Explanation : 

Numeric Lock 3 digit PIN Combinations

All Possible Values for a 3-digit PIN are 753, 752, 751, 742, 741, 731, 642, 641, 631, 531.
So, the maximum number of attempts one need to find out the PIN with certainty is 10.

Explanation : 

X and Y run a 3 km race

The faster runner will cross the slower one when he covers an extra 300 m. Let their speeds be 3 m/sec and 2 m/sec. So, their relative speed = 3 – 2 = 1 m/sec. So, the time taken by the faster runner to cross the slower one = Distance/Relative Speed = 300/1 = 300 seconds.

It basically means that the faster runner will cross the slower one every 300 seconds or 5 minutes. Now, the time taken for the faster racer to complete the entire race = Total Distance/Speed = 3000/3 =1000 seconds.

So, during the entire race, which lasts for 1000 seconds, the faster racer will cross the slower one 3 times – after 300 seconds, 600 seconds, and 900 seconds.

Explanation : 
Let’s assume the initial price of the item to be x.

After a 25% increase in price, the new price will be:
x + 0.25x = 1.25x

Next, the price is decreased by 20%. The new price will be:
0.8(1.25x) = 1x

Finally, the price is increased by 10%. The new price will be:
1.1(1x) = 1.1x

To find the resultant increase in price

we need to find the percentage increase from the initial price x to the final price 1.1x:
Resultant increase = [(1.1x – x) / x] x 100%
= (0.1x / x) x 100%
= 10%

Therefore, the resultant increase in the price is 10%. The correct answer is (b).

Explanation : 

In a tournament of Chess

In a knockout chess tournament like this, each player is eliminated after losing a match. Since there are 150 entrants, only one player remains undefeated at the end of the tournament, and this player is the winner.

To find out how many matches are played, we can think about it in terms of elimination: for every player except the winner, they have to lose exactly one match to be eliminated. Since there are 150 entrants and only one winner, 150 – 1 = 149 players need to be eliminated.

Each elimination comes from one match, so there are 149 matches played in the entire tournament.

Explanation : 

Which number among is smallest:

2^40, 3^21, 4^18, and 8^12

We can also write them as: 2^40, 3^21, 2^36, and 2^36.

We can rewrite 2^36 and 3^21 as:
2^12 and 3^7
4096 > 2187
Hence, 3^21 is the smallest number.

Explanation : 

A bill for 1840

Statement 1: 25 notes of Rs. 50 were used. So, remaining amount = 1840 – (25 × 50) = 1840 – 1250 = Rs. 590.
Even if all the remaining 25 notes are of Rs. 20 denomination, we will only get Rs. 500. So, Statement 1 is definitely incorrect.

Statement 2: 35 notes of Rs. 20 were used. So, remaining amount = 1840 – (35 × 20) = 1840 – 700 = Rs. 1140.
Even if all the remaining 15 notes are of Rs. 50 denomination, we will only get Rs. 750. So, Statement 2 is definitely incorrect.

Statement 3: 20 notes of Rs. 10 were used. So, remaining amount = 1840 – (20 × 10) = 1840 – 200 = Rs. 1640.Even if all the remaining 30 notes are of Rs. 50 denomination, we will only get Rs. 1500. So, Statement 3 is definitely incorrect.

Explanation : 

An Identity Card has the number ABCDEFG

The number ABCDEFG is divisible by 9. Since the sum of the digits of a number divisible by 9 is also divisible by 9, we know that A+B+C+D+E+F+G = 1+2+4+5+7+8+9 = 36.

After deleting the first digit from the right (G), the resulting number ABCDEF is divisible by 6. This means that F must be even (2 or 8) and the sum of the digits (A+B+C+D+E+F) must be divisible by 3.

After deleting two digits from the right (FG), the resulting number ABCDE is divisible by 5. This means that E must be 5.

After deleting three digits from the right (EFG), the resulting number ABCD is divisible by 4. For a number to be divisible by 4, the last two digits must form a number divisible by 4. So, CD must be one of the following pairs: 12, 24, 28, 52, 72, 92.

After deleting four digits from the right (EFGD), the resulting number ABC is divisible by 3. This means that the sum of A, B, and C must be divisible by 3.

After deleting five digits from the right (EFGDC), the resulting number AB is divisible by 2. This means that B must be even (2 or 8).

Now let’s use this information to find the possible value for the sum of the middle three digits (C, D, and E).

Since E is 5, and B is even (2 or 8), we can eliminate the pairs 52 and 92 for CD, as they would require B to be odd. We are left with these possibilities for CD: 12, 24, 28, 72.

Consider CD = 12. Then, A = 7, B = 8, C = 1, D = 2, E = 5, F = 4, and G = 9. The conditions for divisibility by 9, 6, 5, 4, 3, and 2 are satisfied with 7815249. The sum of the middle three digits (C, D, and E) is 1+2+5 = 8.

In this case, the correct answer is (a) 8.