Explanation : 

Smallest non-negative value of the expression

5 – 4 – 3 + 2 × 1 = 0

Explanation : 

Least possible number of cuts

Let’s approach this step-by-step:
1) We need to divide the cube into 64 identical pieces. This means we need to divide each dimension of the cube into 4 equal parts (4 x 4 x 4 = 64).
2) For each dimension, we need 3 cuts to divide it into 4 equal parts.
3) So, we need:
– 3 cuts parallel to the xy-plane
– 3 cuts parallel to the yz-plane
– 3 cuts parallel to the xz-plane
4) In total, this gives us 3 + 3 + 3 = 9 cuts.
Therefore, the least possible number of cuts required to cut a cube into 64 identical pieces is 9.
The correct answer is b) 9

Explanation : 

Prime and Composite Number Analysis

To determine which statements are correct, let’s analyze each one individually:

1. p + c / p – c can be even:

• For this expression to be even, the numerator (p + c) and the denominator (p – c) must both be either even or odd.
• Prime numbers (p) are mostly odd, except for 2.
• Composite numbers (c) can be either even or odd.
• If p is odd and c is even, then:
  • p + c is odd + even = odd
  • p – c is odd – even = odd
  • An odd number divided by an odd number can be even if the result is an integer.
• If p is 2 (even) and c is odd, then:
  • p + c is even + odd = odd
  • p – c is even – odd = odd
  • An odd number divided by an odd number can be even if the result is an integer.

Therefore, it is possible for p + c / p – c to be even.

2. 2p + c can be odd:

For 2p + c to be odd:

• 2p is always even because 2 times any integer is even.
• For the sum to be odd, c must be odd (even + odd = odd).
• Since composite numbers can be odd, this statement is true.

3. pc can be odd:

For the product pc to be odd:

• Both p and c must be odd.
• Since prime numbers (except 2) are mostly odd and composite numbers can be odd, this statement is true.

Given the analysis, all three statements are correct. Therefore, the correct answer is:

(d) 1, 2 and 3

Prime Numbers:

• A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
• In other words, a prime number is a number that can only be divided evenly (without a remainder) by 1 and itself.
• Examples of prime numbers include 2, 3, 5, 7, 11, 13, 17, 19, etc.
• Note that 2 is the only even prime number; all other prime numbers are odd.

Composite Numbers:

• A composite number is a natural number greater than 1 that is not prime, meaning it has more than two positive divisors.
• In other words, a composite number can be divided evenly by numbers other than 1 and itself.
• Examples of composite numbers include 4, 6, 8, 9, 10, 12, 14, 15, etc.
• Composite numbers can be even or odd. For instance, 4 (even) and 9 (odd) are both composite numbers.

Explanation : 

Sum of ABC and DEF

Since the sum of ABC and DEF is 1111, and all the digits are distinct and non-zero, one of the possible combinations for ABC and DEF is 785 and 326.

In the first case, the sum of the digits A+B+C+D+E+F is 7+8+5+3+2+6 = 31

So, in both cases, the sum of the digits is 31. Therefore, the correct answer is (d) 31.

Explanation : 

Let’s take the 3-digit number as XYZ, where X, Y, and Z are its digits.

Statement-1:

The sum of the digits of the number is equal to the product of the digits.

This means X+Y+Z = XYZ. There are multiple 3-digit numbers that satisfy this condition, such as 123, 132, 213, 231, 312, 321, etc.

Statement-2:

The number is divisible by the sum of the digits of the number.

This means XYZ is divisible by (X+Y+Z). Again, there are multiple 3-digit numbers that satisfy this condition, such as 111, 222, 333, etc.

Even if we use both statements together, we still have multiple options for the 3-digit number. For example, both 111 and 222 satisfy both conditions. Therefore, even by using both statements together, we cannot definitively answer the question.

Explanation : 

Analysis of 2-Digit Numbers – pp, qq & rr

The problem is based on the concept of number theory and algebra.

The given condition is that pp, qq and rr are 2-digit numbers where p< q< r and their sum is a 3-digit number ending with zero.

This means that the sum of the digits p, q and r is either 10 or 20, as the sum of their double-digit representations (pp, qq, rr) is a three-digit number ending in zero (either 110 or 220).

For the first statement, we need to find the possible values of p. Since p is the smallest digit and the sum of p, q and r is either 10 or 20, the possible values of p can be 1 or 2 for the sum 10, and 1, 2, 3, 4 or 5 for the sum 20. So, there are 5 possible values for p, making the first statement correct.

For the second statement, we need to find the possible values of q. Since q is the middle digit and the sum of p, q and r is either 10 or 20, the possible values of q can be 2, 3 or 4 for the sum 10, and 6, 7 or 8 for the sum 20. So, there are 6 possible values for q, making the second statement correct.

Therefore, both statements are correct, and the answer is (c) Both 1 and 2.

Explanation : 

The remainder when a number is divided by another number is the difference between the dividend and the largest multiple of the divisor that is less than or equal to the dividend.

In this case, we are given that the

remainder is 31 when 1186 is divided by a certain number.

This means that the number we are looking for must be a divisor of (1186 – 31) = 1155.

The divisors of 1155 are 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155.

However, since the remainder is 31, the divisor must be greater than 31.

So, the possible numbers are 33, 35, 55, 77, 105, 165, 231, 385, 1155.

Hence, there are 9 such natural numbers. So, the correct answer is (d) 9.

Explanation : 

ABCD square. Points on sides AB, CD, BC, DA.

The total number of ways to choose 3 points out of 6 is given by the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and “!” denotes factorial.

So, the total number of triangles is C(6, 3) = 6! / [3!(6-3)!] = 20.