The availability of step-by-step ML Aggarwal Class 12 ISC Solutions Case Study Based Questions Section C can make challenging problems more manageable.

## ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section C

Chapter 1 Application of Calculus in Commerce and Economics

Case-Study Based Questions (Solved)

Case – Study 1 :

A firm has the cost function C(x) = \(\frac{x^3}{3}\) – 7x^{2} + 111x + 50 and demand function x = 100 – p.

.

Based on the above Information, answer the following questions:

(i) The total revenue function is

(a) R(x) = x^{2} – 100x

(b) R(x) = 100x – x^{2}

(c) R(x) = 100 – x

(d) none of these

Solution:

(b) R(x) = 100x – x^{2}

Given C(x) = \(\frac{x^3}{3}\) – 7x^{2} + 111x + 50

and demand function x = 100 – p

⇒ p = 100 – x

∴ Revenue function = R (x)

= px

= (100 – x) x

(ii) The total profit function is

(a) – \(\frac{x^3}{3}\) + 6x^{2} – 11x – 50

(b) \(\frac{x^3}{3}\) – 6x^{2} – 11x + 50

(c) \(\frac{x^3}{3}\) + 6x^{2} – 11x + 50

(d) – \(\frac{x^3}{3}\) – 6x^{2} + 11x + 50

Solution:

(a) – \(\frac{x^3}{3}\) + 6x^{2} – 11x – 50

∴ Total profit function = P (x)

= R (x) – C (x)

= (100 x – x^{2}) – {\(\frac{x^3}{3}\) – 7x^{2} + 111x + 50}

= – \(\frac{x^3}{3}\) + 6x^{2} – 11x – 50

(iii) The value of x for which profit is maximum is

(a) 8

(b) 9

(c) 10

(d) 11

Solution:

(d) 11

\(\frac{d}{d x}\) P(x) = – x^{2} + 12x – 11

and \(\frac{d^2}{d x^2}\) P(x) = – 2x + 12

For maxima / minima,

\(\frac{d}{d x}\) P(x) = 0

⇒ – x^{2} + 12x – 11 = 0

⇒ x^{2} – 12x + 11 = 0

⇒ x = 1, 11

at x = 11 ;

\(\frac{d^2}{d x^2}\) P(x) = – 22 + 12 = – 10 < 0

Thus P(x) is maximum when x = 11.

(iv) The maximum profit is

(a) ₹ 131.11

(b) ₹ 113.31

(c) ₹ 111.33

(d) ₹ 133.11

Solution:

(c) ₹ 111.33

∴ Maximum profit = P (11)

= \(-\frac{11^3}{3}\)

= \(-\frac{1331}{3}\)

= \(\frac{-1331+2178-363-150}{3}\)

= \(\frac{334}{3}\)

= ₹ 111.33

Case – Study 2 :

The average cost function associated with producing and marketing x units of an item is given by

AC = 2x – 11 + \(\frac{50}{x}\).

Based on the above information, answer the following questions:

(i) The total cost function is

(a) C(x) = 2x^{2} – 11x + 50

(b) C(x) = 2 – \(\frac{50}{x^2}\)

(c) C(x) = 2x – \(\frac{50}{x}\)

(d) C(x) = 2x^{2} + 11x – 50

Solution:

(a) C(x) = 2x^{2} – 11x + 50

Given Average cost function = AC

= 2x – 11 + \(\frac{50}{x}\)

∴ total cost function = x AC (x)

= x (2x – 11 + \(\frac{50}{x}\))

= 2x^{2} – 11x + 50

(ii) The marginal cost function is

(a) MC = 4x + 11

(b) MC = 4x – 11

(c) MC = 2 + \(\frac{50}{x^2}\)

(d) MC = \(\frac{100}{x^3}\)

Solution:

(b) MC = 4x – 11

Marginal cost function MC = \(\frac{d}{d x}\) C(x)

= \(\frac{d}{d x}\) [2x^{2} – 11x + 50]

= 4x – 11

(iii) The marginal cost when x = 4 units is

(a) 5

(b) 27

(c) 18

(d) 13

Solution:

(a) 5

When x = 4 ;

MC = 4 × 4 – 11 = ₹ 5

(iv) The range of values of x for which AC ¡s increasing is

(a) x < 5 (b) x > – 5

(c) x < – 5 (d) x > 5

Solution:

(d) x > 5

Now \(\frac{d}{d x}\) AC = 2 – \(\frac{50}{x^2}\)

Now \(\frac{d}{d x}\) AC > 2

⇒ 2 – \(\frac{50}{x^2}\) > 0

⇒ \(\frac{50}{x^2}\) < 2

⇒ x^{2} > 25

⇒ |x| > 5

⇒ x > 5 or x < – 5 (but x > 0)

⇒ x > 5

Case – Study 3:

The marginal cost (in ₹) of a product ¡s given by MC = \(\frac{300}{\sqrt{3 x+25}}\) and the fixed cost is ₹ 5000.

Based on the above information, answer the following questions :

(i) The cost function is

(a) C(x) = 600 \(\sqrt{3 x+25}\) – 100

(b) C(x) = 200 \(\sqrt{3 x+25}\) + 5000

(c) C(x) = 200 \(\sqrt{3 x+25}\) + 4000

(d) C(x) = 200 \(\sqrt{3 x+25}\) – 4000

Solution:

Given MC = \(\frac{300}{\sqrt{3 x+25}}\)

∴ C(x) = ∫ MC(x) dx

= ∫ \(\frac{300}{\sqrt{3 x+25}}\) dx

= \(\frac{300(3 x+25)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right) 3}\) + k

C(x) = 200 \(\sqrt{3 x+25}\) + k ………………(1)

When x = 0,

fixed cost i.e. C (0) = 5000

∴ from(1) ; we have

5000 = 200 × 5 + k

⇒ k = 4000

∴ from (1) ;

C(x) = 200 \(\sqrt{3 x+25}\) + 4000

(ii) The cost of producing 25 units of the product is

(a) ₹ 7000

(b) ₹ 6000

(c) ₹ 8000

(d) ₹ 6500

Solution:

(b) ₹ 6000

∴ required cost of producing 25 units = C(25)

= ₹ 200 \(\sqrt{3 \times 25+25}\) + 4000

= ₹ [200 × 10 + 4000]

= ₹ 6000

(iii) The average cost function is

(a) AC = \(\frac{200 \sqrt{3 x+25}}{x}+\frac{4000}{x}\)

(b) AC = \(\frac{200 \sqrt{3 x+25}}{x}+\frac{5000}{x}\)

(c) AC = \(\frac{100 \sqrt{3 x+25}}{x}+\frac{4000}{x}\)

(d) AC = \(\frac{200 \sqrt{3 x+25}}{x}-\frac{4000}{x}\)

Solution:

(a) AC = \(\frac{200 \sqrt{3 x+25}}{x}+\frac{4000}{x}\)

∴ Average cost function = \(\frac{C(x)}{x}\)

= \(\frac{200 \sqrt{3 x+25}+4000}{x}\)

(iv) The average cost of producing 200 units is

(a) ₹ 5

(b) ₹ 25

(c) ₹ 45

(d) ₹ 50

Solution:

(c) ₹ 45

∴ Required average cost of producing 200 units = AC (200)

= \(\frac{200 \sqrt{3 \times 200+25}+4000}{200}\)

= \(\frac{200 \times 25+4000}{200}\)

= \(\frac{9000}{200}\) = ₹ 45

Chapter 2 Linear Regression

Case – Study Based Questions (Solved)

Case – Study 1 :

For five observations of pairs (x, y) of correlated variables, the following results were obtained :

Σx = 15,

Σy = 18,

Σx^{2} = 55,

Σy^{2} = 74

and Σxy = 58.

Based on the above Information, answer the following questions:

(i) The regression coefficient of x on y i.e. b_{xy} is

(a) \(\frac{2}{5}\)

(b) \(\frac{10}{23}\)

(c) \(-\frac{2}{5}\)

(d) \(-\frac{10}{23}\)

Solution:

(b) \(\frac{10}{23}\)

(ii) The equation of the line of regression of y on x is

(a) 2x – 5y + 12 = 0

(b) 23x – 10y – 13 = 0

(c) 2x + 23y – 113 = 0

(d) 2x – 5y – 12 = 0

Solution:

(a) 2x – 5y + 12 = 0

\(\bar{x}=\frac{\Sigma x}{n}\)

= \(\frac{15}{5}\) = 3 ;

\(\bar{y}=\frac{\Sigma y}{n}\)

= \(\frac{18}{5}\) = 3.6

b_{yx} = \(\frac{n \Sigma x y-\Sigma x \Sigma y}{n \Sigma x^2-(\Sigma x)^2}\)

= \(\frac{5 \times 58-15 \times 18}{5 \times 55-15^2}\)

= \(\frac{20}{50}=\frac{2}{5}\)

Thus repression line ofy on x be given by

⇒ y – \(\vec{y}\) = b (x – \(\vec{x}\))

⇒ \(\left(y-\frac{18}{5}\right)=\frac{2}{5}(x-3)\)

⇒ 5y – 18 = 2x – 6

⇒ 2x – 5y + 12 = 0

(iii) The most likely value of y when x = 8 is

(a) 6.4

(b) 6.5

(c) 5.6

(d) 5.8

Solution:

(c) 5.6

from (1) ;

5y = 2x + 12

⇒ y = \(\frac{1}{5}\) (2x + 12)

∴ y(8) = \(\frac{1}{5}\) (2 × 8 + 12)

= \(\frac{28}{5}\) = 5.6

(iv) The coefficient of correlation between x and y is

(a) 3.41

(b) 0.174

(c) – 0.417

(d) 0.417

Solution:

(d) 0.417

Since b_{xy} = \(\frac{10}{3}\) > 0

and b_{yx} = \(\frac{2}{5}\) > 0

and r = + \(\sqrt{b_{x y} \cdot b_{y x}}\)

= \(\sqrt{\frac{10}{23} \times \frac{2}{5}}\)

= \(\frac{2}{\sqrt{23}}\)

= 0.4170

Case – Study 2 :

For a given bivariate data, the two regression lines x + y – 8 = 0 and 4x + 9y – 57 = 0, and the standard deviation of y is 2.

Based on the above information, answer the following questions :

(i) The line of regression of y on x is

(a) x + y – 80

(b) 4x + 9y – 57 = 0

(c) x + y + 80

(d) 4x + 9y + 57 = 0

Solution:

Let us assume x + y – 8 = 0 be the regression line of y on x.

∴ y = – x + 8

∴ b_{yx} = – 1

Then 4x + 9y – 57 = 0 be a regression line of x on y.

⇒ x = \(-\frac{9}{4} y+\frac{57}{4}\)

∴ b_{xy} = – \(\frac{9}{4}\)

Here b_{xy} . b_{yx} = (- 1) (- \(\frac{9}{4}\))

= \(\frac{9}{4}\) > 1

which is not possible.

Thus our assumption is wrong.

Hence line of regression of y on x be 4x + 9y – 57 = 0

and line of regression of x on y be x + y – 8 = 0

(ii) The coefficient of correlation between x and y is

(a) \(\frac{1}{3}\)

(b) – \(\frac{1}{2}\)

(c) \(\frac{2}{3}\)

(d) \(-\frac{2}{3}\)

Solution:

(d) \(-\frac{2}{3}\)

The regression line of y on x be 4x + 9y – 57 = 0

⇒ 9y = – 4x + 57

⇒ y = \(-\frac{4}{9} x+\frac{57}{9}\)

b_{xy} = – \(\frac{4}{9}\) < 0

and regression line of x on y be x + y – 8 = 0

⇒ x = – y + 8

∴ b_{xy} = – 1 < 0

∴ r = – \(\sqrt{b_{x y} b_{y x}}\)

= \(-\sqrt{-\frac{4}{9} \times(-1)}\)

= \(-\frac{2}{3}\)

[∵ b_{xy}, b_{yx} , 0

⇒ r < 0]

(iii) The variance of x is

(a) 4

(b) 3

(c) 9

(d) 6

Solution:

(c) 9

Given σ_{y} = 2

We know that

b_{xy} = r \(\frac{\sigma_x}{\sigma_y}\)

– 1 = \(-\frac{2}{3} \times \frac{\sigma_x}{2}\)

⇒ σ_{x} = 3

⇒ Var (x) = σ_{x}^{2} = 9

(iv) Covariance between x and y is

(a) – 4

(b) 4

(c) – 9

(d) 9

Solution:

(a) – 4

Cov (x, y) = b_{yx} σ_{x}^{2}

= – \(\frac{4}{9}\) × 9 = – 4

= \(\sqrt{\frac{10}{23} \times \frac{2}{5}}\)

= \(\frac{2}{\sqrt{23}}\)

= 0.4170

Chapter 3 Linear Programming

Case-Study Based Questions (Solved)

Case – Study 1 :

The feasible region for an L.P.P. is shown in the adjoining figure. The line CB is parallel to OA.

Based on the above information, answer the following questions :

(i) The equation of the line OA is

(a) x – 2y = 0

(b) y – 2x = 0

(c) x + 2y = 0

(d) 2x + y = 0

Solution:

(b) y – 2x = 0

The eqn. of line OA be given by

y – 0 = \(\frac{12-0}{6-0}\) (x – 0)

⇒ y = 2x

(ii) The equation of the line BC is

(a) y – 2x = 4

(b) x – 2y = 4

(c) y + 2x = 4

(d) x + 2y = 4

Solution:

(a) y – 2x = 4

The eqn. of line CB be given by

y – 4 = \(\frac{16-4}{6-0}\) (x – 0)

⇒ y – 4 = 2x

⇒ y = 2x + 4

(iii) The constraints for the L.P.P. re

(a) y ≥ 2x, y – 2x ≤ 4, x ≤ 6, x ≥ 0, y ≥ 0

(b) y ≤ 2x, y – 2x ≤ 4, x ≤ 6, x ≥ 0, y ≥ 0

(c) y ≥ 2x, y – 2x ≥ 4, x ≤ 6, x ≥ 0, y ≥ 0

(d) y ≤ 2x, y – 2x ≥ 4, x ≤ 6, x ≥ 0, y ≥ 0

Solution:

(a) y ≥ 2x, y – 2x ≤ 4, x ≤ 6, x ≥ 0, y ≥ 0

Clearly (0, 0) satisfies y – 2x ≤ 4 and (0, 0) satisfies y ≥ 2x.

Also (0, 0) satisfies x ≤ 6.

Thus the associated constraints for given L.P.P is as under:

y – 2x ≤ 4 ;

y ≥ 2x ;

x ≤ 6 ;

x ≥ 0, y ≥ 0

Point B be the point of intersection of lines x = 6 and y – 2x = 4

∴ Coordinates of B are (6, 16).

(iv) The minimum value of the objective function Z = 3x – 4y is

(a) 0

(b) – 16

(c) – 30

(d) – 46

Solution:

(d) – 46

The bounded shaded region OCBAO represents the feasible region with corner points

O (0, 0) ; C (0, 4) ; B (6, 16) and A (6, 12).

Clearly Z_{min} = – 46

and obtained at B (6, 16)

i.e. x = 6

and y = 16.