CATKing Forum · CATKing Forum

21.

If $(a + b \sqrt{3})^{2} = 52 + 30 \sqrt{3}$ , where $a$ and $b$ are natural numbers, then $a + b$ equals

A.
7
B.
10
C.
9
D.
8

Option D is the correct answer.

Video Explanation

Explanatory Answer

Question:
If (a + b√3)^2 = 52 + 30√3, where a and b are natural numbers, find a + b.

Solution:

Expand the left side:
(a + b√3)^2 = a^2 + 2ab√3 + 3b^2

Equate real and irrational parts with the right side 52 + 30√3:

Real part: a^2 + 3b^2 = 52
Irrational: 2ab = 30 ⇒ ab = 15

Find natural-number factor pairs of 15:
(ab = 15) ⇒ (a,b) ∈ {(1,15), (3,5), (5,3), (15,1)}

Check each pair against a^2 + 3b^2 = 52:

(1,15): 1^2 + 3·15^2 = 1 + 675 = 676 ≠ 52
(3,5): 3^2 + 3·5^2 = 9 + 75 = 84 ≠ 52
(5,3): 5^2 + 3·3^2 = 25 + 27 = 52 ✓
(15,1): 15^2 + 3·1^2 = 225 + 3 = 228 ≠ 52

Only (a,b) = (5,3) satisfies both equations.

Therefore:
a + b = 5 + 3 = 8

Answer: 8

22.

In a group of 250 students, the percentage of girls was at least 44% and at most 60%.The rest of the
students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80%
of the girls opted for swimming while 70%of the boys and 60% of the girls opted for running, then the
minimum and maximum possible number of students who opted for both swimming and running, are

A.
72 and 88, respectively
B.
75 and 96, respectively
C.
72 and 80, respectively
D.
75 and 90, respectively

Option C is the correct answer.

Video Explanation

Explanatory Answer

Here, with respect to boys
Swimming + Running = 50% + 70% = 120%
The extra 20% is meant for the boys who opted both swimming and running.
Similarly, with respect to girls
Swimming + Running = 80% + 60% = 140%
The extra 40% is meant for the boys who opted both swimming and running.
In pictorial representation:

So, percentage of people who opted for both swimming and running = 20% boys + 40% girls

The question is "In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are"

Hence, the answer is '72 and 80, respectively'

23.

The air-conditioner (AC) in a large room can be operated either in REGULAR mode or in POWER mode to reduce
the temperature.
If the AC operates in REGULAR mode, then it brings down the temperature inside the room(called inside
temperature) at a constant rate to the set temperature in 1 hour. If it operates in POWER mode, then this is
achieved in 30 minutes.
If the AC is switched off, then the inside temperature rises at a constant rate so as to reach the temperature
outside at the time of switching off in 1 hour.
The temperature outside has been falling at a constant rate from 7 pm onward until 3 am on a particular night.
The following graph shows the inside temperature between 11 pm (23:00)and 2 am (2:00) that night.

The following facts are known about the AC operation that night.
• The AC was turned on for the first time that night at 11 pm (23:00).
• The AC setting was changed (including turning it on/off, and/or setting different temperatures) only at the
beginning of the hour or at 30 minutes after the hour.
• The AC was used in POWER mode for longer duration than in REGULAR mode during this 3-hour period.

21.

How many times the AC must have been turned off between 11:01 pm and 1:59 am?

A.
cannot be determined
B.
2
C.
0
D.
1

Option B is the correct answer.

Video Explanation

Explanatory Answer

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.

22.

What was the temperature outside, in degree Celsius, at 1 am?

Answer : 34

Video Explanation

Explanatory Answer

Following data was captured from the question:

TIME	TEMPERATURE	AC - ON/OFF
23:00	38	ON
23:30	32	ON
00:00	26	OFF
00:30	31	ON
01:00	26	OFF
01:30	30	ON
02:00	28

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.

TIME	TEMPERATURE	AC - ON/OFF	MODE	OUTSIDE TEMP
23:00	38	ON		38
23:30	32	ON		37
00:00	26	OFF	NA	36
00:30	31	ON		35
01:00	26	OFF	NA	34
01:30	30	ON		33
02:00	28			32

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'

23.

What was the temperature outside, in degree Celsius, at 9 pm?

Answer : 42

Video Explanation

Explanatory Answer

Following data was captured from the question:

TIME	TEMPERATURE	AC - ON/OFF
23:00	38	ON
23:30	32	ON
00:00	26	OFF
00:30	31	ON
01:00	26	OFF
01:30	30	ON
02:00	28

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.

TIME	TEMPERATURE	AC - ON/OFF	MODE	OUTSIDE TEMP
23:00	38	ON		38
23:30	32	ON		37
00:00	26	OFF	NA	36
00:30	31	ON		35
01:00	26	OFF	NA	34
01:30	30	ON		33
02:00	28			32

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'

24.

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?

A.
More than 3
B.
Either 2 or 3
C.
Exactly 2
D.
Exactly 3

Option D is the correct answer.

Video Explanation

Explanatory Answer

Following data was captured from the question:

TIME	TEMPERATURE	AC - ON/OFF
23:00	38	ON
23:30	32	ON
00:00	26	OFF
00:30	31	ON
01:00	26	OFF
01:30	30	ON
02:00	28

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.

TIME	TEMPERATURE	AC - ON/OFF	MODE	OUTSIDE TEMP
23:00	38	ON		38
23:30	32	ON		37
00:00	26	OFF	NA	36
00:30	31	ON		35
01:00	26	OFF	NA	34
01:30	30	ON		33
02:00	28			32

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'

25.

What was the maximum difference between temperature outside and inside temperature, in degree Celsius,
between 11:01 pm and 1:59 am?

Answer : 10

Video Explanation

Explanatory Answer

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.

24.

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

21.

Which one among the countries 1 through 8, has the smallest population in 2024?

A.
Country 8
B.
Country 3
C.
Country 7
D.
Country 5

Option A is the correct answer.

Video Explanation

Explanatory Answer

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 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'

22.

The ratio of Country 4’s GDP to Country 5’s GDP in 2026 will be closest to

A.
1.195
B.
0.963
C.
1.314
D.
1.032

Option A is the correct answer.

Video Explanation

Explanatory Answer

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'

23.

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

A.
Country 1
B.
Country 5
C.
Country 7
D.
Country 4

Option A is the correct answer.

Video Explanation

Explanatory Answer

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'

24.

For how many countries among Country 1 through Country 8 will the GDP per capita in 2027 be lower than
that in 2024?

Answer : A

Video Explanation

Explanatory 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.

25.

Comprehension:
Over the top (OTT) subscribers of a platform are segregated into three categories: i) Kid, ii)Elder, and iii) Others.
Some of the subscribers used one app and the others used multiple apps to access the platform. The figure
below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the ‘Kid’ and
‘Elder’ categories.

The following additional facts are known about the numbers of subscribers.
1. The total number of subscribers increased by 10% from 2023 to 2024.
2. In 2024, 1/2 of the subscribers from the ‘Kid’ category and 2/3 of the subscribers from the ‘Elder’ category
subscribers use one app.
3. In 2023, the number of subscribers from the ‘Kid’ category who used multiple apps was the same as the
number of subscribers from the ‘Elder’ category who used one app.
4. 10,000 subscribers from the ‘Kid’ category used one app and 15,000 subscribers from the ‘Elder’ category
used multiple apps in 2023.

21.

How many subscribers belonged to the ‘Others’ category in 2024?

A.
Cannot be determined
B.
65000
C.
55000
D.
45000

Option C is the correct answer.

Video Explanation

Explanatory 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.

22.

What percentage of subscribers in the ‘Kid’ category used multiple apps in 2023?

A.
33.33%
B.
5.00%
C.
25.50%
D.
50.00%

Option A is the correct answer.

Video Explanation

Explanatory 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.
 VIDEO SOLUTION
 VIDEO SOLUTION
Downloaded from Cracku.in For MBA/CAT Courses: 6303239042 68/84
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.

23.

What was the percentage increase in the number of subscribers in the ‘Elder’ categoryfrom 2023 to 2024?

A.
60%
B.
50%
C.
65%
D.
40%

Option C is the correct answer.

Video Explanation

Explanatory 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.

24.

What could be the minimum percentage of subscribers who used multiple apps in 2024?

A.
16.5%
B.
20.0%
C.
10.0%
D.
22.00%

Option B is the correct answer.

Video Explanation

Explanatory 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 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.

26.

The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients, per 100
grams of nutrients in seven food grains. The first column shows the food grain category and the second
column its codename. The table has some missing values.

The following additional facts are known.
1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any
millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the food grains are non-zero multiples of 5.
4. All the missing values of protein, fat and other nutrients amounts (in grams) for all the food grains are nonzero multiples of 4.
5. P1 contained double the amount of protein that M3 contains.

21.

How many foodgrains had a higher amount of carbohydrate per 100 grams ofnutrients than M1?

Answer : 5

Video Explanation

Explanatory Answer

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.

22.

How many grams of protein were there in 100 grams of nutrients in M2?

Answer : 12

Video Explanation

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

23.

How many grams of other nutrients were there in 100 grams of nutrients in M3?

Answer : 24

Video Explanation

Explanatory 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

24.

What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?

Answer : 12

Video Explanation

Explanatory 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.

27.

The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C
and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance
(in km) between two adjacent intersections.Six ATMs are placed at six of the nine road intersections. Each ATM
has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure
indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash
requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed
on the same road.
2. The road distance between the ATM with the second highest cash requirement and the ATM located at the
intersection of R-C and V3 is 12 km.

21.

Which of the following statements is correct?

A.
There is no ATM placed at the (R-C, V2) intersection.
B.
The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.
C.
The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 8 Lakhs.
D.
The cash requirement of the ATM placed at the (R-C, V2) intersection cannot be uniquely determined.

Option B is the correct answer.

Video Explanation

Explanatory Answer

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.

22.

How many ATMs have cash requirements of Rs. 10 Lakhs or more?

Answer : 3

Video Explanation

Explanatory Answer

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.

23.

Which of the following two statements is/are DEFINITELY true?
Statement A: Each of R-A, R-B, and R-C has two ATMs.
Statement B: Each of V1, V2, and V3 has two ATMs.

A.
Only Statement A
B.
Neither Statement A nor Statement B
C.
Only Statement B
D.
Both Statement A and Statement B

Option A is the correct answer.

Video Explanation

Explanatory Answer

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.

24.

What best can be said about the road distance (in km) between the ATMs having the second highest and
the second lowest cash requirements?

A.
4 km
B.
7 km
C.
5 km
D.
Either 4 km or 7 km

Option D is the correct answer.

Video Explanation

Explanatory Answer

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.

25.

What is the number of ATMs whose locations and cash requirements can both be uniquely determined?

Answer : 3

Video Explanation

Explanatory Answer

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.

28.

The passage below is accompanied by four questions. Based on the passage, choose the best answer for
each question.
There is a group in the space community who view the solar system not as an opportunity to expand human
potential but as a nature preserve, forever the provenance of an elite group of scientists and their sanitary
robotic probes. These planetary protection advocates [call] for avoiding “harmful contamination” of celestial
bodies. Under this regime, NASA incurs great expense sterilizing robotic probes in order to prevent the
contamination of entirely theoretical biospheres ...
Transporting bacteria would matter if Mars were the vital world once imagined by astronomers who mistook
optical illusions for canals. Nobody wants to expose Martians to measles, but sadly, robotic exploration reveals
a bleak, rusted landscape, lacking oxygen and flooded with radiation ready to sterilize any Earthly microbes.
Simple life might exist underground, or down at the bottom of a deep canyon, but it has been very hard to find
with robots. . . . The upsides from human exploration and development of Mars clearly outweigh the welfare of
purely speculative Martian fungi ...
The other likely targets of human exploration, development, and settlement, our moon and the asteroids, exist
in a desiccated, radiation-soaked realm of hard vacuum and extreme temperature variations that would kill
nearly anything. It’s also important to note that many international competitors will ignore the demands of
these protection extremists in any case. For example, China recently sent a terrarium to the moon and
germinated a plant seed—with, unsurprisingly, no protest from its own scientific community. In contrast, when it
was recently revealed that a researcher had surreptitiously smuggled super-resilient microscopic tardigrades
aboard the ill-fated Israeli Beresheet lunar probe, a firestorm was unleashed within the space community ...
NASA’s previous human exploration efforts made no serious attempt at sterility, with little notice. As the Mars
expert Robert Zubrin noted in the National Review, U.S. lunar landings did not leave the campsites cleaner than
they found it. Apollo’s bacteria-infested litter included bags of feces. Forcing NASA’s proposed Mars exploration
to do better, scrubbing everything and hauling out all the trash, would destroy NASA’s human exploration budget
and encroach on the agency’s other directorates, too. Getting future astronauts off Mars is enough of a
challenge, without trying to tote weeks of waste along as well.
A reasonable compromise is to continue on the course laid out by the U.S. government and the National
Research Council, which proposed a system of zones on Mars, some for science only, some for habitation, and
some for resource exploitation. This approach minimizes contamination, maximizes scientific exploration ...
Mars presents a stark choice of diverging human futures. We can turn inward, pursuing ever more limited
futures while we await whichever natural or manmade disaster will eradicate our species and life on Earth.
Alternatively, we can choose to propel our biosphere further into the solar system, simultaneously protecting
our home planet and providing a backup plan for the only life we know exists in the universe. Are the lives on
Earth worth less than some hypothetical microbe lurking under Martian rocks?

21.

The author is unlikely to disagree with any of the following EXCEPT:

A.
the proposal for a zonal segregation of the Martian landscape into regions for different purposes.
B.
that while NASA’s earlier missions were not ideal in their approach to space contamination, they likely
did no grave damage.
C.
space contamination should be minimised until the possibility of life on the astronomical body being
explored is ruled out.
D.
the exorbitant costs of continuing to keep the space environment pristine may be unsustainable.

Option C is the correct answer.

Video Explanation

Explanatory Answer

The phrase “unlikely to disagree + EXCEPT” can seem tricky to interpret. In simple terms, the question requires
us to find a statement the author will disagree with. Let us inspect the choices -
Option A: The author supports the proposal for zonal segregation as a reasonable compromise, balancing
scientific exploration with human settlement.
Option B: The author agrees that NASA’s earlier missions did not prioritise contamination but implies they
caused no significant harm.
Option C: This viewpoint reflects a cautious approach to space exploration. The author dismisses concerns
about hypothetical extraterrestrial life as speculative and prioritises human exploration and development over
minimising contamination. Therefore, he’s likely to disagree with this position.
Option D: In the passage, the author argues that the costs of maintaining strict planetary protection measures
are excessive and could undermine future exploration efforts. This is consistent with his stance.
Hence, Option C is the best choice.

22.

The author mentions all of the following reasons to dismiss concerns about contaminating Mars EXCEPT:

A.
the lack of evidence of living organisms on Mars makes possible contamination from earthly microbes a
moot point.
B.
efforts to contain contamination on Mars are likely to be derailed as competitor countries may not
follow similar restrictions.
C.
the use of similar probes on astronomical bodies like the moon have had little effect on the
environment.
D.
earlier explorations have already contaminated pristine space environments.

Option C is the correct answer.

Video Explanation

Explanatory Answer

The passage discusses the debate surrounding planetary protection policies, particularly the concerns about
contaminating Mars with Earth-based microbes. The author argues against these concerns, citing several
reasons why the risk of contamination should not hinder human exploration and development of Mars. These
reasons include:
the lack of evidence for life on Mars (describes Mars as a “bleak, rusted landscape” with no confirmed life)
[Option A]
the disregard for such protocols by international competitors (China’s lenient approach to planetary
protection) [Option B]
the historical precedent of contamination from earlier human missions (Apollo missions left waste on the
Moon) [Option D]
On the other hand, Option C is not presented as a valid reason. The author does not specifically argue that
probes have had “little effect” on the Moon's environment but instead focuses on human waste and
contamination from earlier human missions, not robotic probes.

23.

The author’s overall tone in the first paragraph can be described as

A.
sceptical about the excessive efforts to sanitise planets where life has not yet been proven to exist.
B.
equivocal about the reasons extended by the group of scientists seeking to limit space exploration.
C.
indifferent to the elitism of a few scientists aiming to corner space exploration.
D.
approving of the amount of money NASA spends to restrict the spread of contamination in space.

Option A is the correct answer.

Video Explanation

Explanatory Answer

The first paragraph critiques the stringent planetary protection policies advocated by a group of scientists who
aim to prevent biological contamination of celestial bodies. The author portrays these efforts as excessive,
particularly given the lack of evidence for extraterrestrial life, and highlights the significant financial burden
these measures place on space agencies like NASA. Option A accurately reflects this scepticism, as the author
questions the need to sterilize planets where life has not been proven to exist.
Option B is incorrect because the author is not equivocal (i.e., ambiguous or undecided); instead, he expresses
a clear stance against these strict protocols. Option C is also inaccurate, as the author is not indifferent to
elitism but rather critiques the scientists’ restrictive approach. Similarly, Option D can be eliminated because
the author does not approve of NASA's spending on sterilization but views it as an unnecessary expense.

24.

The contrasting reactions to the Chinese and Israeli “contaminations” of lunar space

A.
are valid as the contamination of the lunar environment from animal sources is far greater than from
plants.
B.
are evidence of China’s reasonable approach towards space contamination.
C.
indicate that national scientists may have different sensitivities to issues of biosphere protection.
D.
reveal global biases prevalent in attitudes towards different countries.

Option C is the correct answer.

Video Explanation

Explanatory Answer

The passage highlights contrasting reactions to two instances of potential contamination of the lunar
environment: China’s germination of a plant seed on the Moon, which elicited little controversy, and Israel’s
accidental release of tardigrades aboard the Beresheet probe, which sparked significant backlash within the
space community. This contrast underscores differences in how national or regional scientific communities
respond to issues of planetary protection. Option C most closely reflects this idea.
The passage does not suggest that contamination from animals is inherently more harmful than from plants, as
suggested in Option A. Similarly, Option B inaccurately implies that the passage endorses China’s approach as
inherently “reasonable,” which it does not. Option D diverges a bit from the discussion by emphasising global
biases against specific countries, but the passage provides no evidence of such biases, focusing instead on
scientific reactions.

29.

The passage given below is followed by four alternate summaries. Choose the option that best captures
the essence of the passage.
Humans have managed to tweak the underlying biology of various plants and animals to produce high-tech
crops and microbes. But regulating these entities is complicated, as the framework of policies and
procedures are outdated and not flexible enough to adapt to emerging technology. The question is whether
regulation will ever be able to keep up with human innovation, to regulate living things, which are apt to be
unpredictable and unique; to capture all the potential risks when new biological entities are introduced, or
when they pass on variations of their genes?

A.
The mercurial nature of biological entities calls for scientists to shape the regulations governing
emerging technology, with regular calibration to handle variations in the field.
B.
A new framework of rules and procedures for regulating the most recent research emerging from
biotechnology is urgently needed, to keep up with this rapidly changing discipline
C.
Current regulation of biotechnology is outdated, but it is debatable if we can create a framework,
imaginative and flexible, to cover all contingencies in this fast-changing area.
D.
The problem with formulating regulation for innovation in the scientific arena it that it is impossible to
imagine the outcomes or risks related to the outcomes of all the research.

Option C is the correct answer.

Video Explanation

Explanatory Answer

The passage discusses the challenges of regulating innovations in biotechnology. It highlights two key issues:
(i) current regulations are outdated and insufficient for new biological advancements, and (ii) the unpredictable
nature of living entities and rapid technological changes make it difficult to design a regulatory framework that
anticipates all risks and contingencies. The core question is whether regulation can ever keep pace with the
unpredictable and unique outcomes of biotechnology. Option C correctly touches upon these ideas.
The other options are either too narrow or miss the essence. Option A focuses on calibration and scientists'
roles, which is not the central concern of the passage. Similarly, Option B highlights the urgency for new rules
but omits the skepticism about achieving comprehensive regulation. Option D overemphasises the
impossibility of imagining all risks without addressing the need for flexibility in regulations.
Hence, Option C is the best choice.

30.

.Five jumbled-up sentences (labelled 1, 2, 3, 4 and 5) related to a topic are given below. Four of them can
be put together to form a coherent paragraph. Identify the odd sentence and key in the number of that
sentence as your answer.
1. To create a synapse, the neuron has specialized structures, often seen as tiny swellings, at its terminal
end of the axon where it stores the chemicals that are emitted to transmit a signal to the next neuron.
2. This fetal warm-up act—the soldering of neural connections before the eyes actually function—is crucial
to the performance of the visual system.
3. The reasons for this paring back of synapses is a mystery, but synaptic pruning is thought to sharpen and
reinforce the “correct” synapses, while removing the weak and unnecessary ones.
4. Neural connections between the eyes and the brain are formed long before birth, establishing the wiring
and the circuitry that allow a child to begin visualizing the world the minute she emerges from the womb.
5. During this rehearsal period, synapses—points of chemical connection—between nerve cells are
generated in great excess, only to be pruned back during later development.

Answer : 1

Video Explanation

Explanatory Answer

A good starting point here would be to link sentences 5 and 3 based on the idea of “pruning.” Sentence 5 talks
about the synaptic development that occurs during the early period described in sentences 4 and 2. It describes
how synapses are initially overproduced, a concept that sets up the next part of the process: synaptic pruning.
Sentence 3 continues this discussion by offering an explanation for why synapses are pruned: to reinforce the
necessary connections and eliminate the weaker ones. This helps to fine-tune the brain’s wiring, ensuring that
the important pathways are retained while the less useful ones are discarded. Together, 5-3 ties in with the
broader purpose of neural wiring presented in Sentence 2: it emphasises that the creation of neural
connections in the fetus is a preparatory stage and that these connections are essential for the future
performance of the visual system. Sentence 4 sets the context of the paragraph: it mentions the formation of
neural connections before birth, specifically focusing on the connections between the eyes and the brain.

Previous Year Papers

Hence, the answer is '72 and 80, respectively'

Hence, the answer is '34'

Hence, the answer is '42'

Hence, the answer is 'Exactly 3'

Hence, the answer is 'Country 8'

Hence, the answer is '1.195'

Hence, the answer is 'Country 1'

To: