Question
Train βAβ can cross a pole in 4 seconds and a 150
metre long platform in 10 seconds. If the ratio of length of train βAβ and train βBβ is 2:5, respectively, then find the time taken by train βBβ to cross a pole with a speed of 25 m/s.Solution
Let the length and speed of the train βAβ be βlβ metre and βsβ m/s, respectively. According to question, l = 4s Also, 10s = 4s + 150 Or, 6s = 150 Or, s = 25 Therefore, length of train βAβ = 4s = 100 metres Length of train βBβ = 150 Γ (5/2) = 250 metres Required time taken = 250 Γ· 25 = 10 second
(1.01) 0 + (2.02) 1 + (2.93) 2 + (4.04) 3 + (5.05) 4 = ?
49.97% ofΒ 2016 β 37.99% of 1050 = ? β 47.98% of 5950
124.88% of 60.101 + 18.09% of 849.87 β 22.12% of 1049.93= ? β 19.93
(√4623.9 + √484.2) – √2303.97 ÷ √1296.4 × √35.98 ÷ √15.99 = ?
√ ({(5.5 × 2.3) × √ (728.91))} = 3(1/7) ÷ ?/28
...111.89 Γ 4.12 β 504.04 Γ· 2.12 = 170.12 + ?
2 (1/4)% of 7999.78 + {49.77% of 899.71} + β144.14 - 20% of 1499.83 = ?
44.78% of 715.62 + 1785% of 42.98 = ?
(1560.23 Γ· 25.98) + (768.32 Γ· 23.9) + 1814.11 = ?
70.14% of 799.95 - 240.12 = ? + 40.17% of 299.95