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Let the length of the train be 'L' metres and speed of the train be 's' m/s. Speed of the man = 14.4 X (5/18) = 4 m/s ATQ; (L/s) = 12 Or, L = 12s ....... (I) Also, {L/(s - 4)} = 15 Or, 12s = 15 X (s - 4) {Using equation (I)} Or, 12s = 15s - 60 Or, 60 = 3s So, s = 20 So, length of the train = 20 X 12 = 240 metres.
16.11 × 9.96 – (238.19 – 64.04 × 2.18) = ?
1242.12 ÷ √530 + 1139.89 ÷ 14.91 = ? + 45.39
24.022 + 6.06 × 36.06 ÷ 24.02 = ? × 5.05
`[(7.99)^2 - (13.001)^2 + (4.01)^3]^2=` ?
64.889% of 399.879 + √? = 54.90% of 799.80 – 44.03% of 400.21
(124.99)² = ?
20.11% of 159.99 + √99.97 ÷ 5.02 = ?