12th Commerce Maths Question Paper 2022 Maharashtra Board Pdf

Maharashtra State Board Class 12th Commerce Maths Question Paper 2022 with Solutions Answers Pdf Download.

Class 12 Commerce Maths Question Paper 2022 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) If A = \(\left[\begin{array}{ll}2 & 3 \\ a & 6 \end{array}\right]\) is a singular matrix, then a = ________ .
(a) 6
(b) -5
(c) 3
(d) 4
(ii)

(iii) The slope of a tangent to the curve y = 3×2 – x + 1 at (1, 3) is ________ .
(a) 5
(b) -5
(c) \(\frac{-1}{5}\)
(d) \(\frac{1}{5}\)
(iv) The order and degree of the differential equation \(\left[1+\left(\frac{d y}{d x}\right)^3\right]^{2 / 3}=8\left(\frac{d^3 y}{d x^3}\right)\) are respectively
(a) 3, 1
(b) 1, 3
(c) 3, 3
(d) 1, 1
(v) The area of the region bounded by the curve y = x², x = 0, x = 3 and X-axis is ________ .
(a) 9 sq. units
(b) \(\frac{26}{3}\) sq. units
(c) \(\frac{52}{3}\) sq. units
(d) 18 sq. units
(vi) \(\int_{-5}^5 \frac{x^7}{x^4+10}\)dx = ________ .
(a) 3, 1
(b) 1, 3
(c) 3, 3
(d) 1, 1
[B] State whether the following statements are true or false (1 mark each): (3)
(i) If f'(x) > 0 for all x ∈ (a, b) then f(x) is decreasing function in the interval (a, b).
(ii) If \(\int \frac{4 e^x-25}{2 e^x-5}\)dx = Ax – 3 log |2e^x – 5| + c. where c is the constant of integration, then A = 5.
(iii) The integrating factor of the differential equation
\(\frac{dy}{dx}+\frac{y}{x}\) = x³ is – x.
[C] Fill in the following blanks: (1 mark each): (3)
(i) If p v q is true, then the truth value of ~p ∧ ~q is ________ .

(iii) y² = (x + c)³ is the general solution of the differential equation ________ .
Solution:
[A]
(i) (d) ∵ Matrix A is a singular matrix.
∴ |A| = 0

∴ order = 3.
degree = 3.

[B] (i) False
If f”(x) > 0 for allx e (a, b), then f(x) is decreasing function in the interval (a, b).

(ii) True
Let 4e^x – 25 = A(2e^x – 5) + B.\(\frac{d}{dx}\) (2e^x – 5)
= 2e^xA- 5A + B(2e^x)
2 × 2e^x – 25 = 2e^x (A + B) – 5A
∴ A + B = 2
and – 25 = – 5A
∴ A = 5
5 + B = 2
∴ B = 2 – 5 = -3.

(iii) False

[C] (i) If p v q is true, then the truth value of ~p ∧ ~q is False.
=> ~(p v q) = ~p ∧ ~q
[By D’Morgan’s law]
∴ ~p ∧ -q = ~(p v q)
= ~(T) = F.

(ii)

⇒ x = A(x + 3) + B(x + 2)
⇒ x = (A + B) x + (3A + 2B)
On equating coefficients of like terms, we get
A + B = 1 …(1)
⇒ B = 1 – A
⇒ B = 1 – (- 2) = 3
⇒ B = 3
and 3A + 2B = 0 …(2)
⇒ 3A + 2(1 – A) = 0
⇒ 3A + 2 – 2A = 0
⇒ A + 2 = 0
⇒ A = – 2

(iii) y² = (x + c)³ is the general solution of the differential equation


This is the required differential equation

Question 2.
[A] Attempt any TWO of the following: (3 marks each): (6)[14]
(i) Write the converse, inverse and contrapositive of the statement, “If 2 + 5 = 10, then 4 + 10 = 20.”
(ii) If x = \(\sqrt{1+u^2}\), y = log (1 + u²), then find \(\frac{dy}{dx}\).
(iii) Find the area between the two curves (parabolas)
y² = 7x and x² = 7y.
[B] Attempt any TWO of the following (4 marks each): (8)
(i) Determine whether the following statement pattern is a tautology, contradiction or contingency:
[(~p ∧ q) ∧ (q ∧ r)] ∧ (~q)
(ii) If ax² + 2hxy + by² = 0, then prove that \(\frac{d^2y}{dx^2}\) = 0.

Solution:
[A] (i) Let p: 2 + 5 = 10;
q: 4 + 10 = 20
Converse: q → p
If 4 + 10 = 20
then 2 + 5 = 10
Inverse: ~p → ~q
If 2 + 5 ≠ 10
then 4 + 10 ≠ 20
Contrapositive: ~q → ~p
If 4 + 10 ≠ 20
then 2 + 5 ≠ 10.

(ii)

(iii) To find the point of intersection of the curves:
Equation of curve is


Hence area between the two curves is \(\frac{49}{3}\) sq units.

[B](i)

Since the entries in the Last column of the above truth table are all false, the given statement is a contradiction.

(ii) ax² + 2 hxy + by² = 0
Differentiating w.r.t. x’,

(iii)

Question 3.
[A] Attempt any TWO of the following quations (3 marks each): (6)[14]

(ii) Divide 20 into two parts, so that their product is maximum.
(iii) Solve the following differential equation
x²y dx – (x³ + y³) dy = 0.
[B] Attempt any ONE of the following: (4)
(i) Find the inverse of the matrix A by using adjoint method,

(ii) Evaluate: \(\int_1^3\)logx dx.
[C] Attempt any ONE of the following questions (Activity): (4)
(i) Complete the following activity to find MPC, MPS, APC and APS, if the expenditure Ec of a person with income I is given as:
E_c = (0.0003) I² + (0.075)|
when I = 1000
(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, complete the following activity to find the number of times the bacteria are increased in 12 hours.
Solution:
[A] (i)


(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) x²y dx – (x³ + y³)dy = 0
⇒ (x³ + y³).dy = x²y.dx

[B]

∴ |A| = 3(0 – 6) + 1(0 + 15) + 1(0 – 0)|
= 18 + 15
= 33 ≠ 0
∴ A^-1 exists.
To find cofactors,
m₁₁ = 0 – 6 = -6
m₁₂ = 0 + 15 = 15
m₁₃ = 0 + 0 = 0
m₂₁ = 6 – 6 = 0
m₂₂ = 18 + 15 = 33
m₂₃ = -18 – 15 = -33
m₃₁ = -1 – 0 = -1
m₃₂ = -3 – 0 = -3
m₃₃ = 0 – 0 = 0
∴ A₁₁ = (-1)² (-6) = -6
A₁₂ = (-1)³.15 = -15
A₁₃ = (-l)⁴.0 = 0
A₂₁ = (-1)³.0 = 0
A₂₂ = (-1)⁴.33 = 33
A₂₃ = (-l)⁵.(-33) = 33
A₃₁ = (-l)⁴.(-1) = -1
A₃₂ = (-l)⁵(-3) = 3
A₃₃ = (-1)⁶.0 = 0
∴ Matrix of co-factors

(ii)

[C] (i) Given E_c = (0.0003)I² + (0.075) I²
We have APC = \(\frac{E_C}{I}\)
= (0.0003).I + 0.075
At I = 1000,
APC = (0.0003) × 1000 + 0.075
= 0.3 + 0.075
∴ APC = 0.375
Now MPC = \(\frac{d(E_C)}{dl}\)
= 2 × (0.0003). I + 0.075
= (0.0006) × I + 0.075
When I = 1000,
MPC = 0.6+ 0.075
= 0.675
At I = 1000, MPS = 1 – MPC
= 1 – 0.675
= 0.325
At I = 1000,
APS = 1 – APC = 1 – 0.375
= 0.625.

(ii) Let N be the number of bacteria present at time t.
∵ The rate of increase is proportional to the number present.
\(\frac{dN}{dt}\) ∝ N
∴ \(\frac{dN}{dt}\) = k.N
where k is the constant of proportionality.
∴ \(\frac{dN}{N}\) = k.dt
Integrating both sides,
∫\(\frac{1}{N}\).dN = k∫dt
log N = k. t + C …(1)
Now at, t = 0, N = N₀
∴ From (1),

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) Banker’s discount
(b) True discount
(c) Banker’s gain
(d) Cash value
(ii) b_xy . b_yx = ________ .
(a) V(X)
(b) σ_x
(c) r²
(d) σ_y²
(iii) The assignment problem is said to be balanced, if it is a ________ .
(a) square matrix
(b) rectangular matrix
(c) row matrix
(d) column matrix
(iv) Price index number by weighted aggregate method is given by ________ .

(v) The following function represents the p.d.f. of a.r.v.X

(c) 4
(d) 3
(vi) If X ~ B(20, \(\frac{1}{10}\)), then E(X) = ________ .
(a) 2
(b) 5
(c) 4
(d) 3
[B] State whether the following statements are true or false: (1 mark each) (3)
(i) If X ~ P(m) with P(X = 1) = P(X = 2) then m = 1.
(ii) Dorbish-Bowley’s Price Index Number is square root of product of Laspeyre’s and Paasche’s Index Numbers.
(iii) To convert maximization type assignment problem into a minimization problem, the smallest element in the matrix is deducted from all elements of matrix.
[C] fill in the blanks (1 mark each): (3)
(i) A wholeseller allows 25% trade discount and 5% cash discount. The net price of an article marked at ₹ 1,600 is ________ .
(ii) For a certain bivariate data on 5 pairs of observations given:
∑x = 20, ∑y = 20, ∑x² = 90, ∑y² = 90, ∑xy = 76 then b_xy = ________ .
(iii) If P₀₁ (L) = 121. P₀₁ (P) = 100,then P₀₁ (F) = ________ .
Solution:
[A] (i) (b)True discount
(ii) (c) r²
(iii) (a) square matrix
(iv) \( \frac{\sum p_1 w}{\sum p_0 w} \times 100\)
(v) (b) f(x) is a p.d.f. of random variable x.

[B](i) False

⇒ m = 2
∴ m ≠ 1
(ii) False
P₀₁(F) = \(\sqrt{P_{01}(L) X P_{01}(P)}\)
Fisher’s Ideal Price. Index Number is square root of product of Laspeyre’s Price. Index number and Paasche’s Price Index Number.

(iii) False
To convert maximization type assignment problem into a minimization problem, all the elements of the matrix are subtracted from the largest element of the matrix.

[C] (i) 25% trade discount on ₹1600.
= \(\frac{25}{100}\) × l600
= ₹400
Invoice Price = 1600 – 400 = ₹1200
5% cash discount on invoice price.

Question 5.
[A] Attempt any TWO of the following questions (3 marks each): (6) [14]
(i) Find the equation of line of regression of y on x for the following data:

(ii) A job production unit has four jobs P, Q, R, S which can be manufactured on each of the four machines I, II, III and IV. The processing cost of each job for each machine is given in the following table:

Find the optimal assignment to minimize the total processing cost.
(iii) In a cattle breeding firm, it is prescribed that the food ration for one animal must contain 14,22 and 1 unit of nutrients A, B and C respectively. Two different kinds of fodder are available. Each unit weight of these two contains the following amounts of these three nutrients:

The cost of fodder 1 is ₹3 per unit and that of fodder 2 is ₹ per unit Formulate the LP.P. to minimize the cost.
[B] Attempt any TWO of the following questions: (4 marks each) (8)
(i) Calculate the cost of living index number for the following data by aggregative expenditure method:

(ii) Five jobs are perofomed first on machine and then on machine M₂. Time taken in hours by each job on each machine is given below:

Determine the optimal sequence of jobs and total elapsed time. Also find the idle time for two machines.
(iii) The probability distribution of a discrete r.v. X is as follows:

(a) Determine the value of k.
(b) Find P(X ≤ 4)
(c) P(2 < X ≤ 4)
(d) P(X ≥ 3)
Solution:
[A] (i) Equation of line of regression of y on x is


(ii) Step 1. Subtract smallest element of each row from every element of that row.

Step 2. Since each column contains element O, column minimization will give us the same matrix.
Step 3. Drawing minimum number of horizontal and vertical lines to cover all zeros.

∵ No. of lines drawn (4) = order of the matrix (4), optimal solution has reached.
Step 4: Assignment:

Job Assignment:

Minimum Processing Cost = ₹99.

(iii) Let x and y be the no. of fodder 1 and fodder 2 respectively. Let Z be the total cost of fodders which has to be minimized.
Z = 3x + 2y
The constraints are as follows:

we get, 2x + y ≥ 14
2x+ 3y ≥ 22
x + y ≥ 1
∴ The required LP.P. is
min Z = 3x + 2y subject to
2x + y ≥ 14
2x + 3y ≥ 22
x + y ≥ 1;
x ≥ 0,
y ≥ 0.

[B](i)

= \(\frac{17.170}{12,000}\) × 100
= 137.36

(ii) The optimal sequence is

∴ Total elapsed time = 41 hrs.
Idle time for machine
M₁ = 41 – 30 = 11 hrs.
Idle time for machine
M₂ = 41 – 36 = 5 hrs.

(iii) (a) ∵ X has probability distribution,
∴ ∑P_i = 1
⇒ k + 2k + 3k + 4k + 5k + 6k = 1
⇒ 21k = 1
⇒ k = \(\frac{1}{21}\)

(b) P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 3k + 4k = 10k
= 10 × \(\frac{1}{21}\) = \(\frac{10}{21}\)

(c) P(2 < X < 4) = P(X = 3)
= 3k = \(\frac{3}{21}=\frac{1}{7}\)

(d) P(X ≥ 3) = 1 – P(X < 3)
= 1 – [P(X = 1) + P(X = 2)]
= 1 – [k + 2k]
= 1 – 3k
= 1 – \(\frac{3}{21}=\frac{18}{21}\)
= \(\frac{6}{7}\)

Question 6.
[A] Attempt any TWO of the following question (3 marks each): (6)[14]
(i) For 50 students of a class, the regression equation of marks in statistics (X) on the marks in accountancy (Y) is 3y – 5x + 180 = 0. The variance of marks in statistics is \({(\frac{9}{16})}^{th}\) of the variance of marks in accountancy. Find the correlation coefficient between marks in two subjects.
(ii) Solve the following LP.P.
Maximize z = 13x + 9y
Subject to 3x + 2y ≤ 12,
x + y ≥ 4,
x ≥ 0,
y ≥ 0.
(iii) Obtain the trend values for the following data using 5 yearly moving averages:

[B] Attempt any ONE of the following questions: (4)
(i) A warehouse valued at ₹40,000 contains goods worth ₹2,40,000. The warehouse is insured against fire for ₹16,000 and the goods to the extent of 90% of their value. Goods worth ₹80,000 are completely destroyed, while the remaining goods are destroyed to 80% of their value due to fire. The damage to the warehouse is to the extent of ₹8,000. Find the total amount that can be claimed under the policy.
(ii) A bill was drawn on 14^th April 2005 for ₹3,500 and was discounted on 6^th July 2005 at 5% p.a. The banker paid ₹3,465 for the bill. Find the period of the bill.
[C] Attempt any ONE of the following questions (Activity): (4)
(i) An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer completely randomly. Complete the following activity to find the probability that,
(a) the student gets 4 or more correct answers.
(b) the student gets less than 4 correct answers.
(ii) Following table shows the amount of sugar production (in lakh tonnes) for the years 1931 to 1941:

Complete the following activity to fit a trend line by method of least squares:
Solution:
[A] (i) 3y – 5x + 180 = 0
∴ 5x = 3 y + 180
∴ x = \(\frac{3}{5}\)y + 36
is the regression equation of x on y.
Comparing with

(ii)

Shaded region is the feasible region ABCA.
Z = 13x + 9y
Z(A) = 0 + 9 x 6 = 54
Z(B) = 13 x 4 + 0 = 52
Z(C) = 0 + 4 x 9 = 36
∴ Max, value of z is 54 at A(0,6).
i.e., when x = 0,
y = 6.
(iii)

[B] (i) Property value of warehouse = ₹40,000
Property value of goods = ₹2,40,000
Policy value of warehouse = ₹16,000
Policy value of goods = 90% × 2,40,000
= \(\frac{90}{100}\) × 2,40,000
= ₹2,16, 000.

Goods worth ₹ 80,000 are completely destroyed.
∴ Loss = 80,000

Value of remaining goods
= 2,40,000 – 80,000
= ₹1,60,000
Loss of remaining goods
= 80% × 1,60,000
= \(\frac{80}{100}\) × 1,60,000 100
= ₹1,28,000

∴ From equations (1), (2) and (3),
Total Claim = 72,000 + 1,15,200 + 3200
= ₹1,90,400.

(ii) Face value of bill = ₹3,500
Cash value = ₹3465
B.D. = F.V. – C.V.
= 3500 – 3465 = ₹35

∴ Legal due date is 73 days after the date of discounting i.e., 6^th July 2005.
July Aug. Sept.
25 31 17 = 73 days
∴ Legal due date = 17^th Sept.2005
∴ Nominal due date = 14^th Sept. 2005
Date of drawing bill = 14^th April 2005
∴ Period of bill = 5 months.

[C] (i) Let x = number of correct Answer
p = Probability of guessing correct Answer

Here n = 5
∴ X ~ B(n,p)
For binomial distribution
p(x) = ⁿC_x.p^x.q^n-x

(a) Probability that student gets 4 or more correct answers

(b) Probability that student gets less than 4 correct answer:
= P(X < 4) = 1 – P(X > 4)
= 1 – \(\frac{1}{64}\)
= \(\frac{63}{64}\)

(ii) Let yt be the trend line represented by the equation


The equation of trend Line becomes,
y_t = a’ + b’u …(1)
The normal equations are
∑y_t = na’ + b’∑u …(2)
∑u. y_t = u’∑u + b’∑u² …(3)
from equation (2), we get
∴ Normal equations are
33 = 11a’ + O.b’
⇒ 11a’= 33
⇒ a’ = 3
From equation (3), we get
44 = a’.0 + llO.b’
⇒ b’= \(\frac{44}{110}\) = 0.4
∴ b’ = 0.4
∴ The equation of the trend line is given by
y_t = 3 + (0.4)u.

Maharashtra Board Class 12 Commerce Maths Previous Year Question Papers