A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more
Answers
Solution
Let a distance be d and speed be x. Since time = distance/speed.
CASE 1→ d/(x + 3) = d/x - 40/60
→ d/x - d/(x + 3) = 2/3
→ (dx + 3d - dx)/(x² + 3x) = 2/3
→ 9d = 2x² + 6x
→ 9d = 2x(x + 3)
CASE 2 → d/(x + 2) = d/x + 40/60
→ d/(x - 2) - d/x = 2/3
→ (dx - dx + 2d)/(x² - 2x) = 2/3
→ 6d = 2x² - 4x
→ 6d = 2x(x - 2)
CASE 1/CASE 2:
→ 9d/6d = 2x(x + 3)/2x(x - 2)
→ 3/2 = (x + 3)/(x - 2)
→ 3x - 6 = 2x + 6
→ x = 12
By substituting values we get
→ 6d = 2(12)(12 - 2)
→ 6d = 240
→ d = 40
Hence distance was of 40 km and original speed was 12 km/h
Answer:
: Solution :
Suppose the original speed = s
The distance = d
The time = t mins
d = s × t
d = (s + 3) × (t - 40)
d = (s - 2) × (t + 40)
= Solving equation 1 & 2
= s × t = (s + 3) × (t - 40)
= s × t = s × t + 3t - 40s - 120
= 3t - 40s = 120
= Solving equation 1 & 3
= s × t = (s - 2) × (t + 40)
= s × t = s × t - 2t + 40s - 80
= 2t + 40s = 80
Solving the two and bold equations :
t = 200 minutes
s = 12km/h
d = 12 × (200/60) = 40
= 40 Km.
distance = 40 Km
original speed = 12 km/h