A man who swims 48 , m/min in still water,swi | vedclass

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Question

(B) $Quantity \, 1$: Let the speed of the current be $x \, m/min$.
The time taken against the current is $T_1 = \frac{200}{48-x}$ and the time taken

Vedclass · Accepted Answer

Quantity $I < $ Quantity $II$

Vedclass · Answer

(B) $Quantity \, 1$: Let the speed of the current be $x \, m/min$.
The time taken against the current is $T_1 = \frac{200}{48-x}$ and the time taken with the current is $T_2 = \frac{200}{48+x}$.
Given $T_1 - T_2 = 10$,so $\frac{200}{48-x} - \frac{200}{48+x} = 10$.
Dividing by $10$: $\frac{20}{48-x} - \frac{20}{48+x} = 1$.
$20(48+x) - 20(48-x) = (48-x)(48+x) \Rightarrow 960 + 20x - 960 + 20x = 2304 - x^2$.
$x^2 + 40x - 2304 = 0$.
Solving the quadratic equation: $(x+72)(x-32) = 0$. Since speed cannot be negative,$x = 32 \, m/min$.
$Quantity \, 2$: The distance covered in $3$ rounds of a circular path is $3 	imes (2 \pi r) = 3 	imes 2 	imes \frac{22}{7} 	imes 49 = 3 	imes 2 	imes 22 	imes 7 = 924 \, m$.
Speed = $\frac{	ext{Distance}}{	ext{Time}} = \frac{924}{14} = 66 \, m/min$.
Comparing the two,$32 < 66$,therefore Quantity $I < $ Quantity $II$.

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