Reasoning Aptitude
MATHEMATICAL OPERATIONS MCQs
H < B -- (i) M ≥ B -- (ii) K + M -- (iii)combining (ii) and (iii), we get:K = M ≥ B => K ≥ B. Hence, neither conclusion II (B = K) or conclusion III (K > B) is true. But, both conclusion I and II together make a complementary pair. Hence, either conclusion II (B = K) or conclusion III (K > B) is true.Again combining all (i), (ii) and (iii) we get K = M ≥ B > H => K > H (conclusion I). Hence conclusion I (K > H) is true.
V ≥ M -- (i) N < V -- (ii) J > N -- (iii)From (i) and (ii), no specific relation between M and N can be established. Hence, conclusion II (M > N) is not necessarily true.Again, from (i), (ii) and (iii), no specific relation between V and J can be established. Hence, conclusion III (V > J) is not necessarily true.
D ≥ N -- (i) N = V -- (ii) W ≤ V -- (iii)combining (i) and (ii), we get:D ≥ N = V => D ≥ N. Hence, conclusion III (V = D) is not necessary true.Again, combining all (i), (ii) and (iii), we getD ≥ N = V ≥ W => D ≥ W. Hence, neither conclusion I (D = W) or conclusion II (W < D) is true. But both conclusion I (D = W) and conclusion II (W
T = R -- (i) R < W -- (ii) W < H -- (iii)combining (ii) and (iii), we getH > W ≥ R => H > T (conclusion II). Hence, conclusion II (H ≥ T) is true but conclusion III (H ≥ T) is not true.
1 8 B 1 2 P 4 M 8 Q 6 = 18 * 12 ÷ 4 - 8 + 6 = 52
V ≥ D, D ≤ K, F > K=> D ≥ K < F => D < F=> D ≥ F => D @ F
T = B, M ≤ Q, J ≥ B=> T ≤ J, M ≤ Q
W ≥ D, D = M, M ≤ F=> D ≤ F => D L F or D = F=> D ≥ F or D = F=> D @ F or D % F
-NA-
M ≤ J, K > J, K < T=> M ≤ J < K < T => M < T=> T > M => T ≤ M => T $ MAlso J < T => J ≥ T => J @ T