Question

Find the odd one out from the options.(ਭਿੰਨ ਵਿਕਲਪ ਚੁਣੋ) *

(100,121)

(25,36)

(81,90)

(64,81)​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

Find the odd one out from the options.

(100,121)

(25,36)

(81,90)

(64,81)

EVALUATION

Here the given sets are

(100,121)

(25,36)

(81,90)

(64,81)

We now try to find a pattern

$\sf{(100,121) = ( {10}^{2}, {11}^{2} ) }$

$\sf{(25,36) = ( {5}^{2}, {6}^{2} ) }$

$\sf{(81,90) = ( {9}^{2}, 90 ) }$

$\sf{(64,81) = ( {8}^{2}, {9}^{2} ) }$

We see that three pairs are in the form

$\sf{ ( {n}^{2}, {(n + 1)}^{2} ) }$

Only one pair (81,90) is not following the pattern

So the odd one is (81,90)

FINAL ANSWER

Hence the correct option is (81,90)

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. if 448A is a multiple of 3 then A =

https://brainly.in/question/38462990

2. using brackets write three pairs of numbers.whose sum is 15. use them to write equalities

https://brainly.in/question/38629103

Answer 2

Given:

(100,121)

(25,36)

(81,90)

(64,81)

To Find:

Find the odd one out from the options.

Solution :

To find the odd one out from the options we need to take out the number that is not related to the rest of the given numbers.

(100, 121)

$= > 100 ,\: 121 = {10}^{2} , {11}^{2}$

(25, 36)

$= > 25 ,\: 36 = {5}^{2} , {6}^{2}$

(81, 90)

$= > 81 ,\: 90 = {9}^{2} , 90 not \: square \: number$

(64, 81)

$= > 64 ,\: 81 = {8}^{2} , {9}^{2}$

we can see that all the pairs are in the form of :

$= > [{n}^{2} , {(n + 1)}^{2}]$

Hence, (81, 90) is the odd one among the given options.