Boats and Streams Questions in English

(C) Let the speed of the boat in still water be $u$ and the speed of the stream be $v$.
Speed along the stream (downstream) is $u + v = 12 \, km/h$.
Speed against the stream (upstream) is $u - v = 8 \, km/h$.
Adding the two equations: $(u + v) + (u - v) = 12 + 8 \implies 2u = 20 \implies u = 10 \, km/h$.
Thus,the speed of the boat in still water is $10 \, km/h$.
Time taken to travel $24 \, km$ in still water is given by $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{24}{10} = 2.4 \, h$.

(B) Let the speed of the stream be $x \text{ km/h}$.
Since boat $A$ moves downstream,its effective speed is $(12 + x) \text{ km/h}$.
Since boat $B$ moves upstream,its effective speed is $(15 - x) \text{ km/h}$.
Let the boats meet after $t$ hours.
The total distance covered by both boats when they meet is equal to the initial distance between them,which is $108 \text{ km}$.
Therefore,$(12 + x)t + (15 - x)t = 108$.
Simplifying the equation: $(12 + x + 15 - x)t = 108$.
$27t = 108$.
$t = \frac{108}{27} = 4 \text{ h}$.

(C) Let the speed of the motorboat be $36x\, km/h$ and the speed of the current be $5x\, km/h$.
The speed of the boat along the current (downstream) is $36x + 5x = 41x\, km/h$.
The time taken to travel downstream is $5\, h\, 10\, min = 5 + \frac{10}{60} = 5 + \frac{1}{6} = \frac{31}{6}\, h$.
Distance covered $= \text{Speed} \times \text{Time} = 41x \times \frac{31}{6} = \frac{1271x}{6}\, km$.
The speed of the boat against the current (upstream) is $36x - 5x = 31x\, km/h$.
Time taken to return (upstream) $= \frac{\text{Distance}}{\text{Upstream Speed}} = \frac{1271x / 6}{31x} = \frac{1271}{6 \times 31} = \frac{41}{6}\, h$.
Converting $\frac{41}{6}\, h$ into hours and minutes: $\frac{41}{6} = 6\, h + \frac{5}{6} \times 60\, min = 6\, h\, 50\, min$.

(D) Speed of man in downstream $= \frac{15\, km}{1\, h} = 15\, km/h$.
Let the speed of the man in still water be $v_m$ and the speed of the current be $v_c = 5\, km/h$.
Downstream speed $= v_m + v_c = 15\, km/h$.
Therefore,$v_m + 5 = 15$,which gives $v_m = 10\, km/h$.
Upstream speed $= v_m - v_c = 10 - 5 = 5\, km/h$.
Time taken to swim upstream for a distance of $15\, km$ is $\text{Time} = \frac{\text{Distance}}{\text{Upstream Speed}} = \frac{15}{5} = 3\, h$.

(C) Let the distance of the place be $x \, km$.
The speed of the boat downstream is $(5 + 3) \, km/h = 8 \, km/h$.
The speed of the boat upstream is $(5 - 3) \, km/h = 2 \, km/h$.
According to the problem,the total time taken for the round trip is $3 \, h$.
Therefore,the time taken to go downstream plus the time taken to return upstream equals $3 \, h$.
$\frac{x}{8} + \frac{x}{2} = 3$
Multiplying by the common denominator $8$:
$x + 4x = 24$
$5x = 24$
$x = \frac{24}{5} = 4.8 \, km$.
Thus,the distance of the place is $4.8 \, km$.

(C) Let the speed of the stream be $x\, km/h$.
The speed of the motorboat in still water is $45\, km/h$.
Speed along the stream (downstream) $= (45 + x)\, km/h$.
Time taken to travel $80\, km$ downstream $= 1\, h\, 20\, min = 1 + \frac{20}{60} = 1 + \frac{1}{3} = \frac{4}{3}\, h$.
Using the formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$:
$45 + x = \frac{80}{4/3} = 80 \times \frac{3}{4} = 60$.
$x = 60 - 45 = 15\, km/h$.
Speed against the stream (upstream) $= (45 - x) = 45 - 15 = 30\, km/h$.
Time taken to cover $80\, km$ upstream $= \frac{\text{Distance}}{\text{Speed}} = \frac{80}{30} = \frac{8}{3}\, h$.
Converting $\frac{8}{3}\, h$ into hours and minutes: $\frac{8}{3}\, h = 2\, h + \frac{2}{3} \times 60\, min = 2\, h\, 40\, min$.

(A) Let the speed of the stream be $x\, km/h$.
The speed of the boat in still water is $10\, km/h$.
Therefore,the upstream speed of the boat is $(10 - x)\, km/h$.
We know that $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.
Given that the boat covers $45\, km$ in $6\, h$ upstream,we have:
$10 - x = \frac{45}{6}$
$10 - x = 7.5$
$x = 10 - 7.5 = 2.5\, km/h$.
Thus,the speed of the stream is $2.5\, km/h$.

(B) Let the distance be $x \, km$.
Speed of the man in still water = $6 \, km/h$.
Speed of the current = $2 \, km/h$.
Speed in downstream = $6 + 2 = 8 \, km/h$.
Speed in upstream = $6 - 2 = 4 \, km/h$.
Time taken in upstream = $\frac{x}{4} \, h$.
Time taken in downstream = $\frac{x}{8} \, h$.
According to the question,the difference in time is $3 \, h$:
$\frac{x}{4} - \frac{x}{8} = 3$
Multiply by $8$ to solve for $x$:
$2x - x = 24$
$x = 24 \, km$.

(A) Let the speed of the sailor in still water be $x$ $km/h$ and the speed of the stream be $y$ $km/h$.
Downstream speed = $x + y = \frac{12}{48/60} = \frac{12}{0.8} = 15$ $km/h$.
Upstream speed = $x - y = \frac{12}{80/60} = \frac{12}{4/3} = 9$ $km/h$.
To find the speed of the sailor in still water $(x)$,we add the two equations:
$(x + y) + (x - y) = 15 + 9$
$2x = 24$
$x = 12$ $km/h$.

(B) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Speed upstream $= (x - y) = \frac{40\, km}{8\, h} = 5\, km/h$.
Speed downstream $= (x + y) = \frac{36\, km}{6\, h} = 6\, km/h$.
To find the speed of the boat in still water $(x)$,we add the two equations:
$(x - y) + (x + y) = 5 + 6$
$2x = 11$
$x = \frac{11}{2} = 5.5\, km/h$.

(C) Let the speed of the boat in still water be $u = 6\, km/h$ and the speed of the stream be $v = 1.5\, km/h$.
The speed downstream is $u + v = 6 + 1.5 = 7.5\, km/h$.
The speed upstream is $u - v = 6 - 1.5 = 4.5\, km/h$.
The distance to the place is $d = 22.5\, km$.
Time taken to go downstream is $t_1 = \frac{d}{u+v} = \frac{22.5}{7.5} = 3\, hours$.
Time taken to come back upstream is $t_2 = \frac{d}{u-v} = \frac{22.5}{4.5} = 5\, hours$.
Total time taken = $t_1 + t_2 = 3 + 5 = 8\, hours$.

(B) Let the speed of the person in still water be $x \text{ km/h}$ and the speed of the stream be $y = 2 \frac{1}{4} = 2.25 \text{ km/h}.$
Downstream speed $= (x + y) \text{ km/h}.$
Upstream speed $= (x - y) \text{ km/h}.$
Let the distance be $d \text{ km}.$
According to the problem,$d = (x + y) \times 6$ and $d = (x - y) \times 9.$
Equating the distances: $6(x + y) = 9(x - y).$
$6x + 6y = 9x - 9y \implies 3x = 15y \implies x = 5y.$
Substituting $y = 2.25 \text{ km/h}:$
$x = 5 \times 2.25 = 11.25 = 11 \frac{1}{4} \text{ km/h}.$

(B) Let the distance between $A$ and $B$ be $x\, km$.
Speed of the boat in still water $= 9\, km/h$.
Speed of the current $= 3\, km/h$.
Upstream speed $= (\text{Speed of boat} - \text{Speed of current}) = 9 - 3 = 6\, km/h$.
Downstream speed $= (\text{Speed of boat} + \text{Speed of current}) = 9 + 3 = 12\, km/h$.
Time taken to travel upstream from $B$ to $A = \frac{x}{6}\, hours$.
Time taken to travel downstream from $A$ to $B = \frac{x}{12}\, hours$.
According to the problem, the total time taken is $3\, hours$:
$\frac{x}{6} + \frac{x}{12} = 3$
To solve for $x$, find a common denominator:
$\frac{2x + x}{12} = 3$
$\frac{3x}{12} = 3$
$\frac{x}{4} = 3$
$x = 12\, km$.
Thus, the distance between $A$ and $B$ is $12\, km$.

(B) Let the speed of the motor boat in still water be $x \, km/h$.
Given,speed of the stream $= 1 \, km/h$.
Speed upstream $= (x - 1) \, km/h$.
Speed downstream $= (x + 1) \, km/h$.
Total time taken for the round trip is $12 \, hours$ for a distance of $35 \, km$ each way.
Time taken upstream $+$ Time taken downstream $= 12$.
$\frac{35}{x - 1} + \frac{35}{x + 1} = 12$.
$35 \left( \frac{x + 1 + x - 1}{x^2 - 1} \right) = 12$.
$35 \left( \frac{2x}{x^2 - 1} \right) = 12$.
$70x = 12(x^2 - 1)$.
$12x^2 - 70x - 12 = 0$.
Dividing by $2$: $6x^2 - 35x - 6 = 0$.
$6x^2 - 36x + x - 6 = 0$.
$6x(x - 6) + 1(x - 6) = 0$.
$(6x + 1)(x - 6) = 0$.
Since speed cannot be negative,$x = 6 \, km/h$.

(C) Let the velocity of the boat in still water be $u \, km/h$ and the velocity of the current be $x \, km/h$.
The speed upstream is $(u - x) \, km/h$ and the speed downstream is $(u + x) \, km/h$.
According to the problem:
$\frac{24}{u-x} + \frac{36}{u+x} = 6$ ... $(1)$
$\frac{36}{u-x} + \frac{24}{u+x} = 6.5$ ... $(2)$
Multiply equation $(1)$ by $1.5$:
$1.5 \times \left( \frac{24}{u-x} + \frac{36}{u+x} \right) = 1.5 \times 6$
$\frac{36}{u-x} + \frac{54}{u+x} = 9$ ... $(3)$
Subtract equation $(2)$ from $(3)$:
$\left( \frac{36}{u-x} + \frac{54}{u+x} \right) - \left( \frac{36}{u-x} + \frac{24}{u+x} \right) = 9 - 6.5$
$\frac{30}{u+x} = 2.5$
$u + x = \frac{30}{2.5} = 12 \, km/h$
Substitute $(u + x) = 12$ into equation $(1)$:
$\frac{24}{u-x} + \frac{36}{12} = 6$
$\frac{24}{u-x} + 3 = 6$
$\frac{24}{u-x} = 3$
$u - x = \frac{24}{3} = 8 \, km/h$
Now,we have:
$u + x = 12$
$u - x = 8$
Subtracting the two equations: $(u + x) - (u - x) = 12 - 8$
$2x = 4$
$x = 2 \, km/h$
Thus,the velocity of the current is $2 \, km/h$.

(B) Let the man's speed in still water be $x\, km/h$.
The speed with the current (downstream) is given by $x + y = 12\, km/h$,where $y$ is the speed of the current.
Given $y = 1.5\, km/h$,we have $x + 1.5 = 12$.
Therefore,$x = 12 - 1.5 = 10.5\, km/h$.
The man's speed against the current (upstream) is given by $x - y$.
Thus,the speed against the current $= 10.5 - 1.5 = 9\, km/h$.

(C) Let the distance to the place be $d \, km$.
Speed downstream $= 5 + 1 = 6 \, km/h$.
Speed upstream $= 5 - 1 = 4 \, km/h$.
Total time taken $= \frac{d}{6} + \frac{d}{4} = 1 \, hour$.
Taking the least common multiple of $6$ and $4$,which is $12$:
$\frac{2d + 3d}{12} = 1$
$\frac{5d}{12} = 1$
$d = \frac{12}{5} = 2.4 \, km$.
Thus,the distance is $2.4 \, km$.

(B) Let the speed of the boat in still water be $u = 6\, km/h$ and the speed of the stream be $v\, km/h$.
The speed of the boat upstream is $(u - v) = (6 - v)\, km/h$.
The speed of the boat downstream is $(u + v) = (6 + v)\, km/h$.
Let the distance be $d$. The time taken to row upstream is $T_{up} = \frac{d}{6 - v}$ and the time taken to row downstream is $T_{down} = \frac{d}{6 + v}$.
According to the problem,the time taken to row upstream is twice the time taken to row downstream:
$T_{up} = 2 \times T_{down}$
$\frac{d}{6 - v} = 2 \times \frac{d}{6 + v}$
Canceling $d$ from both sides:
$\frac{1}{6 - v} = \frac{2}{6 + v}$
$6 + v = 2(6 - v)$
$6 + v = 12 - 2v$
$3v = 6$
$v = 2\, km/h$.
Thus,the rate of the stream is $2\, km/h$.

(A) Let the speed of the man in still water be $u \text{ km/h}$ and the rate of the current be $v \text{ km/h}$.
According to the problem:
$\frac{30}{u-v} + \frac{44}{u+v} = 10$ $...(1)$
$\frac{40}{u-v} + \frac{55}{u+v} = 13$ $...(2)$
Let $x = \frac{1}{u-v}$ and $y = \frac{1}{u+v}$.
$30x + 44y = 10$ $...(3)$
$40x + 55y = 13$ $...(4)$
Multiply $(3)$ by $4$ and $(4)$ by $3$:
$120x + 176y = 40$ $...(5)$
$120x + 165y = 39$ $...(6)$
Subtract $(6)$ from $(5)$:
$11y = 1 \Rightarrow y = \frac{1}{11}$.
So,$u+v = 11$.
Substitute $y = \frac{1}{11}$ into $(3)$:
$30x + 44(\frac{1}{11}) = 10 \Rightarrow 30x + 4 = 10 \Rightarrow 30x = 6 \Rightarrow x = \frac{1}{5}$.
So,$u-v = 5$.
Solving $u+v = 11$ and $u-v = 5$:
$2u = 16 \Rightarrow u = 8 \text{ km/h}$.
$2v = 6 \Rightarrow v = 3 \text{ km/h}$.
Thus,the speed of the man is $8 \text{ km/h}$ and the rate of the current is $3 \text{ km/h}$.

(B) Let the speed of the person in still water be $x$ $km/h$ and the speed of the stream be $y = 3$ $km/h$.
Downstream speed $= (x + 3)$ $km/h$.
Upstream speed $= (x - 3)$ $km/h$.
Since the distance is the same,we have: $6(x + 3) = 9(x - 3)$.
$6x + 18 = 9x - 27$.
$3x = 45$.
$x = 15$ $km/h$.
Alternatively,using the formula: Speed in still water $= y \times \frac{(T_2 + T_1)}{(T_2 - T_1)} = 3 \times \frac{(9 + 6)}{(9 - 6)} = 3 \times \frac{15}{3} = 15$ $km/h$.

(A) Let the speed of the stream be $x$ miles/hour.
Speed of the boat in still water = $10$ miles/hour.
Downstream speed = $(10 + x)$ miles/hour.
Upstream speed = $(10 - x)$ miles/hour.
Time taken to travel $36$ miles upstream = $\frac{36}{10 - x}$ hours.
Time taken to travel $36$ miles downstream = $\frac{36}{10 + x}$ hours.
Given that the difference in time is $90$ minutes,which is $\frac{90}{60} = 1.5$ hours.
Therefore,$\frac{36}{10 - x} - \frac{36}{10 + x} = 1.5$.
$\frac{36(10 + x) - 36(10 - x)}{(10 - x)(10 + x)} = 1.5$.
$\frac{360 + 36x - 360 + 36x}{100 - x^2} = 1.5$.
$\frac{72x}{100 - x^2} = 1.5$.
$72x = 1.5(100 - x^2)$.
$72x = 150 - 1.5x^2$.
$1.5x^2 + 72x - 150 = 0$.
Dividing by $1.5$,we get $x^2 + 48x - 100 = 0$.
$(x + 50)(x - 2) = 0$.
Since speed cannot be negative,$x = 2$ miles/hour.

(B) Let the speed of the stream be $x \text{ km/hr}$.
Speed of the boat in still water = $15 \text{ km/hr}$.
Downstream speed = $(15 + x) \text{ km/hr}$.
Upstream speed = $(15 - x) \text{ km/hr}$.
Total time taken = $4 \text{ hours } 30 \text{ minutes} = 4.5 \text{ hours} = \frac{9}{2} \text{ hours}$.
According to the problem,the total time for the round trip is:
$\frac{30}{15+x} + \frac{30}{15-x} = 4.5$
$\frac{30(15-x) + 30(15+x)}{(15+x)(15-x)} = 4.5$
$\frac{450 - 30x + 450 + 30x}{225 - x^2} = 4.5$
$\frac{900}{225 - x^2} = 4.5$
$900 = 4.5(225 - x^2)$
$200 = 225 - x^2$
$x^2 = 225 - 200 = 25$
$x = 5 \text{ km/hr}$.

(A) Let the distance of the place be $x \text{ km}$.
Speed downstream $= 5 + 1 = 6 \text{ km/hr}$.
Speed upstream $= 5 - 1 = 4 \text{ km/hr}$.
Time taken to travel downstream is $\frac{x}{6} \text{ hours}$ and time taken to travel upstream is $\frac{x}{4} \text{ hours}$.
Given that the total time taken is $1 \text{ hour}$,we have:
$\frac{x}{6} + \frac{x}{4} = 1$
Taking the least common multiple $(LCM)$ of $6$ and $4$,which is $12$:
$\frac{2x + 3x}{12} = 1$
$\frac{5x}{12} = 1$
$x = \frac{12}{5} = 2.4 \text{ km}$.
Therefore,the distance to the place is $2.4 \text{ km}$.

(D) Let the speed of the boat in still water be $x \text{ km/hr}$.
The speed of the stream is $3 \text{ km/hr}$.
Downstream speed $= (x + 3) \text{ km/hr}$.
Upstream speed $= (x - 3) \text{ km/hr}$.
Since the distance covered is the same in both cases,we have:
$\text{Distance} = \text{Speed} \times \text{Time}$
$(x + 3) \times 1 = (x - 3) \times 1.5$
$x + 3 = 1.5x - 4.5$
$1.5x - x = 3 + 4.5$
$0.5x = 7.5$
$x = \frac{7.5}{0.5} = 15 \text{ km/hr}$.
Therefore,the speed of the boat in still water is $15 \text{ km/hr}$.

(C) Let the speed of the boat in still water be $u$ $km/h$ and the speed of the current be $v$ $km/h$.
The time taken to cover a distance $D$ upstream is $8$ $hours$ $48$ $minutes = 8 + \frac{48}{60} = 8 + 0.8 = 8.8$ $hours$.
Upstream speed $= u - v = \frac{D}{8.8} \implies D = 8.8(u - v)$.
The time taken to cover the same distance $D$ downstream is $4$ $hours$.
Downstream speed $= u + v = \frac{D}{4} \implies D = 4(u + v)$.
Equating the two expressions for $D$:
$8.8(u - v) = 4(u + v)$
$2.2(u - v) = u + v$
$2.2u - 2.2v = u + v$
$1.2u = 3.2v$
$\frac{u}{v} = \frac{3.2}{1.2} = \frac{32}{12} = \frac{8}{3}$.
Thus,the ratio of the speed of the boat to the speed of the current is $8:3$.

(C) The speed of the boat in still water is $13\, km/hr$.
The speed of the stream is $4\, km/hr$.
When moving downstream, the effective speed of the boat is the sum of the boat's speed and the stream's speed.
$\text{Downstream speed} = 13\, km/hr + 4\, km/hr = 17\, km/hr$.
To find the time taken to cover a distance of $68\, km$ downstream, we use the formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
$\text{Time} = \frac{68\, km}{17\, km/hr} = 4\, hours$.

(D) The distance covered is $3/4 \text{ km}$.
The time taken is $11 \frac{1}{4} \text{ minutes} = 45/4 \text{ minutes} = (45/4) / 60 \text{ hours} = 3/16 \text{ hours}$.
The upstream speed is $\text{Distance} / \text{Time} = (3/4) / (3/16) = (3/4) \times (16/3) = 4 \text{ km/hr}$.
Since the question asks for the speed of the man in still water but does not provide the speed of the stream,it is assumed that the stream speed is $0$ or the question implies the effective speed against the stream is the speed in still water. Given the options,the calculated upstream speed is $4 \text{ km/hr}$.

(C) Let the speed of the man in still water be $u$ and the speed of the current be $v$.
Given,downstream speed (speed with the current) $= u + v = 15 \, km/hr$.
Given,speed of the current $v = 2.5 \, km/hr$.
Therefore,the speed of the man in still water $u = 15 - 2.5 = 12.5 \, km/hr$.
The speed of the man against the current (upstream speed) is given by $u - v$.
Upstream speed $= 12.5 - 2.5 = 10 \, km/hr$.

(B) Let the speed of the boat in still water be $x\, km/hr$ and the speed of the stream be $y\, km/hr$.
Downstream speed = $\frac{\text{Distance}}{\text{Time}} = \frac{16}{2} = 8\, km/hr$.
Thus,$x + y = 8$ --- $(1)$
Upstream speed = $\frac{\text{Distance}}{\text{Time}} = \frac{16}{4} = 4\, km/hr$.
Thus,$x - y = 4$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$(x + y) + (x - y) = 8 + 4$
$2x = 12$
$x = 6\, km/hr$.
Therefore,the speed of the boat in still water is $6\, km/hr$.

(C) Let the speed of the boat in still water be $x \ km/hr$ and the speed of the stream be $y \ km/hr$.
When the boat goes along the stream (downstream),the effective speed is $(x + y) \ km/hr$.
Given that the boat covers $11 \ km$ in one hour downstream,we have: $x + y = 11 \dots (1)$.
When the boat goes against the stream (upstream),the effective speed is $(x - y) \ km/hr$.
Given that the boat covers $5 \ km$ in one hour upstream,we have: $x - y = 5 \dots (2)$.
Adding equations $(1)$ and $(2)$:
$(x + y) + (x - y) = 11 + 5$
$2x = 16$
$x = 8 \ km/hr$.
Therefore,the speed of the boat in still water is $8 \ km/hr$.

(B) Let the speed of the boat in still water be $u$ $km/hr$ and the speed of the stream be $v$ $km/hr$.
The speed downstream is $(u + v)$ $km/hr$ and the speed upstream is $(u - v)$ $km/hr$.
Let the distance be $d$ $km$.
According to the problem,the time taken to row upstream is twice the time taken to row downstream:
$\frac{d}{u - v} = 2 \times \frac{d}{u + v}$
Canceling $d$ from both sides:
$\frac{1}{u - v} = \frac{2}{u + v}$
Cross-multiplying:
$u + v = 2(u - v)$
$u + v = 2u - 2v$
Rearranging the terms:
$v + 2v = 2u - u$
$3v = u$
Therefore,the ratio of the speed of the boat to the speed of the stream is:
$\frac{u}{v} = \frac{3}{1}$ or $3:1$.

(C) Downstream speed $= \frac{1\, km}{10/60\, hr} = 6\, km/hr$.
Upstream speed $= \frac{2\, km}{1\, hr} = 2\, km/hr$.
Speed of the boat in still water $= \frac{\text{Downstream speed} + \text{Upstream speed}}{2} = \frac{6 + 2}{2} = 4\, km/hr$.
Time taken to travel $5\, km$ in stationary water $= \frac{\text{Distance}}{\text{Speed}} = \frac{5}{4} = 1.25\, hours$.
$1.25\, hours = 1\, hour + 0.25 \times 60\, minutes = 1\, hour\, 15\, minutes$.

(D) Let the speed of the current be $u \text{ miles/hour}$ and the speed of rowing in still water be $x \text{ miles/hour}$.
According to the first condition:
$\frac{12}{x-u} - \frac{12}{x+u} = 6$
$\frac{12(x+u - (x-u))}{x^2 - u^2} = 6$
$\frac{24u}{x^2 - u^2} = 6 \implies x^2 - u^2 = 4u \implies x^2 = u^2 + 4u \quad (1)$
According to the second condition,if the rowing rate is doubled $(2x)$:
$\frac{12}{2x-u} - \frac{12}{2x+u} = 1$
$\frac{12(2x+u - (2x-u))}{4x^2 - u^2} = 1$
$\frac{24u}{4x^2 - u^2} = 1 \implies 4x^2 - u^2 = 24u \implies 4x^2 = u^2 + 24u \quad (2)$
Substitute $x^2$ from $(1)$ into $(2)$:
$4(u^2 + 4u) = u^2 + 24u$
$4u^2 + 16u = u^2 + 24u$
$3u^2 = 8u$
Since $u \neq 0$,$u = \frac{8}{3} = 2 \frac{2}{3} \text{ miles/hour}$.

(B) Let the distance to the place be $x \, km$.
Speed of the man in still water $= 7.5 \, km/hr$.
Speed of the river stream $= 1.5 \, km/hr$.
Upstream speed $= 7.5 - 1.5 = 6 \, km/hr$.
Downstream speed $= 7.5 + 1.5 = 9 \, km/hr$.
The total time taken is $50 \, minutes = \frac{50}{60} \, hours = \frac{5}{6} \, hours$.
Using the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$,we have:
$\frac{x}{6} + \frac{x}{9} = \frac{5}{6}$
Taking the least common multiple $(LCM)$ of $6$ and $9$,which is $18$:
$\frac{3x + 2x}{18} = \frac{5}{6}$
$\frac{5x}{18} = \frac{5}{6}$
Multiplying both sides by $18$:
$5x = \frac{5 \times 18}{6}$
$5x = 15$
$x = 3 \, km$.
Therefore,the place is $3 \, km$ away.

(A) Let the distance between place $A$ and place $B$ be $x\, km.$
For the cyclist:
The speed is constant at $12\, km/hr.$ The total distance covered is $2x\, km.$
Time taken by the cyclist $= \frac{\text{Total Distance}}{\text{Speed}} = \frac{2x}{12} = \frac{x}{6}\, hours.$
For the sailor:
The boat travels downstream from $A$ to $B$ and upstream from $B$ to $A.$
Speed downstream $= 10 + 4 = 14\, km/hr.$
Speed upstream $= 10 - 4 = 6\, km/hr.$
Time taken by the sailor $= \frac{x}{14} + \frac{x}{6} = \frac{3x + 7x}{42} = \frac{10x}{42} = \frac{5x}{21}\, hours.$
Comparing the times:
To compare $\frac{x}{6}$ and $\frac{5x}{21},$ we find a common denominator $(42)$:
$\frac{x}{6} = \frac{7x}{42}$
$\frac{5x}{21} = \frac{10x}{42}$
Since $\frac{7x}{42} < \frac{10x}{42},$ the cyclist takes less time.
Therefore,the cyclist will return to place $A$ first.

(A) Speed of the boat upstream is calculated as distance divided by time.
Time $= 39\, minutes = \frac{39}{60}\, hours = 0.65\, hours$.
Speed upstream $= \frac{13\, km}{0.65\, h} = 20\, km/h$.
Let the speed of the boat in still water be $x\, km/h$.
The speed of the stream is $3\, km/h$.
We know that the upstream speed is given by $(x - \text{speed of stream})$.
Therefore,$x - 3 = 20$.
$x = 20 + 3 = 23\, km/h$.

(B) Let the speed of the boat in still water be $u = 8\, km/h$.
Let the speed of the stream be $v\, km/h$.
The speed of the boat downstream is given by the formula $u + v$.
Given that the downstream speed is $15\, km/h$,we have the equation:
$u + v = 15$
Substituting the value of $u$:
$8 + v = 15$
$v = 15 - 8$
$v = 7\, km/h$.
Therefore,the speed of the stream is $7\, km/h$.

(A) The speed of the man in still water is $v_m = 10 \text{ km/h}$.
The speed of the current is $v_c = 3 \text{ km/h}$.
When moving upstream,the man moves against the direction of the current.
Therefore,the effective speed upstream is given by the formula: $v_{\text{upstream}} = v_m - v_c$.
Substituting the given values: $v_{\text{upstream}} = 10 \text{ km/h} - 3 \text{ km/h} = 7 \text{ km/h}$.
Thus,the correct option is $A$.

(C) Let the speed of the man in still water be $u\, km/h$ and the speed of the stream be $v\, km/h$.
Downstream speed $= u + v = 7\, km/h$.
Upstream speed $= u - v = 3\, km/h$.
To find the speed of the man in still water $(u)$,we add the two equations:
$(u + v) + (u - v) = 7 + 3$
$2u = 10$
$u = 5\, km/h$.
Therefore,the speed of the man in still water is $5\, km/h$.

(B) Let the speed of the swimmer in still water be $u\, km/h$ and the speed of the current be $v\, km/h$.
Speed of the swimmer upstream (against the current) $= u - v = \frac{28}{4} = 7\, km/h$.
Speed of the swimmer downstream (in the direction of the current) $= u + v = \frac{40}{4} = 10\, km/h$.
To find the speed of the current $(v)$,we subtract the upstream speed from the downstream speed and divide by $2$:
$v = \frac{(u + v) - (u - v)}{2} = \frac{10 - 7}{2} = \frac{3}{2} = 1.5\, km/h$.
Thus,the speed of the current is $1.5\, km/h$.

(C) Speed of the boat downstream is calculated as: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1 \, km}{10/60 \, h} = 6 \, km/h$.
Speed of the boat upstream is given as $4 \, km/h$.
Let $v_b$ be the speed of the boat in still water and $v_c$ be the velocity of the current.
Downstream speed: $v_b + v_c = 6 \, km/h$.
Upstream speed: $v_b - v_c = 4 \, km/h$.
Subtracting the two equations: $(v_b + v_c) - (v_b - v_c) = 6 - 4$.
$2v_c = 2$.
$v_c = 1 \, km/h$.
Therefore,the velocity of the current is $1 \, km/h$.

(C) Let the speed of the man in still water be $u \text{ km/h}$ and the speed of the stream be $v \text{ km/h}$.
Given,the speed with the current (downstream speed) is $u + v = 12 \text{ km/h}$.
The speed of the current is $v = 1 \frac{1}{2} = 1.5 \text{ km/h}$.
Substituting the value of $v$ in the equation $u + v = 12$,we get $u + 1.5 = 12$,which implies $u = 10.5 \text{ km/h}$.
The speed against the current (upstream speed) is given by $u - v$.
Therefore,the upstream speed $= 10.5 - 1.5 = 9 \text{ km/h}$.

(A) Let the speed of the boat in still water be $u\, km/h$ and the speed of the stream be $v\, km/h$.
Speed upstream $= (u - v) = \frac{2\, km}{20/60\, h} = 2 \times 3 = 6\, km/h$.
Speed downstream $= (u + v) = \frac{2\, km}{18/60\, h} = 2 \times \frac{60}{18} = \frac{20}{3}\, km/h$.
The rate of the current $v$ is given by the formula: $v = \frac{1}{2} \times (\text{Downstream Speed} - \text{Upstream Speed})$.
$v = \frac{1}{2} \times (\frac{20}{3} - 6) = \frac{1}{2} \times (\frac{20 - 18}{3}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\, km/h$.

(B) Let the speed of the boat in still water be $u\, km/h$ and the speed of the current be $v = 5\, km/h$.
Downstream speed $= u + v = \frac{48\, km}{4\, h} = 12\, km/h$.
Therefore,$u + 5 = 12$,which gives $u = 7\, km/h$.
Upstream speed $= u - v = 7 - 5 = 2\, km/h$.
Time taken to cover $8\, km$ upstream $= \frac{\text{Distance}}{\text{Upstream Speed}} = \frac{8\, km}{2\, km/h} = 4\, hours$.

(B) Let the distance to the point be $d \text{ km}$.
Speed in still water $(u)$ = $10 \text{ km/h}$.
Speed of the river $(v)$ = $4 \text{ km/h}$.
Upstream speed = $u - v = 10 - 4 = 6 \text{ km/h}$.
Downstream speed = $u + v = 10 + 4 = 14 \text{ km/h}$.
Time taken for upstream journey $(t_1)$ = $\frac{d}{6} \text{ hours}$.
Time taken for downstream journey $(t_2)$ = $\frac{d}{14} \text{ hours}$.
Total distance = $2d$.
Total time = $t_1 + t_2 = \frac{d}{6} + \frac{d}{14} = \frac{7d + 3d}{42} = \frac{10d}{42} = \frac{5d}{21} \text{ hours}$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{5d/21} = 2d \times \frac{21}{5d} = \frac{42}{5} = 8 \frac{2}{5} \text{ km/h}$.

(A) Let the distance be $d\, km$.
Speed of the man in still water $(u)$ = $6\, km/h$.
Speed of the stream $(v)$ = $2\, km/h$.
Upstream speed = $u - v = 6 - 2 = 4\, km/h$.
Downstream speed = $u + v = 6 + 2 = 8\, km/h$.
Time taken to travel upstream = $\frac{d}{4}$.
Time taken to travel downstream = $\frac{d}{8}$.
According to the problem,the difference in time is $3\, hours$:
$\frac{d}{4} - \frac{d}{8} = 3$.
Multiplying by $8$ on both sides: $2d - d = 24$.
Therefore,$d = 24\, km$.

(C) Let the speed of the boat in still water be $x$ $km/h$ and the speed of the stream be $y = 3$ $km/h$.
Let the distance be $d$ $km$.
Downstream speed $= (x + 3)$ $km/h$.
Upstream speed $= (x - 3)$ $km/h$.
According to the problem,the time taken downstream is $2$ $hours$ and upstream is $4$ $hours$.
Distance $d = (x + 3) \times 2 = (x - 3) \times 4$.
$2x + 6 = 4x - 12$.
$2x = 18$.
$x = 9$ $km/h$.
Thus,the speed of the boat in still water is $9$ $km/h$.

(B) Let the speed of the boat in still water be $u = 6\, km/h$ and the speed of the stream be $v = 2\, km/h$.
The speed of the boat upstream is $(u - v) = 6 - 2 = 4\, km/h$.
The speed of the boat downstream is $(u + v) = 6 + 2 = 8\, km/h$.
The time taken to travel $32\, km$ upstream is $t_1 = \frac{\text{Distance}}{\text{Upstream Speed}} = \frac{32}{4} = 8\, hours$.
The time taken to travel $32\, km$ downstream is $t_2 = \frac{\text{Distance}}{\text{Downstream Speed}} = \frac{32}{8} = 4\, hours$.
The total journey time is $t = t_1 + t_2 = 8 + 4 = 12\, hours$.

(C) Let the distance between $A$ and $B$ be $d$ $km$.
Speed of boat in still water $(x)$ = $9$ $km/h$.
Speed of current $(y)$ = $3$ $km/h$.
Speed downstream = $x + y = 9 + 3 = 12$ $km/h$.
Speed upstream = $x - y = 9 - 3 = 6$ $km/h$.
Total time taken = Time downstream + Time upstream = $3$ $hrs$.
$\frac{d}{12} + \frac{d}{6} = 3$.
Multiplying by $12$,we get $d + 2d = 36$.
$3d = 36$,so $d = 12$ $km$.

(B) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y = 3\, km/h$.
The speed upstream is $(x - 3)\, km/h$ and the speed downstream is $(x + 3)\, km/h$.
The total time taken is $30\, minutes = 0.5\, hours$.
The time taken for the round trip is given by: $\frac{d}{x-y} + \frac{d}{x+y} = t$.
Substituting the values: $\frac{2}{x-3} + \frac{2}{x+3} = 0.5$.
$\frac{2(x+3) + 2(x-3)}{(x-3)(x+3)} = 0.5$.
$\frac{2x + 6 + 2x - 6}{x^2 - 9} = 0.5$.
$\frac{4x}{x^2 - 9} = 0.5$.
$4x = 0.5(x^2 - 9) \Rightarrow 8x = x^2 - 9$.
$x^2 - 8x - 9 = 0$.
$(x - 9)(x + 1) = 0$.
Since speed cannot be negative,$x = 9\, km/h$.