Master CAT Quant Based DI Questions with practice questions and detailed solutions.
Instructions
Out of 10 countries -- Country 1 through Country 10 -- Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.
• Column 1 gives the country's identity.
• Column 2 gives the country’s GDP as a fraction of the GDP of Country 9.
• Column 3 gives the country’s GDP per capita as a fraction of the GDP per capita of Country10.
• Column 4 gives the country’s annual GDP growth rate.
• Column 5 gives the country’s annual population growth rate.
Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.
Question 1.
Which one among the countries 1, 4, 5, and 7 will have the largest population in 2027?
This table compares the GDP, GDP per capita, and population of eight countries, all of which are referenced using a consistent set of units.
Specifically, the GDP values are all given with the same country as the reference point (The reference point being Country 9)
Similarly, the GDP per capita among the eight countries in the table is given using the same reference point (The reference point being Country 10)
As a result, when comparing two countries from the list in terms of GDP, GDP per capita, or population, we can directly use the values presented in the table, since they are all based on a uniform reference point.
We can compare the population of two countries using the formula, $Population=\dfrac{GDP}{GDP\ Per\ Capita}$
For example, to compare the GDP of Country A and Country B, we can simply use the values listed under "GDP" in the table, as both are referenced using the same country throughout. This eliminates the need for additional conversions or adjustments.
The given question asks us to compare the population of countries between 1, 4, 5, and 7 in the year 2027
Country 1: Population in 2024 will be 0.15/0.41 or 15/41 and the population is decreasing at the rate 0.12%
Country 4: Population in 2024 will be 0.12/0.38 or 12/38 and the population is increasing at the rate of 0.49%
Country 5: Population in 2024 will be 0.1/0.36 or 10/36 and the population is increasing at the rate 0.31%
Country 7: Population in 2024 will be 0.08/0.3 or 8/30 and the population is decreasing at the rate 0.11%
Simplifying and comparing the four fractions in order, $\frac{15}{41}$, $\frac{6}{19}$, $\frac{5}{18}$, $\frac{4}{15}$
The decimal values of the following are: 0.36586, 0.315789, 0.2777, 0.2666
Right away we can eliminate Country 7 since the population is not only the least, but it is decreasing, so there is no chance it will have the highest population in three years time.
For the other three countries,
Country 1: $\left(0.36585\right)\times\ \left(0.9988\right)^3$ this equals 0.36453
Country 4: $\left(0.315789\right)\times\ \left(1.0049\right)^3$ this equals 0.32045
Country 5: $\left(0.2777\right)\times\ \left(1.0031\right)^3$ this equals 0.2803612
Country 7: $\left(0.2666\right)\times\ \left(0.9989\right)^3$ this equals 0.2657211
Hence, Country 1 will have the largest population among these four countries.
Question 2.
The ratio of Country 4’s GDP to Country 5’s GDP in 2026 will be closest to
This table compares the GDP, GDP per capita, and population of eight countries, all of which are referenced using a consistent set of units.
Specifically, the GDP values are all given with the same country as the reference point (The reference point being Country 9)
Similarly, the GDP per capita among the eight countries in the table is given using the same reference point (The reference point being Country 10)
As a result, when comparing two countries from the list in terms of GDP, GDP per capita, or population, we can directly use the values presented in the table, since they are all based on a uniform reference point.
We can compare the population of two countries using the formula, $Population=\dfrac{GDP}{GDP\ Per\ Capita}$
For example, to compare the GDP of Country A and Country B, we can simply use the values listed under "GDP" in the table, as both are referenced using the same country throughout. This eliminates the need for additional conversions or adjustments.
We are given the GDP growth rate and we are told it is constant over the next three years,
Country 4 GDP: 0.12 and the given growth rate is 0.5%
GDP of Country 4 in 2026 will be $0.12\times\ \left(1.005\right)^2$ this equals 0.121203
Country 5 GDP: 0.1 and the given growth rate is 3.2
GDP of Country 5 in 2026 will be $0.1\times\ \left(1.007\right)^2$ this equals 0.1014049
Ratio will be, $\dfrac{0.121203}{0.1065024}$ this equals 1.19523
Hence, the answer is Option A
Question 3.
Which one among the countries 1 through 8, has the smallest population in 2024?
This table compares the GDP, GDP per capita, and population of eight countries, all of which are referenced using a consistent set of units.
Specifically, the GDP values are all given with the same country as the reference point (The reference point being Country 9)
Similarly, the GDP per capita among the eight countries in the table is given using the same reference point (The reference point being Country 10)
As a result, when comparing two countries from the list in terms of GDP, GDP per capita, or population, we can directly use the values presented in the table, since they are all based on a uniform reference point.
We can compare the population of two countries using the formula, $Population=\dfrac{GDP}{GDP\ Per\ Capita}$
For example, to compare the GDP of Country A and Country B, we can simply use the values listed under "GDP" in the table, as both are referenced using the same country throughout. This eliminates the need for additional conversions or adjustments.
In this question we are asked to find which of the countries among the four options has the least population in 2024 (The current year)
Option A: Country 8
GDP=0.07 and GDP Per Capita=0.41
Population=0.07/0.41 or $\frac{7}{41}$
Option B: Country 3
GDP=0.13 and GDP Per Capita=0.02
Population=0.13/0.02 or $\frac{13}{2}$
Option C: Country 7
GDP=0.08 and GDP Per Capita=0.3
Population=0.08/0.3 or $\frac{8}{30}$
Option D: Country 5
GDP=0.1 and GDP Per Capita=0.36
Population=0.1/0.36 or $\frac{10}{36}$
Comparing the four values, we are looking for the smallest values among the four,
This table compares the GDP, GDP per capita, and population of eight countries, all of which are referenced using a consistent set of units.
Specifically, the GDP values are all given with the same country as the reference point (The reference point being Country 9)
Similarly, the GDP per capita among the eight countries in the table is given using the same reference point (The reference point being Country 10)
As a result, when comparing two countries from the list in terms of GDP, GDP per capita, or population, we can directly use the values presented in the table, since they are all based on a uniform reference point.
We can compare the population of two countries using the formula, $Population=\dfrac{GDP}{GDP\ Per\ Capita}$
For example, to compare the GDP of Country A and Country B, we can simply use the values listed under "GDP" in the table, as both are referenced using the same country throughout. This eliminates the need for additional conversions or adjustments.
We are asked to find the number of countries where the GDP per capita is lower in 2027 than it was in 2024.
For the GDP per capita to be lower in the consequent years, the population growth rate has to exceed the GDP growth rate, since GDP per capita is nothing but GDP divided by the population. That means if the population growth rate is lesser than that of the GDP growth rate, the GDP per capita will only increase.
We can clearly see that none of the following countries fall in the scenario where GDP growth rate is lesser than that of the population growth rate. Countries 1, 2 and 7 have actually decreasing population rates thereby definitely increasing the GDP per capita.
The rest of the countries have GDP growth rates larger than the population growth rates.
Hence, we can conclude that, none of the countries will have a smaller GDP per capita in 2027 when compared to 2024.
Instructions
Two students, Amiya and Ramya are the only candidates in an election for the position of class representative. Students will vote based on the intensity level of Amiya’s and Ramya’s campaigns and the type of campaigns they run. Each campaign is said to have a level of 1 if it is a staid campaign and a level of 2 if it is a vigorous campaign. Campaigns can be of two types, they can either focus on issues, or on attacking the other candidate.
If Amiya and Ramya both run campaigns focusing on issues, then
• The percentage of students voting in the election will be 20 times the sum of the levels of campaigning of the two students. For example, if Amiya and Ramya both run vigorous campaigns, then 20 × (2+2)%, that is, 80% of the students will vote in the election.
• Among voting students, the percentage of votes for each candidate will be proportional to the levels of their campaigns. For example, if Amiya runs a staid (i.e., level 1) campaign while Ramya runs a vigorous (i.e., level 2) campaign, then Amiya will receive 1/3 of the votes cast, and Ramya will receive the other 2/3.
The above-mentioned percentages change as follows if at least one of them runs a campaign attacking their opponent.
• If Amiya runs a campaign attacking Ramya and Ramya runs a campaign focusing on issues, then 10% of the students who would have otherwise voted for Amiya will vote for Ramya, and another 10% who would have otherwise voted for Amiya, will not vote at all.
• If Ramya runs a campaign attacking Amiya and Amiya runs a campaign focusing on issues, then 20% of the students who would have otherwise voted for Ramya will vote for Amiya, and another 5% who would have otherwise voted for Ramya, will not vote at all.
• If both run campaigns attacking each other, then 10% of the students who would have otherwise voted for them had they run campaigns focusing on issues, will not vote at all.
Question 5.
What is the maximum possible voting margin with which one of the candidates can win?
We are looking for the minimum possible number of votes that Ramya can get and maximise the number of votes that Amiya can get.
We can borrow the scenario from the previous question where Ramya runs an attacking campaign, and we minimised the number of votes she can get.
To minimise the number of votes, we can have Ramya run a staid campaign to minimise the votes, so minimum intensity, which will get her 20% of the votes if she ran with issues. Now that she is running with attacking, she will loose 20% of the votes to Amiya and 5% of the votes will not vote anymore.
That is a total 25% loss. Remaining votes she will get is 75% of the 20% which will leave her with 15% of the votes.
And to maximise the number of votes Amiya can get, we will have her run an vigorous issues campaign, which will give her 2x20% of the votes, that is 40% of the votes. And since Ramya has been running an attacking campaign, 20% of her votes are transferred to Amiya. 20% of the 20% of the votes which is 4% that were going to Ramya will now go to Amiya. That will bring up Amiya's tally up to 44% leaving Ramya's tally at 15%.
The difference in the votes will be 44-15=29%.
This is the maximum possible vote difference between the two candidates that is possible.
Question 6.
If Ramya runs a campaign attacking Amiya, then what is the minimum percentage of votes that she is guaranteed to get?
We are looking for the minimum possible number of votes that Ramya can get when she runs an attacking campaign.
To minimise the number of votes, we can have Ramya run a staid campaign to minimise the votes, so minimum intensity, which will get her 20% of the votes if she ran with issues. Now that she is running with attacking, she will loose 20% of the votes to Amiya and 5% of the votes will not vote anymore.
That is a total 25% loss. Remaining votes she will get is 75% of the 20% which will leave her with 15% of the votes.
Question 7.
If Amiya runs a campaign focusing on issues, then what is the maximum percentage of votes that she can get?
Amiya runs a campaign on issues, and we need to find the maximum vote share that she can get.
We are trying to maximise the number of voters for Amiya, that means Amiya needs to run a vigorous campaign.
Since, we are trying to increase the vote share for Amiya, we want as many voters as possible transferred from Ramya's share to Amiya's votes.
As the number of votes a candidate receives is proportional to the intensity level of the campaign, we want Ramya also to run a vigorous AND attacking campaign, so that her votes are transferred to Amiya.
In this scenario we have 20x(2+2)% of the people voting, 80% of the people. Of that, if both had ran issues campaign, they would have each received 40% of the votes.
Now, we want Ramya to run an attacking campaign, where 20% of the people that would have voted for her vote for Amiya.
So 20% of 40% of the votes are transferred to Amiya. That is 8% of the votes. And we are also told that, 5% that would have voted for her dont vote, so 5% of 40% dont vote, that is 2% of the voters.
Final Tally is Amiya gets 48% of the votes, and Ramya gets 30% of the votes.
Question 8.
What is the minimum percentage of students who will vote in the election?
We want the minimum vote share, that means both the candidates run staid campaigns.
And, both the campaigns should be attacking, since we see that if one candidate runs an attacking campaign whereas the other candidate runs an issues campaign, a fair number of voters get transferred to the other candidate's vote share.
This points to us to a scenario where both the campaigns are staid and attacking, which is nothing but the scenario described in the previous question.
If both Ramya and Amiya run staid campaigns, the intensity of each staid campaign is 1
So total number of voters that would vote for them if they focused on issues will be 20x(1+1)%=40%
This means if they had both ran regarding issues, then they would get 20% of votes each.
We are told that both of them run attacking campaigns,
And the rule for mutual attacking campaign is 10% of voters who would have voted for each candidate will not vote.
That means 10% of 20% of each candidate will not vote, now that it is a mutually attacking campaign.
That means, each candidate receives 18% of the votes.
Total votes received is 36%, which is the minimum possible.
Question 9.
If both of them run staid campaigns attacking the other, then what percentage of students will vote in the election?
If both Ramya and Amiya run staid campaigns, the intensity of each staid campaign is 1
So total number of voters that would vote for them if they focused on issues will be 20x(1+1)%=40%
This means if they had both ran regarding issues, then they would get 20% of votes each.
We are told that both of them run attacking campaigns,
And the rule for mutual attacking campaign is 10% of voters who would have voted for each candidate will not vote.
That means 10% of 20% of each candidate will not vote, now that it is a mutually attacking campaign.
That means, each candidate receives 18% of the votes.
Total votes received is 36%.
Instructions
The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.
The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:
Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1 => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:
Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1 => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:
The letter B represents 9.
Question 12.
Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?
The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:
Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1 => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:
In all possible cases 7 is already represented by a letter other than D. Hence 7 is the answer.
Question 13.
Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?
The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:
Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1 => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:
From the table it is clear that 6 cannot be represented by G.
Instructions
Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.
There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
Question 14.
What is the total amount of money (in rupees) in the three pouches kept in the first column of the second row?
We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."
If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.
If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.
Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.
It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.
So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.
We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.
We can make the following table:
Similarly, we can find the amount for Column 2.
For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.
For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.
Thus {20,3,4} is the only solution possible.
We can similarly make the following table for the last column.
The total amount of money (in rupees) in the three pouches kept in the first column of the second row=13
Correct answer 13
Instructions
Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.
There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."
If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.
If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.
Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.
It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.
So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.
We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.
We can make the following table:
Similarly, we can find the amount for Column 2.
For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.
For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.
Thus {20,3,4} is the only solution possible.
We can similarly make the following table for the last column.
Answer 8
Instructions
Three countries - Pumpland (P), Xiland (X) and Cheeseland (C) - trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency).
The following terminology is used:
• Trade balance = Exports - Imports
• Total trade = Exports + Imports
• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms
The following information is known.
1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.
2. 40% of exports of X are to P. 22% of imports of P are from X.
3. 90% of exports of C are to P; 4% are to ROW.
4. 12% of exports of ROW are to X, 40% are to P.
5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0. In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively. For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a. For X, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$ $10E\ -\ 10I\ =\ E\ +\ I$ $11I\ =\ 9E$ So, let us assume export to be 11b and import to be 9b. For C, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$ $5I\ -\ 5E\ =\ E\ +\ I$ $2I\ =\ 3E$ So, let us assume export to be 2c and import to be 3c. In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P 22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X Equating both, we get, 4.4b = 2.2a a = 2b. We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0. The total imports for C = 3c = 1200 c = 400 This makes the exports for C = 2c = 800 In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW. 90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P 4% of the exports of C $=\dfrac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW The exports from C to X = 800 - 720 - 32 = 48 In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P. If we assume that Imports of ROW are m and exports of ROW are n. 12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X 40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P Placing all the values in the table, we get, Equating the imports of P and X to the Totals, we get, For P, 4.4b + 720 + 0.4n = 20b 15.6b = 720 + 0.4n --(1) For X, 600 + 48 + 0.12n = 9b 9b = 648 + 0.12n --(2) Solving (1) and (2), we get $\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$ $10108.8+1.872n=6480+3.6n$ $3628.8=1.728n$ $n=2100$ From (1), 15.6b = 720 + 0.4(2100) 15.6b = 1560 b = 100 Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200 Exports of X to ROW = 11b - 4.4b = 6.6b = 660 Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW. The value of exports and imports of ROW to ROW has to be equal. We calculated the value of n to be 2100. The exports of ROW to ROW can be calculated as, Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW. The value of m can be calculated as, m = 200 + 660 + 32 + 1008 = 1900. Filling up the table with all the values, we get, Exports from C to X are 48 IC. Hence, the correct answer is 48.
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0. In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively. For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a. For X, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$ $10E\ -\ 10I\ =\ E\ +\ I$ $11I\ =\ 9E$ So, let us assume export to be 11b and import to be 9b. For C, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$ $5I\ -\ 5E\ =\ E\ +\ I$ $2I\ =\ 3E$ So, let us assume export to be 2c and import to be 3c. In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P 22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X Equating both, we get, 4.4b = 2.2a a = 2b. We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0. The total imports for C = 3c = 1200 c = 400 This makes the exports for C = 2c = 800 In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW. 90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P 4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW The exports from C to X = 800 - 720 - 32 = 48 In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P. If we assume that Imports of ROW are m and exports of ROW are n. 12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X 40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P Placing all the values in the table, we get, Equating the imports of P and X to the Totals, we get, For P, 4.4b + 720 + 0.4n = 20b 15.6b = 720 + 0.4n --(1) For X, 600 + 48 + 0.12n = 9b 9b = 648 + 0.12n --(2) Solving (1) and (2), we get $\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$ $10108.8+1.872n=6480+3.6n$ $3628.8=1.728n$ $n=2100$ From (1), 15.6b = 720 + 0.4(2100) 15.6b = 1560 b = 100 Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200 Exports of X to ROW = 11b - 4.4b = 6.6b = 660 Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW. The value of exports and imports of ROW to ROW has to be equal. We calculated the value of n to be 2100. The exports of ROW to ROW can be calculated as, Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW. The value of m can be calculated as, m = 200 + 660 + 32 + 1008 = 1900. Filling up the table with all the values, we get, The exports from P to ROW 200 IC. Hence, the correct answer is 200.
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0. In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively. For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a. For X, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$ $10E\ -\ 10I\ =\ E\ +\ I$ $11I\ =\ 9E$ So, let us assume export to be 11b and import to be 9b. For C, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$ $5I\ -\ 5E\ =\ E\ +\ I$ $2I\ =\ 3E$ So, let us assume export to be 2c and import to be 3c. In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P 22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X Equating both, we get, 4.4b = 2.2a a = 2b. We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0. The total imports for C = 3c = 1200 c = 400 This makes the exports for C = 2c = 800 In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW. 90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P 4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW The exports from C to X = 800 - 720 - 32 = 48 In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P. If we assume that Imports of ROW are m and exports of ROW are n. 12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X 40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P Placing all the values in the table, we get, Equating the imports of P and X to the Totals, we get, For P, 4.4b + 720 + 0.4n = 20b 15.6b = 720 + 0.4n --(1) For X, 600 + 48 + 0.12n = 9b 9b = 648 + 0.12n --(2) Solving (1) and (2), we get $\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$ $10108.8+1.872n=6480+3.6n$ $3628.8=1.728n$ $n=2100$ From (1), 15.6b = 720 + 0.4(2100) 15.6b = 1560 b = 100 Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200 Exports of X to ROW = 11b - 4.4b = 6.6b = 660 Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW. The value of exports and imports of ROW to ROW has to be equal. We calculated the value of n to be 2100. The exports of ROW to ROW can be calculated as, Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW. The value of m can be calculated as, m = 200 + 660 + 32 + 1008 = 1900. Filling up the table with all the values, we get, The exports from ROW to ROW are 1008 IC. Hence, the correct answer is 1008.
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0. In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively. For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a. For X, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$ $10E\ -\ 10I\ =\ E\ +\ I$ $11I\ =\ 9E$ So, let us assume export to be 11b and import to be 9b. For C, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$ $5I\ -\ 5E\ =\ E\ +\ I$ $2I\ =\ 3E$ So, let us assume export to be 2c and import to be 3c. In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P 22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X Equating both, we get, 4.4b = 2.2a a = 2b. We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0. The total imports for C = 3c = 1200 c = 400 This makes the exports for C = 2c = 800 In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW. 90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P 4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW The exports from C to X = 800 - 720 - 32 = 48 In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P. If we assume that Imports of ROW are m and exports of ROW are n. 12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X 40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P Placing all the values in the table, we get, Equating the imports of P and X to the Totals, we get, For P, 4.4b + 720 + 0.4n = 20b 15.6b = 720 + 0.4n --(1) For X, 600 + 48 + 0.12n = 9b 9b = 648 + 0.12n --(2) Solving (1) and (2), we get $\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$ $10108.8+1.872n=6480+3.6n$ $3628.8=1.728n$ $n=2100$ From (1), 15.6b = 720 + 0.4(2100) 15.6b = 1560 b = 100 Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200 Exports of X to ROW = 11b - 4.4b = 6.6b = 660 Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW. The value of exports and imports of ROW to ROW has to be equal. We calculated the value of n to be 2100. The exports of ROW to ROW can be calculated as, Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW. The value of m can be calculated as, m = 200 + 660 + 32 + 1008 = 1900. Filling up the table with all the values, we get, Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200. Hence, the correct answer is option D.
Question 20.
Which among the countries P, X, and C has/have the least total trade?
In clue 5, we are given that export volumes of P, in IC, to X and C are 600 and 1200, respectively, and P is the only country that exports to C. So, the export from X to C and ROW to C is 0. In clue 1, we are given that normalised trade balances of P, X and C are 0%, 10%, and -20%, respectively. For P, since the normalised trade balance is 0% the exports have to be equal to the imports according to the given formula. So, let us assume them to be 10a and 10a. For X, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ 10$ $10E\ -\ 10I\ =\ E\ +\ I$ $11I\ =\ 9E$ So, let us assume export to be 11b and import to be 9b. For C, $\dfrac{E\ -\ I}{E\ +\ I}\ \times\ 100\ =\ -20$ $5I\ -\ 5E\ =\ E\ +\ I$ $2I\ =\ 3E$ So, let us assume export to be 2c and import to be 3c. In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 40% of the exports of X $=\dfrac{40}{100}\ \times\ 11b\ =\ 4.4b\ =$ exports from X to P 22% of the imports of P $=\dfrac{22}{100}\ \times\ 10a\ =\ 2.2a\ =$ imports to P from X Equating both, we get, 4.4b = 2.2a a = 2b. We are also given that imports for C are only from P, and we can set the rest of the imports for C to 0. The total imports for C = 3c = 1200 c = 400 This makes the exports for C = 2c = 800 In clue 3, we are given that 90% of exports of C are to P and 4% are to ROW. 90% of the exports of C $=\dfrac{90}{100}\ \times\ 2c\ =\ 1.8c\ =1.8\ \times\ 400\ =\ 720\ =\$exports from C to P 4% of the exports of C $=\frac{4}{100}\ \times\ 2c\ =\ 0.08c\ =\ 0.08\ \times\ 400\ =\ 32\ =$ exports from C to ROW The exports from C to X = 800 - 720 - 32 = 48 In clue 4, we are given that 12% of exports of ROW are to X, 40% are to P. If we assume that Imports of ROW are m and exports of ROW are n. 12% of the exports of ROW $=\dfrac{12}{100}\ \times\ n\ =\ 0.12n\ =$ exports from ROW to X 40% of the exports of ROW $=\dfrac{40}{100}\ \times\ n\ =\ 0.4n\ =$ exports from ROW to P Placing all the values in the table, we get, Equating the imports of P and X to the Totals, we get, For P, 4.4b + 720 + 0.4n = 20b 15.6b = 720 + 0.4n --(1) For X, 600 + 48 + 0.12n = 9b 9b = 648 + 0.12n --(2) Solving (1) and (2), we get $\dfrac{648\ +\ 0.12n}{9}\ =\ \dfrac{720\ +\ 0.4n}{15.6}$ $10108.8+1.872n=6480+3.6n$ $3628.8=1.728n$ $n=2100$ From (1), 15.6b = 720 + 0.4(2100) 15.6b = 1560 b = 100 Exports of P to ROW = 20b - 600 - 1200 = 2000 - 1800 = 200 Exports of X to ROW = 11b - 4.4b = 6.6b = 660 Substituting the values of b and n, we can figure out all the values except for Imports and Exports of ROW to ROW. The value of exports and imports of ROW to ROW has to be equal. We calculated the value of n to be 2100. The exports of ROW to ROW can be calculated as, Exports of ROW to ROW = 2100 - 840 - 252 = 1008 = Imports of ROW to ROW. The value of m can be calculated as, m = 200 + 660 + 32 + 1008 = 1900. Filling up the table with all the values, we get, Total trade of P = 2000 + 2000 = 4000 Total trade of X = 900 + 1100 = 2000 Total trade of C = 1200 + 800 = 2000 So, both countries X and C have the least total trade. Hence, the correct answer is option C.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 
Equating the imports of P and X to the Totals, we get,
Exports from C to X are 48 IC.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 
Equating the imports of P and X to the Totals, we get,
The exports from P to ROW 200 IC.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 
Equating the imports of P and X to the Totals, we get,
The exports from ROW to ROW are 1008 IC.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 
Equating the imports of P and X to the Totals, we get,
Trade balance of ROW = Exports - Imports = 2100 - 1900 = 200.
In clue 2, we are given that 40% of exports of X are to P and 22% of imports of P are from X. 
Equating the imports of P and X to the Totals, we get,
Total trade of P = 2000 + 2000 = 4000