The following table shows the distance travelled by a boat moving upstream (in…
2023
The following table shows the distance travelled by a boat moving upstream (in km) during five different days of a week from Monday to Friday, along with the speed of boat (in km/hour), speed of stream (in km/hour) and total time taken (in hours). Some data is missing in the table (indicated as '—') that you are expected to calculate, if required. Based on the data in the table, answer the questions that follow.
Day-wise distance travelled by the boat and other details
Days | Distance Upstream (km) | Speed of Boat (km/hour) | Speed of Stream (km/hour) | Total Time Taken (hours) |
|---|---|---|---|---|
Monday | 320 | — | 4 | — |
Tuesday | — | — | — | 75 |
Wednesday | 270 | — | 6 | — |
Thursday | — | 11 | 7 | — |
Friday | 324 | — | — | 72 |
Note: (1) Distance Upstream = Distance Downstream, if not stated otherwise.
Note: (2) Total Time Taken = Downstream Time + Upstream Time
On Wednesday, if the boat covers half of the upstream distance at its usual still-water speed and the other half at twice that still-water speed, taking 333/4 hours less than the usual upstream journey, what is the total time for the equal-distance upstream and downstream journeys?
Answer: C. 108 hours — ConceptFor a boat with still-water speed b and stream speed s, the upstream speed is b − s and the downstream speed is b + s. Travel time equals distance…
- A.
128 hours
- B.
120 hours
- C.
108 hours
- D.
92 hours
Show answer & explanation
Correct answer: C
Concept
For a boat with still-water speed b and stream speed s, the upstream speed is b − s and the downstream speed is b + s.
Travel time equals distance divided by speed. When equal distances are covered upstream and downstream, the two travel times are calculated separately and then added.
Application
Let b km/h be the boat’s usual still-water speed. On Wednesday, s = 6 km/h and the upstream distance is 270 km, so the usual upstream time is 270/(b − 6) hours.
Half the distance is 135 km. When the boat’s speed is doubled, its still-water speed becomes 2b, so the modified upstream time is 135/(b − 6) + 135/(2b − 6).
The saving is 33.75 hours: 270/(b − 6) − [135/(b − 6) + 135/(2b − 6)] = 33.75.
Therefore, 135[1/(b − 6) − 1/(2b − 6)] = 33.75, which gives b/[(b − 6)(2b − 6)] = 1/4.
Thus 4b = (b − 6)(2b − 6), so b2 − 11b + 18 = 0 and (b − 9)(b − 2) = 0. Since the boat must move faster than the 6 km/h stream, b = 9 km/h.
The upstream speed is 9 − 6 = 3 km/h, while the downstream speed is 9 + 6 = 15 km/h.
Total time = 270/3 + 270/15 = 90 + 18 = 108 hours.
Cross-check
With b = 9, the modified upstream time is 135/3 + 135/12 = 45 + 11.25 = 56.25 hours. The usual upstream time is 90 hours, and 90 − 56.25 = 33.75 hours, exactly the stated saving. Hence the total time is 108 hours.