An increase of speed of 4km/hr on a journey of 32 km reduces the time taken by 4 hours. What's the original speed.
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Let the original speed = v
Thus, the second speed is v + 4
Let the original time = t
Let the second time = T
Since time can be stated as t = s / v (t = time, s = distance, v = velocity),
we can compare the original time with the latter.
T = t - 4
32 / (v + 4) = (32 / v) - 4
Solving the equation
32v = 32(v+4) - 4v(v+4)
32v = 32v + 128 - 4v^2 - 16v
4v^2 + 16v -128 = 0
v^2 + 4v - 32 = 0
(v - 4)(v + 8) = 0
v = -8
v = 4 (since v must not be negative)
So, the original speed is 4.km/hr
Thus, the second speed is v + 4
Let the original time = t
Let the second time = T
Since time can be stated as t = s / v (t = time, s = distance, v = velocity),
we can compare the original time with the latter.
T = t - 4
32 / (v + 4) = (32 / v) - 4
Solving the equation
32v = 32(v+4) - 4v(v+4)
32v = 32v + 128 - 4v^2 - 16v
4v^2 + 16v -128 = 0
v^2 + 4v - 32 = 0
(v - 4)(v + 8) = 0
v = -8
v = 4 (since v must not be negative)
So, the original speed is 4.km/hr
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