Explanation : 

There are three matches (each pair plays once):
1. P vs Q
2. P vs R
3. Q vs R

We know:
• A win gives 2 points, a draw 1 point each, a loss 0 points.
• P’s total = 3 pts, Q’s total = 2 pts, R’s total = 1 pt.
• Each team scored exactly 1 goal in the whole tournament.

Step 1: Find who beat whom (points condition)

– P has 3 points from two games ⇒ P must have one win (2 pts) and one draw (1 pt).
– Q has 2 points ⇒ possible splits are (2+0) or (1+1).
– R has 1 point ⇒ one draw and one loss (1+0).

Try the two ways P can get 3 points:

Case A:
• P beats R (P=2 pts, R=0)
• P draws with Q (P=1 pt, Q=1)
⇒ P total 3, Q so far 1.
Q needs 2 → Q–R must be a draw (Q+1=2, R+1=1).
R’s total = 1 (0 from P, 1 from Q).
This fits perfectly.

Case B:
• P beats Q (P=2, Q=0)
• P draws with R (P=1, R=1)
⇒ P total 3, R so far 1.
Q needs 2 points → Q–R must be a win for Q (Q+2=2, R+0=1).
R’s total would be 1 (1 from P, 0 from Q), OK.
But as we’ll see in Step 2, Case B fails the “one goal each” condition.

So only Case A survives the points check.

Step 2: Distribute exactly one goal per team

Under Case A the results are:
P–R: P wins
P–Q: draw
Q–R: draw

Each team must score exactly 1 goal in total:

1. P has two games (win vs R, draw vs Q) but only 1 goal in all ⇒
• In the draw P–Q, P must have scored 0;
• In P–R, P must have scored its single goal ⇒ score 1–0.

2. R has two games (loss vs P, draw vs Q) and must total 1 goal ⇒
• In P–R (lost 0–1) R scored 0;
• In Q–R draw, R scored its 1 goal ⇒ Q–R is 1–1.

3. Q has two draws (with P and R) and needs 1 goal total ⇒
• In P–Q (draw) Q scored 0;
• In Q–R, Q scored its 1 goal ⇒ 1–1.

Final scores:
P–Q: 0–0
P–R: 1–0
Q–R: 1–1

Check statements:

I. “P vs Q is a 0–0 draw.” True.
II. “R scored 1 goal against Q.” True (in the 1–1 draw).

Therefore both I and II are correct.

Explanation : 

1. We only know exactly one of P, Q, R is the thief.
2. Each makes two statements, but we have no rule linking guilt to lying or innocence to truth-telling. They might lie or tell the truth arbitrarily.
3. We can check each candidate:

• Case P is guilty
– P’s statements (“I did not steal,” “Q stole it”) could both be false (he lies twice) or mixed – there’s no rule forbidding that.
– Q and R’s statements can be assigned true/false arbitrarily to fit any needed pattern.

• Case Q is guilty
– Q’s statements (“R did not steal,” “I did not steal”) could both be false (he lies twice) or otherwise.
– P and R’s statements can again be forced to fit.

• Case R is guilty
– R says “I do not know who did it,” which would then be false (the thief knows his own guilt), but he could simply be lying.
– P and Q’s two statements each can be arranged in truth/lie to avoid contradiction.

4. In every scenario you can assign truth/lie values to each person’s two statements so as to avoid an outright contradiction—because we have not constrained who must lie or tell the truth.

Conclusion: With only the facts given (and no further assumptions about lying patterns), the identity of the thief cannot be determined.

Explanation : 

1. Count digits in the 1-digit numbers (1 to 9):
• 9 numbers × 1 digit = 9 digits

2. Count digits in the 2-digit numbers (10 to 99):
• 90 numbers × 2 digits = 180 digits
• Cumulative total so far = 9 + 180 = 189 digits

3. Since 489 > 189, the 489th digit lies in the 3-digit numbers (100, 101, 102, …).
• Digits to go into the 3-digit block: 489 – 189 = 300

4. Each 3-digit number contributes 3 digits.
• 300 ÷ 3 = 100 exactly, so the 300th digit in this block is the 3rd digit of the 100th 3-digit number.
• The first 3-digit number is 100, so the 100th one is 100 + 100 – 1 = 199.

5. The 3rd (last) digit of 199 is 9.

Therefore, the 489th digit is 9.

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.