Maharashtra HSC Mathematics and Statistics (Commerce) Board Paper March 2025 with Solutions

Q.1 (A) (i) If p : He is intelligent, q : He is strong. Then, symbolic form of statement "It is wrong that, he is intelligent or strong" is :

Q.1 (A) (ii) \(\int\left(x+\frac{1}{x}\right)^3dx=\)

Q.1 (A) (iii) \(\int_2^7\frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}\,dx=\)

Q.1 (A) (iv) The area of the region bounded by the curve \(y=x^2\) and the line \(y=4\) is

Q.1 (A) (v) The order and degree of the differential equation \(\left(\frac{d^2y}{dx^2}\right)^2+\left(\frac{dy}{dx}\right)^2=a^x\) are _____ respectively.

Q.1 (A) (vi) The integrating factor of the differential equation \(\frac{dy}{dx}+\frac{y}{x}=x^3-3\) is

Q.1 (B) (i) State whether true or false: If A is a matrix and K is a constant, then \((KA)^T=KA^T\).

Q.1 (B) (ii) State whether true or false: \(\int\log x\,dx=x\log x+x+c\).

Q.1 (B) (iii) State whether true or false: The differential equation obtained by eliminating arbitrary constants from \(bx+ay=ab\) is \(\frac{d^2y}{dx^2}=0\).

Q.1 (C) (i) The average revenue \(R_A\) is 50 and elasticity of demand \(\eta\) is 5, the marginal revenue \(R_M\) is _____.

Q.1 (C) (ii) \(\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx=\) _____ \(+c\)

Q.1 (C) (iii) If \(f'(x)=x^2+5\) and \(f(0)=-1\) then \(f(x)=\) _____.

Q.2 (A) (i) Write the converse, inverse and contrapositive of the statement "If a triangle is equilateral then it is equiangular".

Q.2 (A) (ii) Find x, y, z if \(\left\{5\begin{bmatrix}0 & 1\\ 1 & 0\\ 1 & 1\end{bmatrix}-\begin{bmatrix}2 & 1\\ 3 & -2\\ 1 & 3\end{bmatrix}\right\}\begin{bmatrix}2\\ 1\end{bmatrix}=\begin{bmatrix}x-1\\ y+1\\ 2z\end{bmatrix}\)

Q.2 (A) (iii) Evaluate : \(\int\frac{1}{x(x^6+1)}dx\)

Q.2 (B) (i) Solve the following equations by the method of inversion :

\(2x-y+z=1\)
\(x+2y+3z=8\)
\(3x+y-4z=1\)

Q.2 (B) (ii) Find MPC, MPS, APC and APS, if the expenditure \(E_c\) of a person with income I is given as : \(E_c=(0.0003)I^2+(0.075)I\); when \(I=1000\)

Q.2 (B) (iii) Evaluate : \(\int_1^2\frac{dx}{x^2+6x+5}\)

Q.3 (A) (i) Find \(\frac{dy}{dx}\) if \(y=(x)^x+(a)^x\)

Q.3 (A) (ii) Find the area of the region bounded by the parabola \(y^2=25x\) and the line \(x=5\).

Q.3 (A) (iii) Find the differential equation by eliminating arbitrary constants from the relation \(y=Ae^{3x}+Be^{-3x}\).

Q.3 (B) (i) Using the truth table, verify \(p\vee(q\wedge r)\equiv(p\vee q)\wedge(p\vee r)\)

Q.3 (B) (ii) If \(x=\frac{4t}{1+t^2}\), \(y=3\left(\frac{1-t^2}{1+t^2}\right)\), then show that \(\frac{dy}{dx}=\frac{-9x}{4y}\).

Q.3 (C) (i) Divide the number 84 into two parts such that the product of one part and square of the other is maximum.

Solution : Let one part be x then the other part will be \(84-x\).

\(\therefore f(x)=\square\)
\(\therefore f'(x)=168x-3x^2\)
For extreme values \(f'(x)=0\)
\(168x-3x^2=0\)
\(\therefore 3x(56-x)=0\)
\(\therefore x=\square\) OR \(\square\)
\(f''(x)=168-6x\)
If \(x=0,\ f''(0)=168-6(0)=168\gt0\)
\(\therefore\) function attains minimum at \(x=0\)
If \(x=56,\ f''(56)=\square\lt0\)
\(\therefore\) function attains maximum at \(x=56\)
\(\therefore\) Two parts of 84 are \(\square\) and \(\square\)

Q.3 (C) (ii) Solve the following differential equation \((x^2-yx^2)dy+(y^2+xy^2)dx=0\)

Solution : Separating the variables, the given equation can be written as :

\(\square\,dy+\square\,dx=0\)
\(\therefore\left(y^{-2}-\frac{1}{y}\right)dy+\left(x^{-2}+\frac{1}{x}\right)dx=0\)
\(\square\,dy-\frac{1}{y}dy+x^{-2}dx+\square\,dx=0\)
Integrating we get,
\(\int y^{-2}dy-\int\frac{1}{y}dy+\int x^{-2}dx+\int\frac{1}{x}dx=0\)
\(\therefore\frac{y^{-1}}{-1}-\square+\frac{x^{-1}}{-1}+\square=c\)
\(-\frac{1}{y}-\frac{1}{x}+\log x-\log y=c\)
\(\log x-\log y=\square+c\)
is the required solution.

Q.4 (A) (i) An agent who gives guarantee to his principal that the party will pay the sale price of goods is called –

Q.4 (A) (ii) In an ordinary annuity, payments or receipts occur at

Q.4 (A) (iii) Moving averages are useful in identifying

Q.4 (A) (iv) If \(P_{01}(L)=90\) and \(P_{01}(P)=40\), then \(P_{01}(D-B)\) is _____.

Q.4 (A) (v) The objective of an assignment problem is to assign

Q.4 (A) (vi) The expected value of the sum of two numbers obtained when two fair dice are rolled is _____.

Q.4 (B) (i) State whether true or false: If \(b_{yx}+b_{xy}=1.30\) and \(r=0.75\) then the given data is inconsistent.

Q.4 (B) (ii) State whether true or false: Cyclic variation can occur several times in a year.

Q.4 (B) (iii) State whether true or false: Cost of living index number is used in calculating purchasing power of money.

Q.4 (C) (i) The amount paid to the holder of the bill after deducting banker's discount is known as _____.

Q.4 (C) (ii) The simplest method of measuring trend of time series is _____.

Q.4 (C) (iii) Quantity index number by weighted aggregate method is given by _____.

Q.5 (A) (i) Compute the appropriate regression equation for the following data : X is the independent variable and Y is the dependent variable.

Q.5 (A) (ii) A company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be at least 5 kg. Cement costs ₹ 20 per kg and sand costs ₹ 6 per kg. Strength consideration dictate that a concrete brick should contain minimum 4 kg of cement and not more than 2 kg of sand. Formulate the L.P.P. for the cost to be minimum.

Q.5 (A) (iii) Find the mean of number of heads in three tosses of a fair coin.

Q.5 (B) (i) Obtain the trend value for the following data using 4-yearly centered moving averages :

Q.5 (B) (ii) Find the sequence that minimizes the total elapsed time to complete the following jobs in the order AB. Find the total elapsed time and idle time for machine B :

Q.5 (B) (iii) Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. Find the probability that :
(a) all the five cards are spades
(b) only 3 cards are spades.

Q.6 (A) (i) A house valued at ₹ 8,00,000 is insured at 75% of its value. If the rate of premium is 0.80%, find the premium paid by the owner of the house. If agent's commission is 9% of the premium, find agent's commission.

Q.6 (A) (ii) Solve the following L.P.P. by graphical method.

Maximize : \(z=4x+6y\)

Subject to : \(3x+2y\le12,\ x+y\ge4,\ x,y\ge0\)

Q.6 (A) (iii) Defects on plywood sheet occur at random with the average of one defect per 50 sq.ft. Find the probability that such a sheet has :
(a) no defect
(b) at least one defect
(use \(e^{-1}=0.3678\))

Q.6 (B) (i) The equations of two regression lines are \(10x-4y=80\) and \(10y-9x=-40\). Find
(a) \(\bar{x}\) and \(\bar{y}\)
(b) \(b_{YX}\) and \(b_{XY}\)
(c) r
(d) If var(Y) = 36, obtain var (X).

Q.6 (B) (ii) Find x if the cost of living index is 150 :

Q.6 (C) (i) A bill of ₹ 18,000 was discounted for ₹ 17,568 at a bank on 25th October 2017. If the rate of interest was 12% p.a. what is the legal due date?

Solution : Given SD = 18,000; CV = 17,568; r = 12% p.a.

Now, BD = \(\square\)
= 18,000 – 17,568
= ₹ 432
Also BD = \(\square\)
\(\therefore 432=\frac{18{,}000\times n\times12}{100}\)
\(n=\frac{432\times100}{18{,}000\times12}\)
\(n=\frac{1}{5}\) years = \(\square\) days

The period for which the discount is deducted is 73 days, which is counted from the date of discounting i.e. 25th October 2017 :

Hence legal due date is \(\square\)

Q.6 (C) (ii) Solve the following assignment problem for minimization :

Solution :

Step-I : Subtract the smallest element of each row from every element of that row

[Note: the Step-I matrix is transcribed exactly as printed. Its column V entries for rows 1 and 2 are misprinted as 4 and 3; after subtracting the row minima they should be 5 and 4. The Step-II matrix below is correct.]

Step-II : Subtract the smallest element of each column from every element of that column :

Step-III : Draw minimum number of lines covering all zeros. [Lines are drawn through rows 2, 3, 4 and column I.] Here minimum number of lines (4) \(\lt\) order of matrix (5).

Step-IV : The smallest uncovered element is 1, which is to be subtracted from all uncovered elements and add it to all elements which lie at the intersection of two lines :

Step-V : Draw minimum number of lines that are required to cover all zeros :

[Lines are drawn through rows 2, 4 and columns I, III.] Here minimum number of lines \(\ne\) order of matrix.

Step-VI : Find the smallest uncovered element (1). Subtract this number from all uncovered elements and add it to all elements which lie at the intersection of two lines :

[Lines are drawn through rows 1, 4, 5 and columns III, IV. The entry in row 4, column IV is printed as 0; by the working it should be 1 (it lies on only one line in Step-V, so it is unchanged).]

Now minimum number of lines = order of matrix. The optimal assignment can be made.

Optimal solution is \(1\to\) I, \(2\to\) IV, \(3\to\square\), \(4\to\square\), \(5\to\) V

Minimum value = \(\square\)