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ATQ, Let the usual time taken by the train be 'T' hours. Using, distance = speed × time We have, 'D' = 90T ----------- (I) And, 'D' = 100 × {T - (15/60)} -------- (II) From equation (I) and (II), We have, 90T = 100 × {T - (15/60)} Or, 90T = 100T - 100 × (15/60) Or, 10T = 25 Or, 'T' = 2.5 On putting the value of 'T' = 2.5 in equation (I), We have 'D' = 90 × 2.5 = 225 km So, the required time = (4 × 225) ÷ 120 = (900/120) = 7.5 hours
24% of 400 × 16% of ? = 384
[{70 + (40 - 22) ÷ 3} ÷ 4] = ?
22 * 6 + 45% of 90 + 65% of 180 = ?
[(36 × 15 ÷ 96 + 19 ÷ 8) × 38] = ?% of 608
36×?² + (25% of 208 +13) = 60% of 2400 + 17×18
(360 - ?)/(25% of 96) = 13
Solve.
15.73 +13.25 +16.73 – 28.71 = 5 ×?
(11/12) × (18/22) × (4/3) + 3 = ?2