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.

Following data was captured from the question:

Since AC is turned OFF at 12:00, it must reach the outside temperature at 12:00 in 1 hour. It rises from 26 degrees to 31 degrees at 12:30 AM (an increase of 5 degrees in 30 minutes). It clearly states that the outside temperature will be 31 + 5 = 36 degree Celsius at 12:00 AM.

Similarly, AC is turned OFF at 1:00 AM, it must reach the outside temperature at 1:00 AM in 1 hour. It rises from 26 degrees to 30 degrees at 1:30 AM (an increase of 4 degrees in 30 minutes). It clearly states that the outside temperature will be 30 + 4 = 34 degree Celsius at 1:00 AM.

Also, temperature outside has been falling at a constant rate from 7 PM onward until 3 AM on a particular night.

The question is "What best can be concluded about the number of times the AC must have either been turned on or the AC temperature setting been altered between 11:01 pm and 1:59 am?"

Solution: We know that the AC got turned ON after 11:01 PM at 12:30 AM and 1:30 AM.

Hence, the answer is 'Exactly 3'

Following data was captured from the question:

Since AC is turned OFF at 12:00, it must reach the outside temperature at 12:00 in 1 hour. It rises from 26 degrees to 31 degrees at 12:30 AM (an increase of 5 degrees in 30 minutes). It clearly states that the outside temperature will be 31 + 5 = 36 degree Celsius at 12:00 AM.

Similarly, AC is turned OFF at 1:00 AM, it must reach the outside temperature at 1:00 AM in 1 hour. It rises from 26 degrees to 30 degrees at 1:30 AM (an increase of 4 degrees in 30 minutes). It clearly states that the outside temperature will be 30 + 4 = 34 degree Celsius at 1:00 AM.

Also, temperature outside has been falling at a constant rate from 7 PM onward until 3 AM on a particular night.

The question is "What was the temperature outside, in degree Celsius, at 9 pm?"

Solution: From the table we can infer that the outside temperature drops 1 degree Celsius for every 30 minutes. Temperature at 23:00 is 38 degrees. Hence at 9 PM it will be 38+1+1+1+1= 42 degrees Celsius.

Hence, the answer is '42'

Following data was captured from the question:

Since AC is turned OFF at 12:00, it must reach the outside temperature at 12:00 in 1 hour. It rises from 26 degrees to 31 degrees at 12:30 AM (an increase of 5 degrees in 30 minutes). It clearly states that the outside temperature will be 31 + 5 = 36 degree Celsius at 12:00 AM.

Similarly, AC is turned OFF at 1:00 AM, it must reach the outside temperature at 1:00 AM in 1 hour. It rises from 26 degrees to 30 degrees at 1:30 AM (an increase of 4 degrees in 30 minutes). It clearly states that the outside temperature will be 30 + 4 = 34 degree Celsius at 1:00 AM.

Also, temperature outside has been falling at a constant rate from 7 PM onward until 3 AM on a particular night.

The question is "What was the temperature outside, in degree Celsius, at 1 am?"

Solution: From the table we can infer that the temperature outside, in degree Celsius, at 1 am is 34.

Hence, the answer is '34'

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.

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.

Country GDP GDP per capita GDP growth rate Population growth rate Country 1 0.15x 0.41y 0.20% -0.12% Country 2 0.14x 0.25y 0.90% -0.41% Country 3 0.13x 0.02y 6.50% 0.70% Country 4 0.12x 0.38y 0.50% 0.49% Country 5 0.10x 0.36y 0.70% 0.31% Country 6 0.08x 0.08y 3.20% 0.61% Country 7 0.08x 0.30y 0.70% -0.11% Country 8 0.07x 0.41y 1.20% 0.71% Country 9 x Country 10 y

The question is "Which one among the countries 1, 4, 5, and 7 will have the largest population in 2027?"

Answer: Population of country 1 in 2027 = * (1 - 0.0012)³ = 0.364 (highest)

Population of country 4 in 2017 = * (1 + 0.0049)³ = 0.320

Population of country 5 in 2017 = * (1 + 0.0031)³ = 0.280

Population of country 7 in 2017 = (1 - 0.0011)³ = 0.261

Hence, the answer is 'Country 1'

Country GDP GDP per capita GDP growth rate Population growth rate Country 1 0.15x 0.41y 0.20% -0.12% Country 2 0.14x 0.25y 0.90% -0.41% Country 3 0.13x 0.02y 6.50% 0.70% Country 4 0.12x 0.38y 0.50% 0.49% Country 5 0.10x 0.36y 0.70% 0.31% Country 6 0.08x 0.08y 3.20% 0.61% Country 7 0.08x 0.30y 0.70% -0.11% Country 8 0.07x 0.41y 1.20% 0.71% Country 9 x Country 10 y

The question is "The ratio of Country 4's GDP to Country 5's GDP in 2026 will be closest to"

Answer: The GDP growth rates, and population growth rates of the countries will remain constant for the next three years.

Country 4’s GDP in 2026 = 0.12x * (1.005) * (1.005) = 0.1212x

Country 5’s GDP in 2026 = 0.10x * (1.007) * (1.007) = 0.1014x

The ratio of Country 4's GDP to Country 5's GDP in 2026 = 1.195.

Hence, the answer is '1.195'

The question is "Which one among the countries 1 through 8, has the smallest population in 2024?"

Answer: GDP per capita = GDP / Population. Therefore, Population of a country = GDP / GDP per capita.

• Population in country 3 = 6.5 x/y
• Population in country 5 = 0.277 x/y
• Population in country 8 = 0.17 x/y (lowest)
• Population in country 7 = 0.266 x/y

Hence, the answer is 'Country 8'

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: 20
Therefore, Option B is the correct answer.

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 65
Therefore, Option C is the correct answer.

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.
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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 33.33 percent
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:
34.C
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.

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.

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

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.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity
being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.

Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and
RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
ATMs that can be uniquely determines are the ATMs with cash 9L, 11L and 12L. Hence the answer is 3.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3
of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity

being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper
and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest
cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on
the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to
12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).
Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to
identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20,
and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not
possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to
either be at (RA, V1) or (RA, V3)
Case 1: 15L ATM is on the intersection (RA, V3)
We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since
the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the
remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the
total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is
already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L
has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.
Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering
the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and

RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other
ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.
Using the two cases, we can answer the given questions.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

Xena coached players numbered 1, 3, 4 and 8
Therefore, Option B is the correct answer

We can determine the ratings of all the players except 1 and 3
Therefore, 6 is the correct answer

The rating of player 6 was 5
Therefore, 5 is the correct answer.

Score of player 7 was 4

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki.
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at
least four players, leaving only one for Zara.
Hence, Yuki trained two people.
Coming to the scores given to the players themselves:
We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating.
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be , hence the repeated score must be 32-28 = 4
Thus, the score of 5 and 7 was 4
Clue 5 says that player 2 got the highest score, which is 7
Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4
getting 2 and player 8 getting 1.
Now, considering clues 2, 3 and 6:
We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4
being trained by the same coach.
Cheer 6 informs us about the average number of players the coaches train.
We know that Yuki trained only two players.
Let's take the average score of Yuki's players to be x
The average of Xena's players would be x/2, and that of Zara's players would be x-2
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2 or 2x.
The total score of Zara's players would be 3x-6 or 2x-4
We have two cases to consider:
Let's say Xena and Zara had three players each.
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6.
2 ×
8 7 = 28

The sum of all these scores would be which should be equal to 32
This would give us the value of x as
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score.
Hence, this must not be the case.
The other possibility:
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.
The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6
Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8
Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven
were scored by player 2. Hence, player two must be under Yuki.
Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach.
Hence, Zara must have gotten 8 through 4+4 with players 5 and 7.
All the remaining scores must be with Xena, adding up to 12, which is the case.
4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement.

The question is "Which among the firms C, D, E, and F had the least amount of R&D spending per employee in 2023?"

Answer:

R&D spending per employee for firm D in 2023 = (^{2400 × 2 × 2}⁄₈₀₀) = 12

R&D spending per employee for firm E in 2023 = (^{3500 × 3 × 3}⁄₁₄₀₀) = 22.5

R&D spending per employee for firm C in 2023 = (^{3000 × 2 × 2}⁄₈₀₀) = 15

R&D spending per employee for firm F in 2023 = (^{3200 × 2 × 2}⁄₁₀₀₀) = 12.8

Firm D has the least amount of R&D spending per employee in 2023.

Hence, the answer is 'Firm D'

The question is "Which among the firms A, C, E, and F had the maximum PAT per employee in 2023?"

Answer:

PAT per Employee for firm C in 2023 = 3000 / 800 = 3.75

PAT per Employee for firm A in 2023 = 3900 / 1300 = 3

PAT per Employee for firm E in 2023 = 3500 / 1400 = 2.5

PAT per Employee for firm F in 2023 = 3200 / 1000 = 3.2

The value is maximum for firm C.

Hence, the answer is 'Firm C'

The question is "The ratio of the amount of money spent by Firm C on R&D in 2019 to that in 2023 is closest to"

Answer:

Amount of money spent by any firm is proportional to areas on the firm’s PAT.

The amount of money spent by Firm C on R&D in 2019 = 2400 * π * 1.5 * 1.5.

The amount of money spent by Firm C on R&D in 2023 = 3000 * π * 1 * 1.

Ratio = (2400 * 1.5 * 1.5) / (3000 * 1 * 1) = 9 / 5.

Hence, the answer is '9 : 5'

The question is "Assume that the annual rate of growth in PAT over the previous year (ARG) remained constant over the years for each of the six firms. Which among the firms A, B, C, and E had the highest ARG? "

Answer:

Since the annual growth rate in PAT is constant, we can compare the ratio changes in the firm’s PAT in 2019 and 2023 and then compare them.

Firm A: 3900 / 3000 = 1.3

Firm B: 3800 / 2800 = 1.35

Firm C: 3000 / 2400 = 1.25

Firm E: 3500 / 2400 = 1.45

Therefore, the highest among these will be for E.

Hence, the answer is 'Firm E'

Day 2

Total number of Buyers in Day 2 = 50.

Daily Average on Day 2 =
(1×5 + 2×10 + 3×5 + 4×20 + 5×10) / 50 = 170 / 50 = 3.4

DAY 1:

Cumulative averages on Day 1 and Day 2 were 3 and 3.1 respectively.

Let the number of Buyers in Day 1 = x.

(3x + 170) / (x + 50) = 3.1

Solving, 3x + 170 = 3.1(x + 50);
3x + 170 = 3.1x + 155;
0.1x = 15;
x = 150.

Therefore, the number of Buyers in Day 1 = 150.

DAY 3:

Total number of Buyers in Day 3 = 100.

The numbers of buyers giving each product rating are non-zero multiples of 10.
Number of buyers giving rating 1 = Number of buyers giving rating 2 = Half of number of buyers giving rating 3.
The modes of the product ratings were 4 and 5.

Sum of the number of buyers = 4A + 2B = 100.

The only possible solution for the equation and B value being the mode is A = 10 and B = 30.

Product rating on Day 3 = (1×10 + 2×10 + 3×20 + 4×30 + 5×30) / 100 = 360 / 100 = 3.6

Cumulative rating on Day 3 = (3×150 + 3.4×50 + 3.6×100) / (150 + 50 + 100) = 980 / 300 = 3.266

Day 2

Total number of Buyers in Day 2 = 50.

Daily Average on Day 2 =
(1×5 + 2×10 + 3×5 + 4×20 + 5×10) / 50 = 170 / 50 = 3.4

DAY 1:

Cumulative averages on Day 1 and Day 2 were 3 and 3.1 respectively.

Let the number of Buyers in Day 1 = x.

(3x + 170) / (x + 50) = 3.1

Solving, 3x + 170 = 3.1(x + 50);
3x + 170 = 3.1x + 155;
0.1x = 15;
x = 150.

Therefore, the number of Buyers in Day 1 = 150.

DAY 3:

Total number of Buyers in Day 3 = 100.

The numbers of buyers giving each product rating are non-zero multiples of 10.
Number of buyers giving rating 1 = Number of buyers giving rating 2 = Half of number of buyers giving rating 3.
The modes of the product ratings were 4 and 5.

Sum of the number of buyers = 4A + 2B = 100.

The only possible solution for the equation and B value being the mode is A = 10 and B = 30.

Product rating on Day 3 = (1×10 + 2×10 + 3×20 + 4×30 + 5×30) / 100 = 360 / 100 = 3.6

Cumulative rating on Day 3 = (3×150 + 3.4×50 + 3.6×100) / (150 + 50 + 100) = 980 / 300 = 3.266

The question is "What is the daily average rating of Day 3?"

Hence, the answer is '3.6'

Day 2

Total number of Buyers in Day 2 = 50.

Daily Average on Day 2 =
(1×5 + 2×10 + 3×5 + 4×20 + 5×10) / 50 = 170 / 50 = 3.4

DAY 1:

Cumulative averages on Day 1 and Day 2 were 3 and 3.1 respectively.

Let the number of Buyers in Day 1 = x.

(3x + 170) / (x + 50) = 3.1

Solving, 3x + 170 = 3.1(x + 50);
3x + 170 = 3.1x + 155;
0.1x = 15;
x = 150.

Therefore, the number of Buyers in Day 1 = 150.

DAY 3:

Total number of Buyers in Day 3 = 100.

The numbers of buyers giving each product rating are non-zero multiples of 10.
Number of buyers giving rating 1 = Number of buyers giving rating 2 = Half of number of buyers giving rating 3.
The modes of the product ratings were 4 and 5.

Sum of the number of buyers = 4A + 2B = 100.

The only possible solution for the equation and B value being the mode is A = 10 and B = 30.

Product rating on Day 3 = (1×10 + 2×10 + 3×20 + 4×30 + 5×30) / 100 = 360 / 100 = 3.6

Cumulative rating on Day 3 = (3×150 + 3.4×50 + 3.6×100) / (150 + 50 + 100) = 980 / 300 = 3.266

The question is "Which of the following is true about the cumulative average ratings of Day 2 and Day 3?"

Answer: Percentage increase from Day 2 to Day 3= ((3.266-3.1))/3.1 *100= 5.37%

Hence, the answer is 'The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.'

Day 2

Total number of Buyers in Day 2 = 50.

Daily Average on Day 2 =
(1×5 + 2×10 + 3×5 + 4×20 + 5×10) / 50 = 170 / 50 = 3.4

DAY 1:

Cumulative averages on Day 1 and Day 2 were 3 and 3.1 respectively.

Let the number of Buyers in Day 1 = x.

(3x + 170) / (x + 50) = 3.1

Solving, 3x + 170 = 3.1(x + 50);
3x + 170 = 3.1x + 155;
0.1x = 15;
x = 150.

Therefore, the number of Buyers in Day 1 = 150.

DAY 3:

Total number of Buyers in Day 3 = 100.

The numbers of buyers giving each product rating are non-zero multiples of 10.
Number of buyers giving rating 1 = Number of buyers giving rating 2 = Half of number of buyers giving rating 3.
The modes of the product ratings were 4 and 5.

Sum of the number of buyers = 4A + 2B = 100.

The only possible solution for the equation and B value being the mode is A = 10 and B = 30.

Product rating on Day 3 = (1×10 + 2×10 + 3×20 + 4×30 + 5×30) / 100 = 360 / 100 = 3.6

Cumulative rating on Day 3 = (3×150 + 3.4×50 + 3.6×100) / (150 + 50 + 100) = 980 / 300 = 3.266

The question is "What is the median of all ratings given on Day 3?"

Answer: No of 1s’= 10; No of 2’s= 10; No of 3’s= 20; No of 4’s= 30; No of 5’s= 30.
Therefore, the median will be average of 50th and 51st term which is (4+4)/2 = 4.

Hence, the answer is '4'

3500
Similar to the previous question, we should try to avoid the hypotenuse

The total path distance would be 300(CD) + 400(DE) + 300(EF) + 200(FK) + 400(KN) + 300(NO) + 150(OP) +
400(PI) + 150(IJ) + 200(JG) + 400(GB) + 300(BC)
Adding up to 3,500 meters
Therefore, 3500 is the correct answer

Counter to the first question, we should minimize the use of those hypotenuse walkways as they reduce the
distance we travel.
The longest possible route would include then travelling through the edges of the rectangles and sqaures, and
not touch any point more than once.
Best case scenario would be us touching each point exactly once.
This can be visualized as:

Since we can only walk along the side of lakes, that drastically reduces the paths we can take.
The diagram below shows the path with the maximum distance travelled.

The path is CD-DE-EF-FK-KL-LM-MN-NK-KJ-JG-GF-FC
(the reverse of this path is also valid)
We can either manually add the lengths or use shorter methods to note that in the path we travel walkways of
length 400 m 4 times (DE, LM, KN, EF), walkways of length 300m 6 times (CD, EF, KL, MN, KJ, GF) and walkways
of length 200 m 2 times (FK, JG)
Giving the total length to be 1600+1800+400 = 3800
Therefore, Option C is the correct answer.

The first thing to realise here is the lengths of the paths.
KN should be equal to LM, giving the length of KO using Pythagoras theorem as 500m
Similarly, the length of HJ=OP=150m and length of GJ=EL=200m, giving the length of HG as 250 m
The shortest path from L to B and then to P (or the other way around would involve) using these hypotenuses
as much as possible instead of the two adjacent sides. The shortest can be visualised as shown below or
multitude of others variations, as there are multiple ways that would make one travel the shortest distance)

The sum of numbers in the 4th column is 10+8+7+1 = 26
Therefore, 26 is the correct answer.

We are given that the numbers keep increasing from left to right (clue 1), and the number keeps decreasing
from top to bottom (clue 2)
The key takeaway from this is that 10 must be placed in Row 1, column 4, as placing it anywhere else would
mean that the number above it, or right to it, must be greater than 10, which is not an option.
Clue 3 says that one is either in the same row or the same column as 10.
The same logic that we used for 10 applies for 1; it must be either in Row 1, column 1, or Row 4, column 4; we
don't know which one yet.
We are given that 2 and 3 are not in the same column or row as 10, meaning that they must occupy two spots
from (Row 2, Col. 2), (Row 2, Col. 3), and (Row 3, Col. 3)
2 and 3 must be present in Row 2, column 2 or Row 3, column 3, as there is no number smaller than it to be in
the cell left to it.
Clue 6 says that 4 and 6 are in the same row; this can be rows 1, 2, or 3.
Clue 5 is a good starting point.
Once we found the position of 10, the only positions possible for 9 are R1C3 or R2C4

If 9 is placed in R2C4, 7 and 8 must be placed in row 1.
Clue 6 says that 4 and 6 are in the same row, but with 7 and 8 in row 1, no rows are left with two spaces. Hence,
9 can not be in R2C4, and the arrangement must be:
7 and 8 must be in column 4, occupying R3C4 and R2C4, respectively.
Hence, 4, 6 must be in row 1:

Placements of all the numbers except 2 and 3 can be determined.
Therefore, 2 is the correct answer