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Parin Sir > Quantitative Aptitude (Maths) > Time And Distance > Examples 21 to 26 of Time and Distance > Examples 21 to 26 of Time and Distance

Examples 21 to 26 of Time and Distance

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(21) A train running at 40 kmph takes 18 seconds to pass a platform. Next, it takes  12 seconds to pass a man walking at the rate of  4 kmph in the same direction. Find the length of  the train and that of  the platform.

Soln. Let, the speed of  the train = 40  kmph   &   the speed of  a man  =  v  =  4  kmph.

Let, the length of  the train  =  x  m  and  that of  the platform  =  z  m.

Time  taken  to  cross a walking a man ( in the same direction )  = \frac{x+y}{u-v}

∴   12   =  \frac{\mathrm{x\ }+0}{\left(40-4\right)\times\frac{5}{18}}

∴   x   =   120  m.

Time taken to pass a platform  = \frac{x+z}{u}

∴   18  =  \frac{120+z}{\mathrm{40}\times\ \frac{5}{\mathrm{18}}}

∴   120  +  z  =  200  m.

∴     z   =   80  m.

The length of  a train = 120 m and that of  a platform = 80 m.

(22) A man starts from Bombay to Delhi, another from Delhi to Bombay at the same time. After passing each other, they complete their journeys in  4 and  9 hours respectively. Find the speed of  the second man if the speed of  the first is 21 km/hr.

Soln. \frac{1^{\mathrm{st}}\mathrm{\ man's\ \ speed} } {2^{\mathrm{nd}}\mathrm{\ man's\ \ speed} }  =  \frac{\sqrt9}{\sqrt4} = \frac{3}{2}

∴ \frac{\mathrm{21}}{2^{\mathrm{nd}}\mathrm{\ man's\ \ speed} }  =  \frac{3}{2}        ∴  2nd  man’s speed  =  14 km/hr.

 

☼ Formula  ( Problems on Boat ) :  

■ If  the speed of  a boat in still water is ‘u’ km/hr

the speed of  the stream is ‘v’ km/hr,

the speed of  the downstream is ‘x’ km/hr

the speed of  the upstream is ‘y’ km/hr, then

\large \mapsto Speed  while traveling with the stream ( down stream )   =  x  =  ( u + v ) km/hr.

\large \mapsto  Speed while traveling against the stream ( up stream )  =  y  =  ( u – v ) km/hr.

\large \mapsto  Speed of  the boat in still water  =   u  =  \frac{x+y}{2}  km/hr

\large \mapsto  Speed of  the river   =  v  =  \frac{x-y}{2}

(22) A man can row upstream at 10 kmph and downstream at 14 kmph. Find man’s rate in still water and rate of  the current.

Soln. Here,  x  = 14 kmph, y  = 10 kmph.

Now,   u

=  \frac{x+y}{2}

=    12  kmph.

Now,  v

=  \frac{x-y}{2}

=    2  kmph.

Hence, the man’s rate in still water  =  12 kmph  &  rate of  the current  =  2  kmph.

(23) A man can row  10 kmph in still water. It takes him thrice as long to row up as to row down the river ( or it takes twice as long to row up again as to row down the river ). Find the rate of  stream.

Soln. Let man’s rate upstream  =  y  kmph  \large \Rightarrow  man’s rate downstream  =  x  =  3y  kmph.

Now,   u  =  \frac{x+y}{2}

∴ 10  =    \frac{\mathrm{3y}+y}{2}   =  2y

∴ y =   5  kmph

∴ 3y =  15  kmph   =   x

∴   v =  \frac{x-y}{2}   =   5   kmph

Hence, the rate of  the stream  =  5  kmph.

(24) In a stream running at 2 kmph, a motor boat goes 3.75 km upstream and back again to the starting point in 150 minutes. Find the speed of  the motor boat in still water.

Soln. Here,  v  =  2,  Total  time =  T  =  T1 +  T2 =  150  min. =  \frac{150}{60}  hrs.  =  \frac{5}{2}  hrs.

i.e.   T1 +  T2  =  \frac{5}{2}

∴    \frac{D}{S_1} + \frac{D}{S_2}  =  \frac{5}{2}

∴    \frac{\mathrm{3.75}}{\mathrm{u\ }-\mathrm{\ 2}} + \frac{\mathrm{3.75}}{\mathrm{u\ }+\mathrm{\ 2}}  =  \frac{5}{2}

∴   u2 – 3u – 4  =  0

∴   ( u – 4 ) ( u + 1 )   =  0

∴   u  =  4  kmph.

Thus, the speed of  the motor boat in still water  =  4  kmph.

(25) A man rows 40 km with the stream and 15 km against the stream taking 5 hours each time. Find his rate per hour in still water and the rate at which the stream flows.

Soln. Speed with the stream   =  x   =  \frac{40}{5} =  8 km/hr      and

Speed against the stream   =   y  =  \frac{15}{5}   =  3   km/hr

∴ Speed of  the man in still water  =   u  =  \frac{x+y}{2}  =  5.5  km/hr.

(26) A man can row 36 km downstream and 24 km upstream in 10 hours. Also, he can row 24 km downstream and 36 km upstream in 13 hours. Find the rate of  current and the speed of  the man in still water.

Soln. Here,   T1 +  T2  =  12

∴    \frac{D}{S_1} + \frac{D}{S_2} =   12

∴    \frac{\mathrm{36}}{x} + \frac{\mathrm{24}}{y} =   12      ––––––  (1)

Here,    T1 +  T2 =  13

∴    \frac{D}{S_1} + \frac{D}{S_2} =   13

∴  \frac{\mathrm{24}}{x} + \frac{\mathrm{36}}{y}  =   13   ––––––––  (2)

From  equations  (1)  and  (2),  we  get  x  =  6  kmph.  and   y  =  4  kmph.

∴   u  =  10  kmph.   And   v  =  2  kmph.

Thus,  the speed of  the man in still water  =  10  kmph  and the rate of  the current  =  2  kmph.

 

Best of Luck

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