A motorcyclist left 6 frac{2}{3} minutes late | vedclass

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(C) Let the usual speed of the motorcyclist be $x \, km/h$.The distance to be covered is $21 \, km$.The time taken at the usual speed is $T_1 = \frac{

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(C) Let the usual speed of the motorcyclist be $x \, km/h$.The distance to be covered is $21 \, km$.The time taken at the usual speed is $T_1 = \frac{21}{x} \, hours$.The time taken at the increased speed is $T_2 = \frac{21}{x+12} \, hours$.The delay is $6 \frac{2}{3} \, minutes = \frac{20}{3} \, minutes = \frac{20}{3 	imes 60} \, hours = \frac{1}{9} \, hours$.According to the problem,the difference in time is $\frac{1}{9} \, hours$:$\frac{21}{x} - \frac{21}{x+12} = \frac{1}{9}$$21 \left( \frac{x+12-x}{x(x+12)} ight) = \frac{1}{9}$$21 \left( \frac{12}{x^2+12x} ight) = \frac{1}{9}$$252 	imes 9 = x^2 + 12x$$x^2 + 12x - 2268 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-12 \pm \sqrt{144 - 4(1)(-2268)}}{2} = \frac{-12 \pm \sqrt{144 + 9072}}{2} = \frac{-12 \pm \sqrt{9216}}{2} = \frac{-12 \pm 96}{2}$Since speed cannot be negative,$x = \frac{84}{2} = 42 \, km/h$.

$A$ motorcyclist left $6 \frac{2}{3}$ minutes later than the scheduled time,but in order to reach his destination $21 \, km$ away on time,he had to increase his speed by $12 \, km/h$ from his usual speed. What is the usual speed (in $km/h$) of the motorcyclist?

Similar Questions

Two men $A$ and $B$ start from a place $P$ walking at $3 \, km/h$ and $3.5 \, km/h$ respectively. How many $km$ will they be apart at the end of $3 \, h$,if they walk in opposite directions (in $.5$)?

$A$ is faster than $B$. $A$ and $B$ each walk $24 \, km$. The sum of their speeds is $7 \, km/hr$ and the sum of the times taken by them is $14 \, hours$. Then $A$'s speed is equal to (in $km/hr$):

$A$ man walks at the rate of $5 \, km/hr$ for $6 \, hours$ and at $4 \, km/hr$ for $12 \, hours$. The average speed of the man (in $km/hr$) is:

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