DSSSB TGT CS Mock 03 – Question 012

Find the odd one out among the four letter-clusters:
A) BGLQ
B) CHMS
C) DIOT
D) EJPU

• EJPU
• DIOT
• CHMS
• BGLQ

Examine alphabetical place values:
– BGLQ: B(2) +5= G(7) +5= L(12) +5= Q(17) [Difference: +5, +5, +5]
– CHMS: C(3) +5= H(8) +5= M(13) +6= S(19) [Difference: +5, +5, +6]
Wait! Let’s check EJPU: E(5) +5= J(10) +6= P(16) +5= U(21).
Let’s check DIOT: D(4) +5= I(9) +6= O(15) +5= T(20).
Notice that in BGLQ all steps are exactly +5 (+5, +5, +5). In EJPU, DIOT, CHMS, the sum of values is different. Specifically, BGLQ is the only strictly uniform arithmetic progression of letters.