Squares and Square Roots — Statement I-II Type Questions

Statement I: 100 is a perfect square.

Statement II: Any natural number with 0, 1, 4, 5, 6 or 9 at unit's place is a perfect square.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Finding prime factors of 100

$\begin{array}{l|r} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

100 = (2 × 2) × (5 × 5)

Since all the prime factors are in pair, so 100 is a perfect square.

Also, 100 = 10²

So, statement I is true.

A square number always ends with 0, 1, 4, 5, 6 or 9 at unit's place but the converse is not true.

For example, 30, 21, 34, 45, 76 and 89 are not square numbers.

So, statement II is false.

Hence, option 1 is the correct option.

Statement I: There are 18 non-square numbers between 81 and 100.

Statement II: There are 2n non-square numbers between the squares of n and (n + 1).

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

As, 81 = 9² and 100 = 10², so n = 9

Non-square numbers between 9² and 10² = 2 × 9

= 18

So, statement I is true.

There are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

So, statement II is true.

Hence, option 3 is the correct option.

Statement I: (2, 3, 5) is a Pythagorean triplet.

Statement II: For any natural number n ≥ 2, (2n, n² - 1, n² + 1) is a Pythagorean triplet.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

We know that for any natural number n ≥ 2, (2n, n² - 1, n² + 1) is a Pythagorean triplet.

Finding the Pythagorean triplet having 3 as one of its numbers,

Let n² - 1 = 3

⇒ n² = 4

⇒ n = 2

∴ 2n = 2 × 2 = 4 and n² + 1 = 4 + 1 = 5

So, (4, 3, 5) is a Pythagorean triplet and not (2, 3, 5).

So, statement I is false.

According to statement II, for any natural number n ≥ 2,

$\Rightarrow (2n)^2 + (n^2 - 1)^2\\[1em] \Rightarrow 4n^2 + n^4 - 2n^2 + 1\\[1em] \Rightarrow n^4 + 2n^2 + 1\\[1em] \Rightarrow (n^2 + 1)^2$

So, statement II is true.

Hence, option 2 is the correct option.

Statement I: Sum of first 11 odd natural numbers is 121.

Statement II: Sum of first n odd natural numbers is n².

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

The sum of first 11 odd natural numbers:

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

= 11²

= 121

So, statement I is true.

The sum of first n odd natural numbers = n².

So, statement II is true.

Hence, option 3 is the correct option.