Question

A person. Can row 8km upstream. And 24 km. Downstream. In 4 hrs he can row 12 km downstream. And 12 km upstream. In 4 hours. Find. The. Speed of the person in still water and also the speed the current

Answer 1

here is the answer. Hope this will help you

Answer 2

$Let \: speed\:of \:the \: current = x \: kmph$

$Speed \: of \: the \: person = y \:kmph$

Case 1 :

$\underline { \blue { Downstream :}}$

$Relative \:speed = ( y + x ) \: kmph$

$Distance \: travelled = 24 \:km$

$Time = t_{1} \: hr$

$t_{1} = \frac{24}{(y+x)} \: ---(1)$

$\underline { \blue { Upstream :}}$

$Relative \:speed = ( y - x ) \: kmph$

$Distance \: travelled = 8 \:km$

$Time = t_{2} \: hr$

$t_{2} = \frac{12}{(y-x)} \: ---(2)$

/* According to the problem given */

$t_{1} + t_{2} = 4 \:hr$

$\implies \frac{24}{y+x} + \frac{8}{y-x} = 4$

/* Divide each term by 4 , we get */

$\implies \frac{6}{y+x} + \frac{2}{y-x} = 1$

$Let \: a = \frac{1}{y+x} \: and \: b = \frac{1}{y-x}$

$\implies 6a + 2b = 1 \: ---(3)$

Case 2:

$\underline { \blue { Downstream :}}$

$Relative \:speed = ( y + x ) \: kmph$

$Distance \: travelled = 12 \:km$

$Time = t_{3} \: hr$

$t_{1} = \frac{12}{(y+x)} \: ---(4)$

$\underline { \blue { Upstream :}}$

$Relative \:speed = ( y - x ) \: kmph$

$Distance \: travelled = 12 \:km$

$Time = t_{4} \: hr$

$t_{4} = \frac{12}{(y-x)} \: ---(5)$

/* According to the problem given */

$t_{3} + t_{4} = 4 \:hr$

$\implies \frac{12}{y+x} + \frac{12}{y-x} = 4$

/* Divide each term by 4 , we get */

$\implies \frac{3}{y+x} + \frac{3}{y-x} = 1$

$Let \: a = \frac{1}{y+x} \: and \: b = \frac{1}{y-x}$

$\implies 3a + 3b = 1 \: ---(6)$

/* From Equations (3) and (6) , we get */

$6a + 2b = 3a + 3b$

$\implies 6a - 3a = 3b - 2b$

$\implies 3a = b\: --(7)$

/* put b = 3a in Equation (6), we get */

$\implies b + 3b = 1$

$\implies 4b = 1$

$\implies b = \frac{1}{4} \: --(8)$

$Put \: b = \frac{1}{4} \: in \: equation (7),\\ we \:get$

$\implies 3a = \frac{1}{4}$

$\implies a = \frac{1}{12}\: --- (9)$

$Now, a = \frac{1}{y+x} = \frac{1}{12}$

$\imples x + y = 12 \: --- (9)$

$b = \frac{1}{y-x} = \frac{1}{4}$

$\implies 4 = y-x$

$\implies y - x = 4 \: ---(10)$

/* Solving equations (9) and (10) ,we get */

$2y = 16$

$\implies y = 8$

/* put y = 8 in equation (9) , we get */

$x + 8 = 12$

$\implies x = 4$

Therefore.,

$\red { speed\:of \:the \: current(x)} \green {=4 \: kmph }$

$\red { speed\:of \:the \: person\: (y)} \green {=8 \: kmph }$

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