Speed Time And Distance - Study Mode

[#211] A train 300 m long is running with a speed of 54 km/hr. In what time it cross a telephone pole ?
Correct Answer

(A) 20 seconds

Explanation

Solution: Here length of pole is considered 0 metre Time will be taken by train to cross the poll : $$eqalign{
& = frac{{{ ext{300 m}}}}{{54 imes frac{5}{{18}}{ ext{ m/s}}}} cr
& = frac{{300}}{{15}} cr
& = 20 cr} $$ Required time = 20 seconds

[#212] One third of a certain journey is covered at the rate of 25 km/hr, one forth at the rate of 30 km/hr and the rest at 50 km/hr. The average speed for the whole journey is :
Correct Answer

(B) $$33frac{1}{3}$$ km/hr

Explanation

Solution: Let the total distance = 1200 km $$eqalign{
& { ext{Total time taken :}} cr
& = frac{{400}}{{25}} + frac{{300}}{{30}} + frac{{500}}{{50}} cr
& = { ext{ }}16 + 10 + 10 cr
& = { ext{ 36 hours}} cr
& herefore { ext{Average speed}} cr
& = frac{{{ ext{Total distance }}}}{{{ ext{Total time}}}} cr
& = frac{{1200}}{{36}} cr
& = 33frac{1}{3}{ ext{ km/hr}} cr} $$

[#213] It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
Correct Answer

(C) 3 : 4

Explanation

Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{speed}},{ ext{of}},{ ext{the}},{ ext{train}},{ ext{be}},x,{ ext{km/hr}} cr
& { ext{and}},{ ext{that}},{ ext{of}},{ ext{the}},{ ext{car}},{ ext{be}},y,{ ext{km/hr}} cr
& { ext{Then}},,frac{{120}}{x} + frac{{480}}{y} = 8 cr
& Rightarrow frac{1}{x} + frac{4}{y} = frac{1}{{15}},.........left( i
ight) cr
& { ext{and}},,frac{{200}}{x} + frac{{400}}{y} = frac{{25}}{3} cr
& Rightarrow frac{1}{x} + frac{2}{y} = frac{1}{{24}},.........left( {ii}
ight) cr
& { ext{Solving}},left( { ext{i}}
ight),{ ext{and}},left( {{ ext{ii}}}
ight){ ext{,}}, cr
& { ext{we}},{ ext{get}},,x = 60,{ ext{and}},y = 80 cr
& herefore { ext{Ratio}},{ ext{of}},{ ext{speeds}} cr
& = 60:80 cr
& = 3:4 cr} $$

[#214] A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
Correct Answer

(C) 16 km

Explanation

Solution: $$eqalign{
& { ext{Let}},{ ext{the}},{ ext{distance}},{ ext{travelled}},{ ext{on}},{ ext{foot}},{ ext{be}},x,km cr
& { ext{Then,}},{ ext{distance}},{ ext{travelled}},{ ext{on}},{ ext{bicycle}} = left( {61 - x}
ight)km cr
& { ext{So}},,frac{x}{4} + frac{{ {61 - x} }}{9} = 9 cr
& Rightarrow 9x + 4left( {61 - x}
ight) = 9 imes 36 cr
& Rightarrow 5x = 80 cr
& Rightarrow x = 16,km cr} $$

[#215] A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
Correct Answer

(D) 40 km

Explanation

Solution: $$eqalign{
& { ext{Let}},{ ext{distance}} = x,km,{ ext{and}},{ ext{usual}},{ ext{rate}} = y,kmph cr
& { ext{Then}},,frac{x}{y} - frac{x}{{y + 3}} = frac{{40}}{{60}} cr
& Rightarrow 2yleft( {y + 3}
ight) = 9x,..........left( i
ight) cr
& { ext{and}},,frac{x}{{y - 2}} - frac{x}{y} = frac{{40}}{{60}} cr
& Rightarrow y(y - 2) = 3x,.............left( {ii}
ight) cr
& { ext{On}},{ ext{dividing}},left( { ext{i}}
ight),{ ext{by}},left( {{ ext{ii}}}
ight){ ext{,}} cr
& { ext{We}},{ ext{get}},x = 40, ext{km} cr} $$