An express train takes 30 min less for a journey of 440 km. if it's usual speed is increased by 8 km/hr . find its usual speed
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distance = 440
let speed = x
time = 440/x
new speed = x+8
a.t.q = 440/x = 440/x+8 + 30/60
440/x = 880+x+8/2x+16
880x+7040 = 880x+x²+8x
x²+8x-7040=0
x²-80x+88x-7040
x(x-80)+88(x-80)
x = 80 and -88
reject -88 answer will be 80
let speed = x
time = 440/x
new speed = x+8
a.t.q = 440/x = 440/x+8 + 30/60
440/x = 880+x+8/2x+16
880x+7040 = 880x+x²+8x
x²+8x-7040=0
x²-80x+88x-7040
x(x-80)+88(x-80)
x = 80 and -88
reject -88 answer will be 80
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