Question
A train service was disrupted due to an accident after a
travel of 30 km. The speed was reduced to four-fifths of the original speed. This caused it a delay of 45 min. If the accident had happened 18 km farther, it would have been running late for 36 min. What was the original speed of the train?Solution
Let the speed of the train be x. We know that. β Speed = Distance/Time Speed after accident = 4/5 of x = 4x/5 According to the question, The time difference in covering 18 km due to change in speed = 45 - 36 = 9 min 18/(4x/5) -(18/x) =9/60 90/4x)-18/x=9/60 β x = 30 km/h β΄ The original speed of the train is 30 Km/h.
√3598 × √(230 ) ÷ √102= ?
15% of 2400 + (β 484 β β 256) = ?
(13)2Β - 3127 Γ· 59 = ? x 4
6269 + 0.25 × 444 + 0.8 × 200 = ? × 15
...(53 + 480 Γ· 4)% of 20 = ?% of 70
Find the simplified value of the following expression:
62 + 122 Γ 5 - {272 + 162 - 422}
(15 Γ 225) Γ· (45 Γ 5) + 480 = ? + 25% of 1240
β [? x 11 + (β 1296)] = 16
11 Γ 25 + 12 Γ 15 + 14 Γ 20 + 15 = ?