CAT 2024 Slot 3 Question Paper

The set's starting point is that the sum of each row must be 100

Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4. 

Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal. 
We have the reference point of P1 at 66

Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95. 
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4. 

Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated. 
Leaving only 80 grams of carb in C1 and 8 grams of protein. 

Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated. 
Taking 85 carbs would give 2 grams of protein, which can be eliminated. 
80 and 70 would also not work for the same reason. 
Leavin has only 75 grams of carbs, leaving 12 grams of protein. 

Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62. 
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid. 

Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.  

Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14. 
So the protein in M2 (and M3) can be 12, 8, 4, 0

The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65

The protein in M3 can be 0, 4, 8 or 12

We are given in clue five that the protein in P1 is double that in M3. 
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values. 

The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3. 

We can see that there were 12 grams of protein in M2.

The set's starting point is that the sum of each row must be 100

Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4. 

Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal. 
We have the reference point of P1 at 66

Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95. 
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4. 

Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated. 
Leaving only 80 grams of carb in C1 and 8 grams of protein. 

Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated. 
Taking 85 carbs would give 2 grams of protein, which can be eliminated. 
80 and 70 would also not work for the same reason. 
Leavin has only 75 grams of carbs, leaving 12 grams of protein. 

Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62. 
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid. 

Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.  

Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14. 
So the protein in M2 (and M3) can be 12, 8, 4, 0

The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65

The protein in M3 can be 0, 4, 8 or 12

We are given in clue five that the protein in P1 is double that in M3. 
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values. 

The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3. 

There were 24 grams of other nutrients in M3. 

The set's starting point is that the sum of each row must be 100

Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4. 

Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal. 
We have the reference point of P1 at 66

Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95. 
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4. 

Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated. 
Leaving only 80 grams of carb in C1 and 8 grams of protein. 

Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated. 
Taking 85 carbs would give 2 grams of protein, which can be eliminated. 
80 and 70 would also not work for the same reason. 
Leavin has only 75 grams of carbs, leaving 12 grams of protein. 

Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62. 
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid. 

Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.  

Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14. 
So the protein in M2 (and M3) can be 12, 8, 4, 0

The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65

The protein in M3 can be 0, 4, 8 or 12

We are given in clue five that the protein in P1 is double that in M3. 
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values. 

The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3. 

The proteins in the foodgrains, when arranged in ascending order, are: 8, 8, 10, 12, 12,14, 16
Giving the median as 12. 

Reading the bar graphs, we can find the distribution of the subscribers as:

Taking clue one into consideration, let's take the total number of subscribers in 2023 as 100x and in 2204 as 110x
This would give the split of subscribers as follows:

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of 55x, 33x used one app. 

Clues 3 and 4 are to be used with each other. 
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps. 
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x - 15000)

Equating these two we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000

The first question asks about the number of others in 2024, which is 55x, or simply 55,000

Therefore, Option C is the correct answer. 

Reading the bar graphs, we can find the distribution of the subscribers as:

Taking clue one into consideration, let's take the total number of subscribers in 2023 as 100x and in 2204 as 110x
This would give the split of subscribers as follows:

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of 55x, 33x used one app.

Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x - 15000)

Equating these two, we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000

There are 15x, that is, 15,000 kids in 2023, of which we are given that 10,000 use one app. 
So, 5,000 use multiple apps. 
The percentage of kids using multiple apps in 2023 would hence be $$\frac{5000}{15000}\times\ 100=\frac{100}{3}=33.33\%$$

Therefore, Option A is the correct answer. 

Reading the bar graphs, we can find the distribution of the subscribers as:

Taking clue one into consideration, let's take the total number of subscribers in 2023 as 100x and in 2204 as 110x
This would give the split of subscribers as follows:

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of 55x, 33x used one app.

Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x - 15000)

Equating these two, we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000

In 2023, there were 20,000 elders; in 2024, there were 33,000 elders. 
The increase in percentage would be $$\frac{33000-20000}{20000}\times\ 100=\frac{13}{20}\times\ 100=65\%$$

Therefore, Option C is the correct answer. 

Reading the bar graphs, we can find the distribution of the subscribers as:

Taking clue one into consideration, let's take the total number of subscribers in 2023 as 100x and in 2204 as 110x
This would give the split of subscribers as follows:

Clue 2 says that half of 22x, which is 11x and 2/3 of 33x, which is 22x, used one app, given that in 2024, out of 55x, 33x used one app.

Clues 3 and 4 are to be used with each other.
Clue 4 says that out of 15x kids, 10,000 used one app, and out of 20x Elders, 15,000 used multiple apps.
Clue 3 says that the kids who used multiple apps (15x - 10000) were the same as Elders who used one app (20x - 15000)

Equating these two, we get:
15x- 10000 = 20x - 15000
5x = 5000
x = 1000

In clue 2, we were given that 33000 kids and elders out of 55000 kids and elders use one app, showing that 22000 use multiple apps. To minimize the total usage of multiple apps, we can assume that all 55000 other subscribers use one app. 

Giving the percentage of multiple app users as: $$\frac{22,000}{110,000}\times\ 100=\frac{100}{5}=20\%$$

Therefore, Option B is the correct answer. 

We are given the following table,

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, 
$$\frac{7}{41}$$, $$\frac{13}{2}$$, $$\frac{8}{30}$$, $$\frac{10}{36}$$

$$\frac{7}{41}$$ is the smallest value among the four, hence Option A or Country 8 is the answer. 

We are given the following table,

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 4 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

We are given the following table,

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 7: $$\left(0.2777\right)\times\ \left(1.0031\right)^3$$ this equals 0.2803612

Hence, Country 1 will have the largest population among these four countries.