Question

In question below three alternative are alike in a certain way but the rest one is different. Find out the odd one and write vorrect answer A. Abyz B.efuv c. ijqr d. mnop
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Answer 1

Answer:

(d) mnop

Step-by-step explanation:

• (a) ABYZ

ABYZ → (B+Y = 2+25 = 27) and (A+Z = 1+26 = 27.)

• (B) EFUV

EFUV → (F+U = 6+21 = 27) and (E+V = 5+22 = 27.)

• (C) ijqr

ijqr→ (J+Q=10+17= 27)  and  ( I+R = 9+16=27)

• (d) mnop

mnop  → (N+O =  14+15 =29 ) and (M+P= 13+16=29)

Answer 2

The odd one out is mnop. (Option d)

Given:

Four combinations: Abyz, efuv, ijqr, mnop

To find:

The odd one out of these combinations

Solution:

We can find the solution by following the steps given below-

We know that the three of these combinations are alike. So, there must be some pattern involved.

Here, the sum of letters in the middle is equal to the sum of letters present at the ends.

In the first option, abyz, we can see

a+z=1+26= 27

and b+y= 2+25= 27

Similarly, in efuv,

e+v= 5+22=27

f+u=6+21=27

Also, in ijqr,

i+r=9+18=27

j+q=10+17=27

But in the last option, mnop,

m+p=13+16=29

n+o=14+15=29

The first three combinations have the same sum and are alike.

Therefore, the odd one out is mnop.