Explanation : 

We have a set of 11 consecutive integers in each case.

Value-I: “Minimum average when the 11 numbers are consecutive integers ≥ –5.”
1. If the 11 numbers are consecutive and all ≥ –5, the smallest possible sequence is
−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5.
2. The average of 11 equally spaced numbers is the middle (6th) term. Here the 6th term is 0.
3. Any shift upward (starting at −4 or higher) would increase every term, hence increase the average.
⇒ Minimum average = 0.
So Value-I = 0.

Value-II: “Minimum product when the 11 numbers are consecutive non-negative integers.”
1. Let the sequence be k, k+1, …, k+10 with k ≥ 0.
2. If k = 0, the sequence is 0, 1, 2, …, 10, and the product is 0.
3. For any k ≥ 1, all terms are positive and the product > 0.
⇒ Minimum product = 0.
So Value-II = 0.

Comparison: Value-I = 0 and Value-II = 0, hence
Answer: (c) Value-I = Value-II.

Explanation : 

We are given p + q = 10, where p and q are integers.

Value-I: Maximum of p·q when p, q are positive integers
1. Positive integers mean p ≥ 1 and q ≥ 1.
2. Since p + q = 10, possible pairs (p,q) are:
(1,9) → 1·9 = 9
(2,8) → 2·8 = 16
(3,7) → 3·7 = 21
(4,6) → 4·6 = 24
(5,5) → 5·5 = 25
(6,4) etc. (symmetric, same products).
3. The maximum product among these is 25 (at p = 5, q = 5).
So Value-I = 25.

Value-II: Maximum of p·q when p ≥ –6, q ≥ –4 (still with p + q = 10)
1. From p + q = 10, we have q = 10 – p.
2. The constraint q ≥ –4 gives 10 – p ≥ –4 ⇒ p ≤ 14.
Combined with p ≥ –6, we get p ∈ {–6, –5, …, 14}.
3. The product is
p·q = p(10 – p) = –p² + 10p,
which is a downward-opening parabola in p.
4. The vertex of –p² + 10p occurs at p = (10)/(2) = 5, giving
p·q = 5·(10 – 5) = 25.
5. Checking the endpoints (p = –6 or p = 14) yields negative products, so the maximum is indeed 25 at p = 5, q = 5.
So Value-II = 25.

Comparison: Value-I = 25 and Value-II = 25, hence
Answer: (c) Value-I = Value-II.

Explanation : 

1. Let X, Y, Z be the runs scored by players X, Y, Z respectively.
We know
X + Y + Z = 37
X/Y = Y/Z.

2. From X/Y = Y/Z ⇒ X·Z = Y².
A standard integer‐solution parametrization is
X = d·r²,
Y = d·r·s,
Z = d·s²,
where d, r, s are positive integers.

3. Then
X + Y + Z = d (r² + r·s + s²) = 37.
Since 37 is prime, the only way to factor is d = 1 and r² + r·s + s² = 37.

4. Solve r² + r·s + s² = 37 in positive integers (r, s):
– Try s = 3 ⇒ r² + 3r + 9 = 37 ⇒ r² + 3r − 28 = 0 ⇒ r = 4 (positive root).
– Or by symmetry r = 3, s = 4 also works (3² + 3·4 + 4² = 9 + 12 + 16 = 37).

5. Thus two possible triples (d = 1):
a) (r, s) = (4, 3) ⇒ (X, Y, Z) = (16, 12, 9) ⇒ X > Y > Z.
b) (r, s) = (3, 4) ⇒ (X, Y, Z) = ( 9, 12,16) ⇒ X < Y < Z.

6. These give different orderings of Value-I (X), Value-II (Y), Value-III (Z), so we cannot decide uniquely which is smallest, middle or largest.

Therefore the data are insufficient to fix the ordering; the correct choice is (d).

Explanation : 

1. Expand the expression:
(p + q)² – 4pq
= p² + 2pq + q² – 4pq
= p² – 2pq + q²
= (p – q)².

2. We ask: Is (p – q)² positive?
– (p – q)² is zero exactly when p = q.
– It is positive whenever p ≠ q.

3. Examine each statement:

Statement I: p < q
→ Then p ≠ q, so (p – q)² > 0.
→ The expression is positive.

Statement II: p > q
→ Again p ≠ q, so (p – q)² > 0.
→ The expression is positive.

Each statement by itself guarantees p ≠ q, hence the expression is positive. Therefore, either statement alone is sufficient to answer the question.

Final choice: (b) The Question can be answered by using either Statement alone.

Explanation : 

Let
Gq₀ = goals scored by Q up to the 80th minute,
Gp₀ = goals scored by P up to the 80th minute.

We know at 80′ Q was ahead by 3 goals:
Gq₀ − Gp₀ = 3 …(1)

Call in the last 10 minutes:
P scores 4 goals (Statement I),
Q scores B goals (unknown).

Thus final tallies are
Gp_f = Gp₀ + 4,
Gq_f = Gq₀ + B.

Statement II says total goals by Q in whole match is 4, so
Gq_f = 4 ⇒ Gq₀ + B = 4. …(2)

From (1): Gp₀ = Gq₀ − 3, so
Gp_f = (Gq₀ − 3) + 4 = Gq₀ + 1.

Then the final goal‐difference is
Gp_f − Gq_f = (Gq₀ + 1) − 4 = Gq₀ − 3.

But from (2) and non‐negativity of B, we get Gq₀ is either 3 or 4 (since Gq₀ + B = 4 and B ≥ 0).

Case 1: Gq₀ = 4, B = 0 ⇒ Gp_f − Gq_f = 1 ⇒ P wins.
Case 2: Gq₀ = 3, B = 1 ⇒ Gp_f − Gq_f = 0 ⇒ It’s a draw.

Since with both statements together P may win or may only draw, we cannot be sure.
Neither statement alone gives a unique answer, and even together they leave two possibilities. Hence (d).

Explanation : 

Assess Statement I alone:
I. “P has two sisters, R and S.”
– This tells us P’s siblings but nothing about Q.
⇒ Statement I alone is insufficient.

Assess Statement II alone:
II. “R’s father is the brother of Q.”
– R’s father is also P’s father (since R is P’s sister).
– So P’s father is the brother of Q ⇒ Q is a sibling of P’s father ⇒ Q is either P’s paternal uncle or paternal aunt.
– But we do not know Q’s gender ⇒ cannot decide uncle vs. aunt.
⇒ Statement II alone is insufficient to pin down the exact relationship.

Combine I and II:
– I only tells us R and S are sisters of P; II tells us Q is a sibling of their father.
– Even together, Q could be male or female ⇒ still “paternal uncle / aunt” ambiguous.
⇒ Both statements together remain insufficient to give a definite single relationship.

Therefore, the question cannot be answered even using any of the statements.
Correct option: (d).

Explanation : 

We need to find digits P,Q,R,S (all nonzero and distinct), with P≤3, Q≤4, such that
  PP × PQ = RRSS
(where “PP” means the two-digit number with both digits P, etc.).

Step 1. List possible values of PP (since P≤3 and nonzero):
 P=1 → PP=11
 P=2 → PP=22
 P=3 → PP=33

Step 2. For each P, try Q (1 through 4, but Q≠P) and compute PP×PQ. We need a 4-digit product of the form RRSS.

1) P=1 → PP=11, PQ=10+Q gives 11×11=121 up to 11×14=154, all 3-digit. No good.

2) P=2 → PP=22, PQ=20+Q gives 22×21=462, 22×23=506, 22×24=528, all 3-digit. No good.

3) P=3 → PP=33, PQ=30+Q:
 33×31 = 1023 (not of form RRSS)
 33×32 = 1056 (no)
 33×34 = 1122 → this is “RRSS” with R=1, S=2, and all digits {3,4,1,2} are distinct.

No other choice works. Hence the only solution is P=3, Q=4, R=1, S=2.

Thus Q=4 and it is uniquely determined from the question itself, without using either statement.

Therefore the correct answer is (d):
“The Question can be answered even without using any of the Statements.”

Explanation : 

1. Find all 1-digit numbers with exactly 4 distinct positive factors.
– List 1-digit numbers 1 through 9 and their factor counts:
1 → {1} (1 factor)
2 → {1,2} (2 factors)
3 → {1,3} (2 factors)
4 → {1,2,4} (3 factors)
5 → {1,5} (2 factors)
6 → {1,2,3,6} (4 factors)
7 → {1,7} (2 factors)
8 → {1,2,4,8} (4 factors)
9 → {1,3,9} (3 factors)
– So the candidates are 6 and 8, and the smallest is 6.

2. Use Statement I (“2 is one of the factors”):
– Both 6 and 8 have 2 as a factor.
– We cannot decide between 6 and 8.
– So Statement I alone is insufficient.

3. Use Statement II (“3 is one of the factors”):
– Among 1-digit numbers with 4 factors, only 6 has 3 as a factor.
– This pins down the answer uniquely to 6.
– Statement II alone is sufficient.

Conclusion: Statement II alone suffices, but Statement I does not. Hence option (a).

Explanation : 

1. Note that 9 ≡ 3 (mod 6).
2. Then for any k ≥ 1,
9^k ≡ 3^k (mod 6).
3. But 3^1 ≡ 3 (mod 6), and multiplying by 3 again gives
3^2 = 9 ≡ 3 (mod 6),
3^3 = 27 ≡ 3 (mod 6),
and so on.
By induction, 3^k ≡ 3 (mod 6) for every k ≥ 1.
4. The sum 9³ + 9⁴ + … + 9¹⁰⁰ has 100 – 3 + 1 = 98 terms, each ≡ 3 (mod 6).
5. Total ≡ 98 × 3 = 294 ≡ 0 (mod 6).

Explanation : 

1. Observe that
64 = 4³ and its code 343 = 7³
216 = 6³ and its code 729 = 9³

2. In each case the cube‐root is increased by 3:
4 → 7 (4 + 3 = 7)
6 → 9 (6 + 3 = 9)

3. Apply the same rule to 512:
512 = 8³
Increase 8 by 3 → 11
Cube it → 11³ = 1331

Hence, 512 is written as 1331.