Boats and Streams Questions in English

(A) Let the speed of the boat in still water be $x \ km/h$.
The speed of the boat upstream is given by $(x - 4) \ km/h$.
Time taken to travel $14 \ km$ upstream is $42 \ minutes$.
Convert time into hours: $42 \ minutes = \frac{42}{60} \ hours = 0.7 \ hours$.
Using the formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we have:
$x - 4 = \frac{14}{0.7}$
$x - 4 = 20$
$x = 24 \ km/h$.
Therefore,the speed of the boat in still water is $24 \ km/h$.

(B) Let the speed of the boat in still water be $u = 7\, km/h$ and the speed of the stream be $v\, km/h$.
The speed of the boat in downstream is given by the formula: $\text{Downstream Speed} = u + v$.
Given that the downstream speed is $10\, km/h$,we have:
$7 + v = 10$
Solving for $v$:
$v = 10 - 7 = 3\, km/h$.
Thus,the speed of the stream is $3\, km/h$.

(C) Let the speed of the boat in still water be $x \text{ km/h}$ and the speed of the stream be $y \text{ km/h}$.
When rowing with the stream (downstream),the effective speed is $(x + y) \text{ km/h}$.
Given,$x + y = 10 \text{ km/h} \quad \dots(1)$
When rowing against the stream (upstream),the effective speed is $(x - y) \text{ km/h}$.
Given,$x - y = 6 \text{ km/h} \quad \dots(2)$
Adding equation $(1)$ and $(2)$:
$(x + y) + (x - y) = 10 + 6$
$2x = 16$
$x = 8 \text{ km/h}$
Therefore,Aditya's speed in still water is $8 \text{ km/h}$.

(D) The speed of a man in still water is $u = 12 \text{ km/h}$.
The speed of the current is $v = 4 \text{ km/h}$.
When moving upstream,the effective speed is the difference between the speed in still water and the speed of the current.
Therefore,the effective speed upstream $= u - v = 12 - 4 = 8 \text{ km/h}$.

(B) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
The speed downstream is given by $x + y = 14\, km/h$.
The speed upstream is given by $x - y = 6\, km/h$.
Adding both equations: $(x + y) + (x - y) = 14 + 6$.
$2x = 20$.
$x = 10\, km/h$.
Therefore,the speed of the boat in still water is $10\, km/h$.

(B) Let the speed of the boat in still water be $x \, km/h$ and the speed of the stream be $y \, km/h$.
The speed downstream is $(x + y) \, km/h$ and the speed upstream is $(x - y) \, km/h$.
According to the problem,the man covers $20 \, km$ downstream in $5 \, hours$:
$x + y = \frac{20}{5} = 4 \, km/h$ $...(i)$
The man covers $10 \, km$ upstream in $5 \, hours$:
$x - y = \frac{10}{5} = 2 \, km/h$ $...(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(x + y) - (x - y) = 4 - 2$
$2y = 2$
$y = 1 \, km/h$
Therefore,the speed of the stream is $1 \, km/h$.

(C) Let the speed of the boat in still water be $x \, km/h$ and the speed of the current be $y \, km/h$.
Given that $y = \frac{1}{4}x$.
Upstream speed $= x - y = x - \frac{1}{4}x = \frac{3}{4}x$.
Time taken to travel $18 \, km$ upstream is $6 \, hours$,so $\frac{18}{\frac{3}{4}x} = 6$.
$18 = 6 \times \frac{3}{4}x \Rightarrow 18 = \frac{18}{4}x \Rightarrow x = 4 \, km/h$.
Thus,the speed of the current $y = \frac{1}{4} \times 4 = 1 \, km/h$.
Downstream speed $= x + y = 4 + 1 = 5 \, km/h$.
Time taken to cover $18 \, km$ downstream $= \frac{18}{5} = 3.6 \, hours$.

(B) Let the speed of the boat in still water be $x \text{ km/h}$ and the speed of the stream be $y \text{ km/h}$.
The speed downstream (with the stream) is given by $x + y = 8 \text{ km/h}$.
The speed upstream (against the stream) is given by $x - y = 4 \text{ km/h}$.
To find the speed of the stream $(y)$,we subtract the second equation from the first:
$(x + y) - (x - y) = 8 - 4$
$2y = 4$
$y = 2 \text{ km/h}$.
Thus,the speed of the current is $2 \text{ km/h}$.

(C) Let the speed of the boat in still water be $x = 4\, km/h$.
Let the speed of the stream be $y = 2\, km/h$.
The formula for the upstream speed of a boat is given by $(x - y)$.
Therefore,the upstream speed $= 4\, km/h - 2\, km/h = 2\, km/h$.

(A) Let the speed of the swimmer in still water be $x = 9\, km/h$.
Let the speed of the river current be $y = 6\, km/h$.
The downstream speed is defined as the sum of the speed of the swimmer in still water and the speed of the river current.
Downstream speed $= x + y = 9 + 6 = 15\, km/h$.

(A) Let the speed of the swimmer in still water be $x \, km/h$ and the speed of the stream be $y \, km/h$.
Given,downstream speed $= x + y = 11 \, km/h$.
Given,speed of the stream $y = 1.5 \, km/h$.
Substituting $y$ in the downstream equation: $x + 1.5 = 11 \Rightarrow x = 11 - 1.5 = 9.5 \, km/h$.
The upstream speed of the swimmer is given by $x - y$.
Therefore,upstream speed $= 9.5 - 1.5 = 8 \, km/h$.

(A) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Along the stream (downstream),the effective speed is $(x + y)\, km/h$.
Given: $1\, km$ in $5\, min = 5/60\, h = 1/12\, h$.
Speed $= \text{Distance} / \text{Time} = 1 / (1/12) = 12\, km/h$.
So,$x + y = 12$ $...(i)$
Against the stream (upstream),the effective speed is $(x - y)\, km/h$.
Given: $6\, km$ in $1\, h$.
Speed $= 6 / 1 = 6\, km/h$.
So,$x - y = 6$ $...(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(x + y) - (x - y) = 12 - 6$
$2y = 6$
$y = 3\, km/h$.
Therefore,the speed of the stream is $3\, km/h$.

(D) Let the speed of the boat in still water be $x \ km/h$ and the speed of the current be $y \ km/h$.
Downstream speed $= (x + y) \ km/h$.
Upstream speed $= (x - y) \ km/h$.
According to the problem:
For downstream: $\frac{60}{x + y} = 10 \Rightarrow x + y = 6$ $...(i)$
For upstream: $\frac{36}{x - y} = 10 \Rightarrow x - y = 3.6$ $...(ii)$
Adding equations $(i)$ and $(ii)$:
$(x + y) + (x - y) = 6 + 3.6$
$2x = 9.6 \Rightarrow x = 4.8 \ km/h$.
Substituting $x = 4.8$ in equation $(i)$:
$4.8 + y = 6 \Rightarrow y = 6 - 4.8 = 1.2 \ km/h$.
Thus,the velocity of the current is $1.2 \ km/h$.

(B) The speed of the boat in still water is $u = 24\, km/h$.
The speed of the stream is $v = 8\, km/h$.
When moving along the stream (downstream),the effective speed of the boat is $u + v = 24 + 8 = 32\, km/h$.
The distance to be covered is $d = 128\, km$.
The time taken is calculated by the formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
Therefore,$\text{Time} = \frac{128}{32} = 4\, h$.

(D) Let the speed of the boat be $x$ $km/h$ and the speed of the stream be $y$ $km/h$.
Speed against the current (upstream) $= x - y = \frac{\text{Distance}}{\text{Time}} = \frac{3/4 \text{ km}}{15/60 \text{ h}} = \frac{3}{4} \times 4 = 3 \text{ km/h}$.
So,$x - y = 3$ ... $(i)$
Speed with the current (downstream) $= x + y = \frac{\text{Distance}}{\text{Time}} = \frac{3/4 \text{ km}}{10/60 \text{ h}} = \frac{3}{4} \times 6 = 4.5 \text{ km/h} = \frac{9}{2} \text{ km/h}$.
So,$x + y = 4.5$ ... $(ii)$
Adding equations $(i)$ and $(ii)$:
$(x - y) + (x + y) = 3 + 4.5$
$2x = 7.5 \Rightarrow x = 3.75 \text{ km/h} = \frac{15}{4} \text{ km/h}$.
Substituting $x$ in equation $(i)$:
$3.75 - y = 3 \Rightarrow y = 0.75 \text{ km/h} = \frac{3}{4} \text{ km/h}$.
The ratio of the speed of the man to the speed of the current is $\frac{x}{y} = \frac{15/4}{3/4} = \frac{15}{3} = \frac{5}{1}$.
Thus,the ratio is $5:1$.

(A) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Downstream speed $= (x + y)\, km/h$.
Upstream speed $= (x - y)\, km/h$.
According to the problem,the boat covers $48\, km$ downstream in $20\, h$:
$x + y = \frac{48}{20} = 2.4\, km/h$ $....(1)$
It takes $4\, h$ more to cover the same distance upstream,so the time taken is $20 + 4 = 24\, h$:
$x - y = \frac{48}{24} = 2\, km/h$ $....(2)$
Adding equations $(1)$ and $(2)$:
$(x + y) + (x - y) = 2.4 + 2$
$2x = 4.4$
$x = 2.2\, km/h$
Thus,the speed of the boat in still water is $2.2\, km/h$.

(C) Let the speed of the man in still water be $x\, km/hr$ and the speed of the stream be $y\, km/hr$.
Downstream speed $= x + y = \frac{\text{Distance}}{\text{Time}} = \frac{15}{3} = 5\, km/hr$ $...(1)$
Upstream speed $= x - y = \frac{\text{Distance}}{\text{Time}} = \frac{5}{2.5} = 2\, km/hr$ $...(2)$
Adding equations $(1)$ and $(2)$:
$(x + y) + (x - y) = 5 + 2$
$2x = 7$
$x = 3.5\, km/hr$
Therefore,the speed of the man in still water is $3.5\, km/hr$.

(A) Let the speed of the current be $y \text{ km/hr}$.
The speed downstream is given by the sum of the speed in still water and the speed of the current,which is $(8 + y) \text{ km/hr}$.
Given that the man rows $44 \text{ km}$ downstream in $4 \text{ hours}$,we have:
$\text{Speed downstream} = \frac{\text{Distance}}{\text{Time}} = \frac{44 \text{ km}}{4 \text{ hr}} = 11 \text{ km/hr}$.
Equating the two expressions for downstream speed:
$8 + y = 11 \Rightarrow y = 3 \text{ km/hr}$.
The speed upstream is given by the difference between the speed in still water and the speed of the current,which is $(8 - y) \text{ km/hr}$.
Substituting $y = 3 \text{ km/hr}$:
$\text{Speed upstream} = 8 - 3 = 5 \text{ km/hr}$.
Time taken to cover $25 \text{ km}$ upstream is:
$\text{Time} = \frac{\text{Distance}}{\text{Speed upstream}} = \frac{25 \text{ km}}{5 \text{ km/hr}} = 5 \text{ hours}$.

(D) Speed of the man in still water $= 20 \text{ km/h}$.
Speed of the stream $= 5 \text{ m/s}$.
To convert the speed of the stream from $\text{m/s}$ to $\text{km/h}$,multiply by $\frac{18}{5}$:
Speed of the stream $= 5 \times \frac{18}{5} = 18 \text{ km/h}$.
The ratio of the speed of the man to the speed of the current is $20:18$.
Simplifying the ratio by dividing both terms by $2$,we get $10:9$.

(A) Let the speed of the man in still water be $x = 9.6 \text{ km/hr}$.
Let the speed of the stream be $y \text{ km/hr}$.
Let the distance to be covered be $d \text{ km}$.
The speed downstream is $(9.6 + y) \text{ km/hr}$ and the speed upstream is $(9.6 - y) \text{ km/hr}$.
According to the problem,the time taken to swim upstream is twice the time taken to swim downstream.
Time taken upstream = $\frac{d}{9.6 - y}$ and time taken downstream = $\frac{d}{9.6 + y}$.
Given: $\frac{d}{9.6 - y} = 2 \times \frac{d}{9.6 + y}$.
Canceling $d$ from both sides: $\frac{1}{9.6 - y} = \frac{2}{9.6 + y}$.
Cross-multiplying: $9.6 + y = 2(9.6 - y)$.
$9.6 + y = 19.2 - 2y$.
$3y = 19.2 - 9.6$.
$3y = 9.6$.
$y = 3.2 \text{ km/hr}$.

(C) Let the speed of the motorboat in still water be $u = 45 \text{ km/h}$ and the speed of the stream be $v \text{ km/h}$.
Time taken to travel $80 \text{ km}$ along the stream (downstream) is $1 \text{ h } 20 \text{ min} = 1 + \frac{20}{60} \text{ h} = 1 + \frac{1}{3} \text{ h} = \frac{4}{3} \text{ h}$.
Downstream speed $= u + v = \frac{\text{Distance}}{\text{Time}} = \frac{80}{4/3} = 80 \times \frac{3}{4} = 60 \text{ km/h}$.
Since $u = 45 \text{ km/h}$,we have $45 + v = 60$,which gives $v = 15 \text{ km/h}$.
Now,the speed against the stream (upstream) is $u - v = 45 - 15 = 30 \text{ km/h}$.
Time taken to cover $80 \text{ km}$ upstream $= \frac{\text{Distance}}{\text{Upstream Speed}} = \frac{80}{30} \text{ h} = \frac{8}{3} \text{ h}$.
Converting to hours and minutes: $\frac{8}{3} \text{ h} = 2 \frac{2}{3} \text{ h} = 2 \text{ h } + (\frac{2}{3} \times 60) \text{ min} = 2 \text{ h } 40 \text{ min}$.

(A) Let the distance be $d\, km$.
Speed of the boat in still water $= 5\, km/h$.
Speed of the stream $= 2\, km/h$.
Speed upstream $= 5 - 2 = 3\, km/h$.
Speed downstream $= 5 + 2 = 7\, km/h$.
Time taken upstream $= \frac{d}{3}$.
Time taken downstream $= \frac{d}{7}$.
According to the problem,the time taken upstream is $2\, h$ more than the time taken downstream:
$\frac{d}{3} = \frac{d}{7} + 2$
$\frac{d}{3} - \frac{d}{7} = 2$
$\frac{7d - 3d}{21} = 2$
$\frac{4d}{21} = 2$
$4d = 42$
$d = 10.5\, km$.

(A) Let the speed of the boat in still water be $u = 10 \text{ km/h}$ and the speed of the current be $v \text{ km/h}$.
The speed downstream is $(10 + v) \text{ km/h}$ and the speed upstream is $(10 - v) \text{ km/h}$.
The total time taken for the round trip is given by the sum of time taken to go downstream and upstream:
$\frac{24}{10 + v} + \frac{24}{10 - v} = 5$
Taking $24$ as a common factor:
$24 \left( \frac{10 - v + 10 + v}{(10 + v)(10 - v)} \right) = 5$
$24 \left( \frac{20}{100 - v^2} \right) = 5$
Simplifying the equation:
$480 = 5(100 - v^2)$
$96 = 100 - v^2$
$v^2 = 4$
$v = 2 \text{ km/h}$ (since speed cannot be negative).
Thus,the speed of the water current is $2 \text{ km/h}$.

(D) Let the speed of the steamer in still water be $x\, km/h$ and the distance between the two ports be $d\, km$.
Given the speed of the stream is $2\, km/h$.
When moving downstream,the effective speed is $(x + 2)\, km/h$. Since the time taken is $4\, h$,we have $d = 4(x + 2) \Rightarrow d = 4x + 8$ ... $(1)$.
When moving upstream,the effective speed is $(x - 2)\, km/h$. Since the time taken is $5\, h$,we have $d = 5(x - 2) \Rightarrow d = 5x - 10$ ... $(2)$.
Equating the two expressions for $d$:
$4x + 8 = 5x - 10$
$x = 18\, km/h$.
Substituting $x = 18$ into equation $(1)$:
$d = 4(18) + 8 = 72 + 8 = 80\, km$.
Thus,the distance between the two ports is $80\, km$.

(B) Let the speed of the boat in still water be $x$ and the speed of the stream be $y$.
Let the distance be $d$.
The speed of the boat with the stream (downstream) is $(x + y)$ and against the stream (upstream) is $(x - y)$.
According to the problem,the time taken to row upstream is twice the time taken to row downstream:
$\frac{d}{x - y} = 2 \times \frac{d}{x + y}$
Canceling $d$ from both sides,we get:
$\frac{1}{x - y} = \frac{2}{x + y}$
$x + y = 2(x - y)$
$x + y = 2x - 2y$
Rearranging the terms,we get:
$x = 3y$
Therefore,the ratio of the speed of the boat in still water to the speed of the stream is $\frac{x}{y} = \frac{3}{1}$,which is $3:1$.

(B) Let the speed of the boat in still water be $x\, km/h$ and the speed of the river current be $v = 4\, km/h$.
The total distance covered is $42\, km$,so the distance for one way is $d = 21\, km$.
The speed downstream is $(x + 4)\, km/h$ and the speed upstream is $(x - 4)\, km/h$.
The time taken for the outward (downstream) journey is $t_1 = \frac{21}{x + 4}$.
The time taken for the return (upstream) journey is $t_2 = \frac{21}{x - 4}$.
According to the problem,the return journey takes $2\, h$ more than the outward journey,so $t_2 - t_1 = 2$.
Substituting the values: $\frac{21}{x - 4} - \frac{21}{x + 4} = 2$.
$21 \left( \frac{(x + 4) - (x - 4)}{(x - 4)(x + 4)} \right) = 2$.
$21 \left( \frac{8}{x^2 - 16} \right) = 2$.
$168 = 2(x^2 - 16) \Rightarrow 84 = x^2 - 16$.
$x^2 = 100 \Rightarrow x = 10\, km/h$.

(C) Let the speed of the motorboat be $v_b = 36x$ and the speed of the current be $v_c = 5x$.
The speed of the motorboat along the current (downstream) is $v_d = v_b + v_c = 36x + 5x = 41x$.
The speed of the motorboat against the current (upstream) is $v_u = v_b - v_c = 36x - 5x = 31x$.
Let the distance be $d$. The time taken to travel downstream is $t_d = 5 \, h \, 10 \, min = 5 + \frac{10}{60} \, h = 5 + \frac{1}{6} \, h = \frac{31}{6} \, h$.
Since $d = v_d \times t_d$,we have $d = 41x \times \frac{31}{6}$.
The time taken to return (upstream) is $t_u = \frac{d}{v_u} = \frac{41x \times \frac{31}{6}}{31x} = \frac{41}{6} \, h$.
Converting $\frac{41}{6} \, h$ into hours and minutes: $\frac{41}{6} \, h = 6 \frac{5}{6} \, h = 6 \, h + (\frac{5}{6} \times 60) \, min = 6 \, h \, 50 \, min$.

(B) Let the speed of the first boat be $5a$ and the speed of the stream be $2a$.
In the second case,let the speed of the stream be $3b$ and the speed of the second boat be $4b$.
Since the speed of the stream is the same in both cases,we equate the stream speeds: $2a = 3b$.
From this,we get $a = \frac{3}{2}b$.
Now,substitute $a$ into the speed of the first boat: $5a = 5 \times (\frac{3}{2}b) = \frac{15}{2}b$.
The ratio of the speed of the first boat to the second boat is $\frac{\frac{15}{2}b}{4b} = \frac{15}{8}$ or $15:8$.

(B) Let the speed of the man in still water be $x = 3.5 \, km/h$ and the speed of the current be $y \, km/h$.
The speed upstream is $(x - y) \, km/h$ and the speed downstream is $(x + y) \, km/h$.
Let the distance be $d$. The time taken to row upstream is $T_u = \frac{d}{x - y}$ and the time taken to row downstream is $T_d = \frac{d}{x + y}$.
According to the problem,$T_u = 2.5 \times T_d$,which is $\frac{d}{x - y} = \frac{5}{2} \times \frac{d}{x + y}$.
Canceling $d$ from both sides,we get $\frac{1}{x - y} = \frac{5}{2(x + y)}$,which implies $2(x + y) = 5(x - y)$.
Expanding the equation: $2x + 2y = 5x - 5y$.
Rearranging the terms: $7y = 3x$.
Substituting $x = 3.5$: $7y = 3 \times 3.5 = 10.5$.
Therefore,$y = \frac{10.5}{7} = 1.5 \, km/h$.

(D) Let the speed of the man in still water be $x = 4 \text{ km/hr}$ and the speed of the current be $y \text{ km/hr}$.
Speed upstream $= (x - y) \text{ km/hr} = (4 - y) \text{ km/hr}$.
Speed downstream $= (x + y) \text{ km/hr} = (4 + y) \text{ km/hr}$.
Let the distance be $d \text{ km}$.
Time taken upstream $T_u = \frac{d}{4 - y}$ and time taken downstream $T_d = \frac{d}{4 + y}$.
According to the problem,$T_u = 3 \times T_d$.
$\frac{d}{4 - y} = 3 \times \frac{d}{4 + y}$.
Canceling $d$ from both sides: $\frac{1}{4 - y} = \frac{3}{4 + y}$.
$4 + y = 3(4 - y) \Rightarrow 4 + y = 12 - 3y$.
$y + 3y = 12 - 4 \Rightarrow 4y = 8$.
$y = 2 \text{ km/hr}$.

(B) Let the speed of the swimmer in still water be $x \, km/hr$ and the speed of the stream be $y = 1.5 \, km/hr$.
The speed downstream is $(x + y) \, km/hr$ and the speed upstream is $(x - y) \, km/hr$.
According to the problem,the distance covered downstream in a given time $t$ is twice the distance covered upstream in the same time $t$.
Distance = Speed $\times$ Time.
So,$(x + y) \times t = 2 \times (x - y) \times t$.
Dividing both sides by $t$,we get $x + y = 2(x - y)$.
$x + 1.5 = 2x - 3$.
$x = 1.5 + 3 = 4.5 \, km/hr$.
Therefore,the speed of the swimmer in still water is $4.5 \, km/hr$.

(D) Let the speed of the man in still water be $x = 8 \text{ km/hr}$ and the speed of the current be $y \text{ km/hr}$.
The speed upstream is given by $(x - y)$.
Given that the man rows $36 \text{ km}$ upstream in $6 \text{ hours}$,we have:
$\frac{36}{x - y} = 6$
$\frac{36}{8 - y} = 6$
$36 = 6(8 - y)$
$36 = 48 - 6y$
$6y = 48 - 36$
$6y = 12$
$y = 2 \text{ km/hr}$.
The speed downstream is given by $(x + y) = 8 + 2 = 10 \text{ km/hr}$.
To find the distance covered downstream in $10 \text{ hours}$:
$\text{Distance} = \text{Speed} \times \text{Time}$
$\text{Distance} = 10 \text{ km/hr} \times 10 \text{ hours} = 100 \text{ km}$.

(D) Let the speed of the boat in still water be $x = 4 \, km/h$ and the speed of the current be $y = 2 \, km/h$.
The speed upstream is $(x - y) = 4 - 2 = 2 \, km/h$.
The time taken to cover distance $d$ upstream is $9 \, hours$.
Using the formula $\text{Distance} = \text{Speed} \times \text{Time}$,we have $d = 2 \times 9 = 18 \, km$.
The speed downstream is $(x + y) = 4 + 2 = 6 \, km/h$.
To find the time $t$ taken to cover the same distance $d$ downstream,we use $t = \frac{d}{\text{Speed downstream}}$.
$t = \frac{18}{6} = 3 \, hours$.

(A) Let the speed of the person in still water be $u = 4 \, km/h$ and the speed of the water current be $v = 2 \, km/h$.
When swimming against the current (upstream),the effective speed of the person is given by $u - v$.
Effective speed $= 4 \, km/h - 2 \, km/h = 2 \, km/h$.
The distance to be covered is $d = 6 \, km$.
The time taken $t$ is calculated using the formula $t = \frac{d}{\text{effective speed}}$.
$t = \frac{6 \, km}{2 \, km/h} = 3 \, hours$.
Therefore,the man will take $3 \, hours$ to swim back against the current.

(A) Let the speed of the man in still water be $x = 4.5 \text{ km/hr}$ and the speed of the river current be $y = 1.5 \text{ km/hr}$.
The speed upstream $(u)$ is given by $x - y = 4.5 - 1.5 = 3 \text{ km/hr}$.
The speed downstream $(v)$ is given by $x + y = 4.5 + 1.5 = 6 \text{ km/hr}$.
The formula for average speed for a round trip is $\text{Average Speed} = \frac{2uv}{u + v}$.
Substituting the values: $\text{Average Speed} = \frac{2 \times 3 \times 6}{3 + 6} = \frac{36}{9} = 4 \text{ km/hr}$.

(D) Let the speed of the man in still water be $x = 15\, km/hr$ and the speed of the river current be $y = 3\, km/hr$.
The speed upstream $(u)$ is $x - y = 15 - 3 = 12\, km/hr$.
The speed downstream $(v)$ is $x + y = 15 + 3 = 18\, km/hr$.
The average speed for a round trip is given by the formula $\text{Average Speed} = \frac{2uv}{u + v}$.
Substituting the values: $\text{Average Speed} = \frac{2 \times 12 \times 18}{12 + 18} = \frac{2 \times 12 \times 18}{30}$.
$\text{Average Speed} = \frac{432}{30} = 14.4\, km/hr$.

(D) Let the distance be $d\, km$.
Speed of the man in still water $(u)$ = $7\, km/hr$.
Speed of the stream $(v)$ = $3\, km/hr$.
Speed downstream = $u + v = 7 + 3 = 10\, km/hr$.
Speed upstream = $u - v = 7 - 3 = 4\, km/hr$.
Time taken to travel downstream = $\frac{d}{10}\, hours$.
Time taken to travel upstream = $\frac{d}{4}\, hours$.
According to the problem,the difference in time is $6\, hours$:
$\frac{d}{4} - \frac{d}{10} = 6$
$\frac{5d - 2d}{20} = 6$
$\frac{3d}{20} = 6$
$3d = 120$
$d = 40\, km$.

(B) Let the distance be $d \text{ km}$.
Speed of the man in still water $u = 9 \text{ km/hr}$.
Speed of the stream $v = 3 \text{ km/hr}$.
Speed downstream $= u + v = 9 + 3 = 12 \text{ km/hr}$.
Speed upstream $= u - v = 9 - 3 = 6 \text{ km/hr}$.
Time taken to travel downstream $t_1 = \frac{d}{12} \text{ hours}$.
Time taken to travel upstream $t_2 = \frac{d}{6} \text{ hours}$.
According to the problem,$t_2 = t_1 + 3$.
$\frac{d}{6} = \frac{d}{12} + 3$.
Multiply by $12$ to clear the denominators: $2d = d + 36$.
$d = 36 \text{ km}$.

(D) Let the distance between $M$ and $N$ be $d$ $km$ and the speed of the stream be $y$ $km/h$.
Given,speed of the boat in still water $u = 4$ $km/h$.
Time taken for downstream journey = $\frac{d}{u+y} = \frac{d}{4+y}$.
Time taken for upstream journey = $\frac{d}{u-y} = \frac{d}{4-y}$.
Total time taken = $\frac{d}{4+y} + \frac{d}{4-y} = 3$.
$\frac{d(4-y) + d(4+y)}{(4+y)(4-y)} = 3$.
$\frac{8d}{16-y^2} = 3$.
$8d = 3(16-y^2)$.
Since there are two variables ($d$ and $y$) and only one equation,the distance $d$ cannot be determined uniquely without knowing the speed of the stream $y$. Thus,the data is inadequate.

(A) Let the distance between $P$ and $Q$ be $d$. Since $Q$ is equidistant from $P$ and $R$,the distance between $Q$ and $R$ is also $d$. The total distance from $P$ to $R$ is $2d$.
Let $x$ be the speed of the boat in still water and $y$ be the speed of the river.
Time taken to row from $P$ to $Q$ and back is $10$ $hours$:
$\frac{d}{x+y} + \frac{d}{x-y} = 10 \quad \dots (1)$
Time taken to row downstream from $P$ to $R$ is $4$ $hours$:
$\frac{2d}{x+y} = 4 \Rightarrow d = 2(x+y) \quad \dots (2)$
From equation $(1)$:
$d \left( \frac{x-y+x+y}{(x+y)(x-y)} \right) = 10 \Rightarrow 2xd = 10(x^2 - y^2) \Rightarrow xd = 5(x^2 - y^2)$
Substitute $d = 2(x+y)$ from equation $(2)$ into the expression:
x($2$(x+y)) = $5$(x+y)(x-y)
Since $x+y \neq 0$,we can divide both sides by $(x+y)$:
$2x = 5(x-y) \Rightarrow 2x = 5x - 5y \Rightarrow 3x = 5y$
Therefore,$\frac{x}{y} = \frac{5}{3}$.

(B) Let the speed of the boat in still water be $x \text{ km/h}$.
Given,speed of the stream $= 4 \text{ km/h}$.
Speed downstream $= (x + 4) \text{ km/h}$ and speed upstream $= (x - 4) \text{ km/h}$.
Time taken to travel $6 \text{ km}$ downstream $= \frac{6}{x + 4} \text{ hours}$.
Time taken to travel $6 \text{ km}$ upstream $= \frac{6}{x - 4} \text{ hours}$.
According to the problem,the total time is $2 \text{ hours}$:
$\frac{6}{x + 4} + \frac{6}{x - 4} = 2$
Divide by $2$: $\frac{3}{x + 4} + \frac{3}{x - 4} = 1$
$3(x - 4) + 3(x + 4) = (x + 4)(x - 4)$
$3x - 12 + 3x + 12 = x^2 - 16$
$6x = x^2 - 16$
$x^2 - 6x - 16 = 0$
$(x - 8)(x + 2) = 0$
Since speed cannot be negative,$x = 8 \text{ km/h}$.

(D) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Downstream speed $= x + y = 15\, km/h$.
Upstream speed $= x - y = 9\, km/h$.
To find the speed of the boat in still water $(x)$,we add the two equations:
$(x + y) + (x - y) = 15 + 9$
$2x = 24$
$x = 12\, km/h$.
Alternatively,the formula is: Speed of the boat in still water $= \frac{1}{2} \times (\text{downstream speed} + \text{upstream speed})$.
$= \frac{1}{2} \times (15 + 9) = \frac{1}{2} \times 24 = 12\, km/h$.

(A) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
According to the problem,the downstream speed is $x + y = \frac{\text{Distance}}{\text{Time}} = \frac{20}{2} = 10\, km/h$ $....(i)$
Similarly,the upstream speed is $x - y = \frac{\text{Distance}}{\text{Time}} = \frac{20}{5} = 4\, km/h$ $....(ii)$
Adding equation $(i)$ and equation $(ii)$:
$(x + y) + (x - y) = 10 + 4$
$2x = 14$
$x = 7\, km/h$
Thus,the speed of the boat in still water is $7\, km/h$.

(B) Let the rate of swimming in still water be $x \text{ km/h}$.
The speed of the boy downstream (along the current) is $(x + 3) \text{ km/h}$.
The speed of the boy upstream (against the current) is $(x - 3) \text{ km/h}$.
Since the time $t$ is fixed,and distance = speed $\times$ time,the problem states that the distance covered downstream is double the distance covered upstream:
$(x + 3) \times t = 2 \times (x - 3) \times t$
Dividing both sides by $t$ (assuming $t \neq 0$):
$x + 3 = 2(x - 3)$
$x + 3 = 2x - 6$
$x = 9 \text{ km/h}$.
Thus,the rate of swimming in still water is $9 \text{ km/h}$.

(A) Let the speed of the boat in still water be $x \text{ km/h}$ and the speed of the stream be $y \text{ km/h}$.
Speed downstream $= x + y = \frac{20 \text{ km}}{1 \text{ h}} = 20 \text{ km/h}$.
Speed upstream $= x - y = \frac{20 \text{ km}}{2 \text{ h}} = 10 \text{ km/h}$.
Adding the two equations: $(x + y) + (x - y) = 20 + 10$.
$2x = 30$.
$x = 15 \text{ km/h}$.
Therefore,the speed of the boat in still water is $15 \text{ km/h}$.

(B) Let the speed of the stream be $x \text{ km/h}$.
Since the speed of the boat in still water is $4$ times the speed of the current,the speed of the boat in still water is $4x \text{ km/h}$.
The speed downstream is $(4x + x) = 5x \text{ km/h}$.
The speed upstream is $(4x - x) = 3x \text{ km/h}$.
Given that the total time taken for the round trip of $30 \text{ km}$ is $8 \text{ h}$,we use the formula $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
$\frac{30}{3x} + \frac{30}{5x} = 8$.
$\frac{10}{x} + \frac{6}{x} = 8$.
$\frac{16}{x} = 8$.
$x = \frac{16}{8} = 2 \text{ km/h}$.
Thus,the speed of the current is $2 \text{ km/h}$.

(A) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Speed downstream $(x + y)$ $= \frac{1\, km}{5\, min} = \frac{1\, km}{(5/60)\, h} = 12\, km/h$.
Speed upstream $(x - y)$ $= \frac{6\, km}{1\, h} = 6\, km/h$.
To find the speed of the stream $(y)$,we subtract the upstream speed from the downstream speed and divide by $2$:
$y = \frac{(x + y) - (x - y)}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3\, km/h$.

(B) Let the speed of the boat in still water be $x \ km/h$ and the speed of the stream be $y \ km/h$.
Downstream speed $= x + y = \frac{18 \ km}{4 \ h} = 4.5 \ km/h$.
Upstream speed $= x - y = \frac{18 \ km}{12 \ h} = 1.5 \ km/h$.
To find the speed of the stream $(y)$,we subtract the upstream speed from the downstream speed and divide by $2$:
$y = \frac{(x + y) - (x - y)}{2} = \frac{4.5 - 1.5}{2} = \frac{3}{2} = 1.5 \ km/h$.

(C) Let the distance to the place be $x \, km$.
Speed of the man in still water $= 5 \, km/h$.
Speed of the current $= 1 \, km/h$.
Speed downstream $= 5 + 1 = 6 \, km/h$.
Speed upstream $= 5 - 1 = 4 \, km/h$.
According to the problem,the total time taken to go and come back is $1 \, h$.
Time taken downstream + Time taken upstream $= 1 \, h$.
$\frac{x}{6} + \frac{x}{4} = 1$.
Taking the least common multiple of $6$ and $4$,which is $12$:
$\frac{2x + 3x}{12} = 1$.
$\frac{5x}{12} = 1$.
$5x = 12$.
$x = \frac{12}{5} = 2.4 \, km$.
Therefore,the distance is $2.4 \, km$.

(C) The speed of the man in still water is $v_m = 4\, km/h$.
The speed of the stream is $v_s = 2\, km/h$.
When swimming upstream,the effective speed is the difference between the man's speed and the stream's speed:
$v_{upstream} = v_m - v_s = 4 - 2 = 2\, km/h$.
The distance to be covered is $d = 10\, km$.
The time taken is given by the formula: $t = \frac{d}{v_{upstream}}$.
$t = \frac{10}{2} = 5\, h$.