# How many numbers lie between squares of the following numbers?

${\text{(i)}}$12 and 13

${\text{(ii)}}$25 and 26

${\text{(iii)}}$ 99 and 100

Last updated date: 22nd Mar 2023

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Answer

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Hint:- Find squares of given numbers.

As, we know that,

Total numbers lying between two numbers a and b,

is given as \[b - a - 1\], where \[b > a\].

So, here we had to find total numbers lying between squares of two numbers.

So, solving all the parts.

${\text{(i)}}$So, a will be \[{(12)^2}\]

\[ \Rightarrow a = {(12)^2} = 144\]

And, b will be \[{(13)^2}\]

\[ \Rightarrow b = {(13)^2} = 169\]

So, total numbers lying between the squares of 12 and 13 will be,

\[ \Rightarrow b - a = 169 - 144 - 1 = 24\]

\[ \Rightarrow \]Hence, the total numbers lying between squares of 12 and 13 is 24.

${\text{(ii)}}$So, a will be \[{(25)^2}\]

\[ \Rightarrow a = {(25)^2} = 625\]

And, b will be \[{(26)^2}\]

\[ \Rightarrow b = {(26)^2} = 676\]

So, total numbers lying between the squares of 25 and 26 will be,

\[ \Rightarrow b - a = 676 - 625 - 1 = 50\]

\[ \Rightarrow \]Hence, the total numbers lying between squares of 25 and 26 is 50.

${\text{(iii)}}$So, a will be \[{(99)^2}\]

\[ \Rightarrow a = {(99)^2} = 9801\]

And, b will be\[{\text{ }}{(100)^2}\]

\[ \Rightarrow b = {(100)^2} = 10000\]

So, total numbers lying between the squares of 99 and 100 will be,

\[ \Rightarrow b - a = 10000 - 9801 - 1 = 198\]

\[ \Rightarrow \]Hence, the total numbers lying between squares of 99 and 100 is 198.

Note:- Whenever we came up with this type of problem then remember that,

for any two numbers a and b such that \[b > a\]. Total numbers lying between

them will be \[b - a - 1\]. But if the given two numbers are consecutive and we

had to find total numbers lying between their squares then we can also directly

say that numbers lying between them is \[2*a\].

As, we know that,

Total numbers lying between two numbers a and b,

is given as \[b - a - 1\], where \[b > a\].

So, here we had to find total numbers lying between squares of two numbers.

So, solving all the parts.

${\text{(i)}}$So, a will be \[{(12)^2}\]

\[ \Rightarrow a = {(12)^2} = 144\]

And, b will be \[{(13)^2}\]

\[ \Rightarrow b = {(13)^2} = 169\]

So, total numbers lying between the squares of 12 and 13 will be,

\[ \Rightarrow b - a = 169 - 144 - 1 = 24\]

\[ \Rightarrow \]Hence, the total numbers lying between squares of 12 and 13 is 24.

${\text{(ii)}}$So, a will be \[{(25)^2}\]

\[ \Rightarrow a = {(25)^2} = 625\]

And, b will be \[{(26)^2}\]

\[ \Rightarrow b = {(26)^2} = 676\]

So, total numbers lying between the squares of 25 and 26 will be,

\[ \Rightarrow b - a = 676 - 625 - 1 = 50\]

\[ \Rightarrow \]Hence, the total numbers lying between squares of 25 and 26 is 50.

${\text{(iii)}}$So, a will be \[{(99)^2}\]

\[ \Rightarrow a = {(99)^2} = 9801\]

And, b will be\[{\text{ }}{(100)^2}\]

\[ \Rightarrow b = {(100)^2} = 10000\]

So, total numbers lying between the squares of 99 and 100 will be,

\[ \Rightarrow b - a = 10000 - 9801 - 1 = 198\]

\[ \Rightarrow \]Hence, the total numbers lying between squares of 99 and 100 is 198.

Note:- Whenever we came up with this type of problem then remember that,

for any two numbers a and b such that \[b > a\]. Total numbers lying between

them will be \[b - a - 1\]. But if the given two numbers are consecutive and we

had to find total numbers lying between their squares then we can also directly

say that numbers lying between them is \[2*a\].

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