Mathematical Economics Questions and Answers

Q: If all the elements in a set X are in a set Y, but all the elements of Y are not in X, then X is :

X is a Proper subset of Y.If every element of X is in Y, X is a subset of Y. If, in addition, Y has some element not in X, the two sets are not equal, and X is a proper subset, written \(X \subset Y\).A union set combines two sets; a power set is the set of all subsets of a set; the empty set has no elements at all. None of these describes the stated relationship.

Q: If y is a function of x, find the mathematical form from its inverse function :

The inverse function is x = f-1(y).If \(y = f(x)\), the inverse function reverses the mapping: it takes a value of \(y\) and returns the \(x\) that produced it. So \(x = f^{-1}(y)\). For example, if \(y = 2x + 3\), then \(x = \frac{y - 3}{2}\).\(y = f(x)\) is the original function. \(y = f^{-1}(x)\) only relabels the variables, and \(x = f(y)\) applies the original function to \(y\), which is not the inverse.

Q: Suppose demand for the oranges is given by the function \(q = 9 - \frac{1}{2}p\). Find the differential coefficient q'(p).

The differential coefficient is \(-\frac{1}{2}\).Differentiate the demand function with respect to price:\[ \begin{aligned} q &= 9 - \tfrac{1}{2}p \\ q'(p) &= -\tfrac{1}{2} \end{aligned} \]The constant 9 drops out, and the slope is negative, as the law of demand requires: quantity falls as price rises. \(\frac{1}{2}\) loses the sign, and 9 is the intercept, not the slope.

Q: Suppose matrix A = \(\begin{bmatrix} 7 & 2 & 3 \\ 4 & 8 & 1 \\ 5 & 12 & 4 \end{bmatrix}\) is multiplied by the identity matrix. What you will get ?

The product is the matrix A itself.The identity matrix \(I\) has 1s on the diagonal and 0s elsewhere. It is the multiplicative identity for matrices: \(AI = IA = A\) for any square matrix \(A\) of the same order, just as multiplying a number by 1 leaves it unchanged.The identity itself would result only if \(A\) were the identity. The zero matrix comes from multiplying by the zero matrix, and the last option is the transpose of \(A\), which multiplication by \(I\) does not produce.

Q: For the production function Q = 20K0.5L0.5 derive K = f(L) when the isoquant Q = 100.

The isoquant is \(K = 25L^{-1}\).Set output at 100 and solve for \(K\):\[ \begin{aligned} 20K^{0.5}L^{0.5} &= 100 \\ (KL)^{0.5} &= 5 \\ KL &= 25 \\ K &= \frac{25}{L} = 25L^{-1} \end{aligned} \]This rectangular hyperbola is the Q = 100 isoquant. The option \(25K^{-1}\) gives \(K\) in terms of itself, and \(25L\) would make the isoquant slope upward, which an isoquant cannot do when both inputs are productive.

Q: Find the determinant of matrix\[ A = \begin{bmatrix} 4 & 6 & 1 \\ 2 & 5 & 2 \\ 9 & 0 & 4 \end{bmatrix}. \]

The determinant is 95.Expand along the first row:\[ \begin{aligned} |A| &= 4(5 \times 4 - 2 \times 0) \\ &\quad - 6(2 \times 4 - 2 \times 9) \\ &\quad + 1(2 \times 0 - 5 \times 9) \\ &= 80 + 60 - 45 = 95 \end{aligned} \]The middle term is \(-6 \times (8 - 18) = +60\). Taking it as \(-60\) gives \(80 - 60 - 45 = -25\), and stopping after the first term gives 80, one of the distractors.

Q: Derive the marginal revenue function for the demand function q = 400 − 0.1p.

The marginal revenue function is 4000 − 20q.First express price in terms of quantity, then find total revenue and differentiate:\[ \begin{aligned} q &= 400 - 0.1p \Rightarrow p = 4000 - 10q \\ TR &= pq = 4000q - 10q^{2} \\ MR &= \frac{d(TR)}{dq} = 4000 - 20q \end{aligned} \]For a straight-line demand curve, MR has the same intercept and twice the slope. Marginal revenue is a function of quantity, so the options in terms of \(p\) are wrong, and \(+20q\) would make MR rise with output.

Q: What is the point elasticity when price is 12 rupees for the demand function P = 60 − 3q.

The point elasticity is − 0.25.At \(P = 12\): \(12 = 60 - 3q\), so \(q = 16\). From \(P = 60 - 3q\), \(q = 20 - \frac{P}{3}\) and \(\frac{dq}{dP} = -\frac{1}{3}\).\[ \begin{aligned} e &= \frac{dq}{dP} \cdot \frac{P}{q} \\ &= -\frac{1}{3} \times \frac{12}{16} = -0.25 \end{aligned} \]Demand is inelastic at this price. The option 0.25 is the same figure without the minus sign that a downward-sloping demand curve gives.

21.

If all the elements in a set X are in a set Y, but all the elements of Y are not in X, then X is :

A.Union set
B.Power set
C.Proper subset
D.Empty set

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

X is a Proper subset of Y.

If every element of X is in Y, X is a subset of Y. If, in addition, Y has some element not in X, the two sets are not equal, and X is a proper subset, written \(X \subset Y\).

A union set combines two sets; a power set is the set of all subsets of a set; the empty set has no elements at all. None of these describes the stated relationship.

Discuss in Forum

22.

If y is a function of x, find the mathematical form from its inverse function :

A.y = f(x)
B.y = f\(^{-1}\)(x)
C.x = f\(^{-1}\)(y)
D.x = f(y)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The inverse function is x = f^-1(y).

If \(y = f(x)\), the inverse function reverses the mapping: it takes a value of \(y\) and returns the \(x\) that produced it. So \(x = f^{-1}(y)\). For example, if \(y = 2x + 3\), then \(x = \frac{y - 3}{2}\).

\(y = f(x)\) is the original function. \(y = f^{-1}(x)\) only relabels the variables, and \(x = f(y)\) applies the original function to \(y\), which is not the inverse.

Discuss in Forum

23.

Suppose demand for the oranges is given by the function \(q = 9 - \frac{1}{2}p\). Find the differential coefficient q'(p).

A.\(-\frac{1}{2}\)
B.\(\frac{1}{2}\)
C.4
D.9

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The differential coefficient is \(-\frac{1}{2}\).

Differentiate the demand function with respect to price:

\[ \begin{aligned} q &= 9 - \tfrac{1}{2}p \\ q'(p) &= -\tfrac{1}{2} \end{aligned} \]

The constant 9 drops out, and the slope is negative, as the law of demand requires: quantity falls as price rises. \(\frac{1}{2}\) loses the sign, and 9 is the intercept, not the slope.

Discuss in Forum

24.

Suppose matrix A = \(\begin{bmatrix} 7 & 2 & 3 \\ 4 & 8 & 1 \\ 5 & 12 & 4 \end{bmatrix}\) is multiplied by the identity matrix. What you will get ?

A.\(\begin{bmatrix} 7 & 2 & 3 \\ 4 & 8 & 1 \\ 5 & 12 & 4 \end{bmatrix}\)
B.\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
C.\(\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
D.\(\begin{bmatrix} 7 & 4 & 5 \\ 2 & 8 & 12 \\ 3 & 1 & 4 \end{bmatrix}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The product is the matrix A itself.

The identity matrix \(I\) has 1s on the diagonal and 0s elsewhere. It is the multiplicative identity for matrices: \(AI = IA = A\) for any square matrix \(A\) of the same order, just as multiplying a number by 1 leaves it unchanged.

The identity itself would result only if \(A\) were the identity. The zero matrix comes from multiplying by the zero matrix, and the last option is the transpose of \(A\), which multiplication by \(I\) does not produce.

Discuss in Forum

25.

For the production function Q = 20K^0.5L^0.5 derive K = f(L) when the isoquant Q = 100.

A.25K\(^{-1}\)
B.25L\(^{-1}\)
C.25K
D.25L

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The isoquant is \(K = 25L^{-1}\).

Set output at 100 and solve for \(K\):

\[ \begin{aligned} 20K^{0.5}L^{0.5} &= 100 \\ (KL)^{0.5} &= 5 \\ KL &= 25 \\ K &= \frac{25}{L} = 25L^{-1} \end{aligned} \]

This rectangular hyperbola is the Q = 100 isoquant. The option \(25K^{-1}\) gives \(K\) in terms of itself, and \(25L\) would make the isoquant slope upward, which an isoquant cannot do when both inputs are productive.

Discuss in Forum

26.

Find the determinant of matrix

\[ A = \begin{bmatrix} 4 & 6 & 1 \\ 2 & 5 & 2 \\ 9 & 0 & 4 \end{bmatrix}. \]

A.31
B.63
C.80
D.95

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The determinant is 95.

Expand along the first row:

\[ \begin{aligned} |A| &= 4(5 \times 4 - 2 \times 0) \\ &\quad - 6(2 \times 4 - 2 \times 9) \\ &\quad + 1(2 \times 0 - 5 \times 9) \\ &= 80 + 60 - 45 = 95 \end{aligned} \]

The middle term is \(-6 \times (8 - 18) = +60\). Taking it as \(-60\) gives \(80 - 60 - 45 = -25\), and stopping after the first term gives 80, one of the distractors.

Discuss in Forum

27.

Derive the marginal revenue function for the demand function q = 400 − 0.1p.

A.4000 − 20q
B.4000 + 20q
C.4000 − 20p
D.4000 + 20p

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The marginal revenue function is 4000 − 20q.

First express price in terms of quantity, then find total revenue and differentiate:

\[ \begin{aligned} q &= 400 - 0.1p \Rightarrow p = 4000 - 10q \\ TR &= pq = 4000q - 10q^{2} \\ MR &= \frac{d(TR)}{dq} = 4000 - 20q \end{aligned} \]

For a straight-line demand curve, MR has the same intercept and twice the slope. Marginal revenue is a function of quantity, so the options in terms of \(p\) are wrong, and \(+20q\) would make MR rise with output.

Discuss in Forum

28.

What is the point elasticity when price is 12 rupees for the demand function P = 60 − 3q.

A.0.25
B.0.5
C.− 0.25
D.− 0.5

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The point elasticity is − 0.25.

At \(P = 12\): \(12 = 60 - 3q\), so \(q = 16\). From \(P = 60 - 3q\), \(q = 20 - \frac{P}{3}\) and \(\frac{dq}{dP} = -\frac{1}{3}\).

\[ \begin{aligned} e &= \frac{dq}{dP} \cdot \frac{P}{q} \\ &= -\frac{1}{3} \times \frac{12}{16} = -0.25 \end{aligned} \]

Demand is inelastic at this price. The option 0.25 is the same figure without the minus sign that a downward-sloping demand curve gives.

Discuss in Forum

29.

Which of the following are not associated with the Leontief's Static Model ?

I. Final demand is a stochastic vector.
II. Reciprocal of the Perron-Frobinius root is the maximum growth rate.
III. Input coefficients are the fixed technological parameters.
IV. \((I - A - RB)^{-1} WL = P\)
Select the correct code from the following :

A.I, II, III
B.I, III, IV
C.I, III
D.I, II, IV

Exam Year(s)

UGC NET - 2012,

Answer: D

The statements not associated with Leontief's static model are I (final demand is a stochastic vector), II (the reciprocal of the Perron-Frobenius root is the maximum growth rate) and IV (the expression with the matrix RB, which belongs to a dynamic model with capital coefficients). So the code is I, II, IV, option D.

Statement III, that input coefficients are fixed technological parameters, is the basic assumption of the static model, so options A, B and C, which include III, are wrong. In the static model the final demand is a given (exogenous) vector, not a random one.

Discuss in Forum

30.

If \(X_{1} = f(P_{1}, P_{2})\) and \(X_{2} = \phi(P_{1}, P_{2})\), then the two commodities are substitutes if

A.\((\delta X_{1}/\delta P_{2}) \lt 0, (\delta X_{2}/\delta P_{1}) \gt 0\)
B.\((\delta X_{1}/\delta P_{2}) \gt 0, (\delta X_{2}/\delta P_{1}) \lt 0\)
C.\((\delta X_{1}/\delta P_{2}) \gt 0, (\delta X_{2}/\delta P_{1}) \gt 0\)
D.\((\delta X_{1}/\delta P_{2}) \lt 0, (\delta X_{2}/\delta P_{1}) \lt 0\)

Exam Year(s)

UGC NET - 2012,

Answer: C

Two commodities are substitutes if a rise in the price of one raises the demand for the other, so that both cross effects are positive: \(\delta X_{1}/\delta P_{2} \gt 0\) and \(\delta X_{2}/\delta P_{1} \gt 0\). This is option C.

If both cross derivatives are negative (option D) the goods are complements. Options A and B give opposite signs for the two effects, which cannot hold for a pair of substitutes, since the relation of substitution is symmetric.

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