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Some laptop are mouse (I) + All mouse are keyboards (A) ⟹ Some laptops are keyboards (I) + No keyboard is a bus (E) ⟹ Some laptops are not buses (O) ⟹ [Probable conclusion] ⟹ Some laptops may be buses (I). All mouse are keyboards (A) + No keyboard is a bus (E) ⟹ No mouse is a bus (E) ⟹ Conversion ⟹ Some buses are not mouses (O). No bus is a pendrive (E) ⟹ Conversion ⟹ Some pendrives are not buses (O) ⟹ [Probable conclusion] ⟹ Some pendrives may be buses (I). By elimination method - By option 1 and 3, no +ve conclusion can be drawn between pendrive and buses. Hence it is eliminated. By option 4, there is no –ve conclusion. Hence it is eliminated. By option 2, the conclusion can be drawn. Hence answer is 2.
14, 27, 40, 53, 67, 79
? + 144.99 – 110.01 = 15.01 × 7.98
? = 540.24 + 1022.97 – 11.992
956.41 of 45.06% = ?
√2025.12 × √256.03 + √399.89 ×√(?) = 33.99 × 40.12
13.232 + 19.98% of 549.99 = ? × 8.99
(44/25) ÷ (154/199.5) × 419.91 = ? – (11.11)3
44.84% of 799.94 + (625.21 ÷ 24.91) – √(224.77) = ?
78% of 1450 + 26² = ? + 1323 ÷ 17