Explanation : 

1. Express P in terms of Q:
P = QQQ = 100·Q + 10·Q + Q = 111·Q

2. Factor 111 and 481:
111 = 3 × 37
481 ÷ 13 = 37, so 481 = 13 × 37

3. Compute HCF(P, 481):
HCF(P, 481) = gcd(111·Q, 481)
= gcd(3·37·Q, 13·37)
= 37 · gcd(3.Q, 13)

Since Q is a digit 1–9, 13 does not divide 3Q, so gcd(3.Q, 13) = 1.
Therefore HCF(P, 481) = 37.

Answer: (c) 37

Explanation : 

1. Observe that 1ⁿ is always 1, whatever n is.
2. So we need only look at 2ⁿ mod 4, then add 1 and reduce mod 4.

– n = 1: 2¹ = 2 ⇒ 2 mod 4 = 2 ⇒ 1 + 2 = 3 mod 4
– n = 2: 2² = 4 ⇒ 4 mod 4 = 0 ⇒ 1 + 0 = 1 mod 4
– n = 3: 2³ = 8 ⇒ 8 mod 4 = 0 ⇒ 1 + 0 = 1 mod 4
– For any n ≥ 2, 2ⁿ is a multiple of 4, so 2ⁿ mod 4 = 0, giving 1 + 0 = 1 mod 4.

3. Hence the only remainders that occur are 3 (when n = 1) and 1 (for all n ≥ 2).

Number of distinct remainders = 2.

Explanation : 

1. Compute each set’s filling or emptying rate (in “tank‐fractions per minute”):

• Set X: 20 pipes fill 0.70 of the tank in 14 min
– Combined rate of X = 0.70 ÷ 14 = 0.05 tank/min
– Rate per X-pipe = 0.05 ÷ 20 = 0.0025 tank/min

• Set Y: 10 pipes fill 3/8 = 0.375 of the tank in 6 min
– Combined rate of Y = 0.375 ÷ 6 = 0.0625 tank/min
– Rate per Y-pipe = 0.0625 ÷ 10 = 0.00625 tank/min

• Set Z: 16 pipes empty 0.5 of the tank in 20 min
– Combined emptying rate of Z = 0.5 ÷ 20 = 0.025 tank/min
– Rate per Z-pipe = 0.025 ÷ 16 = 0.0015625 tank/min

2. In the new scenario:
– Half of X’s pipes are in use ⇒ 10 X-pipes ⇒ filling at 10×0.0025 = 0.025 tank/min
– Half of Y’s pipes are in use ⇒ 5 Y-pipes ⇒ filling at 5×0.00625 = 0.03125 tank/min
– All Z-pipes are open ⇒ 16 Z-pipes emptying at 16×0.0015625 = 0.025 tank/min

3. Net filling rate = (X filling) + (Y filling) – (Z emptying)
= 0.025 + 0.03125 – 0.025 = 0.03125 tank/min

4. Time to fill 50% = required fraction ÷ net rate
= 0.50 ÷ 0.03125 = 16 minutes

Therefore the correct answer is 16 minutes.

Explanation : 

1. Observe the given square is a 17-digit palindrome:
12345678987654321.

2. Recall a well-known fact about “repunit” numbers (numbers consisting only of 1’s):
(111111111)² = 12345678987654321.

3. Thus N = 111 111 111, which clearly has 9 digits.

Explanation : 

1. Let the speed of the tram be v km/h and its length be L metres.

2. Speeds of X and Y in km/h:
X: 3 km/h
Y: 4 km/h

3. When overtaking a walker, the tram’s relative speed (in km/h) is (v − walk_speed).
Time to pass completely (in seconds) × relative speed (in km/h) = length (in km), then convert to metres.

4. For X:
• Relative speed = (v − 3) km/h
• Time = 8 s = 8/3600 h
• Distance (tram length L in km) = (v − 3)·(8/3600)

5. For Y:
• Relative speed = (v − 4) km/h
• Time = 9 s = 9/3600 h
• Distance (same L) = (v − 4)·(9/3600)

6. Equate the two expressions for L:
(v − 3)·(8/3600) = (v − 4)·(9/3600)

⇒ 8(v − 3) = 9(v − 4)
⇒ 8v − 24 = 9v − 36
⇒ 36 − 24 = 9v − 8v
⇒ v = 12 km/h

7. Now find L (use X’s data):
L (in km) = (12 − 3)·(8/3600) = 9·(8/3600) = 72/3600 = 1/50 km
L (in metres) = (1/50)·1000 = 20 m

Therefore, the length of the tram is 20 m.

Explanation : 

1. Map the letters to operations:
P = +
Q = –
R = ×
S = ÷

2. Write down the option QRPS in sequence between the numbers 60, 15, 3, 20, 4:
60 Q 15 R 3 P 20 S 4
→ 60 – 15 × 3 + 20 ÷ 4

3. Apply the usual order of operations (× and ÷ before + and –):
15 × 3 = 45
20 ÷ 4 = 5

So the expression becomes:
60 – 45 + 5

4. Compute left to right:
60 – 45 = 15
15 + 5 = 20

Thus, 60 – 15 × 3 + 20 ÷ 4 = 20, confirming QRPS is correct.

Explanation : 

1. Let t be the time in hours after 5:00 a.m.
2. Since P and Q walk in opposite directions, their relative speed (in rounds per hour) is
 5 + 3 = 8 rounds/hour.
3. They meet each time their combined distance equals an integer number of laps:
 8·t = k , where k = 1, 2, 3, …
so the k-th meeting happens at
 t = k/8 hours after 5:00.

4. We want all meetings between 5:20 a.m. and 7:00 a.m.
• 5:20 a.m. corresponds to t = 20/60 = 1/3 ≃ 0.3333 hours.
• 7:00 a.m. corresponds to t = 2 hours.

So we need
 1/3 ≤ k/8 ≤ 2
Multiply through by 8:
 8/3 ≤ k ≤ 16

• 8/3 ≃ 2.667 ⇒ the smallest integer k is 3.
• The largest k is 16.

5. Thus k = 3, 4, 5, …, 16. How many integers is that?
From 3 up to 16 inclusive there are 16 – 3 + 1 = 14 meetings.

Therefore, they cross each other 14 times between 5:20 and 7:00.

Explanation :

1. Write down the six terms as
a₁ = 24,
a₂ = X,
a₃ = 12,
a₄ = 18,
a₅ = 36,
a₆ = 90.

2. Observe the multipliers from one term to the next (unknown for the first two steps):
a₁ → a₂ multiply by m₁
a₂ → a₃ multiply by m₂
a₃ → a₄ multiply by m₃ = 18/12 = 1.5
a₄ → a₅ multiply by m₄ = 36/18 = 2
a₅ → a₆ multiply by m₅ = 90/36 = 2.5

3. Notice m₃, m₄, m₅ form an arithmetic sequence increasing by 0.5:
m₃ = 1.5 = 3·0.5,
m₄ = 2.0 = 4·0.5,
m₅ = 2.5 = 5·0.5.
It is natural to extend backward to
m₂ = 2·0.5 = 1.0,
m₁ = 1·0.5 = 0.5.

4. Compute a₂ and check a₃:
a₂ = a₁·m₁ = 24·0.5 = 12,
a₃ should be a₂·m₂ = 12·1.0 = 12,
which matches the given third term.

Hence X = a₂ = 12.

Explanation : 

We need the largest integer n such that the product

P = 7 × 343 × 385 × 1000 × 2401 × 77777

is divisible by 35ⁿ. Since 35 = 5·7, 35ⁿ = 5ⁿ·7ⁿ, so n cannot exceed the exponent of 5 in P or the exponent of 7 in P. We compute those exponents term by term.

1. Factor each term by 5 and 7:

• 7 = 7¹
• 343 = 7³
• 385 = 5¹·7¹·11
• 1000 = 2³·5³
• 2401 = 7⁴
• 77777 = 7·11111 = 7¹·41·271

2. Total exponent of 7 in P:

v₇(P) = 1 (from 7)
+ 3 (from 343)
+ 1 (from 385)
+ 0 (from 1000)
+ 4 (from 2401)
+ 1 (from 77777)
= 10

3. Total exponent of 5 in P:

v₅(P) = 0 (from 7)
+ 0 (from 343)
+ 1 (from 385)
+ 3 (from 1000)
+ 0 (from 2401)
+ 0 (from 77777)
= 4

4. Since 35ⁿ = 5ⁿ·7ⁿ, the maximum n is the smaller of v₅(P) and v₇(P):

nₘₐₓ = min(4, 10) = 4.

Therefore, the product is divisible by 35⁴ but not by 35⁵, so the answer is 4.

Explanation : 

1. Mark the blanks with underscores:
AB _ CC _ A _ BCCC _ BBC _ C
There are 5 blanks at positions 3, 6, 8, 13 and 17.

2. Notice the total length of the filled‐in sequence will be 18 letters. If you cut it into six consecutive blocks of 3 letters each, the pattern emerges. Label the blocks by their letter‐positions:

Block 1 = positions 1–3 = A B _
Block 2 = positions 4–6 = C C _
Block 3 = positions 7–9 = A _ B
Block 4 = positions 10–12 = C C C
Block 5 = positions 13–15 = _ B B
Block 6 = positions 16–18 = C _ C

3. The sequence of blocks follows the simple alternating pattern:
ABB, CCC, ABB, CCC, ABB, CCC

4. Fill each blank so each block matches that pattern:

– Block 1 “A B _” must be “A B B” ⇒ blank 3 = B
– Block 2 “C C _” must be “C C C” ⇒ blank 6 = C
– Block 3 “A _ B” must be “A B B” ⇒ blank 8 = B
– Block 4 is already “C C C” (no blank)
– Block 5 “_ B B” must be “A B B” ⇒ blank 13 = A
– Block 6 “C _ C” must be “C C C” ⇒ blank 17 = C

5. Reading the five blanks in order gives B, C, B, A, C – that is option (c).

Thus the completed sequence is
ABB CCC ABB CCC ABB CCC
which clearly shows the alternating ABB/CCC pattern.