Number System: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

We want to maximise x = 20 – y – z, which means minimising y+z. Since y and z must each be a positive integer greater than zero, their smallest possible values are y=1 and z=1, giving y+z=2. So the maximum x = 20 – 1 – 1 = 18. Check: x=18, y=1, z=1 gives x+y+z=20, and y, z are indeed positive integers.

We want to maximise x = 12 – y – z, which means minimising y+z. Since y and z must each be a positive integer greater than zero, their smallest possible values are y=1 and z=1, giving y+z=2. So the maximum x = 12 – 1 – 1 = 10. Check: x=10, y=1, z=1 gives x+y+z=12, and y, z are indeed positive integers.

Let the digits be a and b. Statement I: a+b=11. Statement II: |a-b|=3. Solving both together: a+b=11 and a-b=3 gives a=7, b=4 (the number 74); but a-b=-3 gives a=4, b=7 (the number 47) equally satisfies a+b=11 and |a-b|=3. Both 74 and 47 fit both statements, so even combined, the two statements don’t pin down one unique number.

The key identity here is a² – b² = (a+b)(a-b) — the difference of squares. So (a²-b²)/(a+b) = (a+b)(a-b)/(a+b) = (a-b), as long as a+b≠0. We’re told this quotient equals 13, so a-b = 13 directly. Check: if a-b=13 and a+b=k for any k, then a²-b²=(a+b)(a-b)=13k, and dividing by (a+b)=k gives back 13.

The key identity here is a² – b² = (a+b)(a-b) — the difference of squares. So (a²-b²)/(a+b) = (a+b)(a-b)/(a+b) = (a-b), as long as a+b≠0. We’re told this quotient equals 9, so a-b = 9 directly. Check: if a-b=9 and a+b=k for any k, then a²-b²=(a+b)(a-b)=9k, and dividing by (a+b)=k gives back 9.

Convert each Roman numeral to its value: M = 1000, D = 500, C = 100, L = 50. Now order these values from greatest to least: 1000 > 500 > 100 > 50, i.e. M > D > C > L.

Since 6 = 2 x 3, and 2 and 3 share no common factors, a number is divisible by 6 exactly when it satisfies BOTH the divisibility rule for 2 (last digit even) AND the divisibility rule for 3 (digit sum divisible by 3) at once. Satisfying only one of the two is not enough.

This is the digit-sum rule: add up every digit of the number, and check if that sum is divisible by 3. For example, 123 has digit sum 1+2+3=6, which is divisible by 3, and indeed 123=3×41. This same digit-sum rule also works for divisibility by 9, but not for most other divisors like 4, 7, or 8.

Any number can be split into (everything except the last 2 digits) x 100 + (last 2 digits). Since 100 = 4 x 25 is itself divisible by 4, the first part is always divisible by 4 regardless of its value. So divisibility by 4 depends only on whether the last two digits, taken as their own number, are divisible by 4.

Any number can be split into (everything except the last 3 digits) x 1000 + (last 3 digits). Since 1000 = 8 x 125 is itself divisible by 8, the first part is always divisible by 8 regardless of its value. So divisibility by 8 depends only on whether the last three digits, taken as their own number, are divisible by 8.

The divisibility rule for 11 works differently from the simple digit-sum rules used for 3 or 9. Label the digit positions from the right as odd (1st, 3rd, 5th, …) and even (2nd, 4th, …). Sum the digits at odd positions, sum the digits at even positions, then take the difference. If that difference is 0 or a multiple of 11 (positive or negative), the number is divisible by 11. This alternating-sum test is the standard rule for 11.

This is the zero-product property: if two numbers multiply to give 0, at least one of them must itself be 0 (since any nonzero number times another nonzero number is always nonzero). It does not require BOTH to be zero — for example, a=0, b=5 gives ab=0 with only a zero. So the correct conclusion is a=0, or b=0, or both.

Let x and y both be negative, say x=-3, y=-4 as a test case. xy = (-3)(-4) = 12, positive. x+y = -7, negative (not positive). x-y = -3-(-4) = 1, positive here but this depends on which is more negative, so not always true. x/y = (-3)/(-4) = 0.75, positive (not negative). The rule that a negative times a negative is always positive holds for any two negative numbers, so xy is positive is always true.

Let n be negative, say n = -5 as a test case. Then 2n = -10, -n = 5, n² = 25, and 0 stays 0. Comparing -10, 5, 25, 0: -10 is the smallest. This makes sense generally too: doubling a negative number makes it more negative, while -n and n² both flip to positive. So 2n is always the least.

If n were even, n³ would also be even (even x even x even = even), contradicting that n³ is odd. So n must be odd. Since n is odd, n² = n x n is odd x odd, which (as shown for products of odd numbers) is also odd. So both n and n² are odd.

Let y = 2k for some integer k. Then 5y = 5(2k) = 10k, which is even. 5y + 3 = 10k + 3, and since 10k is even, adding the odd number 3 makes the whole expression odd. Check with y=2: 5(2)+3 = 13, which is odd. So 5y + 3 must be odd.

Let x = 2k+1 for some integer k. Then 3x = 3(2k+1) = 6k+3, which is odd (an odd number times 3 stays odd). 3x – 1 = 6k+3-1 = 6k+2 = 2(3k+1), which is even. Check with x=3: 3(3)-1 = 8, which is even. So 3x – 1 must be even.

Let m = 2a+1 and n = 2b+1 (odd numbers). Product: mn = (2a+1)(2b+1) = 4ab+2a+2b+1 = 2(2ab+a+b)+1, which is odd (2 x integer + 1). Check with an example: 3 x 5 = 15, odd. Meanwhile m+n and m-n are both even (shown in other questions), so they don’t qualify. So the product of two odd numbers is always odd.

Let m = 2a+1 and n = 2b+1 for some integers a, b (the general form of an odd number). Then m+n = 2a+1+2b+1 = 2a+2b+2 = 2(a+b+1), which is divisible by 2, so m+n is always even. Check with an example: m=3, n=5, m+n=8, which is even. So the answer is m + n.

In a 4-digit number, reading left to right, the digits sit in the thousands, hundreds, tens, and units places respectively. So w (leftmost) contributes w x 1000, x contributes x x 100, y contributes y x 10, and z (rightmost) contributes z x 1. Adding these: value = 1000w + 100x + 10y + z.

In a 3-digit number, reading left to right, the leftmost digit is in the hundreds place, the middle digit is in the tens place, and the rightmost digit is in the units place. So digit p (leftmost) contributes p x 100, digit q (middle) contributes q x 10, and digit r (rightmost) contributes r x 1. Adding these: value = 100p + 10q + r.

π = 3.14159265… continues forever without ever settling into a repeating pattern. A number with a non-terminating, non-repeating decimal expansion is, by definition, irrational — not a recurring decimal, not rational, and certainly not a whole number. So π is best classified as an irrational number.

Check each statement. Every natural number (1,2,3,…) is an integer: true. Every natural number is a real number: true, since naturals sit inside reals. Every integer is a rational number (write n as n/1): true. Every real number is a rational number: FALSE, because irrational numbers like √2 or π are real numbers too, but by definition are not rational. So the incorrect statement is the last one.

The real numbers include every rational number and every irrational number — it is the largest of these categories. So any rational number, by definition, is also a real number. It need NOT be a natural number, whole number, or integer though: for example, 1/2 is rational and real, but not a natural number, whole number, or integer. So the only statement true for every rational number is that it’s a real number.

A number is irrational if it cannot be written as a ratio of two integers, i.e. its decimal never terminates and never repeats. Check each option: 22/7 is a ratio of integers, so it’s rational. 0.375 terminates, so it’s rational. 1.5 terminates, so it’s rational. √7 = 2.6457513… never terminates or repeats, since 7 is not a perfect square. So √7 is the irrational one.