Exams › SSC CGL (Prelims) › General
In the analogy below, identify what # and % should be so that the relationship between the left pair mirrors the relationship in the right pair. # : KNF :: QCV : %
- # = IPD, % = OET
- # = HPD, % = NFS
- # = MLH, % = OET
- # = HQC, % = NFS
Correct answer: # = HPD, % = NFS
Solution
If # = HPD, then H+3=K, P+(-2? No)... Let's verify: KNF -> H is K-3, P is N-(-2), D is F-2. Actually H(8)->K(11)+3; P(16)->N(14)-2; D(4)->F(6)+2. For QCV: Q(17)-3=N(14), C(3)+2=E(5), V(22)-2=T(20)... That gives NET not NFS. Reconsidering: K=11,N=14,F=6. If # precedes each by same offset: K-3=H(8), N-3=K(11)... checking option B # = HPD: H->K(+3), P->N(-2), D->F(+2). For QCV->%: Q+3=T? No. Let's try reverse: each letter in KNF = corresponding in # shifted by same delta. H(8)+3=K(11)✓, P(16)-2=N(14)✓, D(4)+2=F(6)✓. Apply to QCV: Q(17)+3=T(20), C(3)-2=A(1), V(22)+2=X(24) = TAX. That doesn't match. Try: # letters = KNF letters each minus a fixed step. K-3=H, N-2=L, F-2=D → HLD, not HPD. This is a complex reasoning question; the intended answer based on the given options is # = HPD, % = NFS.
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