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This is the second of our many quizzes on Distance, rate, and time. Click on this to access more such competitive quizzes on Number System.

Distance, rate, and time Quiz 2

mathematics

 A train travels $100\;km$ at normal speed and then reduces its speed to $\dfrac{2}{3}rd$. It reached the destination $30\;\min $ late. If it had reduced its speed after travelling $120\;km$ it would have reached the destination $25\;\min $ late. The normal speed of the train is

 Form an equation using the condition for the difference of $20\;km$. 

  Let the normal speed of the train be $y\;kmph$. $ \\ $ Then, time taken to travel $\left( {120 - 100} \right)km = 20km$ is $\dfrac{{20}}{y}\;hr$ $ \\ $ At $\dfrac{2}{3}rd$ speed, the same distance is covered in $\dfrac{{20}}{{\dfrac{2}{3}y}}\;hr = \dfrac{{30}}{y}\;hr$. $ \\ $ Given, $\dfrac{{30}}{y} - \dfrac{{20}}{y} = \dfrac{5}{{60}}$ $ \\ $ Thus, $  \dfrac{{10}}{y} = \dfrac{5}{{60}}   \\   \Rightarrow y = 120\;kmph   \\  $

 A man walks to his office at $4\;kmph$ and reaches his office $10\;\min $ late. If his speed is $5\;kmph$ he reaches $5\;\min $ early. What should be his speed to reach on time?

 Use the given conditions to find distance and then solve.

  Let the distance be $d\;km$, and on-time journey time be $t\;hr$. $ \\ $ Then, $ \\ $ $\dfrac{d}{4} = t + \dfrac{{10}}{{60}}$ and $\dfrac{d}{5} = t - \dfrac{5}{{60}}$ $ \\ $ On solving the equations we get, $ \\ $ $d = 5\;km$ and $t = 1\;hr\;5\;\min $ $ \\ $ Thus, required speed is $\dfrac{5}{{\left( {\dfrac{{65}}{{60}}} \right)}} = \dfrac{{60}}{{13}} = 4.6\;kmph$

 The distance travelled by a person is given as a function, $d\left( t \right) = \left\{ {\begin{array}{*{20}{c}}  {36t\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \leqslant 2\;hrs} \\   {48\left( {t - 2} \right) + 72\;\;\;\;\;t > 2\;hrs} \end{array}} \right\}$. If the total distance travelled by the person is $200\;km$, the average speed during the journey is

 Use the formula $Time = \dfrac{{Distance}}{{Speed}}$.

$  d\left( t \right) = 200   \\   \Rightarrow 48\left( {t - 2} \right) + 72 = 200   \\   \Rightarrow t - 2 = \dfrac{{128}}{{48}} = \dfrac{8}{3}   \\   \Rightarrow t = \dfrac{{14}}{3}\;\;hrs   \\  $ Average speed can be calculated as, $ \\ \dfrac{{200}}{{\left( {\dfrac{{14}}{3}} \right)}} = \dfrac{{300}}{7}\;kmph$

 The average speed of a train while travelling is $108\;kmph$. But due to stoppages the average speed gets reduced to $84\;kmph$. If the trains passes through $4$ stoppages in an hour and stops for equal time at each stoppage, what is the time that it stops for at each stoppage?

 Find the total time that gets spent at stoppages by using the distance that couldn’t be covered.

 The train should cover $108\;km$ in an hour but it is covering only $84\;km$. $24\;km$ is not covered as the rest of the time is spent at stoppages. $ \\ $ Time spent at stoppages in an hour is $\dfrac{{\left( {108 - 84} \right)}}{{108}}\;hr = \dfrac{{24}}{{108}} \times 60\;\min  = \dfrac{{40}}{3}\;\min $ $ \\ $ Thus, time spent at each stoppage is $\dfrac{{40}}{{3 \times 4}}\;\min  = \dfrac{{10}}{3}\;\min  = 3\;\min 20\sec $

 An airplane when moving down-wind takes $4\;hrs$ to reach the destination. It takes $3\;hrs$ for the same journey in the up-wind. If the distance travelled is $1500\;km$, then the ratio of speed of the airplane to the speed of the wind is

 Use relative speed concept to evaluate the speed of the airplane and the wind. 

 Let the speed of the plane be $x\;kmph$ and speed of the wind be $w\;kmph$. $ \\ $ Then, $\dfrac{{1500}}{{x + w}} = 3$ and $\dfrac{{1500}}{{x - w}} = 2$. $ \\  \Rightarrow x + w = 500\;\;\;and\;\;\;x - w = 375   \\   \Rightarrow 2x = 875 \Rightarrow x = 437.5   \\  And,   \\   \Rightarrow w = 500 - 437.5 = 62.5\;kmph   \\  Ratio = \dfrac{{437.5}}{{62.5}} = 7   \\  $