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Sia ? 6 years, 5 months ago
Let the speed of the train be x km/hr and that of the car be y km/hr.
Case I Distance covered by train = 250 km.
Distance covered by car = (370 - 250) km = 120 km.
Time taken to cover 250 km by train = {tex}\frac { 250 } { x }{/tex} hours
Time taken to cover 120 km by car = {tex}\frac { 120 } { y }{/tex}hours
Total time taken =4 hours
{tex}\therefore \quad \frac { 250 } { x } + \frac { 120 } { y } = 4 \Rightarrow \frac { 125 } { x } + \frac { 60 } { y } = 2{/tex}.......(i)
Case II Distance covered by train = 130 km.
Distance covered by car = (370 -130) km = 240 km.
Time taken to cover 130 km by train = {tex}\frac { 130 } { x }{/tex} hours
Time taken to cover 240 km by car ={tex} \frac { 240 } { y }{/tex} hours
Total time taken = {tex}4 \frac { 18 } { 60 } \text { hours } = 4 \frac { 3 } { 10 } \text { hours } = \frac { 43 } { 10 }{/tex}hours
{tex}\therefore \quad \frac { 130 } { x } + \frac { 240 } { y } = \frac { 43 } { 10 } \Rightarrow \frac { 1300 } { x } + \frac { 2400 } { y } = 43{/tex}......(ii)
Putting {tex}\frac 1x{/tex}= u and {tex}\frac 1y{/tex}=v, equations (i) and (ii) become
{tex}125u + 60v = 2{/tex} ... (iii) and {tex}1300u + 2400v = 43{/tex}. ... (iv)
On multiplying (iii) by 40 and subtracting (iv) from the result, we get
5000u - 1300v = 80 - 43 {tex}\Rightarrow{/tex} 3700u = 37
{tex}\Rightarrow u = \frac { 37 } { 3700 } = \frac { 1 } { 100 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 100 } \Rightarrow x = 100{/tex}
Putting u = {tex}\frac{1}{100}{/tex} in (iv), we get
{tex}\left( 1300 \times \frac { 1 } { 100 } \right) + 2400 v = 43 \Rightarrow 2400 v = 43 - 13 = 30{/tex}
{tex}\Rightarrow v = \frac { 30 } { 2400 } = \frac { 1 } { 80 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 80 } \Rightarrow y = 80{/tex}
{tex}\therefore{/tex} x = 100 and y = 80.
Hence, the speed of the train is 100 km/hr and that of the car is 80 km/hr
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