Explanation : 

A has some coins

Let’s solve this step by step:

1. Initial Condition:
– A has x coins
– A gives half of coins plus 2 to B

2. After first transfer (A to B):
– A has: x/2 – 2 coins
– B has: x/2 + 2 coins

3. After second transfer (B to C):
– B gives half plus 2 to C
– B has: (x/2 + 2)/2 – 2 = x/4 – 1 coins
– C has: (x/2 + 2)/2 + 2 = x/4 + 3 coins

4. After third transfer (C to D):
– C gives half plus 2 to D
– C has: (x/4 + 3)/2 – 2 = x/8 – 1/2 coins
– D has: (x/4 + 3)/2 + 2 = x/8 + 7/2 coins

5. Given condition:
– D has smallest two-digit number (10)
– Therefore: x/8 + 7/2 = 10

6. Solve for x:
– x/8 + 7/2 = 10
– x/8 = 10 – 7/2
– x/8 = 6.5
– x = 52

Verification:
– A starts with 52 coins
– B gets 28 coins (26 + 2), A keeps 24
– C gets 16 coins (14 + 2), B keeps 12
– D gets 10 coins (8 + 2), C keeps 6

Answer: A had 52 coins initially

Explanation : 

101 plants on a 1.01 km road

Given information:
– Road length: 1.01 km = 1010 meters
– Total plants: 101
– Plants are equally spaced
– First plant at 0 meters
– Last plant at 1010 meters

Solution steps:

1. Find total gaps between plants:
– Total gaps = Total plants – 1
– Total gaps = 101 – 1 = 100 gaps

2. Find distance between consecutive plants:
– Distance = Total length ÷ Number of gaps
– Distance = 1010 ÷ 100 = 10.1 meters

3. Find distance for 5 consecutive plants:
– Number of gaps = Number of plants – 1
– Number of gaps = 5 – 1 = 4 gaps
– Total distance = Distance per gap × Number of gaps
– Total distance = 10.1 × 4 = 40.4 meters

Answer: The distance between 5 consecutive plants is 40.4 meters

Explanation : 

How many 3 digit natural numbers ??

Given conditions:
– Must be a 3-digit number
– All digits must be odd
– No repetition of digits allowed
– Number must be divisible by 5

Solution steps:

1. For divisibility by 5:
– Last digit must be either 0 or 5
– Since only odd digits allowed, last digit must be 5

2. Available odd digits: 1, 3, 5, 7, 9
– 5 is already used in units place
– Remaining digits for other positions: 1, 3, 7, 9

3. Filling other positions:
– Hundreds place: 4 choices (1, 3, 7, 9)
– Tens place: 3 choices (remaining digits after hundreds place)

4. Total possible numbers:
– Using multiplication principle
– 4 × 3 = 12 different numbers

Answer: 12 three-digit numbers possible

Explanation : 

Is x an integer?

Statement-1: x/3 is not an integer

– This tells us x cannot be evenly divided by 3
– Examples: x could be 2/3 (not an integer) or 4 (integer)
– Statement-1 alone is not sufficient

Statement-2: 3x is an integer

– This means x could be an integer or a fraction multiple of 1/3
– Examples: if 3x = 6, then x = 2 (integer)
if 3x = 2, then x = 2/3 (not integer)
– Statement-2 alone is not sufficient

Combined Analysis:
– Statement-1: x cannot be evenly divided by 3
– Statement-2: 3x is an integer
– Together they tell us x must be a fraction multiple of 1/3
– Still cannot determine if x is definitely an integer

Answer: Both Statement-1 and Statement-2 are not sufficient to answer the Question.

Explanation : 

Letters A B C D and E are arranged

The problem asks for the number of arrangements of the letters A, B, C, D and E such that there are exactly two letters between A and E. There are four possible arrangements for A and E: A _ _ E _, _ A _ _ E, E _ _ A _, and _ E _ _ A.

For each of these arrangements, there are 3! = 6 ways to arrange the remaining letters B, C and D. So the total number of arrangements is 4 * 6 = 24.

Explanation : 

In the series aababcabcdabcde , 100th letter ?

Given Series = A (1 Letter) AB (2 Letters) ABC (3 Letters) ABCD (4 Letters) ABCDE (5 Letters) …… so on up to 100th letter.
Sum of first 13 natural numbers = 1 + 2 + 3 + 4 + 5 ….. 13 = n(n+1)/2 = 13×14/2 = 91.
So, if we have to count upto 100th letter, in the last string (100 – 91) = 9 letters would be there.
Which would be – ABCDEFGHI.
So I is the correct answer.

Explanation : 

If 15*14*13

First, let’s compute the prime factorization of 15!:
15! = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Now, let’s break down each of these factors into their prime factors:
15! = (3×5) × (2×7) × 13 × (2^2×3) × 11 × (2×5) × (3^2) × (2^3) × 7 × (2×3) × 5 × (2^2) × 3 × 2
15! = 2^11 × 3^6 × 5^3 × 7^2 × 11 × 13

We want to express this as the product of two integers, 3m and n. Since we want to maximize m, we should try to include as many factors of 3 in 3m as possible. There are 6 factors of 3 in the prime factorization of 15!, so the maximum value of m is 6.

Therefore, the maximum value of m is 6 (option b).

Explanation : 

A B C are three places

One can travel from A to C in 15 different ways. There are two possible routes: directly from A to C or from A to B and then from B to C.

The number of ways to travel directly from A to C is 3 (since there are 3 different roads). The number of ways to travel from A to B and then from B to C is the product of the number of roads between each pair of places: 3 x 4 = 12.

So, in total, there are 3 + 12 = 15 different ways one can travel from A to C using these roads.

Explanation : 

Total number of seconds in x weeks, x days, x hours, x minutes and x seconds is 694861x.

So the correct answer is d. 694861x.

Here’s how you can calculate it: 1 week = 7 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds

So, x weeks = (7 * 24 * 60 * 60) * x seconds x days = (24 * 60 * 60) * x seconds x hours = (60 * 60) * x seconds x minutes = (60) * x seconds

Adding all of these together with the additional x seconds gives us: 604800x + 86400x + 3600x + 60x + x=694861x

Explanation : 

June 2099 – Sunday

Prelims 2022 – 5-JUN-2022 Sunday.

First, calculate the number of years between 2022 and 2099, which is 2099 – 2022 = 77 years.

Determining Number of Leap Years

Next, determine the number of leap years in this period. Leap years are typically every 4 years, but we must exclude centuries that are not divisible by 400. The years from 2022 to 2099 include the following leap years: 2024, 2028, 2032, 2036, 2040, 2044, 2048, 2052, 2056, 2060, 2064, 2068, 2072, 2076, 2080, 2084, 2088, 2092, 2096. That’s a total of 19 leap years.

Calculating Total Number of Odd Days

Each normal year contributes 1 odd day (365 days = 52 weeks + 1 day), and each leap year contributes 2 odd days (366 days = 52 weeks + 2 days).

So, the total number of odd days in 77 years is:

• Odd days from normal years = 58 normal years × 1 odd day = 58 odd days
• Odd days from leap years = 19 leap years × 2 odd days = 38 odd days
• Total odd days = 58 + 38 = 96 odd days

Determining the Day of the Week

Since each week has 7 days, we find the remainder of 96 divided by 7 to get the number of odd days:

96 mod 7 = 5 odd days

Since June 5, 2022, was a Sunday, adding 5 odd days means:

Sunday + 5 days = Friday

Thus, June 5, 2099, will be a Friday.