Explanation : 

Problem: Finding the day of week in July if 12th January is a Sunday

Given:
– 12th January is a Sunday
– Need to determine if 12th July is a Sunday in a leap year

Solution:
1. Count days from 12th January to 12th July:
– Remaining days in January = 31 – 12 = 19 days
– February (leap year) = 29 days
– March = 31 days
– April = 30 days
– May = 31 days
– June = 30 days
– July (till 12th) = 12 days

2. Total days = 19 + 29 + 31 + 30 + 31 + 30 + 12 = 182 days

3. Calculate odd days:
– 182 ÷ 7 = 26 weeks + 0 days
– 0 odd days means same day as start

4. Since 12th January was Sunday:
– 0 odd days means 12th July will also be Sunday

Therefore, in a leap year, 12th July will be a Sunday.

Explanation :

Frog tries to come out of well

Given:
– Well depth = 4.5 meters = 450 cm
– Each jump up = 30 cm
– Each slide down = 15 cm
– Net progress per jump = 30 cm – 15 cm = 15 cm

Solution:
1. Initial calculation:
– Distance to cover = 450 cm
– Net progress per jump = 15 cm
– Number of jumps = 450 ÷ 15 = 30 jumps

2. Important consideration:
– When frog reaches the top, it won’t slide down
– Last jump will give full 30 cm progress, not 15 cm
– This saves one jump

3. Final calculation:
– First 28 jumps: 28 × 15 cm = 420 cm
– Last jump: 30 cm
– Total distance = 420 + 30 = 450 cm
– Total jumps needed = 29

Therefore, the frog needs 29 jumps to come out of the well.

Explanation :

Let’s solve how long it takes two people to complete 40% of work together.

Given:
Person 1:
– Can do 20% (W/5) of work in 8 days
– Rate = (W/5)/8 = W/40 per day

Person 2:
– Can do 25% (W/4) of work in 6 days
– Rate = (W/4)/6 = W/24 per day

Step by step calculations:
1. Combined rate of work = Rate of Person 1 + Rate of Person 2
= W/40 + W/24
= (24W + 40W)/(40 × 24)
= 64W/960
= W/15 per day

2. Time to complete 40% work:
Work = Rate × Time
0.4W = (W/15) × Time
Time = (0.4W) ÷ (W/15)
= 0.4W × (15/W)
= 0.4 × 15
= 6 days

Therefore, working together, they will complete 40% of the work in 6 days.

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.