Explanation : 

1. Understand the notation
• PQR is the three‐digit number 100·P + 10·Q + R
• PS is the two‐digit number 10·P + S
• PPT is the three‐digit number 100·P + 10·P + T = 110·P + T

2. Write the given equation
PQR − PS = PPT
⇒ (100P + 10Q + R) − (10P + S) = 110P + T

3. Simplify the equation
100P + 10Q + R − 10P − S = 110P + T
⇒ 90P + 10Q + R − S = 110P + T
⇒ T = (90P + 10Q + R − S) − 110P
⇒ T = 10Q + R − S − 20P

4. Use the given values and constraints
• Q = 3
• T is a digit, distinct, nonzero, and T < 6 ⇒ 1 ≤ T ≤ 5
• P, Q, R, S, T are all distinct nonzero digits (1–9)

Substitute Q = 3:
T = 10·3 + R − S − 20P = 30 + R − S − 20P

5. Find possible P

We need 1 ≤ T ≤ 5, so
1 ≤ 30 + R − S − 20P ≤ 5

• If P ≥ 2, then 20P ≥ 40 ⇒ 30 + R − S − 20P ≤ 30 + 9 − 1 − 40 = −2, which is < 1.
So P cannot be 2 or more.

• Therefore P = 1 is the only possibility.

6. Solve for R, S when P = 1

For P = 1:
T = 30 + R − S − 20 = 10 + R − S
And we need 1 ≤ T ≤ 5, so
1 ≤ 10 + R − S ≤ 5
⇒ −9 ≤ R − S ≤ −5
⇒ R − S ∈ {−9, −8, −7, −6, −5}

But R and S are digits from 1 to 9, distinct from each other and from P=1, Q=3:

Check each possible difference:

• R − S = −9
Only possibility: R=1, S=10 (impossible, S must be ≤9) ⇒ discard

• R − S = −8
Possibility: R=2, S=10 (impossible) ⇒ discard

• R − S = −7
Possibility: R=2, S=9
T = 10 + 2 − 9 = 3
But T = 3 = Q, not allowed (must be distinct) ⇒ discard

• R − S = −6
Possibility: R=2, S=8
T = 10 + 2 − 8 = 4
Digits now are P=1, Q=3, R=2, S=8, T=4 — all distinct, nonzero, T<6 ⇒ OK

• R − S = −5
Possibilities:
– R=2, S=7 ⇒ T = 10 + 2 − 7 = 5 (digits 1,3,2,7,5 all distinct ⇒ OK)
– R=4, S=9 ⇒ T = 10 + 4 − 9 = 5 (digits 1,3,4,9,5 all distinct ⇒ OK)

7. Count the valid (R, S) pairs
We found exactly three valid pairs:
(R,S) = (2,8), (2,7), (4,9)

Therefore, the number of possible values of (R, S) is 3.

Explanation : 

1. Factor 1001:
1001 = 7 × 11 × 13.

2. Observe:
• Since LCM of the three numbers is 1001, together they must “cover” each prime 7, 11, 13 at least once.
• Since their HCF is 1, no prime can appear in all three numbers—each prime must be missing from at least one of them.

3. Think of each prime “sticker” (7, 11, 13) that you want to stick onto exactly three “boxes” (the three numbers):
– You can stick it on exactly one box (3 choices), or
– You can stick it on exactly two boxes (3 choices).
You cannot use “zero boxes” (would fail LCM) or “all three boxes” (would fail HCF).
So for each prime there are 3 + 3 = 6 valid ways.

4. Primes act independently.
Number of ordered triples (A, B, C) = 6 (ways for 7) × 6 (ways for 11) × 6 (ways for 13) = 216.

5. Even if we treat (A, B, C) as an unordered set {A,B,C}, we divide 216 by at most 3! = 6, giving at least 36 distinct sets.

6. 36 is clearly more than 8.

Therefore n > 8, so the correct answer is More than 8.

Explanation : 

We want primes p and k such that
 p² + k is prime and p² + k < 30.

Step 1. Bound p.
Since k ≥ 2,
 p² + 2 < 30 ⇒ p² < 28 ⇒ p ≤ 5.
So p can be 2, 3, or 5.

Step 2. Test each p:

1. p = 2 ⇒ p² = 4.
Check prime k so that 4 + k is prime < 30:

• k = 2 → 4 + 2 = 6 (not prime)
• k = 3 → 4 + 3 = 7 (prime)
• k = 5 → 4 + 5 = 9 (not prime)
• k = 7 → 4 + 7 = 11 (prime)
• k = 11 → 4 + 11 = 15 (not prime)
• k = 13 → 4 + 13 = 17 (prime)
• k = 17 → 4 + 17 = 21 (not prime)
• k = 19 → 4 + 19 = 23 (prime)
• k = 23 → 4 + 23 = 27 (not prime)

Valid k’s here: 3, 7, 13, 19.

2. p = 3 ⇒ p² = 9.
Check k so that 9 + k is prime < 30:

• k = 2 → 9 + 2 = 11 (prime)
• k = 3 → 12 (no), k = 5 → 14 (no), … all larger k either composite or >30.

Valid k: 2.

3. p = 5 ⇒ p² = 25.
Then 25 + k < 30 ⇒ k < 5; only k = 2 or 3:

• k = 2 → 27 (not prime)
• k = 3 → 28 (not prime)

No solutions for p = 5.

Step 3. Collect all distinct k found:

From p = 2: {3, 7, 13, 19}
From p = 3: {2}

Total distinct k’s = {2, 3, 7, 13, 19}, which is 5 values.

Answer: (b) 5.

Explanation : 

1. We are given the ranges
4 ≤ x ≤ 8
2 ≤ y ≤ 7

2. Find the maximum of (x + y).
• x + y is largest when x and y are as large as possible
• Take x = 8 and y = 7
⇒ max (x + y) = 8 + 7 = 15

3. Find the minimum of (x − y).
• x − y is smallest when x is as small as possible and y is as large as possible
• Take x = 4 and y = 7
⇒ min (x − y) = 4 − 7 = −3

4. Compute the ratio:
(maximum of x + y) / (minimum of x − y)
= 15 ÷ (−3)
= −5

None of the given positive options (6, 15/2, etc.) equals −5, so the correct choice is “None of the above.”

Explanation : 

We want the count of integers from 1 to 100 that are not divisible by any of 2, 3, 5, 7, or 9. Since any multiple of 9 is already a multiple of 3, excluding multiples of 3 covers multiples of 9 as well. So we only need to exclude multiples of 2, 3, 5, and 7.

Let
A₂ = {multiples of 2},
A₃ = {multiples of 3},
A₅ = {multiples of 5},
A₇ = {multiples of 7}.

We apply the Inclusion–Exclusion Principle to find how many are divisible by at least one of these, then subtract from 100.

1. Count of multiples of each single divisor
• |A₂| = ⌊100/2⌋ = 50
• |A₃| = ⌊100/3⌋ = 33
• |A₅| = ⌊100/5⌋ = 20
• |A₇| = ⌊100/7⌋ = 14
Sum₁ = 50 + 33 + 20 + 14 = 117

2. Subtract multiples counted twice (pairwise intersections)
• |A₂∩A₃| = ⌊100/lcm(2,3)=6⌋ = 16
• |A₂∩A₅| = ⌊100/10⌋ = 10
• |A₂∩A₇| = ⌊100/14⌋ = 7
• |A₃∩A₅| = ⌊100/15⌋ = 6
• |A₃∩A₇| = ⌊100/21⌋ = 4
• |A₅∩A₇| = ⌊100/35⌋ = 2
Sum₂ = 16 + 10 + 7 + 6 + 4 + 2 = 45

3. Add back multiples we subtracted too many times (triple intersections)
• |A₂∩A₃∩A₅| = ⌊100/30⌋ = 3
• |A₂∩A₃∩A₇| = ⌊100/42⌋ = 2
• |A₂∩A₅∩A₇| = ⌊100/70⌋ = 1
• |A₃∩A₅∩A₇| = ⌊100/105⌋ = 0
Sum₃ = 3 + 2 + 1 + 0 = 6

Note on Step 3:
After Step 1, a number divisible by, say, 2, 3, and 5 is counted three times.
In Step 2 we subtract it once for each pair (2–3, 2–5, 3–5), so we subtract it three times—bringing its net count to zero.
But in the union it should appear once, so we “add back” each triple‐intersection exactly once.

4. Four-way intersection
lcm(2,3,5,7)=210 > 100 ⇒ 0

By Inclusion–Exclusion, the total divisible by at least one of 2, 3, 5, 7 is
117 (singles)
– 45 (pairs)
+ 6 (triples)
– 0 (quadruple)
= 78.

Therefore the count not divisible by any of 2, 3, 5, 7, 9 is
100 – 78 = 22.

Final Answer: 22.

Explanation : 

1. We want the units digit of the product
P = 1 × 3 × 5 × 7 × 9 × … × 999.

2. Note there are (999 + 1)/2 = 500 odd numbers from 1 to 999.

3. Look at the factors’ last digits; they cycle every five terms:
1, 3, 5, 7, 9, 1, 3, 5, 7, 9, …

4. Compute the product of one cycle modulo 10:
1 × 3 = 3
3 × 5 = 15 → last digit 5
5 × 7 = 35 → last digit 5
5 × 9 = 45 → last digit 5
⇒ Product of any block of five consecutive odd numbers ends in 5.

5. Since there are 500 odd numbers, that is 100 such blocks.
So P ≡ 5¹⁰⁰ (mod 10).
But 5^k always ends in 5 for any k ≥ 1.

6. Therefore, the units digit of P is 5.

Explanation : 

Statement I: “If A < B > C < D > E > F ≥ G = H, then B is always greater than E.”

– From A < B and B > C we know B > C.
– From C < D and D > E we know D > E.
– But there is no direct comparison between B and D, so B could be less than, equal to, or greater than D.
– For example, choose
 C = 0, B = 10, A = –5;
 D = 11, E = 10, F = 9;
 G = H = 0.
Then
 A < B (–5 < 10), B > C (10 > 0), C < D (0 < 11), D > E (11 > 10),
 E > F (10 > 9), F ≥ G (9 ≥ 0), G = H (0 = 0).
Yet here B = E, so B is not greater than E.
– Hence statement I is false.

Statement II: “If P > Q = R > S = T < U = V > W, then S is always less than V.”

– From Q = R > S = T we get S = T and both are strictly less than Q and R.
– From T < U = V we get S < U and U = V ⇒ S < V.
– No matter what values you pick (as long as inequalities hold), S is strictly less than V.
– Hence statement II is always true.

Therefore the correct choice is (b) II only.