Sock Drawer Problem: Minimum Socks to Guarantee a Pair of Same Color

Given:

• 10 black socks
• 10 brown socks
• All socks are mixed up in the drawer

Goal: Find the smallest number of socks to pick without seeing so that at least one pair of same color socks is guaranteed.

Reasoning:

• Worst case: you pick one black sock and one brown sock → no pair yet.
• So after picking 2 socks, you may have no pair.
• Picking one more sock (3rd sock) ensures that at least two socks must be of the same color (by Pigeonhole Principle).

Answer: The smallest number of socks to pick to guarantee at least one pair of the same color is 3.

Solution: Number of Scooters in Second Half of Row

Total vehicles = 36. Let number of cars = $n$, scooters = $\frac{n(n+1)}{2}$.
Solve $n + \frac{n(n+1)}{2} = 36$ gives $n=7$ cars and 28 scooters (total 35 vehicles).
Since total is 36, assume 1 extra scooter added somewhere.

The vehicle sequence (cars + scooters) goes as:
Car1, 1 scooter; Car2, 2 scooters; ... Car7, 7 scooters.

Half of 36 = 18 vehicles (second half is last 18).
Positions 19 to 36 in the sequence contain:
- Last 3 scooters of the 5th group (positions 18-20)
- All 6 scooters of the 6th group (positions 22-27)
- All 7 scooters of the 7th group (positions 29-35)

Counting scooters in these positions:
$2 + 6 + 7 = 15$ scooters in the second half.

Final answer: 15 scooters.

Given: Total questions = 100, Part B = 23 questions (2 marks each).

Let Part A = $a$ questions (1 mark), Part C = $c$ questions (3 marks).

So, $a + 23 + c = 100 \Rightarrow a + c = 77$.
Total marks = $a + 46 + 3c$.
Part A marks ≥ 60% total marks:
$\Rightarrow a \geq 0.6(a + 46 + 3c)$
$\Rightarrow a \geq 0.6a + 27.6 + 1.8c$
$\Rightarrow 0.4a \geq 27.6 + 1.8c$
$\Rightarrow a \geq 69 + 4.5c$.

Using $a = 77 - c$:
$77 - c \geq 69 + 4.5c \Rightarrow 8 \geq 5.5c \Rightarrow c \leq 1.45$.
So, $c = 1$.

Answer: Part C has 1 question.

Distance Difference Between Akash and Sanjay

Let Akash’s Tuesday distance = $x$ km.
Then Akash’s Monday distance = $x - 4$ km.
Sanjay runs same distance on both days = $y$ km.
Sanjay’s Tuesday distance is 5 km more than Akash’s Monday distance:
$$y = (x - 4) + 5 = x + 1$$

Total distance Akash ran in 2 days:
$$(x - 4) + x = 2x - 4$$ Total distance Sanjay ran in 2 days:
$$y + y = 2y = 2(x + 1) = 2x + 2$$

Difference = Sanjay’s total - Akash’s total:
$$(2x + 2) - (2x - 4) = 6 \text{ km}$$

Answer: The difference in total distances covered is 6 km.

Venn Diagram Problem: Students in Elective Courses

Given:

• Total students = 100
• Math (M) = 47
• Economics (E) = 47
• Sociology (S) = 57
• All three (M ∩ E ∩ S) = 7

Objective:

Find the number of students who enrolled in exactly one course.

Step-by-step:

Let’s use the formula for total union of 3 sets:

Let the number of students who enrolled in exactly 2 courses = x
So those who enrolled in all 3 = 7
Let the number of students who enrolled in exactly one course = y

Then, the total number of course enrollments =
1 × y + 2 × x + 3 × 7 = M + E + S = 47 + 47 + 57 = 151

⇒ y + 2x + 21 = 151
⇒ y + 2x = 130 — (1)

Also, total students = 100 = y + x + 7
⇒ y + x = 93 — (2)

Subtracting (2) from (1):

Now, from (2): y + 37 = 93 ⇒ y = 56

✅ Final Answer:

56 students enrolled in exactly one course.

Logical Puzzle: Cat Climbing a Pole

Problem:

A cat climbs a 21-meter pole. Every minute it climbs 3 meters but slips down 1 meter the next minute. How long will it take to reach the top?

Step-by-Step Analysis:

• Every 2 minutes, net climb = 3 - 1 = 2 meters
• In 20 minutes, the cat climbs 10 × 2 = 20 meters
• At the start of the 21st minute, the cat climbs 3 meters and reaches the top (21 meters) before it can slip.

✅ Final Answer: The cat will reach the top in 21 minutes.

Family Age Puzzle: Arjun's Birth Year

Given:

• Arjun is 25 years younger than his mother.
• His brother was born in 1964 and is 35 years younger than their mother.

Step-by-Step Solution:

• Mother’s birth year = 1964 − 35 = 1929
• Arjun’s birth year = 1929 + 25 = 1954

✅ Final Answer: Arjun was born in 1954.