Question

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go
40 km upstream and 55km downstream. Find the speed of the boat in still water and the
speed of the stream.

Answer 1

Given: A Boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream.

Need to find: The speed of the Boat in Still water & the speed of the Stream in Still water?

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❍ Let's say, that the speed of the Boat in Still water be x km/hr and speed of Stream in Still water be y km/hr respectively.

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$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\; :}}$⠀⠀

• The Boat goes 30 km Upstream and 44 km Downstream in 10 hours.
• Downstream = (x + y)
• Upstream = (x – y)

$:\implies\sf\dfrac{30}{x - y} + \dfrac{44}{x + y} = 10 \: \: \: \: \: \: \qquad\bigg\lgroup\sf eq^n \;(1)\bigg\rgroup$

Similarly,

• The Boat goes 40 km Upstream and 55 km Downstream in 13 hours.

$:\implies\sf\dfrac{40}{x - y} + \dfrac{55}{x + y} = 13 \: \: \: \: \: \: \qquad\bigg\lgroup\sf eq^n \;(2)\bigg\rgroup$

$\sf{Let}\begin{cases}\sf{\:\; \: \dfrac{1}{x - y} = \bf{M}}\\ \\ \sf{\;\;\;\dfrac{1}{x + y} = \bf{N}}\end{cases}$

$:\implies\sf30M + 44N = 10 \: \: \: \: \: \: \qquad\bigg\lgroup\sf eq^n \;(3)\bigg\rgroup$

Also,

$:\implies\sf40M + 55N = 13\: \: \: \: \: \: \qquad\bigg\lgroup\sf eq^n \;(4)\bigg\rgroup$

S U B S T I T U T I O N⠀M E T H O D :

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From eqⁿ ( 3 ) :

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$\dashrightarrow\sf 30M + 44N = 10\\\\\\\dashrightarrow\sf 30M = 10 - 44N\\\\\\\dashrightarrow\sf M = \dfrac{10 - 44}{30}$

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» Substituting 'M' in eqⁿ ( 4 ) :

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$\dashrightarrow\sf 40M + 55N = 13$

$\dashrightarrow\sf 40\Bigg\{\dfrac{10 - 44N}{30}\Bigg\} + 55N = 13$

$\dashrightarrow\sf 4\Bigg\{\dfrac{10 - 44N}{3}\Bigg\}+ 55N = 13$

$\dashrightarrow\sf 3 \times 4\Bigg\{ \dfrac{10 - 44N}{3} \Bigg\}+ 55N = 13 \times 3 \: \: \qquad\bigg\lgroup\sf Multiplying\;Both\; Sides\;by\;3\;\bigg\rgroup$

$\dashrightarrow\sf 40 - 176N - 165N = 39$

$\dashrightarrow\sf -11N = -1$

$\dashrightarrow\sf N = \dfrac{1}{11}$

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» Substituting value of 'N' in eqⁿ ( 3 ) :

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$\dashrightarrow\sf 30M + 44N = 10$

$\dashrightarrow\sf 30M + 44\times\Bigg\{\dfrac{1}{11}\Bigg\}=10$

$\dashrightarrow\sf 30M + 4 = 10$

$\dashrightarrow\sf 30M = 10 - 4$

$\dashrightarrow\sf 30M = 6$

$\dashrightarrow\sf M = \cancel\dfrac{6}{30}$

$\dashrightarrow\sf M = \dfrac{1}{5}$

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✇ Now, we know the values of M and N. Also, we've assumed 1/x - y = M and 1/x + y = N. Therefore, we'll substitute values of M and N in these equations.

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$\dashrightarrow\sf M = \dfrac{1}{x - y}$

$\dashrightarrow\sf \dfrac{1}{5} = \dfrac{1}{x - y}$

$\dashrightarrow\sf 5 = x - y$

$\dashrightarrow\sf x - y = 5\: \: \: \: \: \: \qquad\bigg\lgroup\sf eq^n \;(5)\bigg\rgroup$

Similarly,

$\dashrightarrow\sf N = \dfrac{1}{x + y}$

$\dashrightarrow\sf \dfrac{1}{11} = \dfrac{1}{x + y}$

$\dashrightarrow\sf11 = x + y$

$\dashrightarrow\sf x + y =11\: \: \: \: \: \: \qquad\bigg\lgroup\sf eq^n \;(6)\bigg\rgroup$

» Now, from eqⁿ ( 5 ) & ( 6 ) : —

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$\dashrightarrow\sf x - \cancel{~y}+x + \cancel{~y}= 11 + 5$

$\dashrightarrow\sf 2x = 16$

$\dashrightarrow\sf x = \cancel\dfrac{16}{2}$

$\dashrightarrow{\pmb{\sf{x = 8}}}$

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» Similarly, finding the value of y by substituting the value of x in eqⁿ ( 6 ) :

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$\dashrightarrow\sf x + y = 11$

$\dashrightarrow\sf 8 + y = 11$

$\dashrightarrow\sf y = 11 - 8$

$\dashrightarrow{\pmb{\sf{y = 3}}}$

∴ Therefore, the speed of the Boat in Still water is 8km/hr and speed of Stream is 3km/hr respectively.

Answer 2

Given :-

A  boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go  40 km upstream and 55km downstream

To Find :-

Speed of  the boat in still water and the  speed of the stream

Solution :-

Here, first of all we have to assume the speed of boat and stream as variables b and s. Now

In Case 1

Boat goes 30 km upstream and downstream 44 km

$\sf \implies\dfrac{30}{b-s} + \dfrac{44}{b + s} = 10$

In Case 2

Boat goes 40 km upstream and 55 km downstream

$\sf \implies \dfrac{40}{b-s} + \dfrac{55}{b+s} = 13$

Now

Let us again assume that

$\sf \dfrac{1}{b-s} = u \; and\; \dfrac{1}{b+s}=v$

By putting the value

$\sf \implies \dfrac{30}{u} + \dfrac{44}{v} = 10$

$\sf \implies 30u + 44v = 10(..1)$

and

$\sf \implies \dfrac{40}{u} + \dfrac{55}{v} = 13$

$\sf \implies 40u + 55v = 13 (..2)$

Multiply 1 with 4

$\sf \implies 4(30u + 44v) = 4(10)$

$\sf \implies 120u + 176v = 40(..3)$

Multiply 2 with 3

$\sf \implies 3(40u + 55v) = 3(13)$

$\sf \implies 120u + 165v = 39(..4)$

Now

On subtracting we get

$\sf \implies 120u + 165v - 120u - 176v = 40 - 39$

$\sf \implies 165v - 176v = -1$

$\sf \implies -11v = -1$

$\sf \implies v =\dfrac{-1}{-11}$

$\sf \implies v=\dfrac{1}{11}$

Putting the value in above

$\sf \dfrac{1}{b+s}=\dfrac{1}{11}$

$\sf 11 \times 1 = b+s$

$\sf 11=b+s (..5)$

Using 1

$\sf \implies 30u + 44v= 10$

$\sf \implies 30u = 10-44v$

$\sf \implies 30u = 10 - 44\times\dfrac{1}{11}$

$\sf \implies 30u = 10 - 4$

$\sf \implies 30u = 6$

$\sf \implies u =\dfrac{6}{30}$

$\sf \implies u = \dfrac{1}{5}$

Now

$\sf \implies \dfrac{1}{b-s}=\dfrac{1}{5}$

$\sf \implies 1(b-s) = 5$

$\sf \implies b-s=5 (..6)$

add 5 and 6

$\sf \implies b +s+b-s = 11+5$

$\sf \implies b+b=16$

$\sf \implies 2b = 16$

$\sf \implies b=\dfrac{16}{2}$

$\sf \implies b=8$

Using 6

$\sf \implies8 - s = 5$

$\sf \implies -s=5-8$

$\sf \implies -s=-3$

$\sf \implies s=3$

Hence

Speed of the boat in the still water is 8 km/h and speed of stream in downstream is 3 km/h