Maharashtra Board Class 12 Commerce Maths Sample Paper Set 2 with Solutions

Maharashtra State Board Class 12th Commerce Maths Sample Paper Set 2 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Commerce Maths Model Paper Set 2 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.
[A] Choose the correct alternative:
(i) Let p ∧ (q v r) ≡ (p ∧ q) v (p ∧ r). Then this is known as:
(a) Commutative law
(b) Associative law
(c) De-Morgans law
(d) Distributive law
(ii) Ify = e^logX, then \(\frac{dy}{dx}\) =?
(a) \(\frac{e^{\log x}}{x}\)
(b) \(\frac{1}{x}\)
(c) 0
(d) \(\frac{1}{2}\)
(iii) Slope of the tangent to the curve y = 6 – x² at (2, 2) is ……….
(a) 4
(b) -4
(c) -2
(d) -1
(iv)

(v) Using definite integration of circle x² + y² = 25 is ……….
(a) 5 π sq. units
(b) 4 π sq. units
(c) 25 π sq. units
(d) 25 sq. units
(vi) The solution of \(\frac{dy}{dx}\) = 1 is ……….
(a) x + y = c
(b) xy = c
(c) x² + y² = c
(d) y – x = c
[B] State whether each of the following is True or False:
(i) The derivative of a^x is a^x log a.
(ii) A function f(x) is maximum atx = a when f'(o) > 0.
(iii) \(\int_a^b\)f (x)dx = \(\int_a^b\)f(x – a -b)dx.
[C] Fill in the blanks:
(i) A diagonal matrix in which all diagonal elements are same, is called a ………. matrix.

Solution:
[A] (i) (d) Distributive law
(ii) \(\frac{e^{\log x}}{x}\)
(iii) (b) -4
(iv) (c) –\(\int_b^a\)f(x)dx
(v) (c) 25 π sq. units
(vi) (d) y – x = c

[B] (i) True
(ii) False. A function f(x) is maximum at x = a when f(a) <0.
(iii) False, \(\int_b^a\)f(x)dx = \(\int_b^a\)f(a + b – x)dx.

[C] (i) Scalar
(ii) y = x.logx
Differentiating both sides,

(iii)

Question 2.
[A] Attempt any two of the following:
(i) Write the converse, inverse and contrapositive of the following statement: If a man is bachelor, then he is happy.
(ii)

(iii) Determine the maximum and minimum values of the following functions:
f(x)= x² + \(\frac{16}{x}\)
[B] Attempt any two of the following:
(i) The sum of the cost of the Economic book, one Co-operation book and one account book is ₹ 420. The total cost of an Economic book, 2 Co-operation books and an Account book is ₹480. Also the total cost of an Economic book, 3 co-operation book and 2 Account books is ₹ 600. Find the cost of each book using matrix method.
(ii) For the demand function D = 100 – \(\frac{p^2}{2}\). Find the elasticity of demand at p = 10.
(iii) Solve the following differential equations:
x²ydx -(x³ – y³) dy = 0
Solution:
[A] (i) Let p : A man is bachelor
q : He is happy
Then the symbolic form of given statement is p → q
Converse: q → p is the converse of p → q
i.e., if a man is happy, then he is a bachelor.
Inverse: -p → ~q is in Verse of p → q
i.e., If a man is not bachelor, then he is not happy.

Contrapositive : ~q → ~p is the contrapositive of p → q i.e., If a man is not happy, then he is not bachelor.
(ii)

(iii)

∴ f(x) attains minimum value at x = 2
by second derivative test.
∴ Minimum value = f(2) =(2)² + \(\frac{16}{2}\)
= 4 + 8
= 12
∴ The function f(x) has minimum value 12 at x = 2.

[B] (i) Let the cost of one economic book, one co-operation book and one account book be ₹x, ₹y and ₹z respectively
According to the first condition,
x + y + z = 420
According to the second condition,
x + 2 y + z = 480
According to the third condition,
x + 3y + 2z = 600
Matrix form of the above system of equations is


By equality of matrices, we get
x + y + z = 420 …(i)
y = 60
z = 60
Substituting y = 60 and z = 60 in equation (i), we get
x + 60 + 60 = 420
x = 420 – 120 = 300
∴ The cost of one economic book is ₹300, one co-operation book is ₹60 and one account book is ₹60.

(ii) Given, demand function is D = 100 – \(\frac{p^2}{2}\)

Elasticity of demand at p = 10 is 2
Here, η > 0.
∴ The demand is elastic

(iii) x²ydx – (x³ + y³)dy = 0
∴ x²ydx = (x³ + y³)dy

Integrating on both sides, we get

This is the general solution.

Question 3.
[A] Attempt any two of the following:
(i) Show that the following statement pattern is a contingency:
(p → q) ∧ (p → r)
(ii) Evaluate:

(iii) Evaluate the following definite integrals:

[B] Attempt any one of the following:
(i) lf f'(x) = \(\frac{x^2}{2}\) – kx + 1, f(0) = 2 and f(3) = 5, find f(x).
(ii) If the population of a town increases at a rate proportional in the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years, what will be the population in another 20 years?
[C] Attempt any one of the following activities:
(i) Find \(\frac{dy}{dx}\), if y^x = e^x+y
(ii)

Solution:
[A] (i) (p → q) ∧ (p → r)

The truth values in the last column are neither all T nor all F. Hence, it is contingence.

(ii)

(iii)

[B] (i)

(ii) Let x be the population at time ‘t’
∴ \(\frac{dx}{dt}\) = x
∴ \(\frac{dx}{dt}\)= kx, where k is the constant of proportionality
∴ \(\frac{dx}{dt}\) = k.dt
Integrating on both sides, we get

Population after 60 years will be 73482

[C] (i) Given, y^x = e^x+y
Talcing log on both sides, we get,

Differentiating w.r.t x. we get

(ii)

Now comparing the coefficients of x² and constant term, we get

Section – B

Question 4.
[A] Choose the correct alternative:
(i) Which component of the time series refers to erratic time series movements that follow no recognizable or regular pattern:
(a) Trend
(b) Seasonal
(c) Cyclical
(d) Irregular
(ii) |b_xy + b_yz| ≥ ……….
(a) |r|
(b) 2|r|
(c) r
(d) 2r
(iii) Laspeyre’s Price Index Number is given by:

(iv) Feasible region is the set of points which satisfy:
(a) The objective function.
(b) All of the given functions
(c) Some of the given constraints
(d) Only non-negative constraints
(v) The objective of an assignment problems is to assign:
(a) Number of jobs to equal number of persons at maximum cost
(b) Number of jobs to equal number of persons at minimum cost
(c) Only the maximize cost
(d) Only the minimize cost
(vi) If E(x) > Var (x) then X follows ……….
(a) Binomial distribution
(b) Poisson distribution
(c) Normal distribution
(d) None of the above
[B] State whether each of the following is True or False:
(i) Cyclical variation can occur several times in a year.
(ii)

(iii) occurs at the centre of the feasible region.
[C] Fill in the blanks:
(i) ………. component of time series is indicated by periodic variation year after year.
(ii) Fishers Price Index Number is given by ……….
(iii) The region represented by the inequality y ≤ 0 lies in ……….quadrants.
Solution:
[A] (i) (a) Trend
(ii) (b) 2 |r|
(iii) (c) \(\frac{\Sigma p_1 q_0}{\Sigma p_1 q_0} \times 100\)
(iv) (d) AIL of the given functions
(v) (b) Number of jobs to equal number of persons at minimum cost
(vi) (a) Binomial distribution

[B] (i) False
(ii) False.
(iii) False. The optimum value of the objective function of LPP occurs at the corners of the feasible region.

[C] (i) Seasonal
(ii)

(iii) 3^rd and 4^th quadrants.

Question 5.
[A] Attempt any two of the following:
(i) Deepak’s salary was increased from ₹ 4,000 to ₹ 5,000. The sales being the same, due to reduction in the rate of commission from 3% to 2%, his income remained unchanged. Find the sales.
(ii) For the certain bivariate data on 5 pairs of observation given
∑x = 20, ∑y = 20, ∑x² = 90, ∑y² = 90, ∑xy = 76.
Calculate:
(a) cov (x, y)
(b) b_yx and b_xy
(c) r
(iii) Find the sequence that minimizes the total elapsed time to complete the following jobs in the order of AB. Find the total elapsed time and idle time for both the machines.

[B] Attempt any two of the following:
(i) A stock worth ₹ 7,00,000 was insured for ₹ 4,50,000. Fire burnt sotck worth ₹ 3,00,000 completely and damaged the remaining stock to the extent of 7 5% of its value. What amount can be claimed under the policy?
(ii) Find x if Laspeyre’s Price Index Number is same as Paasche’s Price Index Number for the following data:

(iii) The probability distribution of X is as follows:

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% of x.
= 5000 + \(\frac{2x}{100}\) ….(ii)
Income remains unchanged,

∴ Deepak’s sale is ₹ 1,00,000.

(ii) Given, ∑x = 20, ∑y = 20, ∑x² = 90,
∑y² = 90, ∑xy = 76, n = 5.
Now,


Since b_yx and b_xy both are negative.
∴ r is negative
∴ r = -0.4

(iii) Observe that Min (A, B) = 5, corresponds to job VI on machine B and job VII on machine A.
∴ Job VI is placed last and job VII is placed first in sequence

Now, Min (A, B) = 7, corresponds to job I on machine A.
∴ Job I is placed second in sequence

Now, Min (A, B) = 10, corresponds to job IV on machine A as well as on machine B.
∴ Job IV is placed second last in sequence

Now, Min (A, B) = 14, corresponds to job II and job III on machine B and job A on machine A.
These three jobs can be placed in the sequence in order
V-lll-ll or V-ll-III
∴ The optimal sequence can be


∴ We consider the optimal sequence are VII-I-IV-V-III-II-VI.
Total Elapsed Time

∴ Total time elapsed is 91 units
Idle time for machine A = 91-86 = 5 units
Idle time for machine B = 5 + 2 + 3 + 2 + l = 13 units.

[B] (i) Given, stock (property value) = ₹ 7,00,000
Policy value = ₹ 4,50,000
Now, stock worth = ₹ 3,00,000 were burnt completely due to fire.
Remaining stock = Stock value – Burnt stock
= 7,00,000 – 3,00,000
= ₹ 4,00,000
Remaining stock was damaged to 75% of its value.
∴ Damaged amount = 75% of value

∴ Sum of ₹ 3,85,714.30 can be claimed under the policy.

(ii)

From the table,

∴ \(\frac{2 x+18}{3 x+24}=\frac{5}{7}\)
⇒ 14x+ 126 = 15x+ 120
⇒ x = 6.
Hence, x = 6.

(iii) The table gives a probability distribution and therefore
P[X = 0] + P[X = 1) + P [X = 2] + P [X = 3] + P[X = 4] = 1
i.e., 0.1 + k + 2k + 2k + k = 1
i.e„ 6k = 0.9
∴ k = 0.15
(a) k = 0.15
(b) P[X < 2] = P[X = 0] + P[X = 1] = 0.1 + k
= 0.1 + 0.15
= 0.25
(c) P[X > 3] = P[X = 3] + P[X = 4] = 2k + k
= 3k = 3(0.15)
= 0.45
(d) P[1 < X < 4] = P[X = 1] + P[X = 2] + P[X = 3]
= k + 2k + 2k
= 5k = 5(0.15)
= 0.75
(e) P(2) = P[X ≤ 2] = P[X = 0] + P[X = 1] + P[X = 2]
= 0.1 + k + 2k
= 0.1 + 3k
= 0.1 + 0.45
= 0.55

Question 6.
[A] Attempt any two of the following:
(i) A bill drawn on 5^th June for 6 months was discounted at the rate of 5% p.a. on 19^th October. If the cash value of the bill is ₹ 43,500, find face value of the bill.
(ii) In a partially destroyed laboratory record of an analysis of regression data, the following data are Legible:
Variance of X = 9
Regression equations:
8x – 10y + 66 = 0 and 40x – 18y = 214.
Find on the bass of the above information
(a) The mean value of X and Y.
(b) Correlation coefficient between X and Y.
(c) Standard deviation of Y.
(iii) It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.

(a) Verify whether f(x) is a p.d.f.
(b) Find P(0 < x ≤ 1)
(c) Find the probability that X is between the 1 and 3.
[B] Attempt any one of the following:
(i) The following table shows the index of industrial production for the period from 1976 to 1985, using the year 1976 as the base year.

Fit a trend line to the above data by graphical method.
(ii) A company has a team of four salesmen and there are four districts where the company wants to start its business. After taking into account the capabilities of salesmen and the nature of districts, the company estimates that the profit per day in rupees for each salesman in each district is as below:

Find the assignment of salesman to various districts which will yield maximum profit.
[C] Attempt any one of the following activities:
(i) Shraddha wants to invest at most ₹ 25,000 in saving certificates and fixed deposits. She wants to invest at least ₹ 10,000 in saving certificates and at least ₹ 15,000 in fixed deposits. The rate of interest on saving certificates is 5% per annum and that on fixed deposits is 7% per annum. Formulate the above problem as LP.P. to determine maximum yearly income.
(ii) In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.
(Given that e^-0.2 = 0.8187)
Solution:
[A] (I) Given: Date of drawing = 5^th June
Period of the bill = 6 months
∴ Nominal due date = 5^th December
Legal due date = 8^th December
Date of discounting = 19^th October
Cash value (C.V.) = ₹ 43,500 and r = 5%
Now, number of days from the date of discounting to the legal due date are as follows:

∴ Face value of the bill is ₹ 43,800

(ii) Given: σ²_x = 9
∴ σ_x = 3

(a) The two regression equations are

Substituting y = 17 in (i), we get
8x – 10 × 17 = -66
8x = -66 + 170
8x = 104
x = \(\frac{104}{8}\)
∵ The point of intersection of two regression lines

(b) Let 8x – 10y + 66 = 0 be the regression equation of Y on X.
∴ The equation becomes 10y = 8x + 66

Now, the other equation, i.e., 40x – 18y = 214 is the regression equation of x on y

Since b_YX and b_XY both are positive, r is also positive.
∴ r = 0.6
(c)

(iii) (a) Here, f(x) ≥ 0, x e[0, 4]
Now consider,

[B] (i) Taking years on X-axis and index on Y-axis. We plot the points for indices corresponding to the years. Joining these points we get the graph of the given time series. We draw trend line as shown in the figure.

(ii) Step 1: Since it is a maximisation problem, subtract each of the elements in the table from the largest element i.e., 16


Step 2: Row minimum Subtract the smallest element in each row from every element in its row. The matrix obtained is given below:

Step 3: Column minimum here, each column contains element zero.
∴ Matrix obtained by column minimum is same as above matrix.
Step 4: Draw minimum number of vertical and horizontal lines to cover all zeros. First cover all rows and columns which have maximum number of zeroes.

Step 5: From step 4, minimum number of lines covering all the zeroes are 4, which is equal to order of the matrix, i.e„ 4. Hence optimal solution has reached.
∴ Select a row with exactly one zero, enclose that zero in (□) and cross out all zeros in its respective column.
Similarly, examine each row and column and mark the assignment (□).
∴ The matrix obtained is as follows:

Step 6:
The matrix obtained in step 5 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:
∴ The optimal solution is:

∴ The maximum profit = 16 + 15 + 15 + 15 = ₹61

[C] (i) Let x₁: amount (in ₹) invested in saving certificate.
x₂: amount (in ₹) invested in fixed deposits.
x₁ ≥ 0, x₂ ≥ 0
From given conditions x₁ + x₂ ≤ 25,000
She wants to invest at least ₹ 10,000 in saving certificate.
∴ x₁ ≥ 10,000
Shraddha wants to invest at least ₹ 15,000 in fixed deposits.
x₂ ≥ 15,000
Total interest = z = 0.05x₁ + 0.07x₂
Maximize z = 0.05x₁ + 0.07x₂ subject to
x₁ + x₂ ≤ 25,000
x₁ ≥ 10,000
x₂ ≥ 15,000
x₁, x₂ ≥ 0

(ii) Here m = \(\frac{10}{50}\) = 0.2, and hence X ~ P(m) with

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