Question
The Ship's speed in calm water exceeds that of the
current by 20%. After sailing downstream for 4 hours, it traveled upstream for 4 hours with an increased Ship speed of 15 km/h. The Ship covered a distance during downstream travel that is 175% greater than the distance it covered during upstream travel. Determine the original speed of the Ship when moving downstream.Solution
ATQ, Let the speed of the current be ‘a’ km/hr Therefore, speed of the Ship in still water = ‘1.2a’ km/hr Downstream distance travelled by the boat = (a + 1.2a) × 4 = 8.8a km Upstream distance covered by the boat = (1.2a + 15 – x) × 4 = (0.8a + 60) km According to the question, {(8.8a – 0.8a – 60)/(0.8a + 60)} = 1.75 Or, a = (165/6.6) = 25 km/hr Therefore, original downstream speed of the Ship = 2.2a = 55 km/hr
647.1 ÷ ? + 72.3 × 209.81 – 8743.1 = 6404
? (30.03 - 17.98 × 15.99 ÷ 12.01) = 729.03
(699.88% of 32) + (80.44% of 400.23) = ? + (11.67)2
12, 16, ?, 36, 52, 72
(? + 11.86) X 14.89 = 19.89% of 2399.89
- √81.45 + √225.60 + 49.89% of (520.43 + 22.13% of 131.45) = ?
22.03 × 6.97 + 19.01 – 16.02 = ?
P spends 20% of his monthly income in travelling. He spends 25% of his monthly income on household expenses and spends 15% of his monthly income on fami...
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