Explanation : 

1. Let the cuts be made by slicing the cube with planes parallel to its three pairs of faces.
• Suppose we make a₁ cuts along the x–direction, a₂ cuts along y, and a₃ cuts along z.
• Then the total number of small pieces is
(a₁ + 1)·(a₂ + 1)·(a₃ + 1) = 60.
• The total number of cuts is
N = a₁ + a₂ + a₃,
and we want N as small as possible.

2. Factor 60 into three positive integers whose sum is minimal.
Possible triples (a₁ + 1, a₂ + 1, a₃ + 1) and their sums:
1×6×10 → sum = 17
2×5×6 → sum = 13
3×4×5 → sum = 12 ← minimal
Thus take (a₁+1, a₂+1, a₃+1) = (3, 4, 5).
So a₁=2, a₂=3, a₃=4, giving N = 2+3+4 = 9 cuts.
This proves statement I.

3. Count the pieces not painted on any face.
A piece lies entirely interior if it is not on any outer layer.
Along each axis we must avoid the two boundary layers, so interior count =
(a₁+1 − 2)·(a₂+1 − 2)·(a₃+1 − 2)
= (3−2)·(4−2)·(5−2)
= 1·2·3
= 6.
This proves statement II.

Hence both statements I and II are correct.

Explanation : 

Let the original price be p. An increase of k% multiplies the price by (1 + k/100), and a decrease of k% multiplies it by (1 – k/100). The sequence of changes is:

1. Increase by k%: multiplier = (100 + k)/100
2. Decrease by k%: multiplier = (100 – k)/100
3. Increase by k%: multiplier = (100 + k)/100
4. Decrease by k%: multiplier = (100 – k)/100

Overall multiplier = [(100 + k)/100 × (100 – k)/100]²

= [((100)² – k²)/(100)²]²

= (10000 – k²)² / (10000)²

= (10⁴ – k²)² / 10⁸.

Thus the new price q is:
q = p × (10⁴ – k²)² / 10⁸

Rearrange to relate p and q:
p (10⁴ – k²)² = q × 10⁸

Answer: (a) p(10⁴ – k²)² = q × 10⁸

Explanation : 

Let N = total runs scored by Team X (in 20 overs).
Team Y ties this score in “10% less overs” than 20, i.e. in 20 – 0.1·20 = 18 overs.

1. Team Y’s actual run rate r = N runs ÷ 18 overs = N/18.
2. If Team Y’s run rate were 50% higher, it would be
r′ = r + 0.5·r = 1.5·r = 1.5·(N/18) = N/12.
3. At this higher rate r′, the overs needed to score N runs would be
N runs ÷ (N/12 runs per over) = 12 overs.

This matches the statement, but it imposes no restriction on N (it cancels out). Any positive N satisfies the conditions.

Therefore, the value of N cannot be determined from the information given.

Explanation : 

1. Assign each month its number in the calendar year:
January = 1, February = 2, …, December = 12.

2. Observe that to move from term n to term n + 1 we subtract an increasing offset k (starting with k = 0) from the current month number, taking results modulo 12 (with 0 mapped to 12).

3. Verify the pattern:
• Term 1 → Term 2 (k = 0): 1 − 0 = 1 ⇒ January
• Term 2 → Term 3 (k = 1): 1 − 1 = 0 ⇒ 12 ⇒ December
• Term 3 → Term 4 (k = 2): 12 − 2 = 10 ⇒ October
• Term 4 → Term 5 (k = 3): 10 − 3 = 7 ⇒ July → X
• Term 5 → Term 6 (k = 4): 7 − 4 = 3 ⇒ March
• Term 6 → Term 7 (k = 5): 3 − 5 = −2 ⇒ +12 → 10 ⇒ October
• Term 7 → Term 8 (k = 6): 10 − 6 = 4 ⇒ April → Y
• Term 8 → Term 9 (k = 7): 4 − 7 = −3 ⇒ +12 → 9 ⇒ September

4. Conclusion: X = July, Y = April.

Explanation : 

We seek all positive integers p ≤ q ≤ r satisfying
(1) 1/p + 1/q + 1/r = 1.

1. Bound on p.
Since 1/p + 1/q + 1/r ≤ 3/p, we require 3/p ≥ 1 ⇒ p ≤ 3.

2. Case p = 3.
Then 1/3 + 1/q + 1/r = 1 ⇒ 1/q + 1/r = 2/3.
But 1/q + 1/r ≤ 2/q, so 2/q ≥ 2/3 ⇒ q ≤ 3.
With q ≥ p = 3 we get q = 3, and hence 1/r = 2/3 – 1/3 = 1/3 ⇒ r = 3.
Solution: (3, 3, 3), sum = 9.

3. Case p = 2.
Then 1/2 + 1/q + 1/r = 1 ⇒ 1/q + 1/r = 1/2.
Since 1/q + 1/r ≤ 2/q, we need 2/q ≥ 1/2 ⇒ q ≤ 4.
Checking q = 2,3,4:
• q = 2 ⇒ 1/r = 0 (no solution)
• q = 3 ⇒ 1/r = 1/2 – 1/3 = 1/6 ⇒ r = 6 ⇒ (2,3,6), sum = 11
• q = 4 ⇒ 1/r = 1/2 – 1/4 = 1/4 ⇒ r = 4 ⇒ (2,4,4), sum = 10

No other cases arise. Hence the only triples (up to order) are
(3,3,3) with sum 9,
(2,4,4) with sum 10,
(2,3,6) with sum 11.

Therefore there are exactly three distinct values of p + q + r.

Answer: (c) Three.

Explanation : 

We seek all 3-term arithmetic progressions of primes below 20. If p, q, r are in arithmetic progression then

  p − q = q − r ⟹ p + r = 2q.

The primes under 20 are {2, 3, 5, 7, 11, 13, 17, 19}. Testing q in this set and looking for an integer step d>0 with q±d also prime, one finds exactly five progressions (up to order):

1. (3, 5, 7) sum = 15

2. (3, 7, 11) sum = 21

3. (5, 11, 17) sum = 33

4. (3, 11, 19) sum = 33

5. (7, 13, 19) sum = 39

The sums that occur are 15, 21, 33, 39. Hence there are 4 distinct values of p+q+r.

Explanation : 

Step 1: Restate the problem

We seek natural numbers N for which there exist three distinct divisors p, q, r of N satisfying

p + q + r = N.

Step 2: Change to reciprocal form

If p divides N, write p = N/a for some integer a≥1. Similarly q=N/b and r=N/c, with 1≤a<b<c. Then

p+q+r = N becomes

N·(1/a + 1/b + 1/c) = N

⇒ 1/a + 1/b + 1/c = 1.

Step 3: Solve 1/a + 1/b + 1/c = 1 in distinct positive integers

By testing small denominators (or using known Egyptian‐fraction results), the only solution with a<b<c is

a=2, b=3, c=6

because 1/2 + 1/3 + 1/6 = 1.

Step 4: Translate back to p, q, r

From (a,b,c)=(2,3,6) we get

p = N/2, q = N/3, r = N/6.

These are three distinct divisors of N exactly when 6 divides N, and they indeed satisfy p+q+r=N.

Step 5: Count all such N < 50

The positive multiples of 6 below 50 are

6, 12, 18, 24, 30, 36, 42, 48 — a total of 8 numbers.

Therefore, the answer is (c) 8.