CAT 2024 Slot 3 Question Paper

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 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. 

The set's starting point is the conclusion that the temperature rises only when the AC is turned off. 
When the AC is turned off, the temperature rises linearly, reaching the temperature at the time it was turned off in one hour. 

The temperature rises in two instances, 
From 0:00 to 0:30, it rises from 26 to 31, so by 1:00, it would have reached 36 degrees, meaning that the temperature at 0:00 would have been 36 degrees Celsius. 

From 1:00 to 1:30, it rises from 26 to 30, so by 2:00, it would have reached 34 degrees, meaning the temperature at 1:00 would have been 34 degrees Celcius. 

Calculating either of these would have been enough as we have the starting temperature at 23:00 at 38 degree celsius. Since the temperature decreases linearly, we can find the temperature at every instance using another data point. 

Drawing out the temperature outside, we get:

From this alone, we can answer the first question:
The AC was turned off for only two instances between 11:01 pm and 1:59 am:
Once at 0:00 and once at 1:00

Therefore, Option B is the correct answer. 

The set's starting point is the conclusion that the temperature rises only when the AC is turned off.
When the AC is turned off, the temperature rises linearly, reaching the temperature at the time it was turned off in one hour.

The temperature rises in two instances,
From 0:00 to 0:30, it rises from 26 to 31, so by 1:00, it would have reached 36 degrees, meaning that the temperature at 0:00 would have been 36 degrees Celsius.

From 1:00 to 1:30, it rises from 26 to 30, so by 2:00, it would have reached 34 degrees, meaning the temperature at 1:00 would have been 34 degrees Celcius.

Calculating either of these would have been enough as we have the starting temperature at 23:00 at 38 degrees Celsius. Since the temperature decreases linearly, we can find the temperature at every instance using another data point.

Drawing out the temperature outside, we get:

From the table, we can determine that the temperature outside at 1 am was 34 degrees Celsius. 
Therefore, 34 is the correct answer. 

The set's starting point is the conclusion that the temperature rises only when the AC is turned off.
When the AC is turned off, the temperature rises linearly, reaching the temperature at the time it was turned off in one hour.

The temperature rises in two instances,
From 0:00 to 0:30, it rises from 26 to 31, so by 1:00, it would have reached 36 degrees, meaning that the temperature at 0:00 would have been 36 degrees Celsius.

From 1:00 to 1:30, it rises from 26 to 30, so by 2:00, it would have reached 34 degrees, meaning the temperature at 1:00 would have been 34 degrees Celcius.

Calculating either of these would have been enough as we have the starting temperature at 23:00 at 38 degree celsius. Since the temperature decreases linearly, we can find the temperature at every instance using another data point.

Drawing out the temperature outside, we get:

Extrapolating this linear chain of temperature drop (2 degrees drop every hour), we can see that the temperature at 22:00 would be 40 degrees Celsius, and the temperature at 21:00 or 9 pm would be 42 degrees Celsius. 

Therefore, 42 is the correct answer. 

The set's starting point is the conclusion that the temperature rises only when the AC is turned off.
When the AC is turned off, the temperature rises linearly, reaching the temperature at the time it was turned off in one hour.

The temperature rises in two instances,
From 0:00 to 0:30, it rises from 26 to 31, so by 1:00, it would have reached 36 degrees, meaning that the temperature at 0:00 would have been 36 degrees Celsius.

From 1:00 to 1:30, it rises from 26 to 30, so by 2:00, it would have reached 34 degrees, meaning the temperature at 1:00 would have been 34 degrees Celcius.

Calculating either of these would have been enough as we have the starting temperature at 23:00 at 38 degree celsius. Since the temperature decreases linearly, we can find the temperature at every instance using another data point.

Drawing out the temperature outside, we get:

We can see that the AC was turned on at 0:30, 1:30 and 23:00. 
We do not know about the status at 23:30

We are to count the times after 11:01 pm; hence, turning on the AC at 23:00 does not count. 

Hence, the answer should be 2 or 3. 

But in the final clue, we are given that POWER mode was used longer than REGULAR mode. 
Since there are 4 periods of 30 minutes each, the combinations possible are POWER-3, REGULAR-1, and POWER-4, REGULAR-0

If the AC were on REGULAR mode at 23:00, it would have functioned for 1 hour, and the power mode would have also functioned for 1 hour (total), which would go against the final clue. 
Hence, the AC must have been in POWER mode at 23:00, then again in power mode at 23:20, and then again in power mode at 0:30

At 1:30, it could have entered either POWER or REGULAR. 
But there were three instances of activation between 11:01 and 1:59

Hence, Option D would be the correct answer. 

The set's starting point is the conclusion that the temperature rises only when the AC is turned off.
When the AC is turned off, the temperature rises linearly, reaching the temperature at the time it was turned off in one hour.

The temperature rises in two instances,
From 0:00 to 0:30, it rises from 26 to 31, so by 1:00, it would have reached 36 degrees, meaning that the temperature at 0:00 would have been 36 degrees Celsius.

From 1:00 to 1:30, it rises from 26 to 30, so by 2:00, it would have reached 34 degrees, meaning the temperature at 1:00 would have been 34 degrees Celcius.

Calculating either of these would have been enough as we have the starting temperature at 23:00 at 38 degree celsius. Since the temperature decreases linearly, we can find the temperature at every instance using another data point.

Drawing out the temperature outside, we get:

(Side note: We cannot consider the temperatures outside and inside at 23:00 and 2:00 since the time limit does not include those.)

The maximum difference between the elements of the outside and inside is maximum at 0:00 when the outside temperature is 36, and the inside temperature is 26.

Giving the maximum difference as 10

Therefore,10 is the correct answer. 

Total number of students is 250, and we are told that, The percentage of girls was at least 44% and at most 60%.

So the number of girls range from, $$0.44\left(250\right)\le Girls\le0.6\left(250\right)$$
$$110\le Girls\le150$$

Statement 1: 
If 50% of the boys and 80% of the girls opted for swimming, that means if the total number of Boys is B, Girls is G where B+G=250. 
Swimming is: 0.5B+0.8G

Statement 2: 
If 70%of the boys and 60% of the girls opted for running, that means
Running is 0.7B+0.6G

Total number of enrolments for swimming and running together will be
(0.7B+0.6G)+(0.5B+0.8G)=1.2B+1.4G

Using the overlapping principle, where I represents people who have enrolled only for one activity and II represents number of people who have enrolled for two activities. 
We know that, $$I+II=250=B+G$$
$$I+2II=1.2B+1.4G$$

Subtracting the two equations, 
$$II=0.2B+0.4G$$
$$II=0.2\left(B+2G\right)$$
Using B+G=250
$$II=0.2\left(250+G\right)$$

G can at-most be 150 and at least 110. 

So maximum value of II will be $$0.2\left(250+150\right)=80$$

Minimum value of II will be $$0.2\left(250+110\right)=72$$

Opening the square on the left-hand side, we get $$a^2+3b^2+2ab\sqrt{\ 3}$$
Comparing the rational part on both side,s we get: $$a^2+3b^2=52$$
And comparing the irrational par,t we get: $$2ab\sqrt{\ 3}=30\sqrt{\ 3}$$

$$ab=15$$, Since we are given that a and b are natural numbers, the possible values of a and b are (1,15), (3, 5), (5, 3), or (15,1)

Putting these values in the first relation we got, we see that 15 squared would exceed the required value and would not be the case. 
We need not check if a=5, b=3 or a=3, b=5 since the answer would be the same. 

(a=5 and b=3 would satisfy it)

a+b = 5+3 = 8

Therefore, Option B is the correct answer. 

We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84

Let us write $$x+y+z=84$$
x, y, z being the three distinct integers in ascending order. 

If the smallest number is increased by 7 and the largest number is reduced by 10
$$(x+7)+(y)+(z-10)=81$$

New arithmetic mean will be $$\frac{81}{3}=27$$
And this is said to be 2 more than the middle number, meaning
$$27-2=y=25$$

$$x+z=59$$

We are given that difference between the largest and the smallest numbers becomes 64, 
$$(z-10)-(x+7)=64$$
$$z-x=81$$

Adding the two equations we get, $$2z=140$$
$$z=70$$

There are multiple ways of solving such questions involving remainders; one easy way is to look for a power of numerator that leaves a remainder of 1 or 01 when divided by the denominator. 

In this instance, 1000, when divided by 13, leaves a remainder of -1

We can rewrite the numerator as $$\frac{10^{66}\times\ 100}{13}$$

The remainder would be $$\left[\frac{10^{66}}{13}\right]_R\times\ \left[\frac{100}{13}\right]_R$$
$$\left(-1\right)^{22}\times\ 9$$
9

Therefore, Option C is the correct answer.