Question 1

Ashutosh can row at 24 km/h in still water. He takes twice as long to row a fixed distance upstream as he takes to row the same distance downstream. What is the speed of the stream (in km/h)?

Accepted Answer

Introduction / Context:In boats and streams, the effective speed differs depending on whether motion is with the current (downstream) or against it (upstream). When the time relation between upstream and downstream for the same distance is given, we can form an equation directly in terms of the still water speed and the current speed. Given Data / Assumptions: Still water speed v = 24 km/h. For the same one-way distance D, time upstream is twice time downstream. Uniform current and no delays at turns. Concept / Approach:Downstream speed = v + c; upstream speed = v − c. For the same D, time is inversely proportional to speed. Condition: D/(v − c) = 2 * D/(v + c). Cancel D to obtain an equation in v and c. Step-by-Step Solution: D/(v − c) = 2 * D/(v + c) ⇒ (v + c) = 2(v − c).v + c = 2v − 2c ⇒ bring like terms together: c + 2c = 2v − v ⇒ 3c = v.Given v = 24, so c = 24/3 = 8 km/h. Verification / Alternative check:With c = 8: downstream speed = 24 + 8 = 32; upstream speed = 24 − 8 = 16. For distance D, times are D/32 and D/16, whose ratio is 1:2 as required. Why Other Options Are Wrong:4 and 15 or 18 km/h do not satisfy 3c = v when v = 24 and hence do not produce the required 2:1 time ratio. Common Pitfalls:Equating times to the ratio of speeds directly or averaging speeds arithmetically. The time relationship must be set up via reciprocals of speeds. Final Answer:8 km/h

Question 2

A boat travels 12 km in 1 hour in still water (so v = 12 km/h). It takes three times as long to cover the same 12 km against the current. What is the speed of the current (in km/h)?

Accepted Answer

Introduction / Context:This is a direct application of upstream speed determination from a given time factor relative to still water performance. When the upstream time is some multiple of the still-water time for the same distance, we can infer the upstream speed and then the current. Given Data / Assumptions: Still water speed v = 12 km/h. Distance D = 12 km. Upstream time is three times the still-water time for D (which is 1 hour in still water). Concept / Approach:Still-water time for 12 km is 1 h, so the upstream time is 3 h. Therefore, upstream speed = distance / time = 12/3 = 4 km/h. Since upstream speed equals v − c, solve c = v − 4. Step-by-Step Solution: Upstream speed = 12/3 = 4 km/h.v − c = 4 ⇒ c = 12 − 4 = 8 km/h. Verification / Alternative check:With c = 8, downstream speed is 20 km/h and upstream is 4 km/h. To cover 12 km upstream at 4 km/h takes 3 h, which is three times the still-water time of 1 h. Why Other Options Are Wrong:12 km/h would imply zero or negative upstream capability; 6 or 7 km/h do not yield an upstream time exactly three times the still-water time. Common Pitfalls:Mistaking the relation between time and speed; the triple time factor applies to time, not speed, and thus speed is one-third in that leg. Final Answer:8 km/h

Question 3

A boatman takes twice as long to row a fixed distance against the stream as he takes with the stream. Find the ratio of the boat speed in still water to the stream speed.

Accepted Answer

Introduction / Context:Ratio questions in boats and streams often come from time comparisons for equal distances. Because time equals distance divided by speed, a factor-of-two time difference means a specific ratio between the effective speeds, which in turn constrains the relation between the still-water speed and the current. Given Data / Assumptions: D/(v − c) = 2 * D/(v + c) for the same distance D. v is still-water speed; c is current speed. No pauses; steady speeds. Concept / Approach:Cancel D to get (v + c) = 2(v − c). Solve for v in terms of c and then express the ratio v : c. Step-by-Step Solution: v + c = 2v − 2c ⇒ move terms: c + 2c = 2v − v ⇒ 3c = v.Therefore v : c = 3 : 1. Verification / Alternative check:Pick v = 3 and c = 1 (same units). Then downstream speed = 4, upstream speed = 2. For the same D, downstream time is D/4 and upstream time is D/2, which is exactly double. Why Other Options Are Wrong:2:1, 1:2, 1:3 do not satisfy v = 3c and will not produce a twofold time difference for equal distances. Common Pitfalls:Taking 2:1 between speeds instead of times; remember that time ratios invert speed ratios for the same distance. Final Answer:3 : 1

Question 4

A river flows at 4 km/h. A rower goes downstream from the start to a turning point and then rows back upstream to the start. The total distance covered is 42 km (so each leg is 21 km). The upstream journey takes 2 hours more than the downstream jour…

Accepted Answer

Introduction / Context:This round-trip problem provides different times for equal distances in opposite directions, along with a known current. By writing the two leg times and using the given 2-hour difference, we can solve for the still-water speed using a single equation in v. Given Data / Assumptions: Current speed c = 4 km/h. Each leg distance D = 21 km (since total is 42 km). Let v be still-water speed. Upstream time − downstream time = 2 h. Concept / Approach:Downstream time t1 = D/(v + c). Upstream time t2 = D/(v − c). The condition is t2 = t1 + 2. Substitute D and c and solve for v. Step-by-Step Solution: 21/(v − 4) = 21/(v + 4) + 2.Bring terms: 21/(v − 4) − 21/(v + 4) = 2.Left side = 21[(v + 4) − (v − 4)]/((v − 4)(v + 4)) = 168/(v^2 − 16).So 168/(v^2 − 16) = 2 ⇒ v^2 − 16 = 84 ⇒ v^2 = 100 ⇒ v = 10 km/h (positive root). Verification / Alternative check:Downstream speed 14 ⇒ t1 = 21/14 = 1.5 h. Upstream speed 6 ⇒ t2 = 21/6 = 3.5 h. Difference = 2 h matches the condition. Why Other Options Are Wrong:12, 9, 8 km/h do not satisfy the derived equation and will not create a 2-hour time gap over 21 km per leg with c = 4. Common Pitfalls:Using total distance in each time or forgetting to split 42 km into two equal legs. Final Answer:10 km/h

Question 5

A river flows at 5 km/h. A swimmer can swim at 20 km/h in still water. He swims 30 km downstream from point A to point B, then immediately turns and swims 30 km upstream back to A. What is the total time taken?

Accepted Answer

Introduction / Context:When speeds are given for still water and the current, the effective downstream and upstream speeds are v + c and v − c, respectively. For fixed distances on each leg, compute times separately and add them. Given Data / Assumptions: Still water speed v = 20 km/h. Current speed c = 5 km/h. Downstream and upstream distances both equal 30 km. Concept / Approach:Downstream speed = 25 km/h; upstream speed = 15 km/h. Time = distance / speed for each leg. Sum the two times for total duration. Step-by-Step Solution: Downstream time = 30/25 = 1.2 h = 1 h 12 min.Upstream time = 30/15 = 2 h.Total time = 1 h 12 min + 2 h = 3 h 12 min. Verification / Alternative check:Compute in minutes: 1.2 h = 72 min; add 120 min = 192 min = 3 h 12 min, consistent. Why Other Options Are Wrong:2 h 30 min and 3 h 30 min or 3 h 45 min are not equal to 192 min under the given speeds. Common Pitfalls:Using an average of 25 and 15 over the full 60 km; averages of speeds do not apply to mixed legs with different effective speeds. Final Answer:3 h 12 min

Question 6

Ishwar rows a boat. He takes half as much time to go a certain distance downstream as he takes to cover the same distance upstream. What is the ratio of the rate of the boat in still water to the rate of the current?

Accepted Answer

Introduction / Context:Time ratios for equal distances translate into relationships between effective speeds. If one leg takes half the time of the other, then the corresponding speed must be double. This fact helps connect the boat speed in still water and the current speed using v ± c. Given Data / Assumptions: D/(v + c) = (1/2) * D/(v − c). v is still water speed; c is current speed. Same distance both ways; constant speeds. Concept / Approach:Cancel D to obtain 1/(v + c) = (1/2)/(v − c). Cross-multiply to form a linear relation and solve for v in terms of c; then express v : c. Step-by-Step Solution: 2(v − c) = v + c.2v − 2c = v + c ⇒ v = 3c.Therefore v : c = 3 : 1. Verification / Alternative check:Assume v = 3 and c = 1 (arbitrary units). Then downstream speed = 4, upstream speed = 2. Times are D/4 and D/2—indeed, downstream time is half of upstream time. Why Other Options Are Wrong:2:1, 5:1, 7:1 do not maintain the required halving of time for equal distances. Common Pitfalls:Reversing the ratio or confusing whether time or speed is halved or doubled. Remember time and speed are inversely related for a fixed distance. Final Answer:3 : 1

Question 7

A man can row at 7.5 km/h in still water. The river current flows at 1.5 km/h. He rows from the start to a place and returns to the start in a total of 50 minutes. How far is the place from the start (one-way distance, in km)?

Accepted Answer

Introduction / Context:Round-trip time with different downstream and upstream speeds can be used to deduce the one-way distance. This is a straightforward harmonic-like setup using v ± c and a fixed total time. Given Data / Assumptions: v = 7.5 km/h (still water). c = 1.5 km/h (current). Total time T = 50 min = 5/6 h. Let one-way distance be d (km). Concept / Approach:Downstream speed = v + c = 9 km/h. Upstream speed = v − c = 6 km/h. Total time = d/9 + d/6. Set equal to 5/6 and solve for d. Step-by-Step Solution: d/9 + d/6 = 5/6.Compute: (2d + 3d)/18 = 5/6 ⇒ 5d/18 = 5/6.Multiply both sides by 18/5: d = 3 km. Verification / Alternative check:Downstream time = 3/9 = 1/3 h = 20 min; upstream time = 3/6 = 1/2 h = 30 min; total = 50 min as required. Why Other Options Are Wrong:4, 1, or 2 km do not sum to 50 min using speeds 9 and 6 km/h. Common Pitfalls:Using an average speed over the entire loop; correct method is to add segment times with their respective speeds. Final Answer:3 km

Question 8

The ratio of motorboat speed to current speed is 35 : 5 (that is, v : c = 7 : 1). The motorboat goes with the current in 5 h 10 min. How long will it take to come back against the current (choose the nearest option provided)?

Accepted Answer

Introduction / Context:Given the ratio of still water speed to current, we can treat both as multiples of a common unit k. The downstream time gives the downstream distance, which we then use to compute the upstream time over the same distance. Since the options are in coarse 10-minute increments, select the closest. Given Data / Assumptions: v : c = 35 : 5 ⇒ v = 7k and c = k. Downstream time t_down = 5 h 10 min = 31/6 h. Distance downstream equals distance upstream. Concept / Approach:Downstream speed = v + c = 8k; upstream speed = v − c = 6k. Let the one-way distance be D. Then D = (downstream speed) * (downstream time) = 8k * (31/6). Upstream time = D / (6k). Cancel k and compute the value numerically. Step-by-Step Solution: D = 8k * (31/6) = (248/6)k = (124/3)k.t_up = D/(6k) = (124/3)k / (6k) = 124/18 h = 62/9 h.62/9 h ≈ 6.888... h ≈ 6 h 53 min (rounded). Verification / Alternative check:Using exact fractions maintains independence from k; only the ratio matters. The computed upstream time is approximately 6 h 53 min. Why Other Options Are Wrong:5 h 50 min and 6 h are too short; 12 h 10 min is far too long. The nearest provided option to approximately 6 h 53 min is 6 h 50 min. Common Pitfalls:Assuming symmetric times or averaging speeds. Upstream is significantly slower, so time is longer. Final Answer:6 h 50 min

Question 9

Ravi rows in a river where the downstream speed is 9 km/h and the upstream speed is 5 km/h. Find Ravi’s speed in still water and the speed of the current. (Give both in km/h.)

Accepted Answer

Introduction / Context:Problems on boats and streams typically give downstream and upstream speeds to infer the speed in still water (boat speed) and the speed of the current. Downstream means the current helps the boat, upstream means the current opposes it. Given Data / Assumptions: Downstream speed = 9 km/h Upstream speed = 5 km/h Let boat speed in still water be b (km/h) and current speed be c (km/h). Concept / Approach:Use the identities: downstream = b + c and upstream = b − c. Solve the two linear equations to find b and c. Step-by-Step Solution: b + c = 9b − c = 5Add: (b + c) + (b − c) = 9 + 5 ⇒ 2b = 14 ⇒ b = 7Substitute into b + c = 9 ⇒ 7 + c = 9 ⇒ c = 2 Verification / Alternative check:Check: downstream b + c = 7 + 2 = 9 km/h, upstream b − c = 7 − 2 = 5 km/h, which matches the data. Why Other Options Are Wrong:Options with 8 km/h or 3 km/h for the current violate b + c = 9 and b − c = 5 simultaneously; they do not satisfy both equations. Common Pitfalls:Mixing up b and c, or averaging 9 and 5 incorrectly for c. The average (9 + 5)/2 gives b, not c. The half-difference (9 − 5)/2 gives c. Final Answer:Speed in still water = 7 km/h; speed of current = 2 km/h.

Question 10

A boatman rows 1 km in 5 minutes along the stream (downstream), and 6 km in 1 hour against the stream (upstream). Find the speed of the stream (in km/h).

Accepted Answer

Introduction / Context:This question gives downstream and upstream travel rates to deduce the current speed. Converting times to hours is essential. Given Data / Assumptions: Downstream: 1 km in 5 min ⇒ 5 min = 5/60 h = 1/12 h. Upstream: 6 km in 1 h. Downstream speed vd and upstream speed vu are in km/h. Concept / Approach:Compute vd and vu, then use vd = b + c and vu = b − c to find current c = (vd − vu)/2. Step-by-Step Solution: vd = distance/time = 1 / (1/12) = 12 km/hvu = 6 / 1 = 6 km/hc = (vd − vu)/2 = (12 − 6)/2 = 3 km/h Verification / Alternative check:Then b = (vd + vu)/2 = (12 + 6)/2 = 9 km/h; check: b + c = 12 and b − c = 6, consistent with data. Why Other Options Are Wrong:6, 10, 12 km/h do not satisfy both downstream and upstream relations simultaneously when combined with a single boat speed b. Common Pitfalls:Forgetting to convert 5 minutes to hours; misapplying average vs half-difference. Average gives boat speed, not current speed. Final Answer:3 km/h.

Question 11

A man can row at 8 km/h in still water. The river's current is 2 km/h. He rows to a place and back in a total of 32 minutes. How far is that place from the start (one-way distance in km)?

Accepted Answer

Introduction / Context:Round-trip time with different upstream and downstream speeds can be used to determine one-way distance. Here, the still-water speed and current are given, so both leg speeds are known. Given Data / Assumptions: Still-water speed b = 8 km/h Current speed c = 2 km/h Downstream speed vd = b + c = 10 km/h Upstream speed vu = b − c = 6 km/h Total time T = 32 minutes = 32/60 h = 8/15 h Let one-way distance be d (km) Concept / Approach:Total time T = d/vd + d/vu. Solve for d given T, vd, vu. Step-by-Step Solution: T = d/10 + d/6 = d * ( (3 + 5) / 30 ) = d * (8/30) = (4d/15)Set (4d/15) = 8/15 ⇒ 4d = 8 ⇒ d = 2 km Verification / Alternative check:Time downstream = 2/10 = 0.2 h; time upstream = 2/6 ≈ 0.333... h; total ≈ 0.533... h = 32 minutes, matches. Why Other Options Are Wrong:1.5, 2.5, or 3 lead to totals not equal to 32 minutes when substituted into d/vd + d/vu with vd = 10 and vu = 6. Common Pitfalls:Using average speed over unequal legs or averaging 10 and 6. Average speed cannot be used because times per leg differ. Final Answer:2 km.

Question 12

A boat covers 24 km upstream and 36 km downstream in 6 hours, and 36 km upstream with 24 km downstream in 6 hours 30 minutes. Find the speed of the current (in km/h).

Accepted Answer

Introduction / Context:When two travel-time equations are provided for different upstream/downstream mixes, we can solve the system for still-water speed b and current c. Given Data / Assumptions: Trip 1: 24 km upstream + 36 km downstream in 6 h Trip 2: 36 km upstream + 24 km downstream in 6.5 h Upstream speed = b − c; downstream speed = b + c Concept / Approach:Form two equations in b and c using time = distance/speed for each leg and add them. Solve the linear-fractional system. Step-by-Step Solution: 24/(b − c) + 36/(b + c) = 6 … (1)36/(b − c) + 24/(b + c) = 6.5 … (2)Subtract (2) − (1): 12/(b − c) − 12/(b + c) = 0.512 * ( (b + c) − (b − c) ) / ( (b − c)(b + c) ) = 0.5 ⇒ 12 * (2c) / (b^2 − c^2) = 0.524c = 0.5 (b^2 − c^2) ⇒ b^2 − c^2 = 48cAdd (1) and (2): 60/(b − c) + 60/(b + c) = 12.5 ⇒ 60 * 2b / (b^2 − c^2) = 12.5120b / (b^2 − c^2) = 12.5 ⇒ b^2 − c^2 = 120b / 12.5 = 9.6bEquate: 9.6b = 48c ⇒ b = 5cUse b = 5c in b^2 − c^2 = 48c ⇒ 25c^2 − c^2 = 48c ⇒ 24c^2 = 48c ⇒ c = 2 km/h (c ≠ 0) Verification / Alternative check:With c = 2 and b = 10, upstream = 8, downstream = 12. Trip 1 time = 24/8 + 36/12 = 3 + 3 = 6 h; Trip 2 time = 36/8 + 24/12 = 4.5 + 2 = 6.5 h. Why Other Options Are Wrong:1, 1.5, 2.5 do not satisfy both equations concurrently. Common Pitfalls:Attempting to average distances or speeds directly; the harmonic nature of times requires careful algebra, not naive averaging. Final Answer:2 km/h.

Question 13

A boat travels upstream from B to A and then downstream from A back to B in a total of 3 hours. If the boat’s still-water speed is 9 km/h and the current is 3 km/h, find the distance between A and B (in km).

Accepted Answer

Introduction / Context:Round-trip problems use different leg speeds. Knowing still-water speed and current lets us compute each leg's speed and then the distance from total time. Given Data / Assumptions: b = 9 km/h (still water) c = 3 km/h (current) Upstream speed vu = b − c = 6 km/h Downstream speed vd = b + c = 12 km/h Total time = 3 h; one-way distance = d km Concept / Approach:Total time = d/vu + d/vd. Solve for d. Step-by-Step Solution: 3 = d/6 + d/12 = d * (1/6 + 1/12) = d * (1/4)d = 3 * 4 = 12 km Verification / Alternative check:Upstream time = 12/6 = 2 h; downstream time = 12/12 = 1 h; total = 3 h, matches. Why Other Options Are Wrong:4, 6, 8 give totals smaller than 3 h; they do not satisfy the time equation. Common Pitfalls:Using the arithmetic average of speeds to find time. Average speed is not applicable across legs with different speeds unless weighted by time/distance correctly. Final Answer:12 km.

Question 14

A man can swim at 3 km/h in still water. The stream’s speed is 2 km/h. How long will he take to swim to a point 10 km upstream and then return to the start (total time in hours)?

Accepted Answer

Introduction / Context:Swimming problems mirror boat-and-stream logic. The effective speed differs upstream and downstream due to the current. Given Data / Assumptions: Still-water speed b = 3 km/h Current c = 2 km/h Upstream speed vu = b − c = 1 km/h Downstream speed vd = b + c = 5 km/h Upstream distance = 10 km; downstream distance = 10 km Concept / Approach:Total time = time upstream + time downstream = distance/speed per leg. Step-by-Step Solution: Time upstream = 10 / 1 = 10 hTime downstream = 10 / 5 = 2 hTotal time = 12 h Verification / Alternative check:Speeds and distances are straightforward; arithmetic is direct and consistent. Why Other Options Are Wrong:Any alternative value fails to match both leg times computed from the known effective speeds of 1 km/h and 5 km/h. Common Pitfalls:Using average speed across unequal leg speeds; the correct method adds individual times, not averages the speeds. Final Answer:12 hrs.

Question 15

A man rows 3/4 km (0.75 km) against the stream in 11 minutes 15 seconds and returns the same 0.75 km downstream in 7 minutes 30 seconds. Find his speed in still water (in km/h).

Accepted Answer

Introduction / Context:When distances are equal, we can compute separate effective speeds from time and then recover the still-water speed as the average of downstream and upstream speeds. Given Data / Assumptions: Distance each way = 0.75 km Upstream time = 11 min 15 s = 11.25 min = 0.1875 h Downstream time = 7 min 30 s = 7.5 min = 0.125 h Concept / Approach:Compute vu and vd from distance/time. Then still-water speed b = (vd + vu)/2. Step-by-Step Solution: vu = 0.75 / 0.1875 = 4 km/hvd = 0.75 / 0.125 = 6 km/hb = (vd + vu)/2 = (6 + 4)/2 = 5 km/h Verification / Alternative check:Current c = (vd − vu)/2 = (6 − 4)/2 = 1 km/h, so b − c = 4 and b + c = 6 as required. Why Other Options Are Wrong:2, 3, 4 km/h do not average the computed leg speeds and would not reproduce both 11.25 min and 7.5 min times for 0.75 km. Common Pitfalls:Confusing still-water speed with downstream speed; ensure to average vd and vu, not the times. Final Answer:5 km/hr.

Question 16

A man rows at 9 1/3 km/h in still water. He takes three times as long to row a fixed distance upstream as he takes to row the same distance downstream. Find the speed of the current (in km/h).

Accepted Answer

Introduction / Context:Time ratio information links upstream and downstream speeds via reciprocal relations. If upstream time is 3 times downstream time for equal distances, the upstream speed must be one-third of the downstream speed. Given Data / Assumptions: Still-water speed b = 9 1/3 km/h = 28/3 km/h Let current speed be c Upstream speed vu = b − c; downstream speed vd = b + c Given t_up = 3 * t_down ⇒ vu = (1/3) vd Concept / Approach:Use b − c = (1/3)(b + c) to relate b and c and solve for c. Step-by-Step Solution: 3(b − c) = b + c ⇒ 3b − 3c = b + c2b = 4c ⇒ c = b/2c = (28/3) / 2 = 14/3 km/h Verification / Alternative check:vd = b + c = (28/3) + (14/3) = 42/3 = 14; vu = b − c = (28/3) − (14/3) = 14/3. Then t_up/t_down = (distance/vu)/(distance/vd) = vd/vu = 14 / (14/3) = 3, matching the condition. Why Other Options Are Wrong:Other fractions do not equal b/2 when b = 28/3, and they break the 3:1 time ratio. Common Pitfalls:Inverting the 3:1 relationship improperly or averaging speeds instead of using the reciprocal time-speed relation for equal distances. Final Answer:14/3 km/hr.

Question 17

The current of a stream runs at 1 km/h. A motorboat goes 35 km upstream and returns 35 km downstream to the start in a total of 12 hours. Find the boat’s speed in still water (in km/h).

Accepted Answer

Introduction / Context:Round-trip time with a known current allows solving for the still-water speed b using the sum of upstream and downstream times over equal distances. Given Data / Assumptions: Current c = 1 km/h Distance each way = 35 km Total time = 12 h Upstream speed = b − 1; downstream speed = b + 1 Concept / Approach:Set 35/(b − 1) + 35/(b + 1) = 12 and solve the quadratic for b > 0. Step-by-Step Solution: 35/(b − 1) + 35/(b + 1) = 1235 * ( (b + 1) + (b − 1) ) / (b^2 − 1) = 12 ⇒ 35 * (2b) = 12(b^2 − 1)70b = 12b^2 − 12 ⇒ 12b^2 − 70b − 12 = 0Solve ⇒ b = 6 km/h (reject negative root) Verification / Alternative check:Upstream speed 5 km/h ⇒ time 35/5 = 7 h; downstream speed 7 km/h ⇒ time 35/7 = 5 h; total = 12 h. Why Other Options Are Wrong:7, 8, 8.5 do not satisfy the time equation for both legs with c = 1 km/h. Common Pitfalls:Forgetting to set up the rational sum correctly or trying to average speeds arithmetically; the correct method sums times. Final Answer:6 km/hr.

Question 18

A man rows to a place 48 km away and returns in a total of 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. Find the rate of the stream (in km/h).

Accepted Answer

Introduction / Context:Equal-time statements over different distances give a ratio of downstream to upstream speeds. Combine that with round-trip time to deduce the current speed. Given Data / Assumptions: One-way distance d = 48 km Total time = 14 h 4 km downstream takes the same time as 3 km upstream ⇒ vd / vu = 4 / 3 Let b be still-water speed and c the current Concept / Approach:From vd/vu = (b + c)/(b − c) = 4/3 ⇒ b = 7c. Then use time equation d/(b − c) + d/(b + c) = 14 to find c. Step-by-Step Solution: (b + c)/(b − c) = 4/3 ⇒ 3b + 3c = 4b − 4c ⇒ b = 7c14 = 48/(b − c) + 48/(b + c) = 48/(6c) + 48/(8c) = 8/c + 6/c = 14/c14 = 14/c ⇒ c = 1 km/h Verification / Alternative check:b = 7 km/h; times: upstream 48/6 = 8 h; downstream 48/8 = 6 h; total 14 h. Ratio check: vd/vu = 8/6 = 4/3, satisfied. Why Other Options Are Wrong:Other values for c break either the 14 h total or the 4:3 time equivalence ratio. Common Pitfalls:Treating 4:3 as a distance ratio; it is a speed ratio because times are equal. Always convert such statements into a speed ratio relation. Final Answer:1 km/hr.

Question 19

A boat’s speed in still water is 6 km/h and the stream’s speed is 1.5 km/h. The man rows to a place 22.5 km away and returns to the start. Find the total time taken (in hours and minutes).

Accepted Answer

Introduction / Context:Compute leg times with effective upstream and downstream speeds and sum them to get the total duration. Given Data / Assumptions: b = 6 km/h; c = 1.5 km/h Downstream speed vd = 7.5 km/h Upstream speed vu = 4.5 km/h Distance each way = 22.5 km Concept / Approach:Total time = 22.5/7.5 + 22.5/4.5 (hours). Step-by-Step Solution: Downstream time = 22.5 / 7.5 = 3 hUpstream time = 22.5 / 4.5 = 5 hTotal = 3 + 5 = 8 h Verification / Alternative check:All values align with b ± c; no inconsistencies appear. Why Other Options Are Wrong:Any other total would not match both leg computations given the fixed speeds and distance. Common Pitfalls:Averaging the two speeds to get a single speed for the round trip is invalid; use time per leg instead. Final Answer:8 hrs.

Question 20

A boat takes a total of 3 hours to travel from place M to N downstream and then back from N to M upstream. If the boat’s speed in still water is 4 km/h, what is the distance MN? (Assume stream speed is unknown.)

Accepted Answer

Introduction / Context:Data adequacy problems ask whether the given information uniquely determines the answer. Here, we have total round-trip time and still-water speed, but the stream speed is not provided. Given Data / Assumptions: Total time T = 3 h Still-water speed b = 4 km/h Unknown current c Distance each way = d km Concept / Approach:The time equation is T = d/(b + c) + d/(b − c). Without c, multiple (d, c) pairs can satisfy the same T; therefore, d is not uniquely determined. Step-by-Step Reasoning: 3 = d/(4 + c) + d/(4 − c)For different positive c values < 4, one can solve for different d values to satisfy the same total time. Verification / Alternative check:Try c = 1 km/h: then 3 = d/5 + d/3 ⇒ 3 = ( (3d + 5d) / 15 ) ⇒ d = 45/8 = 5.625 km. Try c = 0.5 km/h: 3 = d/4.5 + d/3.5, leading to a different d. Hence non-unique. Why Other Options Are Wrong:Any specific distance assumes a particular c; since c is not given, those values are not guaranteed. Common Pitfalls:Assuming c = 0 or guessing c. A unique distance requires either c or one additional relation. Final Answer:Data inadequate.

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