Word Set Analysis: Meaningful Word Formation

Method: For each word, arrange letters alphabetically and take the third letter from left. Combine these letters to check if they form a meaningful word.

1. SAME, ROOM, BEST, AUTO

• SAME: A, E, M, S → 3rd letter = M
• ROOM: M, O, O, R → 3rd letter = O
• BEST: B, E, S, T → 3rd letter = S
• AUTO: A, O, T, U → 3rd letter = T

Letters: M O S T → "MOST" (Meaningful)

2. GOAT, PEST, WATT, ARMY

• GOAT: A, G, O, T → 3rd letter = O
• PEST: E, P, S, T → 3rd letter = S
• WATT: A, T, T, W → 3rd letter = T
• ARMY: A, M, R, Y → 3rd letter = R

Letters: O S T R → No meaningful English word from these 4 letters

3. MALE, FIND, LOST, THAT

• MALE: A, E, L, M → 3rd letter = L
• FIND: D, F, I, N → 3rd letter = I
• LOST: L, O, S, T → 3rd letter = S
• THAT: A, H, T, T → 3rd letter = T

Letters: L I S T → "LIST" (Meaningful)

4. JUMP, LIME, DUMB, SOME

• JUMP: J, M, P, U → 3rd letter = P
• LIME: E, I, L, M → 3rd letter = L
• DUMB: B, D, M, U → 3rd letter = M
• SOME: E, M, O, S → 3rd letter = O

Letters: P L M O → No meaningful English word

Answer: The word sets that do NOT give a meaningful word are:

• GOAT, PEST, WATT, ARMY
• JUMP, LIME, DUMB, SOME

Statements Given:

• Rajesh will not go to the concert if Rakesh goes.
• Rakesh will go to the concert if his dog barks three times.

Logical Interpretation:

• If Rakesh goes → Rajesh will not go
• If Dog barks 3 times → Rakesh will go

Deduction:

If the dog barks three times, then Rakesh goes. And if Rakesh goes, Rajesh will not go.

✅ Final Answer: If the dog barks three times, Rajesh will not go to the concert.

Minimum Coins Required

Coins available: 2p, 5p, 10p, 25p, 50p

1) For 79 paise: 50p(1) + 25p(1) + 2p(2) = 4 coins

2) For 66 paise: 50p(1) + 10p(1) + 2p(3) = 5 coins

3) For Re 1.01 (101 paise): 50p(1) + 25p(2) + 10p(2) + 2p(3) = 8 coins
Total = 4+5+8=17 coins

Logic Puzzle: Yes and No Types on Kha-Kha Island

Given:

• “Yes” type always asks questions where the truthful answer is Yes.
• “No” type always asks questions where the truthful answer is No.

Scenario:

Kumar asks: "Is at least one of us of type 'No'?"

Analyze:

Let’s consider possibilities:

1. Assume Kumar is “Yes” type:
• Question must have truthful answer = “Yes”.
• So, “At least one of us is ‘No’” is true.
• Therefore, Kevin or Kumar is “No” type.
• Since Kumar is “Yes” type, Kevin must be “No” type.
2. Assume Kumar is “No” type:
• Question must have truthful answer = “No”.
• So, “At least one of us is ‘No’” is false.
• That means neither Kumar nor Kevin is “No” type.
• This contradicts Kumar being “No” type.
• So, this is impossible.

Conclusion:

Kumar is a “Yes” type, and Kevin is a “No” type.

Quick Solution:

Let number of friends planned = $n$

Price $P =$ between 1000 and 1100

Equation:

$\displaystyle \frac{P}{n-2} - \frac{P}{n} = 1 \implies P = \frac{n(n-2)}{2}$

Try $n = 46$:

$P = \frac{46 \times 44}{2} = 1012$ (valid)

Number of friends who actually contributed = $n - 2 = 44$

Caterpillar Climbing Problem

Pole height = 75 inches
Climbs up = 5 inches/day
Slides down = 4 inches/night

Net climb per day (after sliding): 5 - 4 = 1 inch

On the last day, caterpillar reaches the top without sliding back.

So, after n days, it reaches or passes 75 inches.

At the start of the n^th day, it has climbed:

(n - 1) × 1 = n - 1 inches

On the n^th day, it climbs 5 inches:

(n - 1) + 5 ≥ 75

Solving for n:

n - 1 + 5 ≥ 75
n + 4 ≥ 75
n ≥ 71

Answer: The caterpillar reaches the top on the 71^st day.