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All bus are train (A) + Some train are car (I) → Probable conclusion → All bus may be car (A). Hence conclusion I follows. No truck is a bus (E) + All bus are train (A) → Some trains are not truck (O*) → Probable conclusion → All truck may be train (A). Hence conclusion II follows. Some train are car (I) → Probable conclusion → All car may be train (A). Hence conclusion III follows. Alternate Method -
{(22.22 2 + 6.06 3 ) ÷ 15.15} × 3.03 = ? – 55.05 × 4.04
30.05% of 250.05 – 15.15% of 99.99 × 2.02 = ?
[(2/3 of 899.79) + 25% of 500.21] × (√195.77 + 30.03% of 399.79) = ?
(3375)1/3 x 12.11 x 6.97divide; 14.32 = ? + 15.022
142.07 + 21.89 – (128.12 ÷ 7.81 – 12.09 × 6.98) = ?
6106.11 ÷ √? × 55.9 = 3976.21
(9/10 of 3999.79) - √2499.83 + (17.81% of 1199.81) = ?