Explanation : 

1. Assign convenient coordinates to each place, taking east as the +x-direction and north as the +y-direction. For example, let A = (0, 0).
– A is 6 km south of B ⇒ B = (0, 6)
– A is 10 km west of C ⇒ C = (10, 0)
– C is 6 km north of D ⇒ D = (10, –6)
– D is 5 km east of E ⇒ E = (5, –6)
– F is 9 km west of B ⇒ F = (–9, 6)
– F is 12 km north of G ⇒ G = (–9, –6)

2. Compute the lengths of the roads (all align with the axes):
– AB = distance between (0,0) and (0,6) = 6
– AC = distance between (0,0) and (10,0) = 10
– CD = distance between (10,0) and (10,–6) = 6
– DE = distance between (10,–6) and (5,–6) = 5
– BF = distance between (0,6) and (–9,6) = 9
– EG = distance between (5,–6) and (–9,–6) = 14
– FG = distance between (–9,6) and (–9,–6) = 12

3. Two possible routes from D to F along the given roads both give the same total:
• D → C → A → B → F
= DC + CA + AB + BF
= 6 + 10 + 6 + 9
= 31 km

• D → E → G → F
= DE + EG + GF
= 5 + 14 + 12
= 31 km

Hence the distance traveled is 31 km.

Explanation : 

1. List the terms and their differences:
Sequence: 1, 3, 6, 11, 18, X, 42
Differences:
3–1 = 2
6–3 = 3
11–6 = 5
18–11 = 7
X–18 = ?
42–X = ?

2. Observe the differences so far: 2, 3, 5, 7 – these are consecutive primes.

3. The next two primes after 7 are 11 and 13.
So
X – 18 = 11 ⇒ X = 18 + 11 = 29
42 – X = 13 ⇒ 42 – 29 = 13 (checks out)

Thus X = 29.

Answer: (c) 29.

Explanation : 

1. Compute each person’s daily work rate (fraction of the whole work per day):

• X does 1/3 of the work in 6 days ⇒ rate of X = (1/3) ÷ 6 = 1/18
• Y does 1/3 of the work in 8 days ⇒ rate of Y = (1/3) ÷ 8 = 1/24
• Z does 3/4 of the work in 12 days ⇒ rate of Z = (3/4) ÷ 12 = 1/16

2. All three work together for n days. Together they do per day:
1/18 + 1/24 + 1/16
= (8 + 6 + 9) / 144
= 23/144 of the work per day.
In n days they complete 23n/144 of the work.

3. Remaining work after n days = 1 – 23n/144 = (144 – 23n)/144.

4. Then X and Z quit, and Y alone finishes the rest in 8 2/3 days = 26/3 days.
Work done by Y in that time = rate × time = (1/24) × (26/3) = 26/72 = 13/36.

5. Set remaining work = 13/36:

(144 – 23n) / 144 = 13/36
⇒ 144·(13/36) = 144 – 23n
⇒ 4·13 = 144 – 23n
⇒ 52 = 144 – 23n
⇒ 23n = 144 – 52 = 92
⇒ n = 92 / 23 = 4.

Answer: (b) 4

Explanation : 

1. The 5‐digit number is P Q R S T with all digits distinct and T ≠ 0.
2. Given:
• P = 3·T
• Q = R + 3
• S = Q + 4

3. Since P=3T must be a single digit (0–9) and T≠0, the only possibilities for T are 1, 2 or 3:
– T=1 ⇒ P=3
– T=2 ⇒ P=6
– T=3 ⇒ P=9

4. Also Q=R+3 ≤9 ⇒ R≤6, and S=Q+4=R+7 ≤9 ⇒ R≤2. Hence R∈{0,1,2}.

5. Check each T and each R∈{0,1,2}, discarding any repeat digits:

Case A: T=1, P=3
R=0 ⇒ Q=3 (conflicts with P=3) ✗
R=2 ⇒ Q=5, S=9 → digits {3,5,2,9,1} all distinct ✓
→ Number = 35291

Case B: T=2, P=6
R=0 ⇒ Q=3, S=7 → {6,3,0,7,2} distinct ✓ → 63072
R=1 ⇒ Q=4, S=8 → {6,4,1,8,2} distinct ✓ → 64182
R=2 ⇒ Q=5, S=9 → conflicts with T=2 or repeats none but R=2=T? actually R=2 conflicts with T=2 ✗

Case C: T=3, P=9
R=0 ⇒ Q=3 conflicts with T=3 ✗
R=1 ⇒ Q=4, S=8 → {9,4,1,8,3} distinct ✓ → 94183
R=2 ⇒ Q=5, S=9 conflicts with P=9 ✗

6. Valid numbers found: 35291, 63072, 64182, 94183 → total 4.

Answer: (b) Only 4 such numbers.

Explanation : 

1. From “FRANCE → 654321” we get a letter→digit mapping:
F→6, R→5, A→4, N→3, C→2, E→1

2. From “GERMANY → 9158437” the same mapping must hold for letters already seen, and it tells us the new letters’ codes:
G→9, E→1 (matches), R→5, M→8, A→4, N→3, Y→7

3. Now code “YEMEN” by replacing each letter with its digit:
Y → 7
E → 1
M → 8
E → 1
N → 3

So YEMEN → 71813

Answer: (d) 71813

Explanation : 

We notice that each code is the product of the letters’ positions in the alphabet:

• N = 14
• O = 15
• T = 20
• E = 5
• S = 19

Check the given examples:

1. NO → 14 × 15 = 210
2. NOT → 14 × 15 × 20 = 210 × 20 = 4200
3. NOTE → 14 × 15 × 20 × 5 = 4200 × 5 = 21000

Therefore, for NOTES:

NOTES → 14 × 15 × 20 × 5 × 19
= (14 × 15 × 20 × 5) × 19
= 21000 × 19
= 399000

Answer: (a) 399000

Explanation : 

Let’s first restate the given facts in “family‐tree” form:

1. P is the brother of Q and of R.
⇒ P is male, and (in a standard puzzle) P, Q and R share the same two parents.
2. S is R’s mother.
⇒ S is the mother of R (and hence, of P and Q).
3. T is P’s father.
⇒ T is the father of P (and hence, of Q and R).

So the (full) parents of P, Q, R are S (mother) and T (father). Now check each statement:

I. S and T are a couple.
– True (they are mother and father of the same children).

II. Q is T’s son.
– Not necessarily: we know Q is T’s child, but Q’s gender wasn’t given.

III. T is Q’s father.
– True (T is father of P, Q and R).

IV. S is P’s mother.
– True (S is the mother of R, and P is R’s full brother ⇒ same mother).

V. R is T’s daughter.
– Not necessarily: R is T’s child, but R’s gender wasn’t given.

VI. P is S’s son.
– True (P is male and S is the mother of P, Q, R).

Counting the definitely true ones: I, III, IV and VI ⇒ 4 statements.

Answer: (c) Only four.

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.