Maharashtra Board Class 12 Commerce Maths Sample Paper Set 1 with Solutions

Maharashtra State Board Class 12th Commerce Maths Sample Paper Set 1 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Commerce Maths Model Paper Set 1 with Solutions

Time:3 Hrs.
Max. Marks:80

General Instructions:
(i) All questions are compulsory.
(ii) There are 6 questions divided into two sections.
(iii) Write answers of Section-I and Section-ll in the same answer book.
(iv) Use of logarithmic table is allowed. Use of calculator is not allowed.
(v) For LP.P. graph paper is not necessary. Only rough sketch of graph is expected.
(vi) Start answers to each question on a new page.
(vii) For each multiple choice type of questions, it is mandatory to write the correct answer along with its alphabetical eg. (a)………/ (b)………/ (c)………/ (d)………No mark(s) shall be given if’‘ONLY’ the correct answer or the aphabet of the correct answer is written. Only the first attempt will be considered for evaluation.

SECTION – I

Question 1.
1. [A] Choose the correct alternative:
(i) The false statement in the following is:
(a) p ∧ (~p) is a contradiction.
(b) (p → q) ↔ (~q → ~p) is a contradiction.
(c) ~(~p) ↔ p is a tautology.
(d) p v (~p) ↔ p is a tautology.

(v) The integrating factor of \(\frac{dy}{dx}\) + y = e^-x is ………..
(a) x
(b) -x
(c) e^x
(d) e^-x
(vi) If the elasticity of demand η = 1 then demand is:
(a) Constant
(b) Inelastic
(c) Unitary elastic
(d) Elastic
[B] State whether each of the following is True or False:
(i) Conditional of p → q is equivalent to p → ~q.
(ii) Order and degree of a differential equation are always positive integers.
(iii) If f (x) = x \(\frac{1}{x}\), x ∈ R, x ≠ 0 then f(x) is increasing.
[C] Fill in the Blanks:
(i) \(\int \frac{x^2+x-6}{(x-2)(x-1)}\)dx = x + ……….. + c
(ii) \(\int \frac{1}{a^2-x^2} d x=\frac{1}{2 a}\) x ………..
(iii) Using definite integration area of the circle x² + y² = 49 is ……….. .
Solution:
[A] (i) (b) (p → q) ↔ (~q → ~p) is a contradiction.
(ii) (c) \(\frac{1}{4}\)
(iii) (a) \(\frac{1}{2}\)log\(\frac{7}{5}\)
(iv) (a) 3,1
(v) (c) e^x
(vi) (c) Unitary elastic

[B] (i) False
(ii) True
(iii) True

[C]

Question 2.
[A] Attempt any two of the following:
(i) Without using truth table, show that:
~[(p ∧ q)→ ~q] ≡ p ∧ q
(ii) Divide the number 20 into two parts such that their product is maximum.
(iii) If x = a(1 – \(\frac{1}{t}\)).y = a(1+\(\frac{1}{t}\)) then, show that \(\frac{dy}{dx}\) = -1.
[B] Attempt any two of the following:

method.
(ii) Find MPC, APC and APS, if the expenditure E_c of a person with income I is given as
E_c = [(0.0003) I² + (0.075) I], when I = 1000.
(iii) Evaluate the following: \(\int \frac{3 e^x+4}{2 e^x-8}\)dt
Solution:
[A] (i) LH.S. ≡ ~[(p ∧ q) → ~q]
≡ (p ∧ q) ∧ ~(q) (Negation of implication)
≡ (P ∧ q) ∧ q (Negation of a negation)
≡ P ∧ (q ∧ q) (Associative law)
≡ P ∧ q (Identity law)
≡ R.H.S.

(ii) Let one part of 20 be x.
∴ Other part is (20 – x).
∴ Product = x.(20 – x)
which has to be maximised.
∴ f(x) = x.(20 – x)
= 20x – x²
∴ f’(x) = 20 – 2x
f”(x) = – 2 < 0 Let f‘(x) = 0
∴ 20 – 2x = 0
⇒ 2x = 20
⇒ x = 10
and f”(x) = – 2 < 0
∴ By 2^nd derivative test, f is maximum at
x = 10
∴ 20 – x = 20 – 10 = 10.
∴ The required parts of 20 are 10 and 10.

(iii)

[B] (i)


(ii) E_c = [(0.0003)I² + (0.075)I]
MPC = \(\frac{d \mathrm{E}_{\mathrm{C}}}{d \mathrm{~d}}=\frac{d}{d l}\)[(0.0003)I² + (0.075)I]
= (0.0003)(2I) + (0.075)(1)
= (0.0006)1 + 0.075
When I = 1000, then
MPC = (0.0006)(1000) + 0.075
= 0.6 + 0.075
= 0.675
∵ MPC + MPS = 1
MPS = 1 – 0.675
= 0.325
New.APC = \(\frac{E_c}{I}=\frac{\left.(0.0003)\right|^2+(0.075) \mid}{I}\)
= (0.0003)I + 0.075
When I = 1000, then
APC = (0.00083)(1000) + 0.075
= 0.3 + 0.075
= 0.375
∵ APC + APS = 1
∴ APS = 1 – 0.375
= 0.625
Hence, MPC = 0.675, MPS = 0.325
APC = 0.375, APS = 0.625.

(iii)

Question 3.
[A] Attempt any two of the following:
(i) Draw Venn diagram for the following:
(a) No policeman is thief
(b) Some doctors are rich.
(c) Some students are not scholars.
(ii) The rate of growth of population is proportional to the number present If the population doubled in last 25 years and the present population is 1 lac, when will the city have population 4,00,000?
(iii) Evaluate: \(\int_2^5 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{7-x}}\)dx
[B] Attempt any one of the following:
(i) Express the following equations in matrix form and solve them by reduction method:
x + y + z = 1; 2x + 3y + 2z = 2 and x + y + Iz = 4
(ii) Find \(\frac{dy}{dx}\) if x^y = y^x
[C] Attempt any one of the following activities:


Solution:
[A] (i) (a) Let P: The set of all policemen
T: The set of all thieves
U: The set all human beings.

(b) Let D: The set of all doctors
R: The set of humans who are rich
U: The set of all humans

(c) Let S: The set of all students
S’: The set of all scholars
U: The set all all humans

(ii) Let x be the population at time‘t’ years.
∴ \(\frac{dx}{dt}\) ∝ x
∴ \(\frac{dx}{dt}\) = kx,
where k is the constant of proportionality.
∴ \(\frac{dx}{x}\) = kdt
Integrating on both sides, we get
∫\(\frac{dx}{x}\) = ∫kdt
∴ log x = kt + c …(i)
when t = 0, x = 50,000, we have
log(50,000) = k(0) + c
c = log(50,000)
∴ log x = kt + log (50,000)   …(ii) [from (i)}
when t = 25, x = 1,00,000, we have
log(l,00,000) = 25k + log (50,000)
∴ log (2) = 25k
∴ k = \(\frac{1}{25}\)log(2) …(iii)
When x = 4,00,000, we get

∴ t = 75 years.
Thus, the population will be 4,00,000 after 75-25 = 50 years from present date.

(iii)

[B] (i) Matrix form of the given system of equations is:

∴ By equality of matrices, we get
x + y + z = 1 …(i)
y = 0
z = 3
Substituting y and z in (i), we get
x + 0 + 3 = 1
x = – 2
∴ x = – 2, y = 0, z = 3 is the required solution.

(ii) Given x^y = y^x
Taking log on both sides, we get
∴ y log x = x log y
Differentiating w.r.t x, we get

[C] (i)

Section – B

Question 4.
[A] Choose the correct alternative:
(i) Moving averages are useful in identifying:
(a) Seasonal component
(b) Irregular component
(c) Trend component
(d) Cyclical component
(ii) If b_xy > 1 then b_yx is ……….. .
(a) > 1
(b) < 1
(c) > 0
(d) < 0
(iii) If F(x) is distribution function of discrete r.v. x with p.m.f. P(x) = \(\frac{x-1}{3}\); for x = 1, 2, 3 and P(X) = 0, otherwise, then F(4) =
(a) -1
(b) 0
(c) 1
(d) 4
(iv) The optimal value of the objective function is attained at the ……….. of feasible region.
(a) at any point
(b) corner points
(c) middle point
(d) None of these
(v) If E(X) > Var(X) th e n X fo Hows:
(a) Binomial distribution
(b) Poisson distribution
(c) Normal distribution
(d) None of the above
(vi) In sequencing, an optimal path is one that minimizes:
(a) Elapsed time
(b) Idle time
(c) Both (a) and (b)
(d) Ready time.
[B] State whether each of the following is True or False:
(i) Cyclical variation can occur several times in a year.
(ii) Graphical solution set of x < 0, y > 0 in xy system lies in second quadrant.
(iii) IfX ~ B (2, 3) then E(X) = 5:
[C] Fill in the blanks:
(i) ……….. component of time series is indicated by periodic variation year after year.
(ii) If X: is number obtained on upper most face when a fair coin is thrown then E(x) = ……….. .
(iii) Maximisation assignment problem is transformed to minimisation problem by subtracting each entry in the table from the ……….. value of the table.
Solution:
[A] (i) (b) Trend component
(ii) (b) < 1
(iii) (c) 1
(iv) (b) corner points
(v) (a) Binomial distribution
(Vi) (c) Both (a) and (b)

[B] (i) False. Cyclical variation can occur over a long period, usually several years.
(ii) True
(iii) False. If X – B (2,3) then E(X) = 6.

[C] (i) Seasonal
(ii) 3.5
(ii) Largest

Question 5.
[A] Attempt any two of the following:
(i) From the following two regression equations find r, \(\bar{x}\) and \(\bar{y}\).
4y = 9x + 15 and 25x = 4y + 17
(ii) An agent is paid a commission of 7% on cash sales and 5% on credit sales made by him. If on the sale of ₹ 1,02,000 the agent claims a total commission of ₹ 6,420, find his cash sales and credit sales.
(iii) The Price Index Number for year 2004, with respect to year 2000 as base year, is known to be 130. Find the missing numbers in the following table if ∑p₀ = 320

[B] Attempt any two of the following :
(i) The following gives the production of steel (in millions of tones) for years 1976 to 1986.

(a) Fit a trend line to the above data by the graphical method.
(b) Obtain the trend value for the year 1990.
(ii) If ∑p₀q₀ = 120′ ∑p₀q₁ = 160. ∑p₁q₁ =140, and ∑p₁q₀ = 200, find Laspeyre’s, Paasche’s, Dorbish-Bowley’s, and Marshall-Edgeworth’s Price Index Numbers.
(iii) Solve the following problems by graphical method: Maximize z = 4x + 2y subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0.
Solution:
[A] (i) Give regression equations are
4y = 9x + 15
i.e., – 9x + 4y = 15 …(i)
and 25x = 4y + 17
i.e., 25x – 4y = 17 …(ii)
Adding equations (i) and (ii), we get

∴ x = 2
Substituting x = 2 in Q, we get
-9(2) + 4y = 15
4y = + 18 + 15
y = \(\frac{33}{4}\)
y = 8.25
Since the point of intersecting of two regression lines is
(\(\bar{x}\), \(\bar{y}\)) \(\bar{x}\) = 2 and \(\bar{y}\) = 8.25.
Let 4y = 9x + 15 be the regression equation of y on x
∴ The equation become y = \(\frac{9x}{4}+\frac{15}{4}\)
Comparing it with y = b_yx x + a, we get
b_yx = \(\frac{9}{4}\) = 2.25
Now, the other equation, i.e., 25x = 4y + 17 is the regression equation of x of y.
∴ The equation becomes x = \(\frac{4}{25}y+\frac{17}{25}\)
Comparing it with x = b_xyy + a, we get
b_xy = \(\frac{4}{25}\) = 0.16

Since b_yx and b_xy are positive, r is also positive
∴ r = 0.6
∴ \(\bar{x}\) = 2 and \(\bar{y}\) = 8.25 and r = 0.6

(ii) Let the agents cash sales be ₹ x
Commission at 7% on cash sales
= ₹x × \(\frac{7}{100}\) = ₹\(\frac{7x}{100}\) …(i)
Total sales is ₹ 1,02,000
∴ Agents credit sales is ₹ (1,02,000 – x)
Commission at S% on credit sales

The agents cash sales is ₹ 66,000 and his credit sales is ₹ (1,02,000 – 66,000) = ₹ 36,000.

(iii) We first tabulate the given data as shown in the following table:

From the table, we have
∑P₀ = 280 + x, ∑P₁ = 350 + y.
But it is given that ∑P₀ = 320. So that
280 + x = 320
∴ x = 40
Further, using the formula

∴ 416 = 350 + y
∴ y = 416 – 350 = 66
∴ y = 66
x = 40, y = 66

[B] (i) (a)Taking year on X-axis and production on Y-axis, we plot the points for production corresponding to years. Joining these points by straight Lines, we get the graph of the given time series. We draw Line as shown in the figure.

(b) Here, n = 11. We transform year-t to u by taking u = t – 1981.
We construct the following table for calculation:

The equation of trend line is x_t = a’ + b’u
The normal equation are
∑x_t = na’ + b∑u ….(i)
∑ux_t = a’∑_u + b’∑_u² ….(ii)
Here, n = 11, ∑x_t = 62, ∑u = 0,∑u² =110, ∑ux_t = 87
Putting these values in normal equations, we get
62 = lla’ + b'(0) ….(iii)
87 = a'(0) + b'(110) ….(iv)
From equation (3), we get
a’ = \(\frac{62}{11}\) = 5.6364 11
From equation (4), we get 87
b’ = \(\frac{87}{110}\) = 0.7909
Putting a’ = 5.6364 and b’ = 0.7909 in the equation x_t = a’ + b’u, we get the equation of trend line as
x_t = 5.6364 + 0.7909u
Trend for the year 1990:
For t = 1990, u = 1990 – 1981 = 9
Putting u = 9 in x_t = 56364 + 0.7909 u. we get
X₁₉₉₀ = 5.6364 + 0.7909 × 9
= 5.6364 + 7.1181
= 12.7545
Hence, trend value for the year 1990 is 12.7545.

(ii) Given,

Dorbish-Bowley’s Price index Number:

(iii) To find the graphical solution, construct the table as follows:

BHC is the unbounded feasible region with
B ≡ (0.27) and C ≡ (21.0)
Point H is the point of intersection of lines

Objective function, Z = 4x + 24
At B(0, 27), Z(B) = 4 × 0 + 2 × 27 = 54
At H(3, 18), Z(H) = 4 × 3 + 2 × 18 = 48
At C(21, 0), 2(C) = 4 × 21 + 2 × 0 = 84
∴ At H (3, 18) the value of Z is minimum.
Hence Z has minimum value 48, when x = 3, y = 18.

Question 6.
[A] Attempt any two of the following:
(i) The difference between true discount and banker’s discount on a bill due 6 months hence at 4% is ₹ 160. Calculate true discount, banker’s discount and amount of bill.
(ii) Five wagons are available at stations 1, 2, 3, 4 and 5. These are required at 5 stations I, II, III, IV and V. The mileage between various stations are given in the table below. How should the wagons be transported so as to minimize the mileage covered?

(iii) Determine whether each of the following is a probability distribution. Give reasons for your answer.

[B] Attempt any one of the following:
(i) The following results were obtained from records of age (x) and systolic blood pressure (y) of a group of 10 women.

(ii) Find the appropriate regression equation and use it to estimate the blood pressure of a woman with age 47 years.

Find the optimal sequence that minimizes total time required to complete the following jobs in order A-B-C. The processing times are given in hrs.

[C] Attempt any one of the following activities:
(i) For an annuity due, C = ₹ 2000, rate = 16% p.a. compounded quarterly for 1 year

(ii) If X has Poisson distribution with parameter m and P (X = 2) = P (X = 3), then find P(X ≥ 2). Use e^-3 = 0.0497.

Solution:
[A] (i) Let T.D. = ₹ x, n = \(\frac{6}{12}=\frac{1}{2}\) year
B.G. = B.D. – T.D.
= Interest on T.D. for 6 months at 4% p.a.
∴ 160 = x × \(\frac{1}{2}\times\frac{4}{100}=\frac{x}{50}\)
x = 8,000
∴ True discount ₹ = 8,000
B.D. = B.G. + T.D.
= 160 + 8,000 = 8,160
∴ Banker’s discount = ₹ 8,160
B.D. = Interest on F.V. for 6 months at 4% p.a.
Let the face value (F.V.) be y.
∴ B.D. = y × \(\frac{1}{2}\times\frac{4}{100}\)
8,160 = \(\frac{y}{50}\)
y = ₹4,08,000
∴ Amount of bill ₹4,08,000.
(ii) Step 1: Subtract the smallest element is each row form every element of that row.

Step 2 : Subtract the smallest element of each column from every element of that column.

The number of lines covering all zeroes (4) is not equal to order of matrix (5). So solution has not reached.
Step 3 : Therefore, subtract the smallest uncovered element (i) from all uncovered elements and add it to alL elements which lie at the intersection of two unchanged:

The number of lines covering all zeroes is equal to order of matrix.
Step 4: Hence, optimal solution has reached. Therefore, the optimal assignment is made as follows:

The optimal assignment is shown as follows:

The minimum milage covered = 10 + 6 + 4 + 9 + 10 = 39 miles,
(iii) (a) Here, P(x) > 0 for all values of x
∑P(x)= 0.4 + 0.4 + 0.2 = 1
Hence, the given distribution is a probability distribution.
(b) Here, P(x) for x = 3, P(3) = – 0.1 < 0
Probability for an value of x cannot be negative. Hence, the given distribution is not a probability distribution.
(c) Here, P(x) > 0 for all values of x
∑P(x)= 0.1 + 0.6 + 0.3 = 1
Hence, the given distribution is a probability distribution.

[B] (i) Here, we need to find Line of regressoin of y on x, which is given as:

Therefore, regression equation of y on x is
y = 94.3 + 0.9x
Now, the estimate of blood pressure of women
with age 47 years is:
y = 94.3 + 0.9 × 47
= 136.6

(ii) Here Min. (A) = 5, Min. (C) = 3, and Max. (B) = 5.
∵ Min (A) ≥ Max. (B), the problem can be converted into 7 Jobs and 2 machines problems.
Let G = A + B; H = B + C where G, H are two fictious machines.

The following sequence is obtained; by using optimal sequence algorithm.

Total elapsed time is obtained as follows:

Total elapsed time T = 61 hours
Idle time for Machine A = T – Total processing time of Machine A
= 61 – 54
= 7 hours
Idle time for Machine B = T – Total processing time of Machine B
= 61 – 23
= 38 hours
Idle time for Machine C = 15 hours

[C] (i) For on annuity due,
C = ₹ 2000, rate = 16% p.a. compounded quarterly for 1 year.

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