12th Commerce Maths Question Paper 2024 Maharashtra Board Pdf

Maharashtra State Board Class 12th Commerce Maths Question Paper 2024 with Solutions Answers Pdf Download.

Class 12 Commerce Maths Question Paper 2024 Maharashtra State Board 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-1 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.
[A] Select and write the correct answer of the following multiple choice type of questions (1 mark each): (6) [12]
(i) Which of the following is not a statement?
(a) Smoking is injurious to health
(b) 2 + 2 = 4
(c) 2 is the only even prime number
(d) Come here
(ii) If x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 then (y, z) = ____________ .
(a) (-1, 0)
(b) (1, 0)
(C) (1, -1)
(d) (-1, 1)

[B] State whether the following statements are true or false (1 mark each): (3)

(ii) For \(\int \frac{x-1}{(x+1)^3}\)e^xdx = e^xf(x) + c, f(x) = (x + 1)²
(iii) Order and degree of a differential equation are always positive integers.
[C] Fill in the following blanks (1 mark each): (3)
(i) The slope of tangent at any point (a, b) is called as _________
(ii) If f'(x) = \(\frac{1}{x}\) + x and f(l) = \(\frac{5}{2}\) then f(x) = log x + \(\frac{x^2}{2}\) + ________ .
(iii) A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ________ solution.
Solution:
[A]
(i) (d) come here
(ii) (b) (1.0)
Matrix form of equations is

By equality of matrices, we get
x + y + z = 3 ….(1)
y + 2z = 1 ….(2)
2z = 0 ….(3)
z = 0
Putting z = 0 in (2),
y + 0 = 1
∴ y = 1

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

[C] (i) The slope of tangent at any point (a, b) is called as gradient.

(iii) A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called particular solution.

Question 2.
[A] Attempt any TWO of the following questions (3 marks each): (6)[14]
(i) Examine whether the following statement pattern is tautology, a contradiction or a contingency.
~p → (p → ~q)
(ii) Find \(\frac{dy}{dx}\) if, x = e^3t, y = e^{(4t+ 5)}

[B] Attempt any TWO of the following questions (4 marks each): (8)
(i) Consider the following statements
(a) If D is dog, then D is very good.
(b) If D is very good, then D is dog.
(c) If D is not very good, then D is not a dog.
(d) If D is not a dog, then D is not very good.
Identify the pairs of statements having the same meaning. Justify.
(ii) Determine the minimum value of the function: f(x) = 2×3 – 2lx2 + 36x – 20.
(iii) Find the area of the regions bounded by the Line y = – 2x, the X-axis and the lines x = -landx=2.
Solution:
[A](i)

Since the entries in the last column are all T, the given statement is Tautology.

(ii) x = e^3t
Differentiating with respect to ‘f,

[B] (i) Letp : D is dog; q: D is very good.
The given statements can be written in the symbolic form as p → q, q → p, ~q → ~ p, ~ p → ~q respectively.

Since the entries in (i) and (iii) columns are same, they have the same meaning.
Also the entries in (ii) and (iv) columns are same, they have the same meaning.

(ii) f(x) = 2x³ – 21x² + 36x – 20
∴ f'(x) = 6x² – 42x + 36
f”(x) = 12x-42

Let f'(x) = 0
∴ 6x² – 42x + 36 = 0
∴ x² – 7x + 6 = 0
∴ (x – 6)(x – 1) = 0
∴ x = 6 or x = 1.
f”(6) = 72 – 42 = 30 > 0
∴ f(x) has minimum value atx = 6.
f”(1) = 12 – 42 = -30 < 0
f(x) has maximum value at x = 1
∴ f(6) = 2 × (6)³ – 21 × 6² + 36 × 6 – 20
= 432 – 756 + 216 – 20 = – 128
Answer:
f[x) has minimum value -128 at x = 6.

(iii) A = (Area below X-axis) + (Area above X-axis)

Question 3.
[A] Attempt any TWO of the following questions (3 marks each): (6)[14]
(i) Find \(\frac{dy}{dx}\) ify = x^ex
(ii) If f'(x) = 4x³ – 3x² + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
(iii) Obtain the differential equation whose general solution is x³ + y³ = 35ax
[B] Attempt any ONE of the following questions (4 marks each): (4)

(ii) The consumption expenditure E_c of a person with income x, is given by E_c = 0.0006x² + 0.003x. Find average propensity to consume (APC), marginal propensity to consume (MPC) when his income is ₹ 200. Also find his marginal propensity to save (MPS).
[C] Attempt any ONE of the following questions (Activity) (4 marks each):
(i) Complete the following activity:

(ii) The rate of growth of population is proportional to the number of inhabitants. If the population doubles in 25 years and the present population is 1,00,000, when will the city have population 4,00,000?
Solution:
Let ‘P’ be the population at time’t’.
Since rate of growth of population is proportionalto the no. of inhabitants:
\(\frac{dP}{dt}\) ∝ P
∴ Differential equation can be written as \(\frac{dP}{dt}\) = kP
where k is constant of proportionality.
∴ \(\frac{dP}{P}\) = k.dt
On integrating we get
___ = kt + c ……(i)
(i) When t=0,P= 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = ___
∴ log(\(\frac{P}{1,00,000}\)) = kt …….(ii)

(ii) When t = 25, P = 2,00,000
as population doubles in 25 years
from (ii) log 2 = 25k

Solution:
[A] (i) y = x^ex
Taking log on both sides
∴ log y = log x^ex = e^x. log x
Differentiating w.r.t x, we get

(ii) f'(x) = 4x³ – 3x² + 2x + k
∴ f(x) = ∫f'(x)dx
= J(4×3-3×2+2x+k).dx
x4 x3 x2
= 4x-—3x — + 2x—+kx+c 4 3 2
f(x) = x⁴ – x³ + x² + kx + c
f(0) = c = 1
∴ c = 1
f(l) = 1 – 1 + l + k + l
4 = k + 2
∴ k = 2
∴ f(x) = x⁴ – x³ + x² + 2x + 1

(iii) x³ + y³ = 35ax …….(1)
Differentiating w.r.t ‘x’,

This is the required differential equation.

[B] (i)

(ii) E_c = 0.0006x² + 0.003x

When x = 200,
APC = 0.0006 × 200 + 0.003
= 0.12 + 0.003 = 0.123
APS = 1 – A PC
= 1 – 0.123 = 0.877

[C] (i)

(ii) Let ‘P’ be the population at time’t’.
Since rate of growth of population is proportional to the number of inhabitants.

On integrating we get
log P = kt + c ….(i)

(i) When t = 0, P = 1,00,000.
∴ From (i), log 1,00,000 = k. 0 + c
∴ c = log 1,00,000
∴ log(\(\frac{P}{1,00,000}\)) = kt ….(ii)

(ii) When t = 25, P = 2,00,000 as population doubles in 25 years.
∴ From (ii), log 2 = 25k
∴ k = \(\frac{1}{25}\) log2
∴ log(\(\frac{P}{1,00,000}\)) = (\(\frac{1}{25}\) log2).t

(iii) ∴ When P = 4,00,000
log(\(\frac{4,00,000}{1,00,000}\)) = (\(\frac{1}{25}\) log2).t

SECTION-II

Question 4.
[A] Select and write the correct answer of the following multiple choice type of questions (1 mark each): (6) [12]
(i) The difference between- face value and present worth is called __________.
(a) Bankers discount
(b) True discount
(c) Banker’s gain
(d) Cash value
(ii) In an ordinary annuity, payments or receipts occur at __________.
(a) beginning of each period
(b) end of each period
(c) mid of each period
(d) quarterly basis
(iii) b_xy and b_yx are __________.
(a) Independent of change of origin and scale
(b) Independent of change of origin but not of scale
(c) Independent of change of scale but not of origin
(d) Affected by change of origin and scale
(iv) Dorbish-Bowley’s Price Index Number is given by __________.

(v) Objective function of LP.P. is __________.
(a) a constraint
(b) a function to be maximised or minimised
(c) a relation between the decision variables
(d) a feasible region
(vi) To use the Hungarian method, a profit maximization assignment problem requires __________.
(a) Converting all profits to opportunity losses
(b) A dummy person or job
(c) Matrix expansion
(d) Finding the maximum number of Lines to cover all the zeroes in the reduced matrix.
[B] State whether the following statements are true or false (1 mark each): (3)
(i) Broker is an agent who gives a guarantee to seller that the buyer will pay the selling price of goods.
(ii) \(\sum \frac{p_0 q_0}{p_1 q_1} \times 100\) is the Value Index Number by simple aggregate method.
(iii) The optimum value of the objective function of LP.P. occurs at the centre of the feasible region.
[C] Fill in the blanks (1 mark each): (3)
(i) The banker’s discount is always __________ than the true discount.
(ii) The cost of living index number using Weighted Relative Method is given by __________.
(iii) The time interval between starting the first job and completing the last job including the idle time (if any) in a particular order by the given set of machines is called __________.
Solution:
[A] (i) (b)True discount
(ii) (b) end of each period
(iii) (b) Independent of change of origin but not of scale.
(iv)
(v) (b) a function to be maximised or minimised
(vi) (a) Converting all profits to opportunity Losses.

[B] (i) False
A deL credere is an agent who gives a guarantee to seller that the buyer will pay the selling price of goods.

(ii) False
\(\sum \frac{p_0 q_0}{p_1 q_1} \times 100\) is the value index number by simple aggregate method.

(iii) False
The optimum value of the objective function of LP.P. occurs at the corner points of the feasible region.

[C] (i) higher
(ii) \(\frac{\Sigma \mathrm{IW}}{\Sigma \mathrm{~W}}\)
(iii) Total elapsed time.

Question 5.
[A] Attempt any TWO of the following questions (3 marks each): (6) [14]
(i) Deepak’s salary was increased from ₹ 4,000 to ₹ 5,000. The sales being th6 same, due to reduction in the rate of commission from 3% to 2%, his income remains unchanged. Find his sales.
(ii) For a bivariate data, the regression co-efficient of Y on X is 0.4 and the regression co-efficient of X on Y is 0.9. Find the value of variance of Y if variance of X is 9.
(iii) The following table shows the index of industrial production for the period from 1976 to 1985, using the year 1976 as the base year. Obtain the trend values for the following data using 4 yearly centered moving averages:

Years	Index
1976	0
1977	2
1978	3
1979	3
1980	2
1981	4
1982	5
1983	6
1984	7
1985	10

[B] Attempt any TWO of the following questions (4 marks each): (8)
(i) If for the following data, Walsh’s Price Index Number is 150, find ‘x’:

(ii) A toy manufacturing company produces five types of toys. Each toy has to go through three machines A, B and C in the order ABC. The time required in hours for each process is given in the following table:

Find the total elapsed time and also find idle time for machine B.
(iii) A random variable X has the following probability distribution:

Solution:
[A] (i) Let Deepak’s sale be ₹ x.
Salary is ₹ 4000 and rate of commission is 3%.
∴ Deepak’s income = 4000 + 3% of x
3x = 4000 + \(\frac{3x}{100}\) ….(i)
Now salary is ₹ 5000 and rate of income is 2%
∴ Deepak’s income = 5000 + 2% ofx.
= 5000 + \(\frac{2x}{100}\) ….(ii)
Income remains unchanged,

∴ Deepak’s sale is ₹ 1,00,000.

(ii) Given: b_yx = 0.4, b_xy = 0.9
var(x) = 9; var (y) = ?
Solution:

∴ Variance of y is 4.

(iii)

[B] (i)

Walsch’s Price Index number

∴ x = 16.67

(ii) Min (A) = 12; Max (B) = 12, Min (C) = 8
∵ Min (A) ≥ Max (B)
∴ The given problem can be converted into 2 machine problem

Total elapsed time:

Total ideal time for machine B = (102 – 94) + 54
= 8 + 54
= 62 hrs.

(iii) (a) ∵ ∑P(x) = 1, we get
k + 2k + 2k + 3k + k² + 2k² + 7k² + k = l
∴ 10k² + 9k – l = 0
∴ 10k² + 10k – k – l = 0
∴ 10k(k+l)- l(k + l) = 0
∴ (10k – l)(k + l) = 0
∴ k = \(\frac{1}{10}\) or k = -1
But P(x) > 0
∴ k ≠ -1
∴ k = \(\frac{1}{10}\)

(b) P(X < 3) = P(x = 1) + P(x = 2)
= k + 2k = 3k
= 3 × \(\frac{1}{10}\) = \(\frac{3}{10}\)

(c) P(x > 6) = P(x = 7)

Question 6.
[A] Attempt any TWO of the following questions (3 marks each): (6)[14]
(i) The building is insured for 75% of its value. The annual permium at 0.70 percent amounts to ₹2,625. If the building is damaged to the extent of 60% due to fire, how much can be claimed under the policy?
(ii) Three new machines M₁, M₂, M₃ are to be installed in a machine shop. There are four vacant places A, B, C, D. Due to limited space, machine M₂ can not be placed at B. The cost matrix (in hundred ₹) is as follows:

(iii) The eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg in the lot of 10 eggs.
[B] Attempt any ONE of the following questions (4 marks each): (4)
(i) Following table shows the all India infant mortality rates (per ’000) for years 1980 to 2010:
Fit the trend line to the above data by the method of least squares.
(ii) Minimize: z = 6x + 2y
Subject to: x + 2y ≥ 3,
x + 4y ≥ 4,
3x + y ≥ 3,
x ≥ 0, y ≥ 0
[C] Attempt any ONEofthefollowing questions (Activity) (4 marks each): (4)
(i) For a bivariate data \(\bar{x}\) = 10, \(\bar{y}\) = 12, V(X) = 9, σ_y = 4 and r = 0.6
Estimate y when x = 5
Solution:
Line of regression of Y on X is

(ii) If X — P(m) with P(X = 1) = P(X = 2) then find the mean and P(X= 2).
Given e^-2 = 0.1353
Since P(X = 1) = P(X = 2)

Solution:
[A] (0 Let the value of the building be ₹ x.

(ii) This is an unbalanced problem, hence introducing a dummy row of machine M₄ with zero cost.
As machine M₂ cannot be placed at B, a very high cost oo is assigned to the corres element.

Step 1: Row minimization.

Step 2: Column minimization.
∵ Each column contains element zero, column minimization will give us the same matrix.
Step 3: Drawing min. lines to cover all zeros.

∵ No. of lines drawn (4) = order of the matrix (4), optimal assignment can be done.

∴ Optimal solution is

Total cost = ₹ 2800.

(iii) Let X denote number of defective eggs from a lot of 10 eggs.

[B] (i) Here n = 7, Transforming the year t to u by u = \(\frac{t-1995}{5}\)

The equation of trend line is x_t = a’ + b’u
The normal equations are
∑x_t = n.a’ + b’ X0u ….(1)
∑u. x_t = a’∑u + b’∑u² ….(2)
Here n = 7, ∑x_t = 30, ∑u = 0, ∑u² = 28, ∑u.x_t = -44
Substituting the values in normal equations, we get
30 = 7a’ + b'(0)

(ii)

From the graph, the unbounded feasible region is EPQD.
Point P is a point of intersection of lines
3x + -y = 3 …(1)
and x + 2 y = 3 …(2)
Multiplying equation (2) by 3 and subtracting it from (1).

Substituting y = 1.2 in equation (1),
3x + 1.2 =3
∴ 3x = 3 – 1.2 = 1.8
∴ x = \(\frac{1.8}{3}\) = 0.6
∴ P ≡ (0.6,1.2)
Point Q is a point of intersection of
x + 2y = 3 …(3)
and x + 4y = 4 …(4)
Subtracting equation (4) from equation (3)

Substituting y = 0.5 in (3).
x + 1 = 3
∴ x = 3 – 1 = 2
∴ Q ≡ (2, 0.5)

Hence the corner points of unbounded feasible region are E(0,3), P(0.6,1.2), Q(2,0.5) and D(4,0).
Z = 6x + 2y
Z(E) = 0 + 3 × 2 = 6
Z(P) = 6 × 0.6 + 2 × 1.2 = 3.6 + 2.4 = 6
Z(Q) = 6 × 2 + 2 × 0.5 = 12 + 1.0 = 13
Z(D) = 6 × 4 + -0 = 24
∴ Z has minimum value of 6 at E(0, 3) and P(0.6, 1.2)

[C] (i) For a bivariate data \(\bar{x}\) = 10, \(\bar{y}\) = 12, v(x) = 9, σ_y = 4 and r = 0.6
Solution:
Line of regression of Y on X is

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