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Questions:

Multiple Choice Question

5 x 1 = 5

ICT stands for ________

(a)

Information and Communication Technology

(b)

Indian Council for Technology

(c)

International Communication Technology

(d)

Information Course for Technique

_________ of the line represents the ratio of the changes in vertical and horizontal line.

(a)

Slope

(b)

Gradient

(c)

Both 'a' and 'b'

(d)

None of these

The command Ctrl +M is applied for

(a)

Saving

(b)

Copying

(c)

Getting new slide

(d)

Deleting a slide

Differentiation of xⁿ is

(a)

nx^(n-1)

(b)

nx ⁽ⁿ⁺¹⁾

(c)

zero

(d)

one

(y - y₁) = m(x - x₁) gives the

(a)

Slope

(b)

Straight line

(c)

Constant

(d)

Curve

Generic 2

5 x 2 = 10

How to open a word document in MS Word?

Integrate : \(\int { ({ x }^{ 2 } } +x-1)dx\)

Define Supply Function?

What is the slope function Y = 10x⁴when x = 5 =?

10.

Find the average cost function where TC = 60 + 10x + 15x².

Generic 3

4 x 3 = 12

11.

The revenue function and demand function from marginal revenue function. Give the Reason.

12.

The marginal cost function for producing x units is y = 23 + 16x - 3x² and the total cost for producing zero unit is Rs.40. Obtain the total cost function and the average cost function.

13.

The demand and supply curves are given by \({ p }_{ d }=\frac { 16 }{ x+4 } and\ { p }_{ s }=\frac { x }{ 2 } \) Find the consumer's surplus and Producer's surplus at the market equilibrium price.

14.

Illustrate the uses of Mathematical Methods in Economics.

Generic 5

4 x 5 = 20

15.

Given the demand function P_d = 25 - Q² and the supply function p_s = 2Q + 1. Assuming pure competition, find (a) consumers surplus and (b) producers surplus. (P_d = Demand price; P_s = Supply price)

16.

Find the solution of the equation system:
8x₁-2x₂-3x₃=5 ; 15x₁-3x₂+2x₃=15 ; 7x₁-4x₂+x₃=2

17.

Solve by Cramer's rule
\(2{ x }_{ 1 }+3{ x }_{ 2 }=7;\)
\(2{ x }_{ 1 }+{ x }_{ 2 }=5\)

18.

Calculate the elasticity of demand for the demand schedule by using differential calculus method P = 60 - 0.2Q where price is (i) zero, (ii) Rs.20, (iii) Rs.40

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Answers Key:

Multiple Choice Question

5 x 1 = 5

(a)
Information and Communication Technology

(a)
nx^(n-1)

(b)
Straight line

Generic 2

5 x 2 = 10

One can open MSWord from various options.
Click start ⟶ All program ⟶ MS word or Double click the MS word icon from the desktop.

\(\int { ({ x }^{ 2 } } +x-1)dx=\int { { x }^{ 2 } } dx+\int { xdx } -\int { dx } \)
\(=\frac { { x }^{ 2+1 } }{ 2+1 } +\frac { { x }^{ 1+1 } }{ 1+1 } -x+c\)
\(=\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } -x+c\)

Q_s = f (P_x) Where 'Q_s' stands for Quantity supplied of a commodity and P_x is the price of that commodity:

Given function Y = 10x⁴
Slope \(={dy\over dx}\)
=40x3
= 40(5)³
= 40 x 125
Slope is = 5,000

10.

\(\mathrm{AC}=\frac{\mathrm{TC}}{x} \)
\(\mathrm{TC}=60+10 x+15 x^{2} \)
\(\mathrm{AC}=\frac{60}{x}+10+15 x\)

Generic 3

4 x 3 = 12

11.

If R is the total revenue function when the output is x, then marginal revenue.
\(MR=\frac { dR }{ dx } \) integrating with respect to 'x' we get.
Revenue Function: R = ഽ(MR) dx + k
where 'k' is the constant of integration which can be evaluated under given conditions when x = 0, total revenue R = 0.
Demand Functi.on: P = \(\frac { R }{ x } ,x\neq 0\).

12.

Given the marginal cost function y = 23 + 16x - 3x² ; C = 40
Rs.40 is the fixed cost.
We know that
Total Cost function = \(\int { } \) (Marginal cost function) dx + C
TC=\(\int { } \)ydx+C
=\(\int { } \)(23+16x-3x²)dx+c,where C is a constant,
=\(\int { } \)23dx+\(\int { } \)16x dx - \(\int { } \)3x²dx+C
=23x+16\(\left[ \frac { { x }^{ 2 } }{ 2 } \right] \)-3\(\left[ \frac { { x }^{ 3 } }{ 3 } \right] \)+c
TC=23x+8x²-x³+C
c = 40 given
\(\therefore\) TC = 23x + 8x²- x³+ 40
Average cost function=\(\frac { TC }{ x } \)
=23 + 8x - x²+ \(\frac { 40 }{ x } \)

13.

For Market equilibrium
Quantity demanded = Quantity supplied
\(\frac { 16 }{ x+4 } \frac { x }{ 2 } \)
\(x(x+4)=16\times 2\)
\({ x }^{ 2 }+4x=32\)
\({ x }^{ 2 }+4x-32=0\)
\((x-4)(x-8)=0\)
\(x=4;x=-8.\)
\(but\quad x=-8\quad is\text { in admissible.}\)
\(\therefore x=4(i.e.,){ x }_{ 0 }=4\)
\( { p }_{ 0 }{ x }_{ 0 }=\left[ \frac { 16 }{ 4+4 } \right] \times 4\)
\(=\frac { 16 }{ 8 } \times 4\)
\({ p }_{ 0 }{ x }_{ 0 }=8\)
Consumer Surplus,
\(CS=\int _{ 0 }^{ { x }_{ 0 } }{ f(x)dx-{ p }_{ 0 } } { x }_{ 0 }\)
\(=\int _{ 0 }^{ 4 }{ \frac { 16 }{ (x+4) } dx-8 } \)
\(=16[\log { (4+4)-\log { (0+4)]-8 } } \)
\(=16\left[ \log { \left( \frac { 8 }{ 4 } \right) } \right] -8\)
\({ C }_{ s }=16\log { 2-8\quad units } \)
Producer's Surplus:
\({ P }_{ s }={ P }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ { x }_{ 0 } }{ g(x)dx } \)
\(=8-\int _{ 0 }^{ 4 }{ \frac { x }{ 2 } } dx\)
\(=8-\frac { 1 }{ 2 } \int _{ 0 }^{ 4 }{ xdx } \)
\(=8-\frac { 1 }{ 2 } { { { { { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }^{ 4 } }_{ 0 } } } }\)
\(=8-\frac { 1 }{ 4 } [{ \left( 4 \right) }^{ 2 }-(0)]\)
\(=8-\frac { 1 }{ 4 } [16]=8-4\)
\(\therefore { P }_{ s }=4\quad units \)

14.

(i) Mathematical methods help to present the economic problems in a more precise form.
(ii) It helps to explain economic concepts.
(iii) It helps to use a large number of variables in economic analysis.
(iv) To quantify the effect of any economic activity implemented by government.

Generic 5

4 x 5 = 20

15.

For market equilibrium, P_d = P_s
25 - Q² = 2Q + 1
0 = 25 + Q²+ 2Q + 1
0 = -24 + Q²+ 2Q
Q²+ 2Q - 24 = 0
Q² + 6Q - 4Q - 24 = 0
Q(Q + 6) - 4(Q + 6) = 0
(Q + 6)(Q - 4) = 0
So, Q = 4 or Q =-6. Since Q cannot be equal to -6,
Q = 4
When Q = 4, P_d = 25 - 4² = 9;
P₂ = 2(4)+ 1 = 9
Consumer's surplus =\(\int _{ 0 }^{ 4 }{ (25-Q^{ 2 })dQ)dQ-(9\times 4) } \)
=\({ \left[ 25Q-\frac { Q^{ 3 } }{ 3 } \right] }_{ 0 }^{ 4 }\)-36
=\({ \left[ (25)(4)-\frac { 1 }{ 3 } (4)^{ 3 } \right] }\)-(0)-36
=\(\left[ 100-\frac { 64 }{ 3 } \right] \)-(0)- 36 = 42.67
Producer's surplus Ps
(P_s) = (9 x 4) - \(\int _{ 0 }^{ 4 }{ (2Q+1)dQ } \)
= 36 -\(\left[ (Q^{ 2 }+Q \right] _{ 0 }^{ 4 }\)
= 36 - (16 + 4) = 16

16.

The matrix form of the given equation is written as
\(\left[ \begin{matrix} 8 & -2 & -3 \\ 15 & -3 & 2 \\ 7 & -4 & 1 \end{matrix} \right] \left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} -5 \\ 15 \\ 2 \end{matrix} \right]\)
\( \triangle =\left| \begin{matrix} 8 & -2 & -3 \\ 15 & -3 & 2 \\ 7 & -4 & 1 \end{matrix} \right| \)
\(=8\left| \begin{matrix} -3 & 2 \\ -4 & 1 \end{matrix} \right| -(-2)\left| \begin{matrix} 15 & 2 \\ 7 & 1 \end{matrix} \right| +(-3)\left| \begin{matrix} 15 & -3 \\ 7 & -4 \end{matrix} \right| \)
= 8(-3 - (-8» + 2(15 - 14) - 3(-60 + 21)
= 8(5) + 2(1) - 3(-39)
= 40 + 2 + 117= 159   \(\therefore\triangle\)=159
\(\triangle { x }_{ 1 }=\left| \begin{matrix} -5 & -2 & -3 \\ 15 & -3 & 2 \\ 2 & -4 & 1 \end{matrix} \right| \)
= -5(-3 + 8) - (~2) (15 - 4) + (-3) (-60 - (-6»
= -25 + 22 + 162
= 159  \(\therefore\triangle\)x₁=159
\(\triangle { x }_{ 2 }=\left| \begin{matrix} 8 & -5 & -3 \\ 15 & 15 & 2 \\ 7 & 2 & 1 \end{matrix} \right| \)
= 8(15 - 4) - (-5) (15 - 14)+ (-3) (30 - 105)
= 88 + 5 + 225 = 318 \(\therefore\triangle\)=318
\(\triangle { x }_{ 3 }=\left| \begin{matrix} 8 & -2 & -5 \\ 15 & -3 & 15 \\ 7 & -4 & 2 \end{matrix} \right| \)
= 8(-6 + 60) - (-2) (30 - 105) + (-5) (-60 + 21)
= 8 x 54 + 2(-75) - 5(-39)
= 432- 150 + 195 = 477  \(\therefore\triangle\)=477
\({ x }_{ 1 }=\frac { \triangle { x }_{ 1 } }{ \triangle } =\frac { 159 }{ 159 } =1\)
\({ x }_{ 2 }=\frac { \triangle { x }_{ 2 } }{ \triangle } =\frac { 318 }{ 159 } =2\)
\({ x }_{ 3 }=\frac { \triangle { x }_{ 3 } }{ \triangle } =\frac { 477 }{ 159 } =3\)
x₁=1 ; x₂=2 ; x₃=3

17.

The coefficient and the constant terms are given below for the equations
\(\triangle =\left| \begin{matrix} 2 & 3 \\ 2 & 1 \end{matrix} \right| \)
\(\triangle =2-6=-4\)
\(\triangle { x }_{ 1 }=\left| \begin{matrix} 7 & 3 \\ 5 & 1 \end{matrix} \right| \)
\(=7-15=-8\)
\( \triangle { x }_{ 2 }=\left| \begin{matrix} 2 & 7 \\ 2 & 5 \end{matrix} \right| \)
\(=10-14=-4\)
\(By\quad Carmer's\quad rule\)
\( { x }_{ 1 }=\frac { \triangle { x }_{ 1 } }{ \triangle } \)
\(=\frac { -8 }{ -4 } \)
\(=2\)
\({ x }_{ 2 }=\frac { \triangle { x }_{ 2 } }{ \triangle } \)
\(=\frac { -4 }{ -4 } =1\)
\(solution\quad set\)
\(\left( { x }_{ 1 }{ ,x }_{ 2 } \right) =\left\{ 2,1 \right\} \)

18.

i) zero
\(\mathrm{P} =60-0.2 \mathrm{Q} \)
\(0 =60-0.2 \mathrm{Q} \)
\(0.2 \mathrm{Q} =60 \)
\(\mathrm{Q} =\frac{60}{0.2}=\frac{600}{2}=300\)
ii) Rs. 20
\(\mathrm{P} =60-0.2 \mathrm{Q} \)
\(20 =60-0.2 \mathrm{Q} \)
\(0.2 \mathrm{Q} =60-20=40 ; \quad \mathrm{Q}=\frac{40}{0.2}=\frac{400}{2}=200\)
iii) Rs. 40
\(\mathrm{P} =60-0.2 \mathrm{Q} \)
\(40 =60-0.2 \mathrm{Q} \)
\(0.2 \mathrm{Q} =60-40 \)
\(\mathrm{Q} =\frac{20}{0.2}=\frac{200}{2}=100\)