Explanation :

In a class there are three groups a b and c ….

1. Average weight = Total weight ÷ Number of students

2. When students are shifted between groups:
– Total weight of all students remains the same
– Total number of students remains the same
– Students are just reorganized within the same class

Example:
Let’s say in a class of 40 students:
– Total weight = 2000 kg
– Average weight = 2000 ÷ 40 = 50 kg

If students are divided into 2 groups:
– Whether 20-20 students
– Or 15-25 students
– Or any other combination
The class average will still be 50 kg because:
– Total weight is still 2000 kg
– Number of students is still 40

Therefore, shifting students between groups within the same class doesn’t affect the overall average weight.

Explanation :

Average age of a teacher and 3 students is 20

Given information:
– Average age of teacher and 3 students = 20 years
– All students are of same age
– Difference between teacher’s age and each student’s age = 20 years

Solution:
1. Sum of ages = Average age × Number of people
= 20 × 4 = 80 years

2. Let student’s age = x
Then, teacher’s age = x + 20

3. Equation:
3x + (x + 20) = 80
4x + 20 = 80
4x = 60
x = 15

Therefore:
– Each student’s age = 15 years
– Teacher’s age = 15 + 20 = 35 years

Explanation : 

The average score of a batsman ?

The batsman’s performance analysis:

After 50 innings:
– Average score = 46.4
– Total score = 50 × 46.4 = 2320 runs

After 60 innings:
– Average score increased to 49
– Total score = 60 × 49 = 2940 runs

Last 10 innings analysis:
1. Total runs in last 10 innings = Total after 60 innings – Total after 50 innings
= 2940 – 2320 = 620 runs

2. Average score in last 10 innings = Total runs in last 10 innings ÷ Number of innings
= 620 ÷ 10 = 62 runs

Therefore, the batsman’s average score in his last 10 innings was 62 runs.

How many different 5 letter words from DELHI ?

Solution:
1. Since D and I are fixed at the start and end positions, we only need to arrange the remaining three letters (E, L, H) in the middle positions.

2. Formula for arranging n distinct objects = n!
In this case, n = 3 (for letters E, L, H)

3. Calculation:
3! = 3 × 2 × 1 = 6

Therefore, there are 6 different possible arrangements.

Possible words:
– DELHI
– DEHLI
– DHELII
– DHELI
– DLHEI
– DLEHI

Explanation :

Different sums that can be formed with different denominations

There are five denominations in total and we need to choose at least three of them to form a sum. We can use the formula for combinations to calculate the number of ways to choose k items from n items:

Formula: C(n,k) = n! / (k!(n-k)!)
where n is the total number of items and k is the number of items being chosen.

Calculations:
1. Choosing three denominations from five:
C(5,3) = 5! / (3!(5-3)!) = 10

2. Choosing four denominations from five:
C(5,4) = 5! / (4!(5-4)!) = 5

3. Choosing all five denominations:
Only one way possible = 1

Total number of different sums = 10 + 5 + 1 = 16

Explanation :

For what value of n the sum of digits in the number (10^n+1) is 2 ?

Let’s solve step by step.

1) The question asks for what value(s) of n, sum of digits in (10^n + 1) is 2.

2) Testing values systematically:

For n = 0:
10^0 + 1 = 1 + 1 = 2
Sum of digits = 2

For n = 1:
10^1 + 1 = 10 + 1 = 11
Sum of digits = 1 + 1 = 2

For n = 2:
10^2 + 1 = 100 + 1 = 101
Sum of digits = 1 + 0 + 1 = 2

For n = 3:
10^3 + 1 = 1000 + 1 = 1001
Sum of digits = 1 + 0 + 0 + 1 = 2

3) Key observations:
– For n = 0, sum is 2
– For positive integers, number is always in form 1(n zeros)1, so sum is always 2
– For negative integers, not applicable
– For non-integer values, not applicable

4) Checking options:
(a) For n = 0 only – False as works for other values too
(b) For any whole number n – True as works for n = 0 and all positive integers
(c) For any positive integer n only – False as also works for n = 0
(d) For any real number n – False as not applicable for non-integers

The answer is (b).

Note: Whole numbers are non-negative integers, including zero (0, 1, 2, 3, …). They represent whole things without fractions or decimals. Real numbers include all the numbers on the number line: rational numbers (fractions, integers) and irrational numbers (numbers that cannot be expressed as fractions, like √2). They encompass all possible magnitudes and their opposites, describing quantities in the real world.

Explanation :

p-2016=q+2017=r-2018=s+2019

Let’s solve this step by step:

1) From the given equation:
p – 2016 = q + 2017 = r – 2018 = s + 2019

Let this common value be k. Then:

2) The equations can be written as:
p – 2016 = k
q + 2017 = k
r – 2018 = k
s + 2019 = k

3) Solving for each variable:
p = k + 2016
q = k – 2017
r = k + 2018
s = k – 2019

4) Comparing these expressions:
p = k + 2016
q = k – 2017 (less than p)
r = k + 2018 (larger than p)
s = k – 2019 (less than q)

5) Arranging in descending order:
r = k + 2018
p = k + 2016
q = k – 2017
s = k – 2019

For any value of k, r will be the largest as it has the largest constant (2018) being added to k.

Therefore, r is the largest number. The answer is (c) R.

Explanation : 

The passage explains that Article 25 of the Indian Constitution provides for the freedom of conscience and the right to practice, profess, and propagate religion. However, this right is subject to public order, morality, health, and other provisions of Part III of the Constitution, which means that the State has the power to intervene in religious affairs if they threaten public order or morality.

Moreover, the passage points out that the rights to equality before the law and equal protection of laws, which are also fundamental rights guaranteed by the Constitution, take precedence over the right to religious freedom. This means that if any religious practice or belief is in conflict with the principle of equality before law, the State has the power to restrict or prohibit it.

Therefore, the most logical inference from the passage is that religious freedom under the Constitution is open to State intervention, and the State can regulate or restrict religious practices that are deemed to be a threat to public order, morality, health, or the rights of other citizens.

Explanation : 

The critical message of the passage is that the combination of unscientific solid waste management, poor drainage, and dumping of untreated sewage can lead to the spread of water-borne diseases. Therefore, the correct answer is (c) Solid waste management should be integrated with the maintenance of drainage and sewerage networks, as this reflects the need for a comprehensive approach to managing waste, drainage, and sewage to prevent the spread of water-borne diseases. The other options are not directly related to the message of the passage. Option (a) is not mentioned in the passage, option (b) focuses on the resources and legislative authority of urban local bodies rather than the need for integrated management, and option (d) mentions drinking water shortages, which is not directly related to the issue of solid waste management and its impact on health.

Explanation : 

The passage supports both assumptions. It explicitly mentions that researchers use “molecular scissors” to dissect the genome and repair it, which implies that there is no transfer of genes from one plant to another (assumption 1). The passage also mentions that the natural process of creating mutations that enable plants to survive future attacks can effectively be sped up through genome editing, which implies that the chosen genes can be altered precisely in a manner akin to the natural process (assumption 2). Therefore, the correct answer is (c) Both 1 and 2.