Boats and Streams Questions in English

(A) Let the speed of the man in still water be $x \text{ km/h}$ and the speed of the stream be $y = 1.5 \text{ km/h}$.
Downstream speed $= (x + y) \text{ km/h}$.
Upstream speed $= (x - y) \text{ km/h}$.
According to the problem,in a given time $t$,the distance covered downstream is twice the distance covered upstream:
$(x + y)t = 2(x - y)t$
$x + 1.5 = 2(x - 1.5)$
$x + 1.5 = 2x - 3$
$x = 4.5 \text{ km/h}$.
Thus,the speed of the man is $4 \frac{1}{2} \text{ km/h}$.

(D) Speed downstream $= (9 + 3)\, km/h = 12\, km/h$.
Speed upstream $= (9 - 3)\, km/h = 6\, km/h$.
Let the distance $AB = x\, km$.
The time taken for the round trip is the sum of the time taken to travel upstream and downstream.
Time $= \frac{\text{Distance}}{\text{Speed}}$.
Therefore,$\frac{x}{6} + \frac{x}{12} = 3$.
Multiplying by $12$,we get $2x + x = 36$.
$3x = 36$.
$x = 12$.
Thus,the distance $AB = 12\, km$.

(D) Let the man's rowing speed in still water be $x \text{ km/h}$ and the speed of the current be $y \text{ km/h}$.
Speed upstream $= (x - y) \text{ km/h}$.
Speed downstream $= (x + y) \text{ km/h}$.
Given that the time taken for both trips is $5 \text{ hours}$,we have:
$x - y = \frac{12}{5} = 2.4 \text{ km/h}$ (Equation $1$)
$x + y = \frac{28}{5} = 5.6 \text{ km/h}$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$:
$(x + y) - (x - y) = 5.6 - 2.4$
$2y = 3.2$
$y = 1.6 \text{ km/h}$.
Converting $1.6$ to a fraction: $1.6 = \frac{16}{10} = \frac{8}{5} = 1 \frac{3}{5} \text{ km/h}$.

(B) Let the speed in still water be $x \text{ km/h}$ and the speed of the current be $y \text{ km/h}$.
The speed downstream is $(x + y) \text{ km/h}$.
The speed upstream is $(x - y) \text{ km/h}$.
According to the problem,twice the speed downstream is equal to thrice the speed upstream:
$2(x + y) = 3(x - y)$
Expanding the equation:
$2x + 2y = 3x - 3y$
Rearranging the terms to solve for $x$ in terms of $y$:
$2x - 3x = -3y - 2y$
$-x = -5y$
$x = 5y$
Therefore,the ratio of the speed in still water to the speed of the current is:
$\frac{x}{y} = \frac{5}{1}$,which is $5:1$.

(D) Speed upstream $= (3 - 2) \text{ km/h} = 1 \text{ km/h}$.
Speed downstream $= (3 + 2) \text{ km/h} = 5 \text{ km/h}$.
Time taken to travel $10 \text{ km}$ upstream $= \frac{\text{Distance}}{\text{Speed upstream}} = \frac{10}{1} = 10 \text{ hours}$.
Time taken to travel $10 \text{ km}$ downstream $= \frac{\text{Distance}}{\text{Speed downstream}} = \frac{10}{5} = 2 \text{ hours}$.
Total time taken $= 10 + 2 = 12 \text{ hours}$.

(C) Let the speed of the boat in still water be $u \text{ km/h}$ and the speed of the stream be $v \text{ km/h}$.
Then,upstream speed $= (u - v) \text{ km/h}$ and downstream speed $= (u + v) \text{ km/h}$.
According to the problem:
$\frac{24}{u-v} + \frac{36}{u+v} = 6$ --- $(1)$
$\frac{36}{u-v} + \frac{24}{u+v} = 6.5 = \frac{13}{2}$ --- $(2)$
Let $X = \frac{1}{u-v}$ and $Y = \frac{1}{u+v}$.
$24X + 36Y = 6 \Rightarrow 4X + 6Y = 1$ --- $(3)$
$36X + 24Y = \frac{13}{2} \Rightarrow 72X + 48Y = 13$ --- $(4)$
Multiplying $(3)$ by $8$: $32X + 48Y = 8$ --- $(5)$
Subtracting $(5)$ from $(4)$: $40X = 5 \Rightarrow X = \frac{1}{8}$.
Substituting $X$ in $(3)$: $4(\frac{1}{8}) + 6Y = 1 \Rightarrow 0.5 + 6Y = 1 \Rightarrow 6Y = 0.5 \Rightarrow Y = \frac{1}{12}$.
So,$u - v = 8$ and $u + v = 12$.
Adding the two equations: $2u = 20 \Rightarrow u = 10 \text{ km/h}$.
Subtracting the two equations: $2v = 4 \Rightarrow v = 2 \text{ km/h}$.
The velocity of the current is $2 \text{ km/h}$.

(B) Let the speed of the boat in still water be $u\, km/h$ and the speed of the stream be $v\, km/h$.
Upstream speed $= u - v = \frac{2\, km}{1\, h} = 2\, km/h$.
Downstream speed $= u + v = \frac{1\, km}{10\, min} = \frac{1\, km}{1/6\, h} = 6\, km/h$.
Adding the two equations: $(u - v) + (u + v) = 2 + 6 \implies 2u = 8 \implies u = 4\, km/h$.
Thus, the speed of the boat in still water is $4\, km/h$.
Time required to cover $5\, km$ in stationary water $= \frac{\text{Distance}}{\text{Speed}} = \frac{5}{4}\, hours$.
$\frac{5}{4}\, hours = 1\, hour\, 15\, minutes$.

(A) Let the speed of the man in still water be $x \, km/h$ and the speed of the current be $y \, km/h$.
Speed downstream $= (x+y) \, km/h$.
Speed upstream $= (x-y) \, km/h$.
Let the river flow from $P$ to $R$ and $PQ = QR = a$. Then $PR = 2a$.
Given that the man rows from $P$ to $Q$ and returns in $10 \, h$:
$\frac{a}{x+y} + \frac{a}{x-y} = 10$ ---$(1)$
Given that he rows from $P$ to $R$ in $4 \, h$ (downstream):
$\frac{2a}{x+y} = 4 \implies \frac{a}{x+y} = 2$ ---$(2)$
Substituting $(2)$ into $(1)$:
$2 + \frac{a}{x-y} = 10 \implies \frac{a}{x-y} = 8$ ---$(3)$
Dividing $(2)$ by $(3)$:
$\frac{x-y}{x+y} = \frac{2}{8} = \frac{1}{4}$
$4x - 4y = x + y$
$3x = 5y$
$\frac{x}{y} = \frac{5}{3}$
Thus,the ratio is $5:3$.

(C) Let the speed of the motorboat in still water be $x \, km/h$.
The speed upstream is $(x - 2) \, km/h$ and the speed downstream is $(x + 2) \, km/h$.
The total time taken is $55 \, min = \frac{55}{60} \, hours = \frac{11}{12} \, hours$.
According to the problem:
$\frac{10}{x - 2} + \frac{10}{x + 2} = \frac{11}{12}$
$10 \left( \frac{x + 2 + x - 2}{x^2 - 4} \right) = \frac{11}{12}$
$10 \left( \frac{2x}{x^2 - 4} \right) = \frac{11}{12}$
$20x \times 12 = 11(x^2 - 4)$
$240x = 11x^2 - 44$
$11x^2 - 240x - 44 = 0$
$11x^2 - 242x + 2x - 44 = 0$
$11x(x - 22) + 2(x - 22) = 0$
$(11x + 2)(x - 22) = 0$
Since speed cannot be negative,$x = 22 \, km/h$.

(A) Let the upstream speed be $x \text{ km/h}$ and the downstream speed be $y \text{ km/h}$.
According to the problem,we have the equations:
$\frac{30}{x} + \frac{44}{y} = 10$ --- $(1)$
$\frac{40}{x} + \frac{55}{y} = 13$ --- $(2)$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. Then the equations become:
$30u + 44v = 10$ --- $(3)$
$40u + 55v = 13$ --- $(4)$
Multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:
$120u + 176v = 40$
$120u + 165v = 39$
Subtracting the two equations: $11v = 1 \implies v = \frac{1}{11}$.
Substituting $v = \frac{1}{11}$ into equation $(3)$: $30u + 44(\frac{1}{11}) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{1}{5}$.
Thus,$x = 5 \text{ km/h}$ and $y = 11 \text{ km/h}$.
Speed of the man in still water $= \frac{y + x}{2} = \frac{11 + 5}{2} = 8 \text{ km/h}$.
Rate of the current $= \frac{y - x}{2} = \frac{11 - 5}{2} = 3 \text{ km/h}$.

(B) The formula for average speed when the distance covered in both directions is the same is given by $\text{Average Speed} = \frac{2xy}{x+y}$,where $x$ and $y$ are the speeds in the two directions.
Here,$x = 21\, km/h$ and $y = 28\, km/h$.
Substituting these values into the formula:
$\text{Average Speed} = \frac{2 \times 21 \times 28}{21 + 28}$
$\text{Average Speed} = \frac{1176}{49}$
$\text{Average Speed} = 24\, km/h$.

(C) Let the speed of the boat in still water be $x\, km/hr$ and the speed of the stream be $y\, km/hr$.
Upstream speed = $(x - y)\, km/hr$ and downstream speed = $(x + y)\, km/hr$.
Given,time = $\text{distance} / \text{speed}$.
For the first case: $\frac{2}{x-y} + \frac{3}{x+y} = \frac{20}{60} = \frac{1}{3} \implies \frac{6}{x-y} + \frac{9}{x+y} = 1$ ---$(1)$
For the second case: $\frac{7}{x-y} + \frac{2}{x+y} = \frac{53}{60}$ ---$(2)$
Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.
$6u + 9v = 1$ ---$(3)$
$7u + 2v = \frac{53}{60}$ ---$(4)$
Multiplying $(3)$ by $2$ and $(4)$ by $9$:
$12u + 18v = 2$
$63u + 18v = \frac{477}{60} = 7.95$
Subtracting the equations: $51u = 5.95 \implies u = \frac{5.95}{51} = \frac{595}{5100} = \frac{7}{60}$.
So,$x - y = \frac{60}{7}$.
Substituting $u$ in $(3)$: $6(\frac{7}{60}) + 9v = 1 \implies \frac{7}{10} + 9v = 1 \implies 9v = \frac{3}{10} \implies v = \frac{3}{90} = \frac{1}{30}$.
So,$x + y = 30$.
Adding $x - y = \frac{60}{7}$ and $x + y = 30$:
$2x = \frac{60}{7} + 30 = \frac{60 + 210}{7} = \frac{270}{7}$.
$x = \frac{135}{7}\, km/hr$.

(C) Let the speed of the boat in still water be $x \ km/hr$ and the speed of the stream be $y \ km/hr$.
Downstream speed $(D_s)$ = $x + y = \frac{32 \ km}{4 \ hours} = 8 \ km/hr$.
Upstream speed $(U_s)$ = $x - y = \frac{24 \ km}{6 \ hours} = 4 \ km/hr$.
Adding the two equations: $(x + y) + (x - y) = 8 + 4$.
$2x = 12$.
$x = 6 \ km/hr$.
Thus,the speed of the boat in still water is $6 \ km/hr$.

(B) Let the speed of the boat in still water be $x \text{ km/hr}$ and the speed of the stream be $y \text{ km/hr}$.
Downstream speed $= (x+y) \text{ km/hr}$.
Upstream speed $= (x-y) \text{ km/hr}$.
According to the problem:
$\frac{60}{x+y} + \frac{20}{x-y} = 4$ --- $(i)$
$\frac{40}{x+y} + \frac{40}{x-y} = 6$ --- $(ii)$
Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.
$60u + 20v = 4 \Rightarrow 15u + 5v = 1$ --- $(iii)$
$40u + 40v = 6 \Rightarrow 20u + 20v = 3$ --- $(iv)$
Multiplying $(iii)$ by $4$: $60u + 20v = 4$.
Subtracting $(iv)$ from this: $(60u + 20v) - (40u + 40v) = 4 - 6 \Rightarrow 20u - 20v = -2 \Rightarrow 10u - 10v = -1$.
Solving the system,we get $u = \frac{1}{20}$ and $v = \frac{1}{10}$.
$x+y = 20$ and $x-y = 10$.
Adding the two equations: $2x = 30 \Rightarrow x = 15$.
Subtracting: $2y = 10 \Rightarrow y = 5$.
Wait,re-evaluating the system: $15u + 5v = 1$ and $20u + 20v = 3$. From $(iii)$,$5v = 1 - 15u \Rightarrow 20v = 4 - 60u$. Substitute into $(iv)$: $20u + 4 - 60u = 3 \Rightarrow -40u = -1 \Rightarrow u = \frac{1}{40}$.
Then $5v = 1 - 15(\frac{1}{40}) = 1 - \frac{3}{8} = \frac{5}{8} \Rightarrow v = \frac{1}{8}$.
$x+y = 40$ and $x-y = 8$.
$2x = 48 \Rightarrow x = 24$.
$2y = 32 \Rightarrow y = 16$.
The speed of the stream is $16 \text{ km/hr}$.

(A) Let the speed of the boat in still water be $x \ km/hr$ and the speed of the current be $y \ km/hr$.
The speed of the boat along the current (downstream speed) is given by $D_s = x + y = 10 \ km/hr$.
The speed of the boat against the current (upstream speed) is given by $U_s = x - y = 8 \ km/hr$.
To find the speed of the current $(y)$,we subtract the upstream equation from the downstream equation:
$(x + y) - (x - y) = 10 - 8$
$2y = 2$
$y = 1 \ km/hr$.
Therefore,the speed of the current is $1 \ km/hr$.

(D) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the current be $y \, km/hr$.
The speed of the boat along the current (downstream) is given by $D_s = x + y = 14 \, km/hr$.
The speed of the boat against the current (upstream) is given by $U_s = x - y = 8 \, km/hr$.
To find the speed of the current $(y)$,we subtract the upstream equation from the downstream equation:
$(x + y) - (x - y) = 14 - 8$
$2y = 6$
$y = 3 \, km/hr$.
Thus,the speed of the current is $3 \, km/hr$.

(D) Let the speed of the boat in still water be $u$ km/h and the speed of the stream be $v$ km/h. Let the distance between $P$ and $Q$ be $d$ km. Then $PQ = QR = d$ km and $PR = 2d$ km.
The time taken to travel from $P$ to $Q$ and back is given by:
$\frac{d}{u-v} + \frac{d}{u+v} = 12$ hours ---$(1)$
The time taken to travel from $P$ to $R$ (downstream) is:
$\frac{2d}{u+v} = 16 \text{ hours } 40 \text{ minutes} = 16 + \frac{40}{60} = 16 + \frac{2}{3} = \frac{50}{3}$ hours ---$(2)$
From $(2)$,we get $\frac{d}{u+v} = \frac{25}{3}$ hours.
Substitute this into $(1)$:
$\frac{d}{u-v} + \frac{25}{3} = 12$
$\frac{d}{u-v} = 12 - \frac{25}{3} = \frac{36-25}{3} = \frac{11}{3}$ hours.
We need to find the time taken to travel from $R$ to $P$ (upstream),which is $\frac{2d}{u-v}$:
$\frac{2d}{u-v} = 2 \times \frac{11}{3} = \frac{22}{3} = 7 \frac{1}{3}$ hours.

(C) The speed of the boat in still water is $u = 20\, km/hr$.
The speed of the current is $v = 5\, km/hr$.
When the boat travels with the current (downstream),the effective speed is $u + v = 20 + 5 = 25\, km/hr$.
The distance to be covered is $d = 100\, km$.
The time taken is calculated as $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{25} = 4\, hours$.

(A) Downstream speed of the boat $= \frac{1 \text{ km}}{7.5 \text{ minutes}} = \frac{1 \text{ km}}{(7.5 / 60) \text{ hours}} = \frac{60}{7.5} \text{ km/h} = 8 \text{ km/h}$.
Upstream speed of the boat $= 5 \text{ km/h}$.
Speed of the boat in still water $= \frac{1}{2} (\text{Downstream speed} + \text{Upstream speed})$.
Speed of the boat in still water $= \frac{1}{2} (8 + 5) = \frac{13}{2} = 6.5 \text{ km/h}$ or $6 \frac{1}{2} \text{ km/h}$.

(D) Given: Speed of the boat in still water $(S_B)$ = $13 \text{ km/hr}$.
Speed of the stream $(S_C)$ = $4 \text{ km/hr}$.
When the boat travels in the opposite direction (upstream),the effective speed is the difference between the boat's speed and the stream's speed.
Upstream speed = $S_B - S_C = 13 - 4 = 9 \text{ km/hr}$.
Distance to be covered = $63 \text{ km}$.
Time taken = $\frac{\text{Distance}}{\text{Upstream speed}} = \frac{63}{9} = 7 \text{ hours}$.

(B) Speed of boat in still water,$S_B = 6\, km/hr$.
Speed of stream,$S_S = 1.5\, km/hr$.
Downstream speed,$D_S = S_B + S_S = 6 + 1.5 = 7.5\, km/hr$.
Upstream speed,$U_S = S_B - S_S = 6 - 1.5 = 4.5\, km/hr$.
Distance to the place,$D = 22.5\, km$.
Time taken to go downstream,$T_1 = \frac{D}{D_S} = \frac{22.5}{7.5} = 3\, hours$.
Time taken to return upstream,$T_2 = \frac{D}{U_S} = \frac{22.5}{4.5} = 5\, hours$.
Total time taken = $T_1 + T_2 = 3 + 5 = 8\, hours$.

(D) 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 for the downstream journey is $t_1 = \frac{d}{u+v} = \frac{22.5}{7.5} = 3\, hours$.
Time taken for the upstream journey is $t_2 = \frac{d}{u-v} = \frac{22.5}{4.5} = 5\, hours$.
The total time taken is $t_1 + t_2 = 3 + 5 = 8\, hours$.

(D) Let the speed of the boat in still water be $x \text{ km/h}$ and the speed of the current be $y \text{ km/h}$.
Downstream speed $D_S = (x + y) \text{ km/h}$.
Upstream speed $U_S = (x - y) \text{ km/h}$.
Let the distance be $d$ and the time taken downstream be $t$. Then the time taken upstream is $2t$.
Since $\text{Distance} = \text{Speed} \times \text{Time}$,we have:
$d = (x + y) \times t$ and $d = (x - y) \times 2t$.
Equating the two expressions for distance:
$(x + y) \times t = (x - y) \times 2t$
$x + y = 2x - 2y$
$3y = x$
Therefore,the ratio $\frac{x}{y} = \frac{3}{1}$ or $3:1$.

(D) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the stream be $y \, km/hr$.
Upstream speed $= (x - y) \, km/hr$ and downstream speed $= (x + y) \, km/hr$.
According to the problem:
$1) \frac{24}{x-y} + \frac{28}{x+y} = 6$
$2) \frac{30}{x-y} + \frac{21}{x+y} = 6.5 = \frac{13}{2}$
Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.
$24u + 28v = 6 \implies 12u + 14v = 3$ (Equation $i$)
$30u + 21v = \frac{13}{2} \implies 60u + 42v = 13$ (Equation $ii$)
Multiply equation $i$ by $3$: $36u + 42v = 9$ (Equation $iii$)
Subtract equation $iii$ from $ii$: $(60u - 36u) = 13 - 9 \implies 24u = 4 \implies u = \frac{1}{6}$.
Substitute $u = \frac{1}{6}$ in equation $i$: $12(\frac{1}{6}) + 14v = 3 \implies 2 + 14v = 3 \implies 14v = 1 \implies v = \frac{1}{14}$.
So,$x - y = 6$ and $x + y = 14$.
Adding the two equations: $2x = 20 \implies x = 10 \, km/hr$.

(B) Let the speed of the boat in still water be $u$ and the speed of the stream be $v$.
The speed of the boat downstream is $u + v = \frac{12 \text{ km}}{2 \text{ h}} = 6 \text{ km/h}$ (from statement $I$).
The speed of the boat upstream is $u - v = \frac{12 \text{ km}}{4 \text{ h}} = 3 \text{ km/h}$ (from statement $II$).
From statement $III$,$v = \frac{u}{3}$.
To find $u$,we can use statement $I$ and $II$:
$u + v = 6$ and $u - v = 3$.
Adding these equations: $2u = 9 \implies u = 4.5 \text{ km/h}$.
Alternatively,we can use statement $I$ and $III$:
$u + v = 6$ and $v = \frac{u}{3}$.
Substituting $v$ in the first equation: $u + \frac{u}{3} = 6 \implies \frac{4u}{3} = 6 \implies u = \frac{18}{4} = 4.5 \text{ km/h}$.
Thus,statement $I$ and either $II$ or $III$ are sufficient to answer the question.

(B) From statement $I$,the length of the train is $240 \ m$,but the time taken is not given.
Hence,statement $I$ alone is not sufficient.
From statement $II$,the time taken by the train to cross a pole is $24 \ s$,but the length of the train is not given.
Hence,statement $II$ alone is not sufficient.
From statement $III$,the time taken by the train to cross a platform is $48 \ s$,but the lengths of the train and the platform are not given.
Hence,statement $III$ alone is not sufficient.
Now,combining statements $I$ and $II$,we have the distance (length of the train) $= 240 \ m$ and the time taken to cross a pole $= 24 \ s$.
Speed of the train $= \frac{\text{Distance}}{\text{Time}} = \frac{240 \ m}{24 \ s} = 10 \ m/s$.
Therefore,both $I$ and $II$ together are sufficient to answer the question.

(C) Let the speed of the boat in still water be $x\, km/h$ and the speed of the current be $y\, km/h$.
According to the problem:
$\frac{12}{x-y} + \frac{18}{x+y} = 3$ $...(1)$
$\frac{36}{x-y} + \frac{24}{x+y} = \frac{13}{2}$ $...(2)$
Multiply equation $(1)$ by $3$:
$\frac{36}{x-y} + \frac{54}{x+y} = 9$ $...(3)$
Subtract equation $(2)$ from equation $(3)$:
$(\frac{36}{x-y} - \frac{36}{x-y}) + (\frac{54}{x+y} - \frac{24}{x+y}) = 9 - \frac{13}{2}$
$\frac{30}{x+y} = \frac{5}{2}$
$x+y = 12$ $...(4)$
Substitute $x+y = 12$ into equation $(1)$:
$\frac{12}{x-y} + \frac{18}{12} = 3$
$\frac{12}{x-y} + 1.5 = 3$
$\frac{12}{x-y} = 1.5$
$x-y = \frac{12}{1.5} = 8$ $...(5)$
To find the speed of the current $(y)$,subtract equation $(5)$ from equation $(4)$:
$(x+y) - (x-y) = 12 - 8$
$2y = 4$
$y = 2\, km/h$.

(B) Let the required distance be $x\, km$.
The speed of the man in still water is $6\, km/h$ and the speed of the current is $2\, km/h$.
Speed in downstream = $(6 + 2) = 8\, km/h$.
Speed in upstream = $(6 - 2) = 4\, km/h$.
Time taken in upstream = $\frac{x}{4}$ hours.
Time taken in downstream = $\frac{x}{8}$ hours.
According to the problem,the difference in time is $3\, hours$:
$\frac{x}{4} - \frac{x}{8} = 3$
$\frac{2x - x}{8} = 3$
$\frac{x}{8} = 3$
$x = 24\, km$.
Therefore,the distance is $24\, km$.

(A) Let the rate of the stream be $x \, km/h$.
Speed downstream = $(10 + x) \, km/h$.
Speed upstream = $(10 - x) \, km/h$.
According to the problem,the total time taken for the round trip is $20 \, hours$:
$\frac{91}{10 + x} + \frac{91}{10 - x} = 20$
Factor out $91$:
$91 \left( \frac{10 - x + 10 + x}{(10 + x)(10 - x)} \right) = 20$
Simplify the numerator:
$91 \left( \frac{20}{100 - x^2} \right) = 20$
Divide both sides by $20$:
$\frac{91}{100 - x^2} = 1$
$100 - x^2 = 91$
$x^2 = 100 - 91 = 9$
$x = \sqrt{9} = 3 \, km/h$.
Thus,the rate of flow of the river is $3 \, km/h$.

(C) Let the speed of the motorboat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
According to the problem:
$\frac{25}{x-y} + \frac{39}{x+y} = 8$ $...(1)$
$\frac{35}{x-y} + \frac{52}{x+y} = 11$ $...(2)$
Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.
The equations become:
$25u + 39v = 8$ $...(3)$
$35u + 52v = 11$ $...(4)$
Multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:
$100u + 156v = 32$ $...(5)$
$105u + 156v = 33$ $...(6)$
Subtracting $(5)$ from $(6)$:
$5u = 1 \Rightarrow u = \frac{1}{5}$.
Since $u = \frac{1}{x-y}$,we have $x-y = 5$.
Substitute $u = \frac{1}{5}$ into equation $(3)$:
$25(\frac{1}{5}) + 39v = 8$
$5 + 39v = 8 \Rightarrow 39v = 3 \Rightarrow v = \frac{1}{13}$.
Since $v = \frac{1}{x+y}$,we have $x+y = 13$.
Solving $x-y = 5$ and $x+y = 13$:
Adding the two equations: $2x = 18 \Rightarrow x = 9$.
Subtracting the two equations: $2y = 8 \Rightarrow y = 4$.
Thus,the speed of the stream is $4\, km/h$.

(D) Let the speed of the man in still water be $u$ and the speed of the current be $v$.
Speed against the current $(u - v) = \frac{\text{Distance}}{\text{Time}} = \frac{3/4 \text{ km}}{15/60 \text{ h}} = \frac{3}{4} \times 4 = 3 \text{ km/h}$.
Speed with the current $(u + v) = \frac{\text{Distance}}{\text{Time}} = \frac{3/4 \text{ km}}{10/60 \text{ h}} = \frac{3}{4} \times 6 = 4.5 \text{ km/h}$.
Adding the two equations: $(u - v) + (u + v) = 3 + 4.5 \implies 2u = 7.5 \implies u = 3.75 \text{ km/h}$.
Subtracting the first from the second: $(u + v) - (u - v) = 4.5 - 3 \implies 2v = 1.5 \implies v = 0.75 \text{ km/h}$.
The ratio of his speed to that of the current is $u : v = 3.75 : 0.75 = 5 : 1$.

(C) 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 upstream is $(x-y)\, km/h$ and the speed downstream is $(x+y)\, km/h$.
According to the given conditions:
$\frac{25}{x-y} + \frac{39}{x+y} = 8$ $...(i)$
$\frac{35}{x-y} + \frac{52}{x+y} = 11$ $...(ii)$
Let $\frac{1}{x-y} = A$ and $\frac{1}{x+y} = B$.
Then,$25A + 39B = 8$ $...(iii)$
$35A + 52B = 11$ $...(iv)$
To solve for $A$ and $B$,multiply equation $(iii)$ by $4$ and equation $(iv)$ by $3$:
$100A + 156B = 32$ $...(v)$
$105A + 156B = 33$ $...(vi)$
Subtracting $(v)$ from $(vi)$:
$5A = 1 \Rightarrow A = \frac{1}{5}$.
Substituting $A = \frac{1}{5}$ in equation $(iii)$:
$25(\frac{1}{5}) + 39B = 8$
$5 + 39B = 8 \Rightarrow 39B = 3 \Rightarrow B = \frac{1}{13}$.
Now,$x-y = 5$ and $x+y = 13$.
Adding these two equations: $2x = 18 \Rightarrow x = 9$.
Subtracting the first from the second: $2y = 8 \Rightarrow y = 4$.
Thus,the speed of the stream is $4\, km/h$.

(B) Let the total distance between the two points be $d\, km$.
Speed of the boat in still water = $8\, km/h$.
Speed of the stream = $2\, km/h$.
Downstream speed = $8 + 2 = 10\, km/h$.
Usual time taken to cover distance $d$ downstream = $\frac{d}{10}$.
In the actual scenario,the man travels half the distance $(d/2)$ at $10\, km/h$ and the remaining half $(d/2)$ at the speed of the stream $(2\, km/h)$ because the motor failed.
Actual time taken = $\frac{d/2}{10} + \frac{d/2}{2} = \frac{d}{20} + \frac{d}{4} = \frac{d + 5d}{20} = \frac{6d}{20} = \frac{3d}{10}$.
Given that the actual time is $6\, hours$ more than the usual time:
$\frac{3d}{10} - \frac{d}{10} = 6$
$\frac{2d}{10} = 6$
$\frac{d}{5} = 6$
$d = 30\, km$.

(D) Let $d$ be the distance between $A$ and $B$,and $B$ and $C$. Let $x$ be the speed of the swimmer in still water and $y$ be the speed of the stream.
For the round trip between $A$ and $B$: $\frac{d}{x+y} + \frac{d}{x-y} = 10 \dots (1)$
For the downstream trip from $A$ to $C$ (total distance $2d$): $\frac{2d}{x+y} = 8 \Rightarrow d = 4(x+y) \dots (2)$
Substitute $d$ from $(2)$ into $(1)$:
$\frac{4(x+y)}{x+y} + \frac{4(x+y)}{x-y} = 10$
$4 + \frac{4(x+y)}{x-y} = 10$
$\frac{4(x+y)}{x-y} = 6 \Rightarrow \frac{x+y}{x-y} = \frac{6}{4} = \frac{3}{2}$
$2(x+y) = 3(x-y)$
$2x + 2y = 3x - 3y$
$x = 5y$
Therefore,the ratio $\frac{x}{y} = \frac{5}{1}$.

(A) Let the speed of the boat in still water be $x$ and the speed of the current be $y$.
The speed downstream is $(x + y)$ and the speed upstream is $(x - y)$.
According to the problem,$(x + y) = \frac{16}{9}(x - y)$.
Multiplying both sides by $9$,we get $9(x + y) = 16(x - y)$,which simplifies to $9x + 9y = 16x - 16y$.
Rearranging the terms,we get $16x - 9x = 9y + 16y$,which gives $7x = 25y$.
Therefore,the speed of the current $y$ in terms of $x$ is $y = \frac{7x}{25}$.
The required percentage is $\frac{y}{x} \times 100 = \frac{7x/25}{x} \times 100 = \frac{7}{25} \times 100 = 28 \%$.

(B) Let the speed of the man in still water be $u = 20\, km/h$ and the speed of the river flow be $v = 5\, km/h$.
When swimming along the direction of the flow (downstream),the effective speed is $u + v = 20 + 5 = 25\, km/h$.
The time taken to travel from $A$ to $B$ is $t_1 = \frac{Distance}{Speed} = \frac{30}{25} = 1.2\, hours$.
When swimming against the direction of the flow (upstream),the effective speed is $u - v = 20 - 5 = 15\, km/h$.
The time taken to travel from $B$ back to $A$ is $t_2 = \frac{30}{15} = 2\, hours$.
The total time taken is $T = t_1 + t_2 = 1.2 + 2 = 3.2\, hours$.
Converting $0.2\, hours$ to minutes: $0.2 \times 60 = 12\, minutes$.
Thus,the total time is $3\, h\, 12\, min$.

(A) Let the speed of the boat in still water be $x$ $kmph$ and the distance be $d$ $km.$
Given,speed of the stream = $2$ $kmph.$
Downstream speed = $(x + 2)$ $kmph$ and upstream speed = $(x - 2)$ $kmph.$
Time taken downstream = $45$ $minutes = \frac{45}{60} = \frac{3}{4}$ $hours.$
Time taken upstream = $1$ $hour$ $15$ $minutes = \frac{75}{60} = \frac{5}{4}$ $hours.$
Since distance is constant,$d = \text{speed} \times \text{time}.$
$d = (x + 2) \times \frac{3}{4} = (x - 2) \times \frac{5}{4}.$
Multiplying both sides by $4$,we get $3(x + 2) = 5(x - 2).$
$3x + 6 = 5x - 10.$
$2x = 16 \Rightarrow x = 8$ $kmph.$
Thus,the speed of the boat in still water is $8$ $kmph.$

(A) Let the speed of the water scooter in still water be $x \, km/hr$. Given the speed of the stream is $y = 2 \, km/hr$.
The speed upstream is $(x - 2) \, km/hr$ and the speed downstream is $(x + 2) \, km/hr$.
The total time taken for the round trip is $55 \, minutes$,which is $\frac{55}{60} = \frac{11}{12} \, hours$.
The equation for the total time is: $\frac{10}{x - 2} + \frac{10}{x + 2} = \frac{11}{12}$.
Simplifying the equation: $10 \left( \frac{x + 2 + x - 2}{(x - 2)(x + 2)} \right) = \frac{11}{12} \Rightarrow \frac{20x}{x^2 - 4} = \frac{11}{12}$.
Cross-multiplying gives: $240x = 11(x^2 - 4) \Rightarrow 11x^2 - 240x - 44 = 0$.
Factoring the quadratic equation: $11x^2 - 242x + 2x - 44 = 0 \Rightarrow 11x(x - 22) + 2(x - 22) = 0$.
This gives $(11x + 2)(x - 22) = 0$. Since speed cannot be negative,$x = 22 \, km/hr$.

(A) Let the speed of the man in still water be $x\, km/h$ and the speed of the current 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:
$\frac{55}{x+y} + \frac{40}{x-y} = 13 \quad ...(1)$
$\frac{44}{x+y} + \frac{30}{x-y} = 10 \quad ...(2)$
Let $v = x+y$ and $u = x-y$. The equations become:
$\frac{55}{v} + \frac{40}{u} = 13 \quad ...(3)$
$\frac{44}{v} + \frac{30}{u} = 10 \quad ...(4)$
Multiply equation $(3)$ by $3$ and equation $(4)$ by $4$:
$\frac{165}{v} + \frac{120}{u} = 39 \quad ...(5)$
$\frac{176}{v} + \frac{120}{u} = 40 \quad ...(6)$
Subtracting $(5)$ from $(6)$:
$\frac{11}{v} = 1 \Rightarrow v = 11$.
Substitute $v=11$ into $(4)$:
$\frac{44}{11} + \frac{30}{u} = 10 \Rightarrow 4 + \frac{30}{u} = 10 \Rightarrow \frac{30}{u} = 6 \Rightarrow u = 5$.
Now,$x+y = 11$ and $x-y = 5$.
Adding these equations: $2x = 16 \Rightarrow x = 8$.
Subtracting these equations: $2y = 6 \Rightarrow y = 3$.
Thus,the speed of the man in still water is $8\, km/h$ and the speed of the current is $3\, km/h$.

(C) Let the distance to the destination be $d \text{ km}$.
Speed of the boat in still water $= 10 \text{ km/h}$.
Speed of the stream $= 4 \text{ km/h}$.
Speed downstream $= 10 + 4 = 14 \text{ km/h}$.
Speed upstream $= 10 - 4 = 6 \text{ km/h}$.
The total time taken is the sum of time taken downstream and time taken upstream: $\frac{d}{14} + \frac{d}{6} = 5$.
Taking the least common multiple of $14$ and $6$,which is $42$:
$\frac{3d + 7d}{42} = 5$
$\frac{10d}{42} = 5$
$10d = 5 \times 42$
$10d = 210$
$d = 21 \text{ km}$.

(A) Let $u$ be the upstream speed and $v$ be the downstream speed in $\text{km/h}$.
From the given information:
$\frac{24}{u} + \frac{36}{v} = 6 \quad \dots (i)$
$\frac{36}{u} + \frac{24}{v} = \frac{13}{2} \quad \dots (ii)$
Multiply equation $(i)$ by $3$ and equation $(ii)$ by $2$:
$\frac{72}{u} + \frac{108}{v} = 18 \quad \dots (iii)$
$\frac{72}{u} + \frac{48}{v} = 13 \quad \dots (iv)$
Subtracting $(iv)$ from $(iii)$:
$\frac{60}{v} = 5 \implies v = 12 \text{ km/h}$.
Substitute $v = 12$ into $(i)$:
$\frac{24}{u} + \frac{36}{12} = 6 \implies \frac{24}{u} + 3 = 6 \implies \frac{24}{u} = 3 \implies u = 8 \text{ km/h}$.
Let $x$ be the speed of the boat in still water and $y$ be the velocity of the current.
$x - y = 8$ and $x + y = 12$.
Adding the two equations: $2x = 20 \implies x = 10 \text{ km/h}$.
Subtracting the two equations: $2y = 4 \implies y = 2 \text{ km/h}$.
Thus,the velocity of the current is $2 \text{ km/h}$.

(C) Let the speed of the boat in still water be $x = 10\, km/hr$ and the speed of the stream be $y\, km/hr$.
The speed downstream is $(x + y)\, km/hr$ and the speed upstream is $(x - y)\, km/hr$.
According to the problem,the time taken to travel $26\, km$ downstream is equal to the time taken to travel $14\, km$ upstream:
$\frac{26}{x + y} = \frac{14}{x - y}$
Substitute $x = 10$:
$\frac{26}{10 + y} = \frac{14}{10 - y}$
Simplify the equation:
$26(10 - y) = 14(10 + y)$
$260 - 26y = 140 + 14y$
$260 - 140 = 14y + 26y$
$120 = 40y$
$y = \frac{120}{40} = 3\, km/hr$.
Thus,the speed of the stream is $3\, km/hr$.

(A) Let the speed of the motorboat in still water be $x \ km/h$.
Given the speed of the stream $y = 2 \ km/h$.
The speed upstream is $(x - 2) \ km/h$ and the speed downstream is $(x + 2) \ km/h$.
The total time taken is $33 \ minutes = \frac{33}{60} \ hours = \frac{11}{20} \ hours$.
According to the problem: $\frac{6}{x-2} + \frac{6}{x+2} = \frac{11}{20}$.
$6 \left( \frac{x+2+x-2}{x^2-4} \right) = \frac{11}{20} \Rightarrow 6 \left( \frac{2x}{x^2-4} \right) = \frac{11}{20}$.
$12x \times 20 = 11(x^2 - 4) \Rightarrow 240x = 11x^2 - 44$.
$11x^2 - 240x - 44 = 0$.
$11x^2 - 242x + 2x - 44 = 0 \Rightarrow 11x(x - 22) + 2(x - 22) = 0$.
$(x - 22)(11x + 2) = 0$.
Since speed cannot be negative,$x = 22 \ km/h$.

(C) Let the speed of the man in still water be $x = 9 \frac{1}{3} = \frac{28}{3} \text{ km/h}$ and the speed of the current be $y \text{ km/h}$.
The speed upstream is $(x - y) \text{ km/h}$ and the speed downstream is $(x + y) \text{ km/h}$.
Let the distance be $d$. According to the problem,the time taken to row upstream is thrice the time taken to row downstream:
$\frac{d}{x - y} = 3 \times \frac{d}{x + y}$
Canceling $d$ from both sides:
$\frac{1}{x - y} = \frac{3}{x + y}$
$x + y = 3(x - y)$
$x + y = 3x - 3y$
$4y = 2x$
$y = \frac{x}{2}$
Substituting $x = \frac{28}{3}$:
$y = \frac{28/3}{2} = \frac{14}{3} = 4 \frac{2}{3} \text{ km/h}$.

(B) Let the speed of the man in still water be $x = 18 \ km/h$ and the speed of the stream be $y \ km/h$.
The speed downstream is $(18 + y) \ km/h$ and the speed upstream is $(18 - y) \ km/h$.
Let the distance be $d$. The time taken to row downstream is $t_1 = \frac{d}{18 + y}$ and the time taken to row upstream is $t_2 = \frac{d}{18 - y}$.
According to the problem,the time taken to row upstream is thrice the time taken to row downstream:
$t_2 = 3 \times t_1$
$\frac{d}{18 - y} = 3 \times \frac{d}{18 + y}$
Canceling $d$ from both sides:
$\frac{1}{18 - y} = \frac{3}{18 + y}$
$18 + y = 3(18 - y)$
$18 + y = 54 - 3y$
$4y = 36$
$y = 9 \ km/h$.
Thus,the rate of the stream is $9 \ km/h$.

(D) Let the speed of Rohit in still water be $x$ $km/hr$ and the speed of the stream be $y = 5$ $km/hr$.
Let the distance be $d$ $km$.
Downstream speed = $(x + y)$ $km/hr$.
Upstream speed = $(x - y)$ $km/hr$.
Given that time taken downstream is $8$ $hours$: $d = 8(x + 5) = 8x + 40$ $...(i)$.
Given that time taken upstream is $12$ $hours$: $d = 12(x - 5) = 12x - 60$ $...(ii)$.
Equating $(i)$ and $(ii)$:
$8x + 40 = 12x - 60$
$4x = 100$
$x = 25$ $km/hr$.
Thus,the speed of Rohit in still water is $25$ $km/hr$.

(A) Let the speed of the boat in still water be $x \; km/hr$ and the speed of the river current be $y \; km/hr$.
Downstream speed $= x + y$.
Time taken downstream $= 3 \; hours \; 45 \; minutes = 3 + \frac{45}{60} = 3 + \frac{3}{4} = \frac{15}{4} \; hours$.
Distance $= 15 \; km$.
Using $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we get $x + y = \frac{15}{15/4} = 4 \; km/hr \; \dots (i)$.
Upstream speed $= x - y$.
Time taken upstream $= 2 \; hours \; 30 \; minutes = 2 + \frac{30}{60} = 2.5 = \frac{5}{2} \; hours$.
Distance $= 5 \; km$.
Using $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$,we get $x - y = \frac{5}{5/2} = 2 \; km/hr \; \dots (ii)$.
Subtracting equation $(ii)$ from equation $(i)$:
$(x + y) - (x - y) = 4 - 2$
$2y = 2$
$y = 1 \; km/hr$.
Thus,the speed of the river current is $1 \; km/hr$.

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

(D) Let the speed of the boat in still water be $x = 9 \text{ km/h}$ and the speed of the stream be $y = 1.5 \text{ km/h}$.
The speed downstream is $(x + y) = 9 + 1.5 = 10.5 \text{ km/h}$.
The speed upstream is $(x - y) = 9 - 1.5 = 7.5 \text{ km/h}$.
The time taken to travel $105 \text{ km}$ downstream is $T_1 = \frac{\text{Distance}}{\text{Speed downstream}} = \frac{105}{10.5} = 10 \text{ hours}$.
The time taken to travel $105 \text{ km}$ upstream is $T_2 = \frac{\text{Distance}}{\text{Speed upstream}} = \frac{105}{7.5} = 14 \text{ hours}$.
The total time taken is $T = T_1 + T_2 = 10 + 14 = 24 \text{ hours}$.

$(A)$ Given: Speed of boat in still water $(u) = 9\, km/h$, Speed of stream $(v) = 1.5\, km/h$.
Downstream speed $= u + v = 9 + 1.5 = 10.5\, km/h$.
Upstream speed $= u - v = 9 - 1.5 = 7.5\, km/h$.
Distance $(d) = 105\, km$.
Time taken to travel downstream $= \frac{d}{u+v} = \frac{105}{10.5} = 10\, \text{hours}$.
Time taken to travel upstream $= \frac{d}{u-v} = \frac{105}{7.5} = 14\, \text{hours}$.
Total time taken $= 10 + 14 = 24\, \text{hours}$.