1
Question
A train take a 2 hours less for a journey of 300 km . If the speed of the is increased by 5 km / hr from its usual speed , find the usual speed of the train.
A train take a 2 hours less for a journey of 300 km . If the speed of the is increased by 5 km / hr from its usual speed , find the usual speed of the train.
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Solution
Let x = normal speed
Increased speed =x+5
Distance to travel=300
usual time =distance/usual speed=300/x---(1)
New time =distance/increased speed
=300/x+5 ----(2)
Question says that new time is 2 hour less than usual time.
Ie, new time =old time-2
new time +2=old time---(3)
From equation (1) and (3) ,we will get the formula for new time and old time.
[new time=300/x+5 and old time =300/x]
Substitute these formula in equation (3).
(300/(x+5))+2=300/x
. . . combine left side into a single fraction
(300+2x+2×5)/(x+5) =300/x
(310 + 2x) / (x + 5) = 300 / x
. . . cross multiply
310x + 2x^2 = 300x + 1500
2x^2 + 10x - 1500 = 0
x^2 + 5x - 750 = 0
Solve this quadratic equation.
(x - 25) (x + 30) = 0
x = -30 <== invalid negative time
x = 25 km / h <===== answer
Increased speed =x+5
Distance to travel=300
usual time =distance/usual speed=300/x---(1)
New time =distance/increased speed
=300/x+5 ----(2)
Question says that new time is 2 hour less than usual time.
Ie, new time =old time-2
new time +2=old time---(3)
From equation (1) and (3) ,we will get the formula for new time and old time.
[new time=300/x+5 and old time =300/x]
Substitute these formula in equation (3).
(300/(x+5))+2=300/x
. . . combine left side into a single fraction
(300+2x+2×5)/(x+5) =300/x
(310 + 2x) / (x + 5) = 300 / x
. . . cross multiply
310x + 2x^2 = 300x + 1500
2x^2 + 10x - 1500 = 0
x^2 + 5x - 750 = 0
Solve this quadratic equation.
(x - 25) (x + 30) = 0
x = -30 <== invalid negative time
x = 25 km / h <===== answer
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