Explanation : 

1. Let the two natural numbers be n and n + 10.
2. Any number divisible by 5 has the form 5k. We want those 5k strictly between n and n + 10:
n < 5k < n + 10
Divide by 5:
n/5 < k < n/5 + 2
This is an interval of length 2 on the real line, so it can contain either one or two integers k.
3. Examples:
• If n = 5, then the numbers between 5 and 15 are 6,7,…,14. Only 10 = 5·2 lies between – that’s 1 multiple of 5.
• If n = 3, then the numbers between 3 and 13 are 4,5,…,12. Here 5 = 5·1 and 10 = 5·2 lie between – that’s 2 multiples of 5.
4. Conclusion: depending on n, there may be either 1 or 2 such numbers, so “only one” and “only two” are both false, and there is always at least one. Hence the correct statement is that you can have more than one such number.

Explanation : 

Let 0 < x < 1.

1. Check I: x² > x³
– Since x > 0, we can divide both sides of x² > x³ by x² to get
1 > x.
– And indeed 1 > x holds for all x in (0,1).
⇒ Statement I is true.

2. Check II: x > √x
– Since x > 0, divide both sides of x > √x by √x to get
√x > 1.
– That is equivalent to x > 1, which contradicts x < 1.
– In fact, for 0 < x < 1 we have √x > x, not the other way around.
⇒ Statement II is false.

Therefore, only statement I is correct.

Explanation : 

1. Given that the average of p, q and r is k, so
(p + q + r)/3 = k
⇒ p + q + r = 3k.

2. We are also told that “p is as much more than the average as q is less than the average,” i.e.
p – k = k – q
⇒ p + q = 2k.

3. Substitute p + q = 2k into p + q + r = 3k:
2k + r = 3k
⇒ r = 3k – 2k = k.

Answer: (a) k.

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.”