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## ML Aggarwal Class 12 Maths Solutions Section C Chapter 1 Application of Calculus in Commerce and Economics MCQs

Choose the correct answer from the given four options in questions (1 to 14) :

Question 1.

For manufacturing a certain item, the fixed cost is ₹ 9000 and the variable cost of producing each unit is ₹ 30. The average cost of producing 60 units is

(a) 150

(b) 180

(c) 240

(d) 120

Solution:

(b) 180

fixed cost = ₹ 9000

∴ variable cost of x items @ ₹ 30 = ₹ 30x

∴ Total cost function C (x) = 30x + 9000

Thus, average cost function = \(\frac{C(x)}{x}\)

= \(\frac{30 x+9000}{x}\)

= 30 + \(\frac{9000}{x}\)

∴ Average cost of producing 60 items = 30 + \(\frac{9000}{60}\) = ₹ 180

Question 2.

If the cost function of a certain commodity is C(x) = 2000 + 50x – \(\frac{1}{5}\) x^{2}, then the average cost of producing 5 units is

(a) ₹ 45

(b) ₹ 450

(c) ₹ 449

(d) ₹ 2245

Solution:

(c) ₹ 449

Given C(x) = 2000 + 50x – \(\frac{x^2}{5}\)

∴ Average cost function = AC (x)

= \(\frac{C(x)}{x}\)

= \(\frac{2000}{x}+50-\frac{x}{5}\)

∴AC (5) = \(\frac{2000}{5}+50-\frac{5}{5}\)

= 400 + 50 – 1 = ₹ 449

Question 3.

If the demand function is x = \(\frac{24-2 p}{3}\), where x is the number of units produced and p is the price per unit, then the revenue function R (x) is

(a) R(x) = 12x – \(\frac{3}{2}\) x^{2}

(b) R(x) = 8x – \(\frac{2}{3}\) x^{2}

(c) R(x) = \(\frac{2}{3}\) x^{2} – 12x

(d) R(x) = 24x + \(\frac{3}{2}\) x^{2}

Solution:

(a) R(x) = 12x – \(\frac{3}{2}\) x^{2}

Given demand function x = \(\frac{24-2 p}{3}\)

⇒ 3x = 24 – 2p

⇒ p = \(\frac{24-3 x}{2}\)

∴ Revenue function = R (x) = px

= x [12 – \(\frac{3}{2}\) x]

= 12x – \(\frac{3}{2}\) x^{2}

Question 4.

If the selling price of a commodity is fxed at ₹ 45 and the cost function is C (x) = 30x + 240, then the breakeven point is

(a) x = 10

(b) x = 12

(c) x = 15

(d) x = 16

Solution:

(d) x = 16

S.P of x commodities @ ₹ 45 = ₹ 45x

and C(x) = 30x + 240

∴ profit function P (x) = S.P – C (x)

= 45x – (30x + 240)

= 15x – 240

At breakeven point, P (x) = 0

⇒ 15x – 240 = 0

⇒ x = \(\frac{240}{15}\) = 16

Question 5.

The fixed cost of a product is ₹ 1800 and the variable cost per unit is ₹ 55. If the demand function is p (x) = 400 – 15x, then the breakeven values are at

(a) x = 10, 15

(b) x = 8, 15

(c) x = 8, 20

(d) x = 12, 15

Solution:

(b) x = 8, 15

Given fixed cost = ₹ 1800

variable cost = ₹ 55x

∴ Total cost function C (x) = ₹ (1800 + 55x)

and R(x) = xp(x)

= x (400 – 15x)

∴ profit function P (x) = R (x) – C (x)

= 400x – 15x^{2} – 1800 – 55x

= – 15x^{2} + 345x – 1800

For breakeven point, P (x) = 0

⇒ – 15x^{2} + 345x – 1800 = 0

⇒ x^{2} – 23x + 120 = 0

⇒ (x – 8) (x – 15) = 0

⇒ x = 8, 15

Question 6.

The demand function of a monopolist is given by x = 100 – 4p. The quantity at which MR (marginal revenue) = 0 will be

(a) 25

(b) 10

(c) 50

(d) 40

Solution:

(c) 50

Given demand function p (x) = \(\frac{100-x}{4}\)

∴ Revenue function R (x) = xp (x)

= \(\frac{100 x-x^2}{4}\)

∴ MR = \(\frac{d}{d x}\) R(x)

= \(\frac{d}{d x}\left[\frac{100 x-x^2}{4}\right]\)

= \(\frac{1}{4}\) [100 – 2x]

Now MR = 0

⇒ \(\frac{100-2 x}{4}\) = 0

⇒ x = 50

Question 7.

If the total cost of producing x units of a commodity is given by C(x) = \(\frac{1}{3}\) x^{3} + x^{2} – 15x + 3000 then the marginal cost when x = 5 is

(a) ₹ 25

(b) ₹ 20

(c) ₹ 30

(d) ₹ 50

Solution:

(b) ₹ 20

Given C(x) = \(\frac{1}{3}\) x^{3} + x^{2} – 15x + 3000

∴ MC = \(\frac{d}{d x}\) C(x)

= x^{2} + 2x – 15

at x = 5 ;

MC(x) = 5^{2} + 2 × 5 – 15

= 25 + 10 – 15 = 20

Question 8.

If the total cost function is given by C(x) = 10x – 7x^{2} + 3x^{3}, then the marginal average cost function (MAC) is given by

(a)10 – 14x + 9x^{2}

(b)10 – 7x + 3x^{2}

(c) – 7 + 6x

(d) – 14 + 18x

Solution:

(c) – 7 + 6x

Given C(x) = 10x – 7x^{2} + 3x^{3}

AC(x) = \(\frac{C(x)}{x}\)

= \(\frac{10 x-7 x^2+3 x^3}{x}\)

= 10 – 7x + 3x^{2}

∴ MAC (x) = \(\frac{d}{d x}\) {AC(x)}

= – 7 + 6x

Question 9.

If the total cost function fora production of x units of a commodity is given by \(\frac{3}{4}\) x^{2} – 7x + 27, then the number of units produced for which MC = AC is

(a) 4

(b) 6

(c) 9

(d) 36

Solution:

(b) 6

Given C(x) = \(\frac{3}{4}\) x^{2} – 7x + 27

∴ AC = \(\frac{C(x)}{x}\)

= \(\frac{3}{4} x-7+\frac{27}{x}\)

and MC = \(\frac{d}{d x}\) C(x)

= \(\frac{3}{2}\) x – 7

Now MC = AC

⇒ \(\frac{3}{2}\) x – 7 = \(\frac{3}{4}\) x – 7 + \(\frac{27}{x}\)

⇒ \(\frac{3}{4} \dot{x}=\frac{27}{x}\)

⇒ x^{2} = 36

⇒ x = 6 (∵ x > 0)

Question 10.

If the total cost function of producing x units of a commodity is given by 360 – 12x + 2x^{2}, then the level of output at which the total cost is minimum is

(a) 24

(b) 12

(c) 6

(d) 3

Solution:

(d) 3

Given total cost function C(x) = 360 – 12x + 2x^{2}

∴ \(\frac{d \mathrm{C}(x)}{d x}\) = – 12 + 4x

For maxima / minima,

\(\frac{d \mathrm{C}}{d x}\) = 0

⇒ – 12 + 4 × 0

⇒ x = 3

and \(\frac{d^2}{d x^2}\) C(x) = 4 > 0

∴ C (x) is minimum when x = 3.

Question 11.

If the demand function is p (x) = 20 – \(\frac{x}{2}\), then the marginal revenue when x = 10 is

(a) ₹ 5

(b) ₹ 10

(c) ₹ 15

(d) ₹ 150

Solution:

(b) ₹ 10

Given demand function p (x) = 20 – \(\frac{x}{2}\)

∴ R(x) = px

= (20 – \(\frac{x}{2}\)) x

= 20 x – \(\frac{x^2}{2}\)

Thus MR = \(\frac{d}{d x}\) R(x)

= 20 – x

∴ MR (10) = 20 – 10 = ₹ 10

Question 12.

If the demand function for a product is p = \(\frac{80-x}{4}\), where x is the number of units and p is the price per unit, then the value of x for which the revenue will be maximum is

(a) 40

(b) 20

(c) 10

(d) 80

Solution:

(a) 40

Given P = \(\frac{80-x}{4}\)

∴ R(x) = xp

= x (\(\frac{80-x}{4}\))

∴ \(\frac{d \mathrm{R}}{d x}=\frac{1}{4}(80-2 x)\)

For maxima / minima,

\(\frac{d \mathrm{R}}{d x}\) = 0

⇒ \(\frac{1}{4}\) (80 – 2x) = 0

⇒ x = 40

\(\frac{d^2 \mathrm{R}}{d x^2}=-\frac{1}{2}\) < 0 at x = 40

Thus R(x) is maximum when x = 40.

Question 13.

If the marginal cost function of a product is given by MC= 10 – 4x + 3x^{2} and fixed cost is ₹ 500, then the cost function is

(a) 10x – 2x^{2} + x^{3}

(b) 500+ 10x- 2x^{2} + x^{3}

(c) – 4 + 6x

(d) 500 + 10x – 8x^{2} + 9x^{3}

Solution:

(b) 500+ 10x- 2x^{2} + x^{3}

Given MC(x) = 10 – 4x + 3x^{2}

∴ C(x) = ∫ MC(x) dx

= ∫ (10 – 4x + 3x^{2}) dx

= 10x – 2x^{2} + x^{3} + K …………….(1)

When x = 0

∴ C(0) = fixed cost = ₹ 500

∴ 500 = K

∴ from (1) ;

C(x) = 10x – 2x^{2} + x^{3} + 500

Question 14.

If the marginal revenue function of a commodity is MR = 2x – 9x^{2}, then the revenue function is

(a) 2x^{2} – 9x^{3}

(b) 2 – 18x

(c) x23x3

(d) 18+x2 – 3×3

Solution:

(a) 2x^{2} – 9x^{3}

Given MR (x) = 2x – 9x^{2}

∴ R(x) = ∫ MR(x) dx

= ∫ (2x – 9x^{2}) dx

= x^{2} – 3x^{3} + K

When x = 0

⇒ R(0) = 0

∴ K = 0

Thus, R(x) = x^{2} – 3x^{3}