on March 31, 2020, was ₹ 80,000.
(iv) AdditionaL Government Securities worth ₹ 30,000 were purchased on March 31, 2021. (5)
Answer:


Working Note:
Interest on 8% Government Securities for year 2020-21 = ₹ 80,000 x \(\frac{8}{100}\) = ₹ 6,400
Accrued Interest on 8% Government Securities = ₹ 6,400 – ₹ 5,400 = ₹ 1,000

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) Decrease in outstanding employees benefits by ₹ 3000
(B) Increase in Current Investment by ₹ 6,000. (2)
Answer:
(A) Income and Expenditure Account
(B) No Flow. [Marking Scheme]
Explanation:
(A) Decrease in outstanding employee benefits by ₹ 3,000 will result in outflow of cash while preparing Cash Flow Statement.
(B) Increase in Current Investment by ₹ 6,000 will result in no flow of cash, while preparing Cash Flow Statement.

Question 11.
From the following details provided by Kumud Ltd., prepare a Comparative Statement of Profit & Loss for the year ended 31st March 2021:

OR
From the following Balance Sheets of Vinayak Ltd. as at 31st March 2021, Prepare a Common” size Balance Sheet.
Vinayak Ltd.
Balance Sheet as on 31st March, 2021 (3)

Answer:
Comparative Statement Profit and Loss For the year ended 31st March 2021

OR
Common Size Balance Sheet of Vinayak Ltd. as at 31st March 2020 and 2021

Explanation: Kumud Ltd.
Comparative Statement of Profit and Loss for the year ended 31st March, 2021

Common Size Balance Sheet of Vinayak Ltd. as at 31st March 2020 and 2021

Question 12.
On the basis of information given by Aradhana Ltd., prepare Cash Flow Statement for the year ending 31st March 2021:
Aradhana Ltd.
Balance Sheet as on 31st March 2021


Additional Information:
(I) Debentures were redeemed on 1st April 2020.
(ii) Tax paid during the year ₹ 2,80,000 (5)
Answer:




Working Notes:
(1) Calculation of Net Profit before Tax and Extraordinary Items:

(2) Calculation of Interest on Debentures Paid:
Interest on Debentures = 10% of ₹ 2,00,000 = ₹ 20,000
(3)

Students can access the CBSE Sample Papers for Class 12 Computer Science 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 Computer Science Term 2 Set 6 with Solutions

Maximum Marks: 35
Time: 2 hours

Instructions:

• The question paper is divided into 3 sections – A, B and C.
• Section A, consists of 7 questions (1-7). Each question carries 2 marks.
• Section B, consists of 3 questions (8-10). Each question carries 3 marks.
• Section C, consists of 3 questions (11-13). Each question carries 4 marks.
• Internal choices have been given for question numbers 7, 8 and 12.

Section – A
(Each question carries 2 Marks)

Question 1.
Evaluate following postfix expression using a stack and show content of the stack after each step.
8, 2, +, 9, +, 2, 8,*,+
Answer:
Given, postfix expression is 8, 2, +, 9, +, 2, 8, *, +

Question 2.
(i) Expand the following
(a) HTTP
(b) VoIP
Answer:
(a) HTTP—HyperText Transfer Protocol
(b) VoIP—Voice over Internet Protocol

(ii) What is the name of the network topology in which there are bi-directional links between each possible node?
Answer:
Mesh topology

Question 3.
What do you mean by referential integrity?
Answer:
A referential integrity is a system of rules that a DBMS uses to ensure that relationships between records in related tables are valid and that users do not accidently delete or change related data. We can set the referential integrity when all of the following conditions are met

• The matching field from the primary table is a primary key or has a unique value.
• The related fields have the same data types.
• Both tables belong to the same database.

Question 4.
Write the code to create the connection in which database’s name is Python, name of host, user and password can be taken by user. Also, print that connection.
Answer:

import mysql. connector
mycon = mysql . connector, connect (
host = "localhost",
user = "test",
passwd = "testData",
database = "Python")
print(mycon)

Question 5.
Write the output of the queries (i) to (iv) based on the table TEACHER given below

(i) SELECT TeacherlD, DOJ FROM TEACHER WHERE Departments’Computer’ OR Department=’Math’;
Answer:

(ii) SELECT DISTINCT Department FROM TEACHER;
Answer:

(iii) SELECT Department, COUNT!*), MIN(Salary) FROM TEACHER GROUP BY Department HAVING COUNT(*)>1:
Answer:

(iv) SELECT SUM (Salary) FROM TEACHER WHERE Gender = “M”;
Answer:

Question 6.
(i) Which aggregate function returns the count of all rows in a specified table?
Answer:
COUNT()

(ii) Which keyword can be used to return only different values in a particular column or a whole table?
Answer:
DISTINCT

Question 7.
Observe a table ITEM and answer the following questions.

(i) Write names of columns can be considered for a candidate key.
(ii) Write an SQL query to display all the items where Qty is more than 20.
Or
(i) Display Itemno, Name and Qty in descending order of the Qty from table ITEM.
(ii) Identify the primary key. Also, give the reason to choose primary key of particular field.
Answer:
(i) Itemno, Name are most appropriate for making candidate key.
(ii) SELECT * FROM ITEM WHERE Qty>20;
Or
(i) SELECT Itemno, Name, Qty FROM ITEM ORDER BY Qty DESC;
(ii) Itemno should be made the primary key as it uniquely identifies each record of the table.

Section B
(Each question carries 3 Marks)

Question 8.
A list has containing 5 integers. To create a program with separate user defined functions to perform following tasks:

• Insert odd numbers
• Remove and display the content of the stack For example,

Sample input
LI = [15, 26, 72, 53, 44]

Or

A linear stack called Status contains the following information
(i) Phone number of Employee
(ii) Name of Employee
Write the method to perform the operation Pop to pop an object from the stack and to release the memory.
Answer:

LI = [15, 26, 72, 53, 44]
def Insert (E, LI):
E.append (LI)
def Delete ( E):
if E! = [ ]:
return E. pop ( )
else :
return None
Stack1 = [ ]
for i in [ ] :
if i% 2! = E:
Insert (Stack1, i) while True:
if Stack1 ! = [ ]:
print(Delete(Stackl), end = " ”)
else:
break
Or
def Pop_element(Status, Top):
Slen = len(Status)
if (Slen < = 0):
print (“Status is empty”)
else:
phone_no*, emp_name = Status. pop( )
Top = Top - 1
print (“Phone number % s and Name %sdeleted” % (phone_no, emp_name))
return Top

Question 9.
(i) Which method is used to create a connection between the MySQL database and Python?
Answer:
Connect

(ii) What are different type of commands category is provided by SQL?
Answer:
SQL provides different type of commands used for different purposes

• DDL Data Definition Language called as DDL, used to define all the commands which are related with the definition of a relation. Such as – CREATE TABLE, ALTER TABLE, etc.
• DCL Data Control Language called as DCL, used to control the access of the users inside the database by using some set of SQL commands. Such as – GRANT and REVOKE.
• DML Data Manipulation Language called as DML, used to define all the SQL commands required for data manipulation is table. Such as – INSERT, DELETE, etc.
• TCL Transaction Control Language called as TCL, used to define all the commands deals with transactions in the database. Such as – COMMIT, ROLLBACK, etc.

Question 10.
Create a database University in MySQL. Also, create a table STUDENT with given constraints and data types for each column.
Rollno – integer and primary key,
Sname – character max length 20, cannot be empty DOB – date type,
Phoneno – character max 10
Answer:

CREATE DATABASE University:
CREATE TABLE STUDENT (
Rollno INT,
Sname VARCHAR(20) NOT NULL,
DOB DATE,
Phoneno CHAR(10),
PRIMARY KEY(Rol1 no));

Section-C
(Each question carries 4 Marks)

Question 11.
Write SQL queries for (i) to (iv) based on tables TEACHER and POSTING.

(i) Display number of male and female teachers.
Answer:
SELECT Gender, COUNT(*) FROM TEACHER GROUP BY Gender;

(ii) Display TeacherlD and Tname working in Delhi.
Answer:
SELECT Teacherld, Tname, City
FROM TEACHER, POSTING WHERE
TEACHER.Department=P0STING.
Department AND
POSTING.City=‘Delhi’;

(iii) Display Tname, DOJ and Gender whose Salary is less than 45000.
Answer:
SELECT Tname, D0J, Gender FROM TEACHER WHERE Salary<45000;

(iv) Display number of teachers’ department wise.
Answer:
SELECT Department, COUNT(*) FROM TEACHER GROUP BY Department;

Question 12.
(i) Explain the “distributed computing power can be achieved” using networking.
Or
Define advantages and disadvantages of fibre optical cables.
Answer:
(i) In computer networks, we can distribute tasks across multiple computers throughout the network, by breaking complex problems into hundreds or thousands of smaller operations, which are then parcelled out to individual computers. Each computer in the network performs its operations on its own portion of the larger problem and return its result. Then, all these results are gathered in such a form that impact as the solution of the complex problem and finally are used for the further task.

Or

There are some advantages of fibre optical cable

• These are highly suitable for harsh industrial environment. These are immune to noise caused by electrical and magnetic fields.
• It can support dramatically higher bandwidths (and hence data rates) than either twisted pair or co-axial cable.

There are some disadvantages of fibre optical cable

• These are very difficult to install and maintain.
• They require more protection around the cable than copper cables.
• Propagation of light is uni-directional. For bi-directional communication, two fibres are needed. These are relatively more expensive than other guided media.

(ii) Compare and contrast the wireless transmission media- bluetooth and infrared.
Answer:
Differences between bluetooth and infrared are as follows

Question 13.
A university is setting a network between the buildings. The University has 4 blocks – Admin(A), Technology(T), Workshop(W) and HR(H).

Distance between wings are as follows
A to T -100 m
A to W – 200 m
A to H – 300 m
T to H – 400 m
T to W -100 m
W to H – 350 m
Number of computers installed at each building
A – 15
T – 100
W – 30
H – 10
(i) Suggest most suitable place to install a server.
Answer:
Most suitable building is “Technology” block to install a server because it has maximum number of computers.

(ii) Suggest a layout to connect these blocks for wired channel.
Answer:
Layout connection

(iii) Which device will be suggested by you to install in each block for efficient connection of all computers within block?
Answer:
Switch is a perfect device for making connection between computers within a block.

(iv) Suggest the placement of repeater.
Answer:
Repeater can be placed when distance between two buildings are more than 70 metres. Total 4 repeaters can be placed.