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

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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 – \(\frac{1}{5}\) x², 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 – \(\frac{3}{2}\) x²
(b) R(x) = 8x – \(\frac{2}{3}\) x²
(c) R(x) = \(\frac{2}{3}\) x² – 12x
(d) R(x) = 24x + \(\frac{3}{2}\) x²
Solution:
(a) R(x) = 12x – \(\frac{3}{2}\) x²

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 – \(\frac{3}{2}\) x²

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 – 15x² – 1800 – 55x
= – 15x² + 345x – 1800
For breakeven point, P (x) = 0
⇒ – 15x² + 345x – 1800 = 0
⇒ x² – 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) = \(\frac{1}{3}\) x³ + x² – 15x + 3000 then the marginal cost when x = 5 is
(a) ₹ 25
(b) ₹ 20
(c) ₹ 30
(d) ₹ 50
Solution:
(b) ₹ 20

Given C(x) = \(\frac{1}{3}\) x³ + x² – 15x + 3000
∴ MC = \(\frac{d}{d x}\) C(x)
= x² + 2x – 15
at x = 5 ;
MC(x) = 5² + 2 × 5 – 15
= 25 + 10 – 15 = 20

Question 8.
If the total cost function is given by C(x) = 10x – 7x² + 3x³, then the marginal average cost function (MAC) is given by
(a)10 – 14x + 9x²
(b)10 – 7x + 3x²
(c) – 7 + 6x
(d) – 14 + 18x
Solution:
(c) – 7 + 6x

Given C(x) = 10x – 7x² + 3x³
AC(x) = \(\frac{C(x)}{x}\)
= \(\frac{10 x-7 x^2+3 x^3}{x}\)
= 10 – 7x + 3x²
∴ 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 \(\frac{3}{4}\) x² – 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) = \(\frac{3}{4}\) x² – 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 + \(\frac{27}{x}\)
⇒ \(\frac{3}{4} \dot{x}=\frac{27}{x}\)
⇒ x² = 36
⇒ x = 6 (∵ x > 0)

Question 10.
If the total cost function of producing x units of a commodity is given by 360 – 12x + 2x², 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 + 2x²
∴ \(\frac{d \mathrm{C}(x)}{d x}\) = – 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 + 3x² and fixed cost is ₹ 500, then the cost function is
(a) 10x – 2x² + x³
(b) 500+ 10x- 2x² + x³
(c) – 4 + 6x
(d) 500 + 10x – 8x² + 9x³
Solution:
(b) 500+ 10x- 2x² + x³

Given MC(x) = 10 – 4x + 3x²
∴ C(x) = ∫ MC(x) dx
= ∫ (10 – 4x + 3x²) dx
= 10x – 2x² + x³ + K …………….(1)
When x = 0
∴ C(0) = fixed cost = ₹ 500
∴ 500 = K
∴ from (1) ;
C(x) = 10x – 2x² + x³ + 500

Question 14.
If the marginal revenue function of a commodity is MR = 2x – 9x², then the revenue function is
(a) 2x² – 9x³
(b) 2 – 18x
(c) x23x3
(d) 18+x2 – 3×3
Solution:
(a) 2x² – 9x³

Given MR (x) = 2x – 9x²
∴ R(x) = ∫ MR(x) dx
= ∫ (2x – 9x²) dx
= x² – 3x³ + K
When x = 0
⇒ R(0) = 0
∴ K = 0
Thus, R(x) = x² – 3x³