12th Commerce Maths Question Paper 2023 Maharashtra Board Pdf

Maharashtra State Board Class 12th Commerce Maths Question Paper 2023 with Solutions Answers Pdf Download.

Class 12 Commerce Maths Question Paper 2023 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) The dual statement of the statement (p v g) ∧ (r v s) is _________.
(a) (p ∧ g) ∧ (r ∧ S)
(C) (p v q) v (r v s)
(b) (p ∧ q) v (r ∧ S)
(d) (r v s) ∧ (p v q)
(ii) If y = x. loqx then \(\frac{dy}{dx}\) = _________.
(a) 1
(b) \(\frac{1}{x}\)
(c) logx
(d) 1 + log x
(iii) Ify = 2x² + a² + 2², then \(\frac{dy}{dx}\) = _________.
(a) 4x
(b) 4x + 2a
(c) 4x + 4
(d) 2x
(iv) A function f is said to be increasing at a point c if _________.
(a) f'(c) = 0
(b) f(c) > 0
(c) f'(c) < 0
(d) f(c) = l
(v) \(\int_0^2\)e^x dx = _________.
(a) e – 1
(b) 1 – e
(c) e² – 1
(d) 1 – e²
(vi) The integrating factor of \(\frac{dy}{dx}\) + y = e^-x is _________.
(a) e^x
(b) e^-x
(c) x
(d) -x
[B] State whether the following statements are true or false (1 mark each): (3)
(i) The derivative of f(y) = a^x is x. a^x-1 where a is constant
(ii) ∫(1 – x)^-2 dx = (l – x)^-1 + c
(iii) The degree of the differential equation
\(\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3\) = a^x is 3.
[C] Fill in the following blanks: (1 mark each):
(i) Converse of the statement q → p is _________.
(ii) If y = (loqx)² then \(\frac{dy}{dx}\) = _________.
(iii) If 0 < η < 1 then the demand is _________.
Solution:
[A] (i) (b) (p ∧ g) v (r ∧ s)

(ii) (d) 1 + log x
y = x.logx

(iii) (a) 4x
y = 2x² + a² + 2²
∴ \(\frac{dy}{dx}\) = 2 × 2x + 0 + 0 = 4x

(iv) (b) f'(c) > 0
(v) (c) e² – 1

[B] (i) False
f'(x) = a^x
f'(x) a^x.log a

(ii) True
∫(1 – x)^-2.dx
= \(\frac{(1-x)^{-2+1}}{(-2+1) \times(-1)}\) + c
= \(=\frac{(1-x)^{-1}}{(-1)(-1)}\)+c = (1-x)^-1 + c
(iii) False
The degree of D.E \(\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3\) = a^x is 2.

[C] (i) Converse of the statement q → p is p → q.

(iii) If 0 < η < 1 then the demand is relatively inelastic.

Question 2.
[A] Attempt any TWO of the following questions: (3 marks each): (6)[14]
(i) Construct the truth table for the statement pattern
(p ∧ ~g) ↔ (g → p)
(ii) If x⁵. y⁷ = (x + y)¹² then show that \(\frac{dy}{dx}=\frac{y}{x}\)
(iii) Evaluate: ∫\(\frac{1}{7+6x-x^2}\)dx
[B] Attempt any TWO of the following questions (4 marks each): (8)
(i) Solve the following equations by reduction method:
x + 2y + z = 8, 2x + 3y – z = 11, 3x – y – 2z = 5
(ii) For manufacturing x units, labour cost is ₹(150 – 54x) and processing cost is ₹(x²). Price of each unit is P = 10800 – 4x². Find the values of x for which
(a) Total cost is decreasing
(b) Revenue increasing
(iii) Evaluate: \(\int_1^3 \frac{\sqrt{x+5}}{\sqrt{x+5}+\sqrt{9-x}}\)
Solution:
[A] (i)

(ii) x⁵.y⁷ =(x + y)¹²
Taking log on both sides,
log (x⁵.y⁷) = log (x + y)¹²
∴ log x⁵ + log y⁷ = 12. log (x + y)
∴ 5. log x + 7 Log y = 12. log(x + y)
Differentiating w.r.t, V, on both sides

Hence proved

[B] (i) The given equations can be written in matrix form as:

∴ By equating of matrices, we get
x + 2 y + z =8 …(1)
-y – 3z =-5 …(2)
16z = 16 …(3)
from (3), z = 1
putting z = 1 in equation (2),
-y – 3 = -5
∴ -y = -2
∴ y = 2
Substituting z = 1, y = 2 in eqn. (1),
x + 4 + l = 8
x = 3
∴ x = 3, y = 2, z = 1 is the required solution

(ii) (a) Total cost = labour cost + processing cost
∴ c = 150 – 54x + x²
= x² – 54x + 150
∴ \(\frac{dc}{dx}\) = 2x – 54.
Total cost is decreasing.
∴ \(\frac{dc}{dx}\) < 0
2x – 54 < 0
∴ 2x < 54
∴ x < 27
∴ Total cost is decreasing if x < 27.

(b) Revenue, R = p.x
∴ R = (10800 – 4x²)x
= 10800 x – 4x³
∴ \(\frac{dR}{dx}\) = 10800 – 12x²
∴ Revenue is increasing \(\frac{dR}{dx}\) >0 dx
∴ 10800 – 12x² > 0
∴ 10800 > 12x²
∴ x² < 900
∴ x < 30 …. ∵ x > 0
∴ Revenue is increasing, if x < 30

(iii)

Question 3.
[A] Attempt any TWO of the following questions (3 marks each): (6)[14]
(i) Write the negation of each of the following statements:
(a) ∃n ∈ N, (n² + 2) is odd number.
(b) Some continuous function are differentiable.
(c) (p → q) v (p → r)
(ii) A rod of 108 meters long is bent to form rectangle. Find its dimensions if the area of rectangle is maximum.
(iii) Evaluate: ∫x³.log xdx
[B] Attempt any ONE of the following questions: (4)
(i) Find the area of the region bounded by the parabola y² = 4x and the line x = 3,
(ii) Find the particular solution of the differential equation (x – y²x) dx – (y + x²y) dy = 0 when x = 2, y = 0.
[C] Attempt any ONE of the following questions (Activity): (4)

Final (AB)^-1 by adjoint method.
Solution:

(ii) In a certain culture of bacteria, the rate of increase is proportional to the number present If it is found . that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
solution
Let N be the number of bacteria present at time’t’. Since the rate of increase of N is proportional to N, the differential equation can be written as:

when t = 0, N = No where No is initial number of bacteria,

∴ Bacteria are increased 8 times in 12 hours.
Solution:
[A] (i) (a) ∀n ∈ N,(n² + 2) is not odd number
(b) All continuous functions are not differentiable.
(c) ~[(p → q) v (p → r)]
= ~(p→q)∧~(p→r)
≡ (p ∧ ~q) ∧ (p ∧ ~r)

(ii) Let x and y be the length and breadth of a rectangle respectively
∴ 2(x + y) = 108
∴ x + y = 54
∴ y = 54-x
Now, Area = x. y
= x(54 – x)
∴ f(x) = 54x – x²
∴ f(x) = 54 – 2x
f'(x) = -2
Let f(x) = 0
∴ 54 – 2x = 0
2x = 54 ∴ x = 27
f'(x) = – 2 < 0
∴ f(x) is maximum at x = 27
∴ y = 54 – x = 54 – 27 = 27
∴ Area is maximum at x = 27, y = 27.
∴ Dimensions of the rectangle are 27m × 27m
∴ It is a square

(iii)

[B] (i) Area of region bounded by parabola
= area of region OABO
= 2(area of region OACO)

(ii) (x – y²x).dx – (y + x²y)dy = 0
x(1 – y²).dx – y( 1 + x²).dy = 0

∴ log |(1 + x²).(l – y²)| = log c
∴ (1 + x²)(l – y²) = c
This is the general solution
When x = 2, y = 0, we get
(1 + 4) (1-0) = c
The particular solution is
(1 + x²)(l – y²) = 5

[C] (i)


(ii) Let N be the number of bacteria present at time ‘t’. Since the rate of increase of N is proportional to N, the differential equation can be written as,
\(\frac{dN}{dt}\) ∝ N
∴ \(\frac{dN}{dt}\) = k.N
where k is the constant of proportionality.
∴ \(\frac{dN}{dt}\) = k.dt
Integrating both sides,
∴ ∫\(\frac{1}{N}\).dN = k∫1.dt
∴ log N = k.t + C
∴ When t = 0, N = No where No. is initial number of bacteria.
∴ log No. = k × 0 + c
∴ c = log No. …(1)
Also when t = 4, N = 2 No.
∴ log (2 No) = k × 4 + log No. ….from (1)

∴ Bacteria are increased 8 times in 12 hours.

Question 4.
[A] Select and write the correct answer of the following multiple choice type of questions (1 mark each): (6) [12]
(i) The sum due is also called as _________.
(a) Face value
(b) Present value
(c) Cash value
(d) True discount
(ii) _________ is a series of cash flows over a limited period of time.
(a) Policy value
(b) Annuity
(c) Present value
(d) Future value
(iii) b_yx is _________.
(a) Regression coefficient of y on x
(b) Regression coefficient of x on y
(c) Correlation coefficient between x and y
(d) Covariance between x and y
(iv) The complicated but efficient method of measuring trend of time series is _________.
(a) graphical method
(b) method of moving average
(c) method of least squares
(d) method of addition
(v) Quantity index number by Weighted Aggregate Method is given by _________.

(vi) If the corner points of the feasible region are (0, 0), (3, 0), (2, 1) and (0, 7/3), then maximum value of z = 4x + 5y is _________.
(a) 12
(b) 13
(c) \(\frac{35}{3}\)
(d) 0
[B] State whether the following statements are true or false (1 mark each): (3)
(i) The bankers discount is also called true discount.
(ii) Laspeyre’s Price Index Number uses current year’s quantities as weights.
(iii) In an assignment problem, if number of columns is greater than number of rows, then dummy column is added.
[C] Fill in the blanks (1 mark each): (3)
(i) The date by which the buyer is legally allowed to pay the amount is known as _________.
(ii) Walsch’s Price Index Number is givne by _________:
(iii) Graphical solution set of the inequations x ≥ 0 and y ≤ 0 lies in _________ quadrant.
Solution:
[A] (i) (a) Face value
(ii) (b) Annuity
(iii) (a) Regression coefficient of y on x
(iv) (c) method of least squares
(v) (d) \(\frac{\sum q_1 w}{\sum q_0 w}\) × 100
(vi) (b) 13
z( 0,0) = 0
z(3,0) = 4 × 3 + 5 × 0 = 12
z(2,1) = 4 × 2 + 5 × 1 = 13
z (0,\(\frac{7}{3}\)) = 4 × 0 + 5 × \(\frac{7}{3}\) = \(\frac{35}{3}\) = 11.66
∴ Max.z = 13

[B] (i) False
(ii) False
(iii) False
[C] (i) Legal due date
(ii) \(\frac{\sum p_1 \sqrt{q_0 q_1}}{\sum p_0 \sqrt{q_0 q_1}}\) × 100
(iii) 4^th

Question 5.
[A] Attempt any TWO of the following questions (3 marks each): (6) [14]
(i) An agent places insurance for ₹4,00,000 on life of a person. The .{Sremium is to be paid annually at the rate of ₹35 per thousand per annum. Find the agent’s commission at 15% on the premium.
(ii) For a bivariate data:

Find: (a) b_yx, (b) b_yx and (c) Correlation coefficient between x and y.
(iii) The following table shows gross capital information (in Crore ₹) for years 1966 to 1975:

Obtain trend values using 5-yearly moving values.
[B] Attempt any TWO of the following questions: (4 marks each) (8)
(i) Solve the following LPP by graphical method:
Maximize: z = 4x × 6y
Subjectto 3x + 2y ≤ 12, x + y ≥ 4.
x ≥ 0, y ≥ 0.
(ii) A marketing manager has list of salesmen and territories. Considering the travelling cost of the salesmen and the nature of the territory, the marketing manager estimates the total of cost per month (in thousand repees) for each salesman in each territory. Suppose these amounts are as follows:

Find the assignment os salesman to territories that wilL result in minimum cost.
(iii) A random variable X has the following probability distribution:

Solution:
[A] (i) Policy value = ₹4,00,000
Rate of premium = ₹35 per thousand p.a.
Rate of commission = 15%

= \(\frac{-5}{12}\) (∵ b_yx, b_xy < 0)

(iii)

[B] (i)

Region ABD is the feasible region with corner points
A(4, 0), B(0, 6) and D(0,4)
Now z = 4x + 6y
∴ z(A) = 4 × 4 + 6 × 0 = 16
z(B) = 4 × 0 + 6 × 6 = 36
z(D) = 4 × 0 + 6 × 4 = 24
∴ z has maximum value 36 at x = 0, y = 6

(ii) In the given problem, no of rows < no. of columns, hence it is an unbalanced assignment problem. It can be balance by introducing a dummy salesman E with zero cost in each territory.

Step 1: Subtract the smallest element of each row from every element of that row.

Step 2: Since all column minimums are zero, no need to subtract anything from columns.
Step 4: Drawing minimum number of horizontal and vertical lines, to cover all zeros.
Step 5: Since number of lines drawn (4) order of the matrix, subtract the smallest uncovered element (4) from all uncovered elements and add it to the elements at the intersection of two lines.
∴ we get,


∴ Minimum cost = ₹39,000

(iii) (a) For a probability distribution,.
∑P(x) = 1
∴ k + 2k + 2k + 3k + k² + 2k² + 7k² + k = l
∴ 9k + 10k² = l
10k² + 9k – l = 0
∴ 10k² + 10k – k – l = 0
10k(k+l)-l(k+l) = 1
(k+l)(10k-l) = 1
k = -l or l0k = l
k = -1 or k = \(\frac{1}{10}\)
k = -1 is not possible ∴ k = \(\frac{1}{10}\)

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

(c) P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7)
= k² + 2k² + 7k² + k
= 10k² + k
= 10 × \(\frac{1}{100}+\frac{1}{10}\)
= \(\frac{2}{10}=\frac{1}{5}\)

Question 6.
[A] Attempt any TWO of the following questions (3 marks each): (6)[14]
(i) An agent was paid ₹88,000 as commission on the sales of computers at the rate of 12.5%. if the price of each computer was ₹32.000, how many computers did he sell?
(ii) The publisher of a magazine wants to determine the rate of increase in the number of subscribers. The following table shows the subscription information for eight consecutive years:

Fit a trend line by graphical method.
(iii) If x follows Poisson distribution such that P(x = 1) = 0.4 and P(x = 2) = 0.2, find variance of x.
[B] Attempt any ONE of the following questions: (4)
(i) From the following data, find the regression equation of y on x and estimate y when x = 4:

(ii) Calculate Marshall – Edgeworth’s price index number for the following data:

[C] Attempt any ONE of the following questions (Activity): (4)
(i) Six jobs are performed on Machines M₁ and M₂ respectively. Time in hours taken by each job on each machine is given below:

Determine the optimal sequence of jobs and find total elapsed time. Also find the idle time for machines M₁ and M₁.
Solution:
Given jobs can be arranged in optimal sequence as,


Total Elapsed time = _____ hrs.
Idle time for Machine M₁ = 43 – 42 = 1 hour
Idel time for Machine M₂ = _____ hrs.
(ii) A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of getting at least two success.
Solution: A pair of dice is thrown 3 times
∴ n = 3
Let x = number of successes (doublets)
p = probability of success (doublets)
∴ p = ___, q = ___
∴ x ~ B(n, p)
P(x) = ⁿC_xp^xq^n-x
Probability of getting at least two success means x ≥ 2.
∴ P(x ≥ 2) = P(x = 2) + P(x = 3)
= ___ + ___
= \(\frac{2}{27}\)
Solution:
[A] (i) Price of each computer = ₹32,000
Rate of commission = 12.5%
∴ Commission on each computer = 12.5% × 32,000
= \(\frac{12.5}{100}\) × 32,000 100
= ₹4,000
Total commission = ₹32,000
∴ Number of computers sold

= \(\frac{88,000}{4,000}\) = 22
∴ An agent sold 22 computers.

(ii)

(iii) x follows Poisson distribution.

[B] (i)


∴ Regression equation of Y on X is
y = (-1) + 2x
∴ y = 2x – 1
when x = 4, y = 2 × 4 – 1 = 7

(ii)

From table, ∑p₀q₀ = 728.∑p₀q₁ = 1056,
∑p₁q₀ = 1032, ∑p₁q₁ = 1484.
Marshall – Edyeworth’s price index numbers is given by

[C] (i) Given jobs can be arranged in optimal sequence as

Total elapsed time = 43 hrs.
Idle time for machine m₁ = 43 – 42 = 1 hr.
Idle time for machine m₂ = 2 + 4 = 6 hrs.

(ii) A pair of dice is thrown 3 times
∴ n = 3
Let x = numberer of success (doublets)
p = probality of sucess (doublets)

probability of getting atleast 2 success means x ≥ 2.
∴ p(x ≥ 2) = p(x = 2) + p(x = 3)

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