Squares and Square Roots Test - 8

The square of a number ending with 5, say k5, can be calculated as

k(k + 1) × 100 + 5².

∴ 215² = 21(21 + 1) × 100 + 25

= (21 × 22) × 100 + 5²

⇒ a = 21

Similarly, 405² = 40(40 + 1) × 100 + 5²

⇒ b = 40

Thus, the value of (a × b) = 21 × 40 is 840.

The correct answer is C.

We know that Pythagorean triplets can be found by using the general form 2m, m² − 1, m² + 1, where m is a natural number greater than 1. Among these members, m² + 1 is the largest.

∴ m² + 1 = 65

⇒ m² = 65 − 1

⇒ m² = 64

∴ m = 8

⇒ 2m = 2 × 8 = 16

m² − 1 = 8² − 1 = 64 − 1 = 63

Thus, the smallest number of the required Pythagorean triplet is 16.

The correct answer is B.

We know that a perfect square can never end with the digits 2, 3, 7, or 8. Also, the number of zeroes at the end of a perfect square is always even.

It can be seen that 64000 has three zeroes at its end.

Thus, the number 64000 cannot be a perfect square.

The correct answer is D.

We can write 69 as (60 + 9).

∴ 69² = (60 + 9)² = (60 + 9) × (60 + 9)

= 60 × (60 + 9) + 9 × (60 + 9)

= 60² + 60 × 9 + 9 × 60 + 9²

= 3600 + 540 + 540 + 81

Thus, the value of 69² is (3600 + 540 + 540 + 81).

The correct answer is C.

We know that if a number has 3 or 7 at the unit's place, then its square ends in 9.

Thus, the unit's digit of the square of the number 465d27 is 9.

The correct answer is D.

If a natural number m can be expressed as n × n = n², where n is a natural number, then m is said to be a perfect square.

It can be seen that there are five perfect squares between 35 and 111, which are:

6² = 36, 7² = 49, 8² = 64, 9² = 81,10² = 100

Thus, there are 5 perfect squares between the numbers 35 and 111.

The correct answer is C.

The sum of first n odd natural numbers is n².

For example:

1 [one odd number] = 1 = 1²

1 + 3 [Sum of first two odd numbers] = 4 = 2²

1 + 3 + 5 [Sum of first three odd numbers] = 9 = 3²

1 + 3 + 5 + 7 [Sum of first four odd numbers] = 16 = 4² and so on.

Thus, the sum of first 26 odd numbers = 26² = 676.

The correct answer is C.

We know that if a number has 2 or 8 at the unit's place, then its square ends in 4 and if a number has 1 or 9 at the unit's place, then its square ends in 1.

∴ 608x8² has 4 at the unit's place and 29y09² has 1 at the unit's place.

Thus, the expression (608x8² − 29y09²) has (4 − 1) = 3 at its unit's place.

The correct answer is C.

We start by prime factorizing the number 6468. The number 6468 can be factorized as:

∴ 6468 = 2 × 2 × 3 × 7 × 7 × 11

It can be seen that the prime factors 3 and 11 do not occur in pairs.

Therefore, if we divide 6468 by (3 × 11) = 33, we get 2 × 2 × 7 × 7 = 196, which is a perfect square.

Thus, the required number is 33.

Hence, the correct answer is option D.

We know that (a − 1) × (a + 1) = a² − 1 where a is a natural number.

The given expression (416 × 418 = x² − 1) can be expressed as:

(417 − 1) × (417 + 1) = x² − 1

∴ x = 417

Thus, the value of x is 417.

The correct answer is C.

We start by calculating the LCM of 20, 28 and 70.

2	20, 28, 70
2	10, 14, 35
5	5, 7, 35
7	1, 7, 7
	1, 1, 1

∴ L.C.M (20, 28, 70) = 2 × 2 × 5 × 7 = 140

It can be seen that in the prime factorization of 140, 5 and 7 do not appear in pairs. Therefore, when we multiply 140 by 5 and 7, we get a square number.

∴ 140 × 5 × 7 = 2 × 2 × 5 × 5 × 7 × 7 = 4900

Thus, the smallest square number which is a multiple of 20, 28 and 70 is 4900.

Hence, the correct answer is option C.

We know that if the unit's digit of a square number is 9, then the unit's digit of its square root is either 3 or 7.

3 × 3 = 9

7 × 7 = 49

Thus, the possible unit's digit of the square root of the square number 32x489 is 3 or 7.

The correct answer is B.