A current changes a boat's speed relative to the bank. A race compares distance covered by competitors over the same time. Both are relative-motion problems: define the frame of reference and the event being compared before calculating.
1. Boat speed and stream speed
Let b be boat speed in still water and s be stream speed, with b>s for ordinary upstream travel. Downstream speed is b+s; upstream speed is b−s. Given both observed speeds, b=(downstream+upstream)/2 and s=(downstream−upstream)/2. The arithmetic mean and half-difference recover the two components.
Worked example 1. A boat moves downstream at 18 km/h and upstream at 10 km/h. Its still-water speed is (18+10)/2=14 km/h; stream speed is (18−10)/2=4 km/h. Check: 14+4=18 and 14−4=10.
2. Time over equal or unequal legs
For a downstream leg of distance d, time is d/(b+s). For the same upstream distance, time is d/(b−s), which is longer. A round trip's average speed is total distance divided by total time. For equal distances, the harmonic-mean formula gives 2(b+s)(b−s)/[(b+s)+(b−s)] = (b²−s²)/b. It is less than b when s is positive.
Worked example 2. With still-water speed 12 km/h and stream 3 km/h, a 45 km downstream trip takes 45/15=3 hours; the same trip upstream takes 45/9=5 hours. Round-trip average speed is 90/8=11.25 km/h, not 12 km/h.
3. Rafts and floating objects
A free-floating raft moves with the stream, so its speed relative to the bank is s. An upstream-moving boat approaches it at relative speed (b−s)+s=b, if they move towards each other along the river. A downstream-moving boat gains on a raft at (b+s)−s=b. These cancellations are useful, but draw the directions first to avoid using the wrong sign.
Questions involving a boat turning back after dropping an object can often be solved with relative motion: the boat's motion relative to the water is symmetric before and after the turn if its engine speed is unchanged. Write distances in the bank frame if the wording is complex; never invoke the symmetry without checking the conditions.
4. Race margins
When A and B run at constant speeds over the same interval, distances are in the ratio of speeds. If A beats B by 20 m in a 100 m race, A covers 100 m while B covers 80 m in the same time, so A:B speeds are 5:4. A's lead in metres is a distance difference at A's finish time, not a time gap.
Worked example 3. In a 200 m race, A finishes when B has covered 160 m. Their speed ratio is 200:160=5:4. In a 300 m race at unchanged speeds, B has covered 240 m when A finishes; A wins by 60 m.
5. Head starts and dead heats
If the faster runner gives a head start, distinguish a head start in distance from a delay in starting time. A 20 m head start in a 100 m race means the slower runner has 80 m left while the faster has 100 m. A 5-second head start gives the slower runner five extra seconds of travel, which requires a time equation.
Worked example 4. A runs at 10 m/s and B at 8 m/s. In a 100 m race, B begins 20 m ahead. A needs 10 seconds for 100 m; B needs 80/8=10 seconds for the remaining distance. They finish together.
6. Error checks
Downstream speed must exceed upstream speed for a positive stream. Still-water speed must lie between the two observed speeds. A stronger runner's finishing distance cannot be less than the official race distance. Check whether “beats by 10 seconds” or “beats by 10 metres” is being asked: they require different conversions.
Recall before practice
1. Recover boat and stream speed from upstream and downstream observations. 2. Explain why a round-trip average speed can be below still-water speed. 3. Convert a race winning margin into a speed ratio. 4. Distinguish a distance head start from a time head start.
Chapter 14 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. In a 2000 m race, A beats B by 12 m or 3 seconds. What is the speed of B?
A. 0.01 m/s B. 666.67 m/s C. 5 m/s D. 4 m/s
2. In a game of 150 points, A can give B 15 points. How many points can A give B in a game of 450 points?
A. 45 points B. 15 points C. 50 points D. 315 points
3. A swimmer takes twice as long to go upstream as to come downstream over the same distance. If the speed of the stream is 6 km/h, find the speed in still water.
A. 12 km/h B. 18 km/h C. 8 km/h D. 9 km/h
4. In a 1500 m race, P beats Q by 18 m or 6 seconds. What is the speed of Q?
A. 4 m/s B. 3 m/s C. 0.01 m/s D. 250 m/s
5. The ratio of the downstream speed to the upstream speed of a motorboat is 5 : 3. What is the ratio of the speed in still water to the speed of the stream?
A. 8 : 3 B. 1 : 4 C. 5 : 3 D. 4 : 1
6. In a 800 m race, P beats Q by 8 m or 4 seconds. What is the speed of Q?
A. 3 m/s B. 0.01 m/s C. 200 m/s D. 2 m/s
7. A boatman can travel at 6 km/h in still water. If the speed of the current is 2 km/h, how much time will it take to cover 8 km upstream?
A. 1 hour B. 1 hour 20 minutes C. 2 hours 30 minutes D. 2 hours
8. A swimmer travels at 14 km/h downstream and 4 km/h upstream. What is the speed in still water?
A. 18 km/h B. 10 km/h C. 9 km/h D. 5 km/h
9. In a 400 m race, A beats B by 60 m or 15 seconds. What is the speed of B?
A. 26.67 m/s B. 5 m/s C. 4 m/s D. 0.15 m/s
10. A motorboat takes four times as long to go upstream as to come downstream over the same distance. If the speed in still water is 20 km/h, find the speed of the stream.
A. 12 km/h B. 16 km/h C. 5 km/h D. 15 km/h
11. A boatman can travel at 24 km/h in still water. If the speed of the current is 6 km/h, how much time will it take to cover 72 km upstream?
A. 4 hours B. 4 hours 30 minutes C. 3 hours D. 2 hours 24 minutes
12. The ratio of the downstream speed to the upstream speed of a boatman is 5 : 4. What is the ratio of the speed in still water to the speed of the stream?
A. 1 : 9 B. 9 : 1 C. 9 : 2 D. 5 : 4
13. In a 200 m race, A beats B by 10 m and B beats C by 20 m. By how many metres does A beat C?
A. 30 m B. 31 m C. 10 m D. 29 m
14. The speeds of P and Q are in the ratio 8 : 7. In a 800 m race, how much start should P give Q so that the race ends in a dead heat?
A. 110 m B. 90 m C. 100 m D. 700 m
15. In a 2000 m race, P beats Q by 40 m and Q beats R by 200 m. By how many metres does P beat R?
A. 160 m B. 244 m C. 236 m D. 240 m
16. A swimmer covers 34 km downstream in 2 hours, while a piece of wood takes 8 hours 30 minutes to float the same 34 km with the current. What is the speed of the boat in still water?
A. 9 km/h B. 17 km/h C. 21 km/h D. 13 km/h
17. The speed of a boatman in still water is 11 km/h and the stream flows at 5 km/h. How much more time does it take to go 48 km upstream than to come the same distance downstream?
A. 11 hours B. 8 hours C. 5 hours D. 8 hours 44 minutes
18. A boatman takes four times as long to go upstream as to come downstream over the same distance. If the speed of the stream is 6 km/h, find the speed in still water.
A. 8 km/h B. 10 km/h C. 24 km/h D. 30 km/h
19. The speed of a boatman in still water is 15 km/h and the stream flows at 3 km/h. How long will it take to go 72 km upstream and come back?
A. 10 hours B. 9 hours 36 minutes C. 2 hours D. 6 hours
20. In a 100 m race, A beats B by 20 m or 5 seconds. In how much time does B complete the race?
A. 5 s B. 25 s C. 30 s D. 20 s
21. In a 200 m race, B can give A a start of 40 m and C can give B a start of 45 m. What start can C give A in the same race?
A. 5 m B. 76 m C. 85 m D. 94 m
22. A runs 100 m in 15 s and B runs the same distance in 20 s. By how many metres does A beat B?
A. 25 m B. 75 m C. 5 m D. 33.33 m
23. Two boats P and Q are 200 km apart on a river flowing at 6 km/h. P starts downstream at 23 km/h (still water) and Q starts upstream at 27 km/h (still water) at the same time, towards each other. After how long do they meet?
A. 4 hours B. 5 hours 16 minutes C. 50 hours D. 3 hours 14 minutes
24. The speed of a motorboat in still water is 4 km/h and the stream flows at 2 km/h. How much more time does it take to go 36 km upstream than to come the same distance downstream?
A. 6 hours B. 12 hours C. 18 hours D. 24 hours
25. Anita can row 85 km downstream in 5 hours and 11 km upstream in 1 hour. What is the speed of the stream?
A. 14 km/h B. 6 km/h C. 3 km/h D. 16 km/h
26. In a 1000 m race, Q can give P a start of 175 m and R can give Q a start of 80 m. What start can R give P in the same race?
A. 255 m B. 95 m C. 241 m D. 269 m
27. A boatman can row at 16 km/h in still water in a stream flowing at 4 km/h. How long will it take to go 75 km downstream and return to the starting point?
A. 10 hours B. 8 hours 26 minutes C. 6 hours 15 minutes D. 9 hours 23 minutes
28. The speeds of P and Q are in the ratio 4 : 3. In a 200 m race, how much start should P give Q so that the race ends in a dead heat?
A. 50 m B. 60 m C. 40 m D. 150 m
29. A swimmer covers 36 km downstream and 12 km upstream in 5 hours. If the speed of the stream is 3 km/h, find the speed of the boat in still water.
A. 12 km/h B. 6 km/h C. 11 km/h D. 9 km/h
30. In a 400 m race, Q can give P a start of 60 m and R can give Q a start of 100 m. What start can R give P in the same race?
A. 160 m B. 145 m C. 175 m D. 40 m
31. Two boats P and Q are 184 km apart on a river flowing at 5 km/h. P starts downstream at 22 km/h (still water) and Q starts upstream at 24 km/h (still water) at the same time, towards each other. After how long do they meet?
A. 3 hours 17 minutes B. 5 hours 7 minutes C. 4 hours D. 92 hours
32. In a 800 m race, A beats B by 160 m or 20 seconds. In how much time does B complete the race?
A. 5 s B. 80 s C. 120 s D. 100 s
33. The speeds of A and B are in the ratio 3 : 2. By how many metres will A beat B in a 1500 m race?
A. 500 m B. 1000 m C. 750 m D. 510 m
34. A runs 1500 m in 12 min 18 s and B runs the same distance in 12 min 30 s. By how many metres does A beat B?
A. 12 m B. 1476 m C. 24.39 m D. 24 m
35. In a 800 m race, Q can give P a start of 120 m and R can give Q a start of 80 m. What start can R give P in the same race?
A. 188 m B. 212 m C. 40 m D. 200 m
36. A boat covers 57 km downstream and 9 km upstream in 4 hours. If the speed of the stream is 5 km/h, find the speed of the boat in still water.
A. 14 km/h B. 16 km/h C. 19 km/h D. 9 km/h
37. A boatman covers 36 km downstream in 4 hours, while a piece of wood takes 12 hours to float the same 36 km with the current. What is the speed of the boat in still water?
A. 3 km/h B. 6 km/h C. 9 km/h D. 12 km/h
38. The speed of a boat in still water is 8 km/h and the stream flows at 1 km/h. How much more time does it take to go 63 km upstream than to come the same distance downstream?
A. 15 hours 45 minutes B. 2 hours C. 9 hours D. 16 hours
39. In a 1500 m race, A beats B by 20 m or 4 seconds. In how much time does A complete the race?
A. 296 s B. 300 s C. 75 s D. 304 s
40. In a 1000 m race, P beats Q by 60 m and Q beats R by 50 m. By how many metres does P beat R?
A. 110 m B. 10 m C. 107 m D. 113 m
41. Boat A leaves a jetty and moves downstream at 6 km/h in still water. 2 hours later boat B leaves the same jetty downstream at 9 km/h in still water. If the stream flows at 3 km/h, how long after starting does B catch up with A?
A. 8 hours B. 6 hours C. 4 hours D. 1 hour 30 minutes
42. Two boats P and Q are 34 km apart on a river flowing at 4 km/h. P moves downstream at 9 km/h in still water and Q moves upstream at 8 km/h in still water, starting simultaneously towards each other. How far from P's starting point do they meet?
A. 18 km B. 8 km C. 17 km D. 26 km
43. In a 1500 m race, A beats B by 32 m or 8 seconds, and B beats C by 375 m. By how many seconds does A beat C?
A. 134 s B. 125 s C. 133 s D. 132 s
44. In a 1000 m race, P beats Q by 25 m or 5 seconds, and Q beats R by 200 m. By how many seconds does P beat R?
A. 54 s B. 56 s C. 55 s D. 50 s
45. In a 500 m race, P gives Q a start of 90 m and still beats Q by 50 m. What is the ratio of the speeds of P and Q?
A. 25 : 18 B. 50 : 41 C. 10 : 9 D. 41 : 36
46. In a 100 m race, A beats B by 25 m and A beats C by 40 m. In a 100 m race between B and C, by how many metres will B beat C?
A. 25 m B. 65 m C. 20 m D. 15 m
47. In a 2000 m race, P gives Q a start of 90 m and still beats Q by 60 m. What is the ratio of the speeds of P and Q?
A. 40 : 37 B. 191 : 185 C. 200 : 191 D. 100 : 97
48. A boatman whose speed in still water is 10 km/h takes 6 hours more to go 45 km upstream than to go the same distance downstream. What is the speed of the stream?
A. 4 km/h B. 10 km/h C. 6 km/h D. 5 km/h
49. Boat A leaves a jetty and moves downstream at 15 km/h in still water. 1 hour later boat B leaves the same jetty downstream at 21 km/h in still water. If the stream flows at 3 km/h, how long after starting does B catch up with A?
A. 2 hours 30 minutes B. 45 minutes C. 4 hours D. 3 hours
50. A boatman goes 84 km downstream in 7 hours and returns upstream in 21 hours. How long would a piece of wood take to drift 84 km down the same stream?
A. 10 hours 30 minutes B. 14 hours C. 28 hours D. 21 hours
Chapter 14 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. D. "Beats by 12 m or 3 s" means B covers the last 12 m in 3 s. Speed of B = 12/3 = 4 m/s. (Dividing the whole 2000 m by 3 s, 666.67 m/s, is the common error.)
APAR26-14-29 | Loser's speed from 'x m or t s' | Easy
2. A. When A scores 150, B scores 150 − 15 = 135. The ratio A : B = 150 : 135 stays fixed, so when A scores 450, B scores 450 × 135/150 = 405. Points A can give = 450 − 405 = 45. (Keeping 15 points regardless of game size is the trap.)
APAR26-14-31 | Points in a game of a different size | Easy
3. B. Time ratio up : down = 2 : 1 ⇒ speed ratio down : up = 2 : 1. Let downstream = 2x, upstream = x. Still water = (2x + x)/2 = 1.5x, stream = (2x − x)/2 = 0.5x, so b : s = 3 : 1. With s = 6, x-unit = 6/1 = 6, giving b = 3 × 6 = 18 km/h.
APAR26-14-02 | Upstream time is a multiple of downstream time | Easy
4. B. "Beats by 18 m or 6 s" means Q covers the last 18 m in 6 s. Speed of Q = 18/6 = 3 m/s. (Dividing the whole 1500 m by 6 s, 250 m/s, is the common error.)
APAR26-14-32 | Loser's speed from 'x m or t s' | Easy
5. D. Let downstream = 10k and upstream = 6k. Still water = (10k + 6k)/2 = 8k, stream = (10k − 6k)/2 = 2k. Ratio = 8k : 2k = 4 : 1. (Rule: D : U = p : q ⇒ b : s = (p + q) : (p − q).)
APAR26-14-03 | Speed ratio from downstream : upstream | Easy
6. D. "Beats by 8 m or 4 s" means Q covers the last 8 m in 4 s. Speed of Q = 8/4 = 2 m/s. (Dividing the whole 800 m by 4 s, 200 m/s, is the common error.)
APAR26-14-30 | Loser's speed from 'x m or t s' | Easy
7. D. Upstream speed = 6 − 2 = 4 km/h. Time = distance/speed = 8/4 = 2 hours = 2 hours. (Using the still-water speed gives 1.33 h, which ignores the current.)
APAR26-14-01 | Time for one leg | Easy
8. C. Speed in still water = (downstream + upstream)/2 = (14 + 4)/2 = 9 km/h; speed of stream = (downstream − upstream)/2 = (14 − 4)/2 = 5 km/h. Required answer: 9 km/h. (Taking the bare difference 10 as the stream speed is the usual slip.)
APAR26-14-07 | Still-water and stream speed from up/down speeds | Easy
9. C. "Beats by 60 m or 15 s" means B covers the last 60 m in 15 s. Speed of B = 60/15 = 4 m/s. (Dividing the whole 400 m by 15 s, 26.67 m/s, is the common error.)
APAR26-14-27 | Loser's speed from 'x m or t s' | Easy
10. A. Time ratio up : down = 4 : 1 ⇒ speed ratio down : up = 4 : 1. Let downstream = 4x, upstream = x. Still water = (4x + x)/2 = 2.5x, stream = (4x − x)/2 = 1.5x, so b : s = 5 : 3. With b = 20, x-unit = 20/5 = 4, giving s = 3 × 4 = 12 km/h.
APAR26-14-04 | Upstream time is a multiple of downstream time | Easy
11. A. Upstream speed = 24 − 6 = 18 km/h. Time = distance/speed = 72/18 = 4 hours = 4 hours. (Using the still-water speed gives 3 h, which ignores the current.)
APAR26-14-06 | Time for one leg | Easy
12. B. Let downstream = 10k and upstream = 8k. Still water = (10k + 8k)/2 = 9k, stream = (10k − 8k)/2 = 1k. Ratio = 9k : 1k = 9 : 1. (Rule: D : U = p : q ⇒ b : s = (p + q) : (p − q).)
APAR26-14-05 | Speed ratio from downstream : upstream | Easy
13. D. When A runs 200 m, B runs 190 m. When B runs 200 m, C runs 180 m, so when B runs 190 m, C runs 190 × 180/200 = 171 m. Hence A beats C by 200 − 171 = 29 m. (Simply adding the margins, 10 + 20 = 30 m, is the trap.)
APAR26-14-28 | Chained winning margins | Easy
14. C. In the time P runs 800 m, Q runs 800 × 7/8 = 700 m. For a dead heat Q must start 800 − 700 = 100 m ahead. (Formula: L(a − b)/a; using L(a − b)/b = 114.29 m is the usual slip.)
APAR26-14-26 | Start for a dead heat | Easy
15. C. When P runs 2000 m, Q runs 1960 m. When Q runs 2000 m, R runs 1800 m, so when Q runs 1960 m, R runs 1960 × 1800/2000 = 1764 m. Hence P beats R by 2000 − 1764 = 236 m. (Simply adding the margins, 40 + 200 = 240 m, is the trap.)
APAR26-14-34 | Chained winning margins | Medium
16. D. A floating object moves at the speed of the stream: s = 34/8.5 = 4 km/h. Downstream speed = 34/2 = 17 km/h. Still-water speed = downstream − stream = 17 − 4 = 13 km/h. (Do not halve the downstream speed; 17/2 is not the still-water speed here.)
APAR26-14-13 | Still-water speed from a floating object | Medium
17. C. Downstream speed = 11 + 5 = 16 km/h, upstream speed = 11 − 5 = 6 km/h. Time downstream = 48/16 = 3 h, time upstream = 48/6 = 8 h. Difference = 8 − 3 = 5 h = 5 hours. (2d/b = 8.73 h is the trap answer.)
APAR26-14-14 | Difference of upstream and downstream times | Medium
18. B. Time ratio up : down = 4 : 1 ⇒ speed ratio down : up = 4 : 1. Let downstream = 4x, upstream = x. Still water = (4x + x)/2 = 2.5x, stream = (4x − x)/2 = 1.5x, so b : s = 5 : 3. With s = 6, x-unit = 6/3 = 2, giving b = 5 × 2 = 10 km/h.
APAR26-14-17 | Upstream time is a multiple of downstream time | Medium
19. A. Downstream speed = 15 + 3 = 18 km/h, upstream speed = 15 − 3 = 12 km/h. Time downstream = 72/18 = 4 h, time upstream = 72/12 = 6 h. Total = 4 + 6 = 10 h = 10 hours. (2d/b = 9.6 h is the trap answer.)
APAR26-14-18 | Total time up and down | Medium
20. B. B runs 20 m in 5 s, so B's speed = 20/5 = 4 m/s and B's time for 100 m = 100/4 = 25 s. A finishes 5 s earlier: 25 − 5 = 20 s. Required time = 25 s. (Subtracting 5 s gives A's time, 20 s, not B's.)
APAR26-14-42 | Finishing time from 'x m or t s' | Medium
21. B. B : A = 200 : 160 and C : B = 200 : 155. Multiplying, C : A = 200 × 200 : 160 × 155, so when C runs 200 m, A runs 160 × 155/200 = 124 m. Start = 200 − 124 = 76 m. (Adding the starts gives 85 m, which is wrong.)
APAR26-14-40 | Chained head starts | Medium
22. A. A finishes 20 − 15 = 5 s before B. In those 5 s, B (speed 100/20 = 5 m/s) is still 5 × 5 = 25 m short of the finish. So A beats B by 25 m. (Using A's speed for the gap, 33.33 m, is wrong: the gap is what B has left to run.)
APAR26-14-35 | Winning distance from two timings | Medium
23. A. P's effective speed = 23 + 6 = 29 km/h, Q's = 27 − 6 = 21 km/h. Relative speed = 29 + 21 = 50 km/h — the stream cancels out when boats move in opposite directions. Time = 200/50 = 4 hours.
APAR26-14-15 | Boats meeting from opposite points | Medium
24. B. Downstream speed = 4 + 2 = 6 km/h, upstream speed = 4 − 2 = 2 km/h. Time downstream = 36/6 = 6 h, time upstream = 36/2 = 18 h. Difference = 18 − 6 = 12 h = 12 hours. (2d/b = 18 h is the trap answer.)
APAR26-14-10 | Difference of upstream and downstream times | Medium
25. C. Downstream speed = 85/5 = 17 km/h, upstream speed = 11/1 = 11 km/h. Still-water speed = (17 + 11)/2 = 14 km/h and stream speed = (17 − 11)/2 = 3 km/h. Answer: 3 km/h. (Averaging total distance over total time, 16 km/h, is wrong.)
APAR26-14-12 | Speeds from two timed legs | Medium
26. C. Q : P = 1000 : 825 and R : Q = 1000 : 920. Multiplying, R : P = 1000 × 1000 : 825 × 920, so when R runs 1000 m, P runs 825 × 920/1000 = 759 m. Start = 1000 − 759 = 241 m. (Adding the starts gives 255 m, which is wrong.)
APAR26-14-43 | Chained head starts | Medium
27. A. Downstream speed = 16 + 4 = 20 km/h, upstream speed = 16 − 4 = 12 km/h. Total time = 75/20 + 75/12 = 3.75 + 6.25 = 10 hours = 10 hours. (Using 2d/b = 9.38 h ignores the current.)
APAR26-14-16 | Round-trip time | Medium
28. A. In the time P runs 200 m, Q runs 200 × 3/4 = 150 m. For a dead heat Q must start 200 − 150 = 50 m ahead. (Formula: L(a − b)/a; using L(a − b)/b = 66.67 m is the usual slip.)
APAR26-14-37 | Start for a dead heat | Medium
29. D. Let still-water speed = x km/h. Then 36/(x + 3) + 12/(x − 3) = 5. Clearing denominators: 5x² − 48x + 27 = 0, whose admissible root is x = 9. Check: 36/12 + 12/6 = 3 + 2 = 5 ✓. (Total distance/total time = 9.6 km/h is a trap.)
APAR26-14-08 | Still-water speed from a mixed trip | Medium
30. B. Q : P = 400 : 340 and R : Q = 400 : 300. Multiplying, R : P = 400 × 400 : 340 × 300, so when R runs 400 m, P runs 340 × 300/400 = 255 m. Start = 400 − 255 = 145 m. (Adding the starts gives 160 m, which is wrong.)
APAR26-14-41 | Chained head starts | Medium
31. C. P's effective speed = 22 + 5 = 27 km/h, Q's = 24 − 5 = 19 km/h. Relative speed = 27 + 19 = 46 km/h — the stream cancels out when boats move in opposite directions. Time = 184/46 = 4 hours.
APAR26-14-11 | Boats meeting from opposite points | Medium
32. D. B runs 160 m in 20 s, so B's speed = 160/20 = 8 m/s and B's time for 800 m = 800/8 = 100 s. A finishes 20 s earlier: 100 − 20 = 80 s. Required time = 100 s. (Subtracting 20 s gives A's time, 80 s, not B's.)
APAR26-14-45 | Finishing time from 'x m or t s' | Medium
33. A. In the time A runs 1500 m, B runs 1500 × 2/3 = 1000 m. So A wins by 1500 − 1000 = 500 m. (Formula: L(a − b)/a; using L(a − b)/b = 750 m is the usual slip.)
APAR26-14-39 | Winning margin from speed ratio | Medium
34. D. A finishes 750 − 738 = 12 s before B. In those 12 s, B (speed 1500/750 = 2 m/s) is still 2 × 12 = 24 m short of the finish. So A beats B by 24 m. (Using A's speed for the gap, 24.39 m, is wrong: the gap is what B has left to run.)
APAR26-14-44 | Winning distance from two timings | Medium
35. A. Q : P = 800 : 680 and R : Q = 800 : 720. Multiplying, R : P = 800 × 800 : 680 × 720, so when R runs 800 m, P runs 680 × 720/800 = 612 m. Start = 800 − 612 = 188 m. (Adding the starts gives 200 m, which is wrong.)
APAR26-14-33 | Chained head starts | Medium
36. A. Let still-water speed = x km/h. Then 57/(x + 5) + 9/(x − 5) = 4. Clearing denominators: 4x² − 66x + 140 = 0, whose admissible root is x = 14. Check: 57/19 + 9/9 = 3 + 1 = 4 ✓. (Total distance/total time = 16.5 km/h is a trap.)
APAR26-14-19 | Still-water speed from a mixed trip | Medium
37. B. A floating object moves at the speed of the stream: s = 36/12 = 3 km/h. Downstream speed = 36/4 = 9 km/h. Still-water speed = downstream − stream = 9 − 3 = 6 km/h. (Do not halve the downstream speed; 9/2 is not the still-water speed here.)
APAR26-14-09 | Still-water speed from a floating object | Medium
38. B. Downstream speed = 8 + 1 = 9 km/h, upstream speed = 8 − 1 = 7 km/h. Time downstream = 63/9 = 7 h, time upstream = 63/7 = 9 h. Difference = 9 − 7 = 2 h = 2 hours. (2d/b = 15.75 h is the trap answer.)
APAR26-14-20 | Difference of upstream and downstream times | Medium
39. A. B runs 20 m in 4 s, so B's speed = 20/4 = 5 m/s and B's time for 1500 m = 1500/5 = 300 s. A finishes 4 s earlier: 300 − 4 = 296 s. Required time = 296 s. (Forgetting to subtract the 4 s gives 300 s.)
APAR26-14-38 | Finishing time from 'x m or t s' | Medium
40. C. When P runs 1000 m, Q runs 940 m. When Q runs 1000 m, R runs 950 m, so when Q runs 940 m, R runs 940 × 950/1000 = 893 m. Hence P beats R by 1000 − 893 = 107 m. (Simply adding the margins, 60 + 50 = 110 m, is the trap.)
APAR26-14-36 | Chained winning margins | Medium
41. B. Effective speeds: A = 6 + 3 = 9 km/h, B = 9 + 3 = 12 km/h. In 2 h, A is 9 × 2 = 18 km ahead. B gains 12 − 9 = 3 km/h (same direction ⇒ the stream cancels). Time = 18/3 = 6 hours. (Ignoring the stream in A's lead gives 4 h.)
APAR26-14-25 | Delayed chase downstream | Difficult
42. D. Relative speed = (9 + 4) + (8 − 4) = 17 km/h, so they meet after 34/17 = 2 h. In 2 h, P travels 2 × (9 + 4) = 26 km downstream. (Using 2 × 9 = 18 km forgets the current that carries P.)
APAR26-14-23 | Meeting point of two boats | Difficult
43. C. B's speed = 32/8 = 4 m/s, so B takes 1500/4 = 375 s and A takes 375 − 8 = 367 s. When B finishes (375 s), C has run 1500 − 375 = 1125 m, so C's speed = 1125/375 = 3 m/s and C's time = 1500/3 = 500 s. A beats C by 500 − 367 = 133 s.
APAR26-14-49 | Time margin across three runners | Difficult
44. C. Q's speed = 25/5 = 5 m/s, so Q takes 1000/5 = 200 s and P takes 200 − 5 = 195 s. When Q finishes (200 s), R has run 1000 − 200 = 800 m, so R's speed = 800/200 = 4 m/s and R's time = 1000/4 = 250 s. P beats R by 250 − 195 = 55 s.
APAR26-14-48 | Time margin across three runners | Difficult
45. A. P runs the full 500 m. Q starts 90 m ahead and is still 50 m short at the finish, so Q runs 500 − 90 − 50 = 360 m in the same time. Speed ratio P : Q = 500 : 360 = 25 : 18. (Ignoring the 50 m margin gives 50 : 41.)
APAR26-14-50 | Speed ratio from start and margin | Difficult
46. C. When A runs 100 m, B runs 75 m and C runs 60 m. So while B runs 75 m, C runs 60 m; while B runs 100 m, C runs 60 × 100/75 = 80 m. B beats C by 100 − 80 = 20 m. (Taking the difference 40 − 25 = 15 m ignores that B still has 25 m to run.)
APAR26-14-46 | Margin between the two losers | Difficult
47. A. P runs the full 2000 m. Q starts 90 m ahead and is still 60 m short at the finish, so Q runs 2000 − 90 − 60 = 1850 m in the same time. Speed ratio P : Q = 2000 : 1850 = 40 : 37. (Ignoring the 60 m margin gives 200 : 191.)
APAR26-14-47 | Speed ratio from start and margin | Difficult
48. D. Let stream speed = x km/h. 45/(10 − x) − 45/(10 + x) = 6 ⇒ 45 × 2x/(100 − x²) = 6 ⇒ 90x = 6(100 − x²) ⇒ 6x² + 90x − 600 = 0. Check x = 5: 45/5 − 45/15 = 9 − 3 = 6 ✓. Stream speed = 5 km/h.
APAR26-14-22 | Stream speed from extra upstream time | Difficult
49. D. Effective speeds: A = 15 + 3 = 18 km/h, B = 21 + 3 = 24 km/h. In 1 h, A is 18 × 1 = 18 km ahead. B gains 24 − 18 = 6 km/h (same direction ⇒ the stream cancels). Time = 18/6 = 3 hours. (Ignoring the stream in A's lead gives 2.5 h.)
APAR26-14-21 | Delayed chase downstream | Difficult
50. D. Downstream speed = 84/7 = 12 km/h, upstream speed = 84/21 = 4 km/h. Stream speed = (12 − 4)/2 = 4 km/h, and a floating object moves at exactly this speed. Drift time = 84/4 = 21 h = 21 hours. (Using D − U = 8 km/h without halving gives 10.5 h.)
APAR26-14-24 | Drift time of a floating object | Difficult