Math, asked by vanshdevgan73, 21 days ago

find the odd one out from the option
CFI
RUX
LOR
SVZ​

Answers

Answered by shanthakabbur
2

Answer:

suz

Step-by-step explanation:

suz

+3 +3

C → F → I

+3 +3

U → X → A

+3 +3

R → U → X

+3 +3

L → O → R

But,

+3 +4

S → V → Z

Therefore, ‘SVZ’ does not belong to the group.

Answered by RvChaudharY50
0

Given :- Find the odd one out from the option :-

A) CFI

B) RUX

C) LOR

D) SVZ

Solution :-

we know that, 1A, 2B, 3C, 4D, 5E, 6F, 7G, 8H, 9I, 10J, 11K, 12L, 13M, 14N, 15O, 16P, 17Q, 18R, 19S, 20T, 21 U, 22V, 23W, 24X, 25Y and 26Z.

so, checking positions of all alphabets of given options we get,

A) CFI

→ C = 3

→ F = 6

→ I = 9

  • All alphabets are multiple of 3 or difference between them is 3 .

B) RUX

→ R = 18

→ U = 21

→ X = 24

  • All alphabets are multiple of 3 or difference between them is 3 .

C) LOR

→ L = 12

→ O = 15

→ R = 18

  • All alphabets are multiple of 3 or difference between them is 3 .

D) SVZ

→ S = 19

→ V = 22

→ Z = 26

  • All alphabets are not multiple of 3 or difference between them is not 3 .(22 - 19 = 3 but 26 - 22 = 4)

therefore, we can conclude that, (D) SVZ is odd one out from the options.

Learn more :-

Find the odd one out from the options.(ਭਿੰਨ ਵਿਕਲਪ ਚੁਣੋ) * 289 1331 2197 343

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