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Gnom [1K]
3 years ago
9

Pakistan’s GDP in 2010, using the official exchange rate on 1/1/2010, is equal to $300 billion USD ($300,000,000,000). When 2010

GDP is converted using the PPP-implied exchange rate on 1/1/2010, the USD value of the GDP increases to $500 billion ($500,000,000,000). On 1/1/2010, is the Pakistani Rupee overvalued or undervalued against the USD? Explain your answer.
Business
1 answer:
docker41 [41]3 years ago
5 0

Answer:

Undervalued

Explanation:

The PPP exchange rate is the implicit exchange rate, so that everywhere, one dollar has the same purchasing power. In general, this exchange rate is different from the exchange rate on the market.

Because the same nominal GDP translates to a higher real GDP by using the PPP exchange rate, one Pakistan Rupee must be valued more in terms of U.S. dollars than in contexts of the market exchange rate under the PPP exchange rate. The Pakistan Rupee is therefore worth less than its true value in the economy, i.e., undervalued.

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Suppose that output (Y ) in an economy is given by the following aggregate production function: Yt = Kt + Nt where Kt is capital
shusha [124]

Answer:

Check the explanation

Explanation:

Yt = Kt + Nt

Taking output per worker, we divide by Nt

Yt/Nt = Kt/Nt + 1

yt = kt + 1

where yt is output per worker and kt is capital per worker.

a) With population being constant, savings rate s and depreciation rate δ.

ΔKt = It - δKt

dividing by Nt, we get

ΔKt/Nt = It/Nt - δKt/Nt ..... [1]

for kt = Kt/Nt, taking derivative

d(kt)/dt = d(Kt/Nt)/dt ... since Nt is a constant, we have

d(kt)/dt = d(Kt/Nt)/dt = (dKt/dt)/Nt = ΔKt/Nt = It/Nt - δKt/Nt = it - δkt

thus, Capital accumulation Δkt = i – δkt

In steady state, Δkt = 0

That is I – δkt = 0

S = I means that I = s.yt

Thus, s.yt – δkt = 0

Then kt* = s/δ(yt) = s(kt+1)/(δ )

kt*= skt/(δ) + s/(δ)

kt* - skt*/(δ) = s/(δ)

kt*(1- s/(δ) = s/(δ)

kt*((δ - s)/(δ) = s/(δ)

kt*(δ-s)) = s

kt* = s/(δ -s)

capital per worker is given by kt*

b) with population growth rate of n,

d(kt)/dt = d(Kt/Nt)/dt =

= \frac{\frac{dKt}{dt}Nt - \frac{dNt}{dt}Kt}{N^{2}t}

= \frac{dKt/dt}{Nt} - \frac{dNt/dt}{Nt}.\frac{Kt}{Nt}

= ΔKt/Nt - n.kt

because (dNt/dt)/Nt = growth rate of population = n and Kt/Nt = kt (capital per worker)

so, d(kt)/dt = ΔKt/Nt - n.kt

Δkt = ΔKt/Nt - n.kt = It/Nt - δKt/Nt - n.kt ......(from [1])

Δkt = it - δkt - n.kt

at steady state Δkt = it - δkt - n.kt = 0

s.yt - (δ + n)kt = 0........... since it = s.yt

kt* = s.yt/(δ + n) =s(kt+1)/(δ + n)

kt*= skt/(δ + n) + s/(δ + n)

kt* - skt*/(δ + n) = s/(δ + n)

kt*(1- s/(δ + n)) = s/(δ + n)

kt*((δ + n - s)/(δ + n)) = s/(δ + n)

kt*(δ + n -s)) = s

kt* = s/(δ + n -s)

.... is the steady state level of capital per worker with population growth rate of n.

3. a) capital per worker. in steady state Δkt = 0 therefore, growth rate of kt is zero

b) output per worker, yt = kt + 1

g(yt) = g(kt) = 0

since capital per worker is not growing, output per worker also does not grow.

c)capital.

kt* = s/(δ + n -s)

Kt*/Nt = s/(δ + n -s)

Kt* = sNt/(δ + n -s)

taking derivative with respect to t.

d(Kt*)/dt = s/(δ + n -s). dNt/dt

(dNt/dt)/N =n (population growth rate)

so dNt/dt = n.Nt

d(Kt*)/dt = s/(δ + n -s).n.Nt

dividing by Kt*

(d(Kt*)/dt)/Kt* = s/(δ + n -s).n.Nt/Kt* = sn/(δ + n -s). (Nt/Kt)

\frac{sn}{\delta +n-s}.\frac{Nt}{Kt}

using K/N = k

\frac{s}{\delta +n-s}.\frac{n}{kt}

plugging the value of kt*

\frac{sn}{\delta +n-s}.\frac{(\delta + n -s)}{s}

n

thus, Capital K grows at rate n

d) Yt = Kt + Nt

dYt/dt = dKt/dt + dNt/dt = s/(δ + n -s).n.Nt + n.Nt

using d(Kt*)/dt = s/(δ + n -s).n.Nt from previous part and that (dNt/dt)/N =n

dYt/dt = n.Nt(s/(δ + n -s) + 1) = n.Nt(s+ δ + n -s)/(δ + n -s) = n.Nt((δ + n)/(δ + n -s)

dYt/dt = n.Nt((δ + n)/(δ + n -s)

dividing by Yt

g(Yt) = n.(δ + n)/(δ + n -s).Nt/Yt

since Yt/Nt = yt

g(Yt) = n.(δ + n)/(δ + n -s) (1/yt)

at kt* = s/(δ + n -s), yt* = kt* + 1

so yt* = s/(δ + n -s) + 1 = (s + δ + n -s)/(δ + n -s) = (δ + n)/(δ + n -s)

thus, g(Yt) = n.(δ + n)/(δ + n -s) (1/yt) =  n.(δ + n)/(δ + n -s) ((δ + n -s)/(δ + n)) = n

therefore, in steady state Yt grows at rate n.

5 0
3 years ago
Wall Drugs offered an incentive stock option plan to its employees. On January 1, 2021, options were granted for 75,000 $1 par c
valkas [14]

Answer:

the total compensation cost is $75,000

Explanation:

The computation of the total compensation cost for this plan is shown below:

Total compensation cost = option granted × fair value of each option

total compensation cost = 75000 × $1

total compensation cost = $75,000

Here to determined the total compensation cost we simply multiplied the option granted with the fair value of each option so that the correct amount could come

Therefore the total compensation cost is $75,000

4 0
3 years ago
Cogswell cola purchased a machine for $237,500. The firm paid another $5,750 for delivery and installation. In addition the firm
Novay_Z [31]

Based on the cost of purchasing the machine and the delivery and installation fees, the initial outlay is $243,250

<h3>How much is the initial outlay?</h3>

This can be found as:

= Cost of purchasing machine + Installation and delivery cost

Solving gives:

= 237,500 + 5,750

= $243,250

Find out more on fixed asset capitalization at brainly.com/question/25355478

#SPJ1

3 0
2 years ago
The following note transactions occurred during the year for Towell Company: Nov. 10 Towell issued a 90-day, 9% note payable for
Pani-rosa [81]

Answer: See explanation

Explanation:

The general journal entries necessary to adjust the interest accounts at December 31 will be:

1. December 31:

Debit: Interest Expenses = $8,000 × 9% × 51/ 360 = $102

Credit: Interest payable = $102

(To accrue interest expenses for the note issued on November 10).

2. December 31:

Debit: Interest Expenses = $12,000 × 10% ×30/360 = $120

Credit: Interest payable = $120

(To accrue interest expenses for the note issued on December 1)

3. December 31:

Debit: Interest Expenses = $12,000 × 10% × 11/360 = $36.67

Credit: Interest payable = $36.67

(To accrue interest expenses for the note issued on December 20).

3 0
2 years ago
Differential Analysis for a Discontinued Product A condensed income statement by product line for Healthy Beverage Inc. indicate
Varvara68 [4.7K]

Answer:

Healthy Beverage Inc.

a) Differential Analysis

1) Continue Fruit Cola (Alt. 1)

Sales                            $12,750,000

Cost of goods sold         8,500,000

Gross profit                  $4,250,000

Operating expenses      6,000,000

Loss from operations ($1,750,000)

2) Discontinue Fruit Cola (Alt. 2)

Differential Effect on Income (Alternative 2):

Fixed costs:

Cost of goods sold        $2,125,000

Operating expenses          900,000

Income (Loss)               ($3,025,000)

b. Should Fruit Cola be retained ?

The production and sale of the Fruit Cola should be continued.  Discontinuing it would not save the company the incurrence of the fixed cost.

Explanation:

Differential analysis is a managerial accounting technique for analyzing the different costs and benefits that would arise from alternative solutions to a particular problem.

In the above scenario, discontinuing the production and sale of Fruit Cola would not save the company the fixed costs, so the product should be continued.  It is not the product that is causing the net loss but allocated fixed costs.  Fixed cost is a sunk cost that is not relevant in differential analysis type of decision making.

5 0
2 years ago
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