Question

The distance b/w A & B is 211 km.
One boat starts moving from point A
towards B in downstream at 7:45am
After one hour another boat starts
from the point B towards A.At
12:45 pm both boats will meet.
It speed of boat first and second
is 26 km/hr and 18 km/hr de
in still water then find the speed
of Current​

Answer 1

$Distance \: between \:A \:and \:B = 211 \:km$

$Let \: speed \: of \:Current = x \:kmph$

$\underline { \blue { Case \: 1 : }}$

$\pink { ( Moving \: downstream )}$

$Speed \:of \:the \: first \:boat = 26 \: kmph$

$Relative \:speed \:of \:the \:boat (s_{1})\\ = (26+x) \:kmph$

$Time \:to \:meet \: second \:boat = 5\:hours$

$\blue { ( From \: 7.45 \:am \:to \: 12.45 \:pm = 5 hr )}$

$Distance \: travelled (d_{1}) = speed \times Time \\= (26+x)5 \:km \: ---(1)$

$\underline { \blue { Case \: 2: }}$

$\pink { ( Moving \: Upstream )}$

$Speed \:of \:the \: Second \:boat = 18 \: kmph$

$Relative \:speed \:of \:the \:boat (s_{2})\\ = (18-x) \:kmph$

$Time \:to \:meet \: second \:boat = 4\:hours$

$\blue { ( From \: 8.45 \:am \:to \: 12.45 \:pm = 4 hr )}$

$Distance \: travelled ( d_{2})= speed \times Time \\= (18-x)4 \:km \: ---(2)$

/* According to the problem given */

$d_{1} + d_{2} = 211 \:km$

$\implies (26+x)5 + (18-x)4 = 211$

$\implies 130 + 5x + 72 - 4x = 211$

$\implies x + 202 = 211$

$\implies x = 211 - 202$

$\implies x = 9 \:kmph$

Therefore.,

$\red { Speed \:of \:the \: Current } \green { = 9 \:kmph }$

•••♪