Maharashtra Board Class 12 Maths Sample Paper Set 7 with Solutions

Maharashtra State Board Class 12th Maths Sample Paper Set 7 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Maths Model Paper Set 7 with Solutions

Time : 3 Hours
Total Marks: 80

General Instructions:

1. All questions are compulsory.
2. There are 6 questions divided into two sections.
3. Write answers of Section I and Section II in the same answer book.
4. Use of logarithmic tables is allowed. Use of calculator is not allowed.
5. For LPP graph paper is not necessary. Only rough sketch of graph is expected.
6. Start answer to each question on a new page.
7. For each multiple choice type of question, it is mandatory to write the correct answer along with its alphabet eg. (a) ………. /(b) ……… /(c) ……../ (d) …….. No mark(s) shall be given if “ONLY” the correct answer or the alphabet of the correct answer is written. Only the first attempt will be considered for evaluation.

Section – I

i. Area of the region bounded by the curve x² = y, the X-axis and the lines x = 1 and x = 3 is ___.
(A) \(\frac{26}{3}\) sq. units
(B) \(\frac{3}{26}\) sq. units
(C) 26 sq. units
(D) 3 sq. units
Solution:
(A) \(\frac{26}{3}\) sq.units [1 Mark]

Explanation:
Required area = \(\int_1^3 y d x\)
= \(\int_1^3 x^2 d x\)
= \(\left[\frac{x^3}{3}\right]_1^3\)
= \(\frac{1}{3}\)(3³ – 1³)
= \(\frac{1}{3}\)(27 – 2)
= \(\frac{26}{3}\) sq. units

ii. \(\int \frac{\mathrm{d} x}{\left(x-x^2\right)}\) =
(A) log x – log (1 – x) + c
(B) log (1 – x²) + c
(C) -log x + log (1 – x) + c
(D) log(x – x²) + c
Solution:
(A) log x – log (1 – x) + c [1 Mark]

Explanation:

iii. The demand function is P = 25 + 4D – 2D². The marginal revenue function is
(A) R_m = 4 – 4D
(B) R_m = \(\frac{25}{\mathrm{D}}\) + 4 – 2D
(C) R_m = 25D + 4D² – 2D³
(D) R_m = 25 + 8D – 6D²
Solution:
(D) R_m = 25 + 8D – 6D² [1 Mark]

Explanation:
R_m = \(\frac{d}{d D}\)(p.D.) = \(\frac{d}{d D}\)(25D + 4D² – 2D³) = 25 + 8D – 6D²

iv. The order and degree of the differential equation \(\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}\) = a²(\(\frac{d^2 y}{d x^2}\))² are respectively
(A) 2.2
(B) 3, 2
(C) 2, 4
(D) 6, 4
Solution:
(C) 2, 4

Explanation:

v. If y = log\(\left(\frac{\mathrm{e}^x}{x^2}\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) = ?
(A) \(\frac{2-x}{x}\)
(B) \(\frac{x-2}{x}\)
(C) \(\frac{\mathrm{e}-x}{\mathrm{e} x}\)
(D) \(\frac{x-e}{e x}\)
Solution:
(B) \(\frac{x-2}{x}\) [1 Mark]

Explanation:
log = log\(\left(\frac{e^x}{x^2}\right)\) = log(e^x) – log(x²)
= x log e – 2 log x
= x(1) – 2 log x
∴ y = x – 2 log x
Differentiating both sides w.r.t.x, we get
\(\frac{\mathrm{dy}}{\mathrm{~d} x}\) = 1 – 2\(\left(\frac{1}{x}\right)\) = \(\frac{x-2}{x}\)

vi. \(\int_2^3 \frac{\mathrm{~d} x}{x^2-x}\) =
(A) log \(\left(\frac{2}{3}\right)\)
(B) log \(\left(\frac{1}{4}\right)\)
(C) log \(\left(\frac{4}{3}\right)\)
(D) log \(\left(\frac{8}{3}\right)\)
Solution:
(C) log \(\left(\frac{4}{3}\right)\) [1 Mark]

Explanation:
\(\int_2^3 \frac{d x}{x^2-x}\) = \(\int_2^3 \frac{d x}{x(x-1)}\) = \(\int_2^3\left[\frac{1}{x-1}-\frac{1}{x}\right] \mathrm{d} x\)
= \([\log (x-1)]_2^3\) – \([\log x]_2^3\)
= [log 2 – log 1] – [log 3 – log 2]
= log 2 – log 3 + log 2
= log \(\left(\frac{4}{3}\right)\)

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

i. The derivative of x^m.yⁿ = (x + y)^m+n is \(\frac{x}{y}\)
Solution:
False

Justification:

x^myⁿ = (x + y)^{m + n}
Taking log on both sides, we get
m log x+ n log y (m + n) log (x+ y)
∴ \(\frac{m}{n}+\frac{n}{y} \frac{d y}{d x}\) = \(\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right)\)
∴ \(\left(\frac{m}{n}-\frac{m+n}{x+y}\right)\) = \(\left(\frac{m+n}{x+y}-\frac{n}{y}\right) \frac{d y}{d x}\)
∴ \(\frac{\mathrm{dy}}{\mathrm{~d} x}\) = \(\frac{y}{x}\)

ii. The demand function is D = 50 – p. When the price is 10, the price elasticity of demand is 1.
Solution:
False

Justification:

Price elasticity of demand = \(\frac{-p}{D} \frac{d D}{d p}\) = \(\frac{-P}{50-p}(-1)\) = \(\frac{p}{50-p}\)
At p = 10, \(\frac{p}{50-p}\) = \(\frac{10}{40}\) = 0.25

iii. If y = a + \(\frac{\mathrm{a}}{x}\), then the differential equation is of the form x² \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) – x \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) – y = 0
Solution:
False

Justification:

y = a + \(\frac{a}{x}\)
∴ \(\frac{d y}{d x}\) = \(\frac{-a}{x^2}\)
a = -x²\(\frac{\mathrm{d} y}{\mathrm{~d} x}\)
∴ ⇒ y = -x²\(\frac{d y}{d x}\) – x\(\frac{d y}{d x}\)
i.e., x²\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) + x\(\frac{\mathrm{d} y}{\mathrm{~d} x}\) + y = 0

(C) Fill in the following blanks: (1 mark each)

i. If f(x) = \(\frac{7}{x}\) – 3, x ∈ R, x ≠ 0 then f “(x) is x _____
Solution:
14x^-3

Explanation:
f(x) = \(\frac{7}{x}\) – 3
∴ f'(x) = \(\frac{-7}{x^2}\)
∴ f”(x) = \(\frac{14}{x^3}\) = 14x^-3

ii. If y = x.log x, then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) = _____
Solution:
\(\frac{1}{x}\)

Explanation:
y = x.logx
Differentiating both sides w.r.t. x,we get
\(\frac{d y}{d x}\) = x. \(\frac{\mathrm{d}}{\mathrm{dx}}\)(loq x) + log x. \(\frac{d}{d x}\)(x)
= x.\(\frac{1}{x}\) + log x.(1)
∴ \(\frac{d y}{d x}\) = 1 + log x
Again differentiating both sides w.r.t. x, we get
\(\frac{d^2 y}{d x^2}\) = 0 + \(\frac{1}{x}\) = \(\frac{1}{x}\)

iii. Matrix B = \(\left[\begin{array}{ccc}
0 & 3 & 1 \\
-3 & 0 & -4 \\
p & 4 & 0
\end{array}\right]\) is skew symmetric, then the value of p is ____
Solution:
-1

Explanation:
Matrix B is skew symmetric.
∴ B = -B^T
∴ \(\left[\begin{array}{ccc}
0 & 3 & 1 \\
-3 & 0 & -4 \\
p & 4 & 0
\end{array}\right]\) = \(\left[\begin{array}{ccc}
0 & 3 & -p \\
-3 & 0 & -4 \\
-1 & 4 & 0
\end{array}\right]\)
∴ p = -1

Question 2.(A)
Attempt any TWO of the following: (3 marks each) [6]

i. If f ‘(x) = x² + 5 and f(0) = -1, then find the value of f(x).
Solution:
f'(x) = x² + 5 ……[Given]
f(x) = ∫f'(x) . dx
= ∫(x² + 5) . dx
= ∫x²dx + 5∫dx
∴ f (x) = \(\frac{x^3}{3}\) + 5x + c …….(i)
Now, f(0) = -1 ……[Given]
∴ \(\frac{(0)^3}{3}\) + 5(0) + c = -1
∴ c = -1
Substituting c = – 1 in (i), we get
f(x) = \(\frac{x^3}{3}\) + 5x – 1

ii. If the demand function is D = 50 – 3p – p². Find the elasticity of demand at p = 2. Comment on the
result.
Solution:

iii. Construct the matrix A = [a_jj]_3×3, where a_ij = i – j. State whether A is symmetric or skew- symmetric.
Solution:

(B) Attempt any TWO of the following: (4 marks each) [8]

i. Find the population of a city at any time t given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.
Solution:
Let ‘x’ be the population of the city at time ‘t’.
∴ \(\frac{\mathrm{d} x}{\mathrm{dt}}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k ¡s the constant of proportionality.
∴ \(\frac{\mathrm{d} x}{x}\) = k dt
Integrating on both sides, we get
\(\int \frac{d x}{x}\) = \(k \int \mathrm{~d} t\)
∴ log x = kt + c …… (i)
When t = 0, x = 30000
∴ log (30000) = k(0) + c
∴ c = log (30000)
∴ log x = kt + log (30000) ……(ii)[From (i)]
When t = 40, x = 40000
∴ log (40000) = k(40) + log (30000)
∴ log (40000) – log (30000) = 40k
∴ log \(\left(\frac{40000}{30000}\right)\) = 40k

ii. Determine the truth value of the following statements.

a. 4 + 5 = 7 or 9 – 2 = 5
Solution:
Let p : 4 + 5 = 7
q : 9 – 2 = 5
The truth valves of p and q are F and F respectively. The given statement in symbolic form is p ∨ q.
∴ p ∨ q ≡ F ∨ F ≡ F
∴ Truth value of the given statement is F.

b. If 9 > 1 then x² – 2x + 1 = 0 for x = 1
Solution:
Let p : 9 > 1
q : x² – 2x + 1 = 0 for x = 1
The truth values of p and q are T and T respectively.
The given statement in symbolic form is p → q.
∴ p → q ≡ T → T ≡ T
∴ Truth value of the given statement is T

c. x + y = 0 is the equation of a straight line if and only if y² = 4x is the equation of the parabola.
Solution:
Let p : x + y = 0 is the equation of a straight line.
q : y² = 4x is the equation of the parabola.
The truth values of p and q are T and T respectively.
The given statement in symbolic form is p ↔ q.
∴ p ↔ q ≡ ≡ T ↔ T ≡ T
∴ Truth value of the given statement is T

d. It is not true that 2 + 3 = 6 or 12 + 3 = 5
Solution:
Let p : 2 + 3 = 6
q : 12 + 3 = 5
The truth values of p and q are F andF respectively.
The given statement ¡n symbolic form is ~(p ∨ q).
∴ ~(p ∨ q) ≡ (F ∨ F) ≡ ~F ≡ T
∴ Truth value of the given statement is T

iii. Evaluate : \(\int_2^7 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}} \mathrm{dx}\)
Solution:

Question 3.
(A) Attempt any TWO of the following questions: (3 marks each)

i. Find the area of the region bounded by y = x², the X-axis and x = 1, x = 4.
Solution:

ii. Evaluate: \(\int \frac{1}{7+6 x-x^2} \mathrm{~d} x\)
Solution:

iii. Determine the maximum and minimum values of the given function.
f(x) = x² + \(\frac{16}{x}\)
Solution:
f(x) = x² + \(\frac{16}{x}\)
∴ f ‘(x) = 2x – \(\frac{16}{x^2}\) and f“(x) = 2 + \(\frac{32}{x^3}\)
Consider, f ‘(x) = 0
∴ 2x – \(\frac{16}{x^2}\) = 0
∴ 2x = \(\frac{16}{x^2}\)
∴ x³ = 8
∴ x = 2
For x = 2,
f”(2) = 2 + \(\frac{32}{2^3}\) = 2 + \(\frac{32}{8}\) = 2 + 4 = 6 > 0
∴ f(x) attains minimum value at x = 2.
∴ Minimum value = f(2) = (2)² + \(\frac{16}{(2)}\) = 4 + 8 = 12
∴ The function f(x) has minimum value 12 at x = 2.

(B) Attempt any ONE of the following:

i. If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
a. A^T + 4B^T
b. 5A^T – 5B^T
Solution:

ii. Solve: y² dx + (xy + x²)dy = 0
Solution:

(C) Attempt any ONE of the following questions (Activity):

i. If p ↔ q and p → q both are true, then find the truth values of the following with the help of activity.
a. p ∨ q
b. p ∧ q
p ↔ q and p → q are true if p and q have truth values T, T or F, F.
p ∨ q
If both p and q are true, then p ∨ q ≡ ___
p ∧ q
If both p and q are false, then p ∧ q ≡ ____
If both p and q are true, then p ∧ q ≡ ____
Solution:
p ↔ q and p → q are true if p and q have truth values T, T or F, F.
p ∨ q
If both p and q are true, then p ∨ q ≡ T ∨ T ≡ T
If both p and q are false, then p ∨ q ≡ F ∨ F ≡ F
p ∧ q
If both p and q are true, then p ∧ q ≡ T ∧ T ≡ T
If both p and q are false, then p ∧ q ≡ F ∧ F = F

ii. Complete the following activity to find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\), if y = x^{(log x} + 10^x.
Solution:

Let y = x^{(log x)} + 10^x
Let u = x^{log x}, v = 10^x
∴ y = u + v ………. (i)
u = x^{log x}
Taking logarithm of both sides, we get
log u = log x^{log x}
∴ log u = log x. log x
∴ log u = (log x)²
Differentiating w.r.t. x, we get

Section – II

Question 4. (A)
Select and write the correct answer of the following multiple choice type of questions (1 mark each)

i. An agent who is given the possession of goods to be sold is known as
(A) Factor
(B) Broker
(C) Auctioneer
(D) Del Credere Agent
Solution:
(A) Factor

ii. Time-series analysis is based on the assumption that ____.
(A) random error terms are normally distributed.
(B) the variable to be forecast and other independent variables are correlated.
(C) past patterns in the variable to be forecast will continue unchanged into the future.
(D) the data do not exhibit a trend.
Solution:
(C) past patterns in the variable to be forecast will continue unchanged into the future.

iii. The probability distribution of a discrete random variable X is given below:

Then k =
(A) 8
(B) 32
(C) 16
(D) 48
Solution:
(B) 32

Explanation:

As \(\frac{5}{k}\) + \(\frac{7}{k}\) + \(\frac{9}{k}\) + \(\frac{11}{k}\) = 1, we get
\(\frac{32}{k}\) = 1 ⇒ k = 32

iv. The cost of Living Index Number using Aggregate Expenditure Method is given by

Solution:
(A) \(\frac{\sum \mathrm{p}_1 q_0}{\sum \mathrm{p}_0 q_0}\) × 100

v. The half plane represented by 4x + 3y ≥ 24 contains the point
(A) (0, 0)
(B) (2, 2)
(C) (3, 4)
(D) (1, 1)
Solution:
(C) (3, 4)

Explanation:

Only (3, 4) satisfies the given inequality.

vi. Moving averages are useful in identifying ____.
(A) seasonal component
(B) irregular component
(C) trend component
(D) cyclical component
Solution:
(C) trend component

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

i. The difference between the banker’s discount and true discount is called sum due.
Solution:
False

ii. The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.
Solution:
True

iii. The optimum value of the objective function of LPP occurs at the center of the feasible region.
Solution:
False

(C) Fill in the blanks: (1 mark each) [3]

i. Trade discount is allowed on the ____ price.
Solution:
Catalogue/list

ii. The simplest method of measuring trend of time series is ____
Solution:
graphical

iii. In an assignment problem, a solution having ________ total cost is an optimum solution.
Solution:
zero

Question 5.
(A) Attempt any TWO of the following question: (3 marks each) [6]

i. The time (in hours) required to perform the printing and binding operations (in that order) for each book is given in the following table:

Book	I	II	III	IV	V
Printing Machine M1	3	7	4	5	7
Printing Machine M2	6	2	7	3	4

Find the sequence that minimizes the total elapsed time (in hours) to complete the work.
Also find the minimum elapsed time T and idle times for two machines.
Solution:

Here, Min {M11, Mi2) = 2,
which corresponds to book II of machine 2
∴ Book II is processed last
i.e.,

Now, the remaining books are I, III, IV and V
Here, Min {M_i1, M_i2} = 3,
which corresponds to book I of machine 1 and book IV of machine 2
∴ book I is processed first and book IV is processed fourth before book II
i.e.,

again, the remaining books are III and V
Here, Min {M_i1, M_i2} = 4,
which corresponds to book III of machine 1 and book V of machine 2
∴ book III is processed second after book I and book V is processed third before book IV
i.e.,

this the optimal sequence for completing a task.
The minimum elapsed time T (i.e., total processing time) can be computed as follows:

∴ From the table,
minimum elapsed (processing) time T = 28 hrs.
idle time for machine M₂ = 6 hrs.
idle time for machine M₁ = T – (sum of the processing time of the five books on M₁)
= 28 – 26
= 2 hrs.

ii. Calculate Dorbish-Bowley’s Price Index Number for the following data.

Solution:

iii. Following is the probability distribution of a r.v.X.

Find the probability that

a. X is positive.
Solution:
P(X is positive) = P(X = 1 or X = 2 or X = 3)
= P(X = 1) + P(X = 2 ) + P(X = 3)
= 0.25 + 0.15 + 0.10 = 0.50

b. X is non-negative
Solution:
P(X is non-negative) = P(X = 0 or X = 1 or X = 2 or X = 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.20 + 0. 25 + 0.15 + 0.10 = 0.70

c. X is odd.
Solution:
P(X is odd) = P(X = -3 or X = -1 or X = 1 or X = 3)
= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(B) Attempt any TWO of the following questions : (4 marks each) [8]

i. The two regression equations are 5x – 6y + 90 = 0 and 15x – 8y – 130 = 0. Find \(\bar{x}\), \(\bar{y}\), r.
Solution:
Given, the two regression equations are
5x – 6y + 90 = 0
i.e., 5x – 6y = -90 ……(i)
and 15x – 8y – 130 = 0
i.e., 15x – 8y = 130 …….(ii)
By (i) × 3 – (ii), we get

Substituting y = 40 in (i), we get
5x – 6(40) = -90
∴ 5x – 240 = -90
∴ 5x = -90 + 240 = 150
∴ x = 30
Since the point of intersection of two regression lines is (\(\bar{x}\), \(\bar{y}\)),
∴ \(\bar{x}\) = 30 and \(\bar{y}\) = 40
Now let 5x – 6y + 90 = 0 be the regression equation of Y on X
The equation becomes 6Y = 5X + 90
i.e., Y = \(\frac{5}{6}\)X + \(\frac{90}{6}\)
Comparing it with Y = b_YX X + a, we get
∴ b_YX = \(\frac{5}{6}\)
Now, other equation 15x – 8y – 130 – 0 be the regression equation of X on Y.
∴ The equation becomes 15X = 8Y + 130
i.e., X = \(\frac{8}{6}\)Y + \(\frac{130}{15}\)
Comparing it with X = b_XY Y + a’, we get
∴ b_XY = \(\frac{8}{15}\)
∴ r = ±\(\sqrt{b_{x y} \cdot b_{y x}}\) = ±\(\sqrt{\frac{8}{15} \cdot \frac{5}{6}}\) = ±\(\sqrt{\frac{4}{9}}\) = ±\(\frac{2}{3}\)
Since b_YX and b_XY both are positive,
r is positive.
∴ r = \(\frac{2}{3}\)

ii. Five different machines can do any of the five required jobs, with different profits resulting from each assignment as shown below:

Find the optimal assignment schedule.
Solution:
Step 1:
Since it is a maximization problem, subtract each of the elements in the table from the largest element, i.e., 62

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
Subtract the smallest element in each column of assignment matrix obtained in step 2 from every element in its column.

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 zeros.

Step 5:
From step 4, minimum number of lines covering all the zeros are 4, which is less than order of matrix, i.e., 5.
∴ Select smallest element from all the uncovered elements, i.e., 4 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.

Step 6:
Draw minimum number of vertical and horizontal lines to cover all zeros.

Step 7:
From step 6, minimum number of lines covering all the zeros are 5, which is equal to order of the matrix, i.e., 5.
∴ 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 8:
The matrix obtained in step 7 contains exactly one assignment for each row and column.
∴ Optimal assignment schedule is as follows:

∴ Total maximum profit = 40 + 36 + 40 + 36 + 62
= ₹ 214

iii. A property valued at ₹ 7,00,000 is insured to the extent of ₹ 5,60,000 at \(\left(\frac{5}{8}\right)^{\text {th }}\)% less 20%. Calculate the saving made in the premium. Find the amount of loss that the owner must hear.
including premium, if the property is damaged to the extent of 40% of its value.
Solution:
Given, Property value = ₹ 7,00,000,
Policy value = ₹ 5,60,000,
Rate of premium = \(\frac{5}{8}\) % less 20%.
Now, for the rate of premium = \(\frac{5}{8}\)%
Amount of premium = \(\frac{5}{8}\)% of policy value
= \(\frac{5}{8}\) × \(\frac{1}{100}\) × 5,60,000
= ₹ 3,500
For the rate of premium = \(\frac{5}{8}\) % less 20%
= (\(\frac{5}{8}\) – 20% of \(\frac{5}{8}\))
= (\(\frac{5}{8}\) – \(\frac{20}{100}\) × \(\frac{5}{8}\))
= (\(\frac{5}{8}\) – \(\frac{1}{8}\))% = \(\frac{4}{8}\)% = \(\frac{1}{2}\)%
Amount of premium = \(\frac{1}{2}\)% of policy value
= \(\frac{1}{2}\) × \(\frac{1}{100}\) × 5,60,000
= ₹ 2,800
∴ Savings made in the premium – premium on \(\frac{5}{8}\)% – premium on \(\frac{1}{2}\)%
= 3,500 – 2,800 = ₹ 700
∴ Savings made in the premium is ₹ 700.
Now, the property is damaged to the extent of 40% of its value.
∴ Value of damaged property (loss) = 40% of property value
= \(\frac{40}{100}\) × 7,00,000 = ₹ 2,80,000
Since, Claim \(=\frac{\text { Policy value }}{\text { Property value }}\) × Loss = \(\frac{5,60,000}{7,00,000}\) × 2,80,000
∴ Claim = ₹ 2,24,000
∴ Loss borne by the owner including premuim = (Loss – Claim) + Premium on \(\frac{1}{2}\)%
= (2,80,000 – 2,24,000) + 2,800
= 56,000 + 2,800
= ₹ 58,800
∴ Loss borne by the owner including the premium is ₹ 58,800.

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

i. After deducting commission at 7\(\frac{1}{2}\)% on first ₹ 50,000 and 5% on balance of sales made by
him, an agent remits ₹ 93,750 to his principal. Find the value of goods sold by him.
Solution:
Agent earns commission at 7\(\frac{1}{2}\) % i.e. 7.5% on first ₹ 50,000 sales and on balance, at 5%.
Let total sales be ‘x’.
Commission earned on first ₹ 50,000 at 7.5% = \(\frac{7.5}{100}\) × 50,000 = ₹ 3,750
Commission earned on balance at 5% = (x – 50,000) × \(\frac{5}{100}\)
Total commission earned = Total sales – Amount remitted
= x – 93,750
∴ (x – 93,750) = 3,750 + \(\frac{5}{100}\) × (x – 50,000)
∴ (x – 93,750) × 100 = 3,750 × 100 + 5(x – 50,000)
∴ 100x – 93,75,000 = 3,75,000 + 5x – 2,50,000
∴ 100x – 5x = 1,25,000 + 93,75,000
∴ 95x = 95,00,000
∴ x = \(\frac{95,00,000}{95}\)
∴ x = 1,00,000
∴ The value of goods sold by the agent is ₹ 1,00,000.

ii. Calculate Marshall-Edgeworth’s Price Index Number for the following data.

Solution:

iii.

Solution:
P(X ≤ 2) = P(X = 0 or X = 1 or X = 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{{ }^5 C_0}{2^5}\) + \(\frac{{ }^5 C_1}{2^5}\) + \(\frac{{ }^5 C_2}{2^5}\)
= \(\frac{{ }^5 C_0+{ }^5 C_1+{ }^5 C_2}{2^5}\)
= \(\frac{1+5+10}{2^5}\)
= \(\frac{16}{32}\) = \(\frac{1}{2}\) …… (i)
P(X ≥ 3) = P (X = 3 or X = 4 or X = 5)
= P(X = 3) + P(X = 4) + P(X = 5)
= \(\frac{{ }^5 C_3}{2^5}\) + \(\frac{{ }^5 C_4}{2^5}\) + \(\frac{{ }^5 C_5}{2^5}\)
= \(\frac{10+5+1}{2^5}\)
= \(\frac{16}{32}\)
= \(\frac{1}{2}\) ….. (ii)
From (i) and (ii), we get
P(X ≤ 2) = P(X ≥ 3)

(B) Attempt any ONE of the following questions:

i. A person makes two types of gift items A and B requiring the services of a cutter and a finisher. Gift item A requires 4 hours of cutter’s time and 2 hours of finisher’s time. B requires 2 hours of cutters time, 4 hours of finishers time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75 and on gift item B is ₹ 125. Assuming that the person can sell all the items produced, determine how many gift items of each type should be make every month to obtain the best returns?
Solution:
Let x gift items of type A and y gift items of type B be produced by the person.
∴ Total profit Z = 75x + 125y
This is the objective function to be maximized.
The given information can be tabulated as shown below:

∴ The constraints are 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
∴ Given problem can be formulated as
Maximize Z = 75x + 125y
Subject to, 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
To draw feasible region, construct table as follows:

Shaded portion OABC is the feasible region, whose vertices are O ≡ (0, 0), A ≡ (52, 0), B and C ≡ (0, 38).
B is the point of intersection of the lines 4x + 2y = 208
i.e. 2x + y = 104 and 2x + 4y = 152

Solving the above equations,
we get
B ≡ (44, 16)
Here, the objective function is
Z = 75x + 125y
∴ Z at O(0, 0) = 75(0) + 125(0) = 0
Z at A(52, 0) = 75(52) + 125(0) = 3900
Z at B(44, 16) = 75(44) + 125(16) = 5300
Z at C(0, 38) = 75(0) + 125(38) = 4750
∴ Z is maximum, when x = 44, y = 16
Thus, a person should make 44 gift items of type A and 16 gift items of type B every month to obtain the best returns of ₹ 5300.

ii. Fit a trend line to the given data by the method of least squares.

Solution:
In the given problem, n = 8 (even), two middle t – values are 1980 and 1981, h = 1
u \(=\frac{t-\text { mean of two middle values }}{\frac{h}{2}}\) = \(\frac{t-1980.5}{\frac{1}{2}}\) = 2(t – 1980.5)
We obtain the following table.

From (i), a’ = \(\frac{39}{8}\) = 4.875
From (ii), b’ = \(\frac{79}{168}\) = 0.4702
∴ The equation of the trend line is y_t = a’ + b’u
i.e., y_t = 4.875 + 0.4702 u, where u = 2(t – 1980.5)

(C) Attempt any ONE of the following questions (Activity):

i. It is observed that it rains on 10 days out of 30 days. Complete the following activity to find the probability that it rains on at most 2 days of a week.
Solution:
Let X denote the number of days it rains in a week.
P(it rains) = p = ____
∴ q = ____
Here, n = ____
∴ X ~ B(n, p)
The p.m.f. of X is given by
P(X = x) = ⁿC_xp^xq^n-x
P(it rains at most 2 days of a week) = P(X ≤ 2)
= ____
Solution:
Let X denote the number of days it rains in a week.
P(it rains) = p = \(\frac{10}{30}\) = \(\frac{1}{3}\)
Here, n = 7
∴ X ~ B(n, p)
The p.m.f. of X is given by
P(X = x) = ⁿC_x p^x q^{n – x}

ii.
Complete the following activity:

Solution:

Maharashtra Board Class 12 Maths Previous Year Question Papers