Start learning 50% faster. Sign in now
ATQ, Let the length of the train = ‘y’ metres Then, distance covered by the train in 10 seconds = ‘y’ metres Distance covered by the train in 14 seconds = (y + 112) metres So, distance covered by the train in (14 – 10) seconds = y + 112 – y = 112 metres So, speed of the train = (112/4) = 28 m/s So, length of the train = 28 × 10 = 280 metres Relative speed of the bike with respect to the train = 33 – 28 = 5 m/s So, time taken by the bike to cross the train = (280/5) = 56 seconds
(23.99)2– (17.99)2+ (1378.88 + 44.88) ÷ ? = 607.998
8.992 + (5.01 × 4.98) + ? = 224.03
14.12 × 21.98 + 25.22% of 195.99 = ? × 50.9
? = 49.99² ÷ (1.98⁵ + 8.01 × 89.91) + 75.15% of (263.89 × 49.11)
10.10% of 999.99 + 14.14 × 21.21 - 250.25 = ?
960.11 ÷ 23.98 × 5.14 – 177.9 = √?
? = 65.78² ÷ (5.01⁵ + 7.02 × 33.33) + 33.33% of (290.88 × 23.09)
? = 49.97% of 38.09% of 1998.95