A train moving with uniform speed for a certain distance takes 6 hours less if its speed is increased by 6 km/hour. It would have taken 6 hours more had its speed been decreased by 4 km/hour. Find the distance travelled and the speed of the train.
Step-by-step solution
Idea: Let the speed be v km/h and the time t h, so distance = vt. The distance is the same in all three situations, which gives two equations; the vt terms cancel and leave linear equations.
- Distance = vt. Faster: (v + 6)(t − 6) = vt ⇒ vt − 6v + 6t − 36 = vt ⇒ −6v + 6t = 36 ⇒ t − v = 6 … (1).1½ marks
- Slower: (v − 4)(t + 6) = vt ⇒ vt + 6v − 4t − 24 = vt ⇒ 6v − 4t = 24 ⇒ 3v − 2t = 12 … (2).1½ marks
- From (1), t = v + 6. In (2): 3v − 2v − 12 = 12 ⇒ v = 24, so t = 30.1 mark
- Distance = 24 × 30 = 720 km; speed = 24 km/h.1 mark
Check: 30 km/h × 24 h = 720 ✓ (6 hours less); 20 km/h × 36 h = 720 ✓ (6 hours more).
Answer to write in the exam
Let speed = v km/h, time = t h; distance = vt
(v + 6)(t − 6) = vt ⇒ t − v = 6 … (1)
(v − 4)(t + 6) = vt ⇒ 3v − 2t = 12 … (2)
t = v + 6 in (2): v = 24, t = 30
∴ Speed = 24 km/h, distance = 720 km
Common mistakes that cost marks
- Expanding (v + 6)(t − 6) wrongly, e.g. forgetting the −36 term.
- Giving only the speed and not the distance.
- Using distance = speed ÷ time.
How this can come in the exam
A car covers a fixed distance. If it went 10 km/h faster it would take 2 hours less; if 10 km/h slower, 3 hours more.
(i) Write the two equations in v and t. (ii) Find v and t. (iii) Find the distance.
Show answer
(i) (v + 10)(t − 2) = vt ⇒ 10t − 2v = 20 ⇒ 5t − v = 10; (v − 10)(t + 3) = vt ⇒ 3v − 10t = 30 (1 mark). (ii) v = 5t − 10 ⇒ 15t − 30 − 10t = 30 ⇒ t = 12, v = 50 (2 marks). (iii) 600 km (1 mark).Try one yourself
A cyclist covers a distance. At 2 km/h faster she takes 1 hour less; at 2 km/h slower, 2 hours more. Find her speed and time.
Show answer
(v + 2)(t − 1) = vt ⇒ 2t − v = 2; (v − 2)(t + 2) = vt ⇒ v − t = 2 ⇒ v = 6 km/h, t = 4 h (24 km).
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