Olympiad Test: Squares And Square Roots - UPSC

Yes the correct option is A because..... 23×23=529.....

Since the square of an odd natural number is odd and that of an even number is an even number.

∴    (i)  The square of 431 is an odd number

(∵  431 is an odd number)

(ii)   The square of 2826 is an even nnumber.

(∵ 2826 is an even number)

(iii) The square of 7779 is an odd number

(∵ 7779 is an odd number)

(iv)  The square of 131 is an odd nnumber.

(∵ 82004 is an odd number)

Squares of odd numbers are always odd

81 is the square of 9..... 9 × 9 = 81

One digit will be 6 because 1234
4, is at he last and 4² = 16. 
So, 6 is at the last.

Square of 400 = 160000

Therefore, the number of zeroes is 4

Since 35,32,39 are not perfect squares . hence the only perfect square between 30 and 40 is 36

The squares that have 1 and 9 at their unit place, ends with 1. 

IN this case, 81² will end in 1

Between 9² and 10² 

Here, n = 9 and n + 1 =10

 :. Natural number between 9² and 10²  are (2 x n) or 2 x 9, i.e. 18. hence,correct option is (c).

Between 11² and 12²

Here, n = 11 and n + 1 = 12

 ∴ Natural numbers between 11² and 12² are (2 × n) or 2 × 11, i.e. 22.

The question should be a²+ac+bc+ab
=a(a+c)+b(a+c)=(a+c)(a+b)

The number of non square numbers
between n² and ( n + 1 )² is 2n
Here,
n = 12,
n + 1 = 13
Therefore ,
Number of natural numbers lie
between 12² and 13² = 2 × 12
= 24

Square root of 390625=625.Hence its a 3 digit number

Therefore, the square root is 80

By prime factorisation we have 90=2*3*3*5, so the least number by which 90 should be multiplied is 2*5=10

By Prime factorisation, we have 48=2*2*2*2*3. We have two pairs of 2 but no pair of 3. Hence 48 must be divided by 3 to make it a perfect square.

The smallest number to be added to 80 so as to obtain a perfect square number is 1.

 80+1=81 and the square root of 81 is 9,so 81 becomes a perfect square.

625 is square root of 25 and one's digit of 25 is 5

Yes as the the smallest 3 digit no. is 100 and it is the square of 10 so also a perfect square.

We know that the largest 3 digit number is 999. So we will find out the square root of 999 by long division method and check whether it is a perfect square or not.
We will see that 999 is not a perfect square as it leaves 38 as remainder.

So if 38 will be added to 999 then it will become a 4 digit number. Therefore to find the largest 3 digit perfect square we will subtract 38 from 999. 

999 - 38 = 961.

Hence 961 isthe largest 3 digit perfect square whose square root is 31.

By division method we find the root of 9999 and we get the remainder 198.So subtracting 198 from 9999 we get 9801 as the greatest 4-digit perfect square.

Apply division method on 1000, gives 3*3=9 leaves remainder 1 .Then get the two zeros, which makes it 100 then you find the next number ie,62*2=124 and then subtract 100 from 124 you get 24 which is to be added in 1000. Hence it’s 1024

The square root of 25600 is 160. Hence, number of digits is 3. 

The square root of 1296 is 36. 

Hence, number of digits os 2

The square root of 12.25 is 3.5

We know area of a square =a^2 given area =441cm^2 so a^2=441cm2 a^2=(21cm)^2 so a=21...length of square =21cm

We have (AB)²+(BC)² =( AC)² which is (AC)² = 36+64 . Hence AC = 10cm.

If we subtract 4 from the given number, then it becomes 625. 

625 is a square of 25. 

Hence, the answer is 4

LCM of 6,9 and 15 is 90. So any multiple of 90 will be divisible by 6,9 and 15. The only square number in the options is 900, so 900 is the smallest square number which is divisible by 6,9 and 15.

Required square of 1.2 is 1.44 

13 is the square root of 169.

Here let us consider ABCD is a rectangle given in the figure.

AD = 3 cm (Breadth)

DC = 4 cm (Length)

To Find:

AC (Diagonal)

We know that, Triangle ADC is a right triangle, right angled at D. So by using Pythagorus Theorem

Length^2 + Breadth^2 = Hypotenuse^2     (diagonal)

DC^2 + AD^2 = AC^2

=> 4^2 + 3^2 = AC^2

=> 16 + 9 = AC^2

=> 25 = AC^2

=> AC = √25 = 5

Therefore the length of the diagonal AC is 5 cm. 

By using Pythagoras theorem
H² = P² + B²
Third side = = 4cm

Squares of all integers are known as perfect squares. All perfect squares end in 1, 4, 5, 6, 9 or 00 (i.e. Even number of zeros). Therefore, a number that ends in 2, 3, 7 or 8 is not a perfect square.

We know that 40² = 1600

So we must start finding squares of numbers less than 40

Take 39 first

40² = 1600, :. 1600 -1500 = 100

39² = 1521,  :.1521 -1500 = 21

38²= 1444, .: 1500 -1444 = 56 

Thus nearest perfect square is 39².

But there is no option of difference of 21.

so we'll go the next available nearest possible that is 1444 i.e 38² 

So answer is 1500-1444 = 56