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Let’s assume the speed of car is ‘8a’. The speed of a car is 60% more than the speed of a bus. 8a = (100+60)% of speed of a bus 160% of speed of a bus = 8a 1.6 x speed of a bus = 8a speed of a bus = 5a The speed of train is 2.4 times the difference between the speed of bus and car. speed of train = 2.4x(8a-5a) = 2.4x3a = 7.2a The train can cover (d-250) km distance in (t-3.5) hours. 7.2a = (d-250)/(t-3.5) a = (d-250)/7.2(t-3.5) Eq.(1) The car covers (d+140) km distance in (t-2) hours. 8a = (d+140)/(t-2) a = (d+140)/8(t-2) Eq.(2) The distance travelled by bus is 1500 km in which it will take (t+4) hours. 5a = 1500/(t+4) a = 300/(t+4) Eq.(3) So Eq.(1) = Eq.(2) (d-250)/7.2(t-3.5) = (d+140)/8(t-2) (d-250)/0.9(t-3.5) = (d+140)/(t-2) (d-250)/(0.9t-3.15) = (d+140)/(t-2) d = (376t−941)/(0.1t+1.15) Eq.(4) So Eq.(1) = Eq.(3) (d-250)/7.2(t-3.5) = 300/(t+4) (d-250)/(t-3.5) = 2160/(t+4) d = (2410t−6560)/(t+4) Eq.(5) So Eq.(4) = Eq.(5) (376t−941)/(0.1t+1.15) = (2410t−6560)/(t+4) After solving the above equation, t = 8, 3.5 But if t = 3.5, then the time taken by train to cover distance will be zero which is not possible. So t = 8 . Put the value of ‘t’ in Eq.(4). d = (376x8−941)/(0.1x8+1.15) = (3008−941)/(0.8+1.15) = 2067/1.95 = 1060 (i) The value of ‘t’ is an odd number. The above given statement is not true. Because the value of ‘t’ cannot be an odd number. (ii) The value of ‘d’ is a three digit number. The above given statement is not true. Because the value of ‘d’ is not a three digit number. (iii) The ratio between the distance travelled by bus and train is (6t+2) : (3t+3) respectively. distance travelled by bus and train = (6t+2) : (3t+3) = 5ax(t+4) : 7.2ax(t-3.5) Put the value of ‘t’ in the above equation. (6x8+2) : (3x8+3) = 5x(8+4) : 7.2x(8-3.5) (48+2) : (24+3) = 5x12 : 7.2x4.5 50 : 27 = 50 : 27 The above given statement is true. Because the above given expression is satisfied.
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