Question 1

Which of the following statements describes the behaviour of the sequence of numbers which satisfy   with initial condition, Y0 = - 1?&#10;A)Uniform convergence&#10;B)Oscillatory convergence&#10;C)None of these&#10;D)Oscillatory divergence&#10;E)Uniform divergence

Accepted Answer

A

Question 2

The value of an economic variable over time satisfies the relation

Accepted Answer

In order to find the equilibrium value of R, we need to take the limit as t approaches infinity, since in the long run, the effects of any transitory shocks or disturbances will disappear, and R will settle down to its steady state or equilibrium value.

As t approaches infinity, the second term in the expression for R goes to zero, since e^(-0.4t) approaches zero faster than any power of t.

Thus, we have:

lim_{t->infinity} R(t) = lim_{t->infinity} [6/(1+t^2) + 3e^(-0.4t)]
= 6/0 + 3*0 (using L'Hopital's rule or simply observing that the denominator goes to infinity faster than the numerator, so the fraction goes to zero)
= 0

Therefore, the equilibrium value of R is 0. Note that this does not imply that R is always zero or that it never fluctuates around zero over time, but only that in the long run, on average, it tends towards this value.

Question 3

Which of the following difference equations represents the reduced form of the market model

Accepted Answer

We start by substituting the expression for Pt from the third equation into the first two equations to obtain:
QSt = a(Pt-1 - e(QSt-1 - QDt-1)) - b
QDt = -c(Pt-1 - e(QSt-1 - QDt-1)) + d
Rearranging the equations, we get:
QSt + aeQSt-1 - acQDt-1 = (1-a)ePt-1 - b 
QDt - ceQSt-1 + cdQDt-1 = -c ePt-1 + d
Solving for Pt-1, we get:
Pt-1 = (QSt + aeQSt-1 - acQDt-1 + b)/(1-a e - c e)
Substituting this expression for Pt-1 into the third equation, we get:
Pt = (1-a e - c e)Pt-1 + e(QSt-1 - QDt-1)
Substituting the expression we derived for Pt-1 above, we get:
Pt = (1-a e - c e)((QSt + aeQSt-1 - acQDt-1 + b)/(1-a e - c e)) + e(QSt-1 - QDt-1)
Simplifying this expression yields:
Pt = (1 - a e - c e ) P t − 1 + e ( b + d − a e Q D t − 1 + ( a e − c e ) Q S t − 1 )
Comparing this to the answer choices, we see that the correct option is (E).

Question 4

A principal of $60 is invested. The value, I(t), of investment, t days later satisfies the differential equation   Find the value of the investment after 27 days, correct to two decimal places.&#10;A)$301.38&#10;B)$97.65&#10;C)$176.34&#10;D)$234.98&#10;E)$281.96

Accepted Answer

The differential equation is solved using an integrating factor or separation of variables. Solving it gives the function for I(t), which is then evaluated at t = 27 to find the investment value.

Question 5

Consider the market model

Accepted Answer

We start by substituting the equation for Qs and Qd into the third equation and solve for dP/dt.

\begin{aligned} \frac { d P } { d t } &= 0.4 \left( Q _ { D } - Q _ { S } ight) \ &= 0.4 \left( ( - 2 P + 13 ) - ( 4 P - 3 ) ight) \ &= - 0.8 P + 4.4 \end{aligned}

This is a separable differential equation that we can solve using integration.

\begin{aligned} \frac { d P } { d t } &= - 0.8 P + 4.4 \ \frac { d P } { - 0.8 P + 4.4 } &= d t \ \int \frac { 1 } { - 0.8 P + 4.4 } d P &= \int d t \ - \frac { 1 } { 0.8 } \ln | - 0.8 P + 4.4 | &= t + C \ \ln | 4.4 - 0.8 P | &= - 0.8 t + C \ | 4.4 - 0.8 P | &= e ^ { - 0.8 t + C } \ 4.4 - 0.8 P &= \pm e ^ { - 0.8 t + C } \ P &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { - 0.8 t + C } \end{aligned}

To determine the value of the constant C, we use the initial condition P(0) = 2.

\begin{aligned} P ( 0 ) &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { C } \ 2 &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { C } \ e ^ { C } &= \frac { 9 } { 4 } \ C &= \ln \frac { 9 } { 4 } \end{aligned}

Substituting this value of C back into our equation for P, we get

\begin{aligned} P &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { - 0.8 t + \ln \frac { 9 } { 4 } } \ &= \frac { 23 } { 8 } - \frac { 1 } { 2 } \cdot \frac { 9 } { 4 } e ^ { - 0.8 t } \ &= \frac { 23 } { 8 } - \frac { 9 } { 8 } e ^ { - 0.8 t } \end{aligned}

To find Qd, we substitute this expression for P into the equation for Qd.

\begin{aligned} Q _ { D } &= - 2 P + 13 \ &= - 2 \left( \frac { 23 } { 8 } - \frac { 9 } { 8 } e ^ { - 0.8 t } ight) + 13 \ &= \frac { 1 } { 4 } e ^ { - 0.8 t } + \frac { 11 } { 2 } \end{aligned}

Thus the final answer is (B).

Question 6

Which of the following is a particular solution of the differential equation  &#10;A)0.6t2 - 0.448&#10;B)0.6t2 + 0.24t - 0.448&#10;C)0.6t2 + 0.448&#10;D)0.6t2 - 0.24t + 0.448&#10;E)0.6t2 + 0.24t + 0.448

Accepted Answer

The answer of Which of the following is a particular...

Question 7

Consider the two- sector model Yt = Ct + It&#10;Ct = 0.75Yt - 1 + 400&#10;It = 200&#10;Find the value of C2, given that Y0 = 4,000.&#10;A)3,600&#10;B)3,200&#10;C)2,400&#10;D)2,700&#10;E)3,100

Accepted Answer

Using the two-sector model equation, we can solve for Y1: 
Y1 = C1 + I1 = 0.75Y0 - 1 + 400 + 200 = 0.75(4,000) + 599 = 3,599. 
Then, we can use the consumption function to solve for C1: 
C1 = 0.75Y1 - 1 + 400 = 0.75(3,599) + 399 = 2,999. 
Finally, we can use the consumption function again to solve for C2: 
C2 = 0.75Y2 - 1 + 400 = 0.75(Y1 - I1) - 1 + 400 
= 0.75(3,599 - 200) - 1 + 400 
= 3,101.5 
Rounding to the nearest hundred gives us a value of 3,100 for C2.

Question 8

Which of the following gives the solution of the difference equation Yt = bYt - 1 + c&#10;With initial condition, Y0 = a?

Accepted Answer

The correct solution to the given difference equation $Y_t = bY_{t-1} + c$ with initial condition $Y_0 = a$ is found by solving the difference equation iteratively or using the method of characteristic equations. The solution involves both the homogeneous part ($a b^t$, which accounts for the initial condition and the behavior of the system without the constant term c) and the particular solution ($\frac{c(1 - b^t)}{1 - b}$, which accounts for the constant term c). The combination of these two parts gives the complete solution, which correctly matches option B.