Question
Two identical trains A and B running in opposite directions at same speed take 2 minutes to cross each other completely. The number of bogies of A are increased from 12 to 16. How much more time would they now require to cross each other?
Answer: Option A
Let the length of each train be x meters and let the speed of each of them by y m/sec
Then, $$\frac{{{\text{2x}}}}{{2{\text{y}}}}$$ = 120
⇒ $$\frac{{{\text{x}}}}{{{\text{y}}}}$$ = 120 . . . . . . . (i)
New length of train A $$ = \left( {\frac{{16}}{{12}}{\text{x}}} \right){\text{m}} = \left( {\frac{{4{\text{x}}}}{3}} \right){\text{m}}$$
∴ Time taken by trains to cross each other
$$\eqalign{
& = \left( {\frac{{{\text{x}} + \frac{{4{\text{x}}}}{3}}}{{2{\text{y}}}}} \right){\text{sec}} \cr
& = \frac{{7{\text{x}}}}{{6{\text{y}}}} \cr
& = \frac{7}{6} \times \frac{{\text{x}}}{{\text{y}}} \cr
& = \left( {\frac{7}{6} \times 120} \right){\text{sec}} \cr
& = 140\,{\text{sec}} \cr} $$
Hence, difference in times taken
= (140 - 120) sec
= 20 sec
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Let the length of each train be x meters and let the speed of each of them by y m/sec
Then, $$\frac{{{\text{2x}}}}{{2{\text{y}}}}$$ = 120
⇒ $$\frac{{{\text{x}}}}{{{\text{y}}}}$$ = 120 . . . . . . . (i)
New length of train A $$ = \left( {\frac{{16}}{{12}}{\text{x}}} \right){\text{m}} = \left( {\frac{{4{\text{x}}}}{3}} \right){\text{m}}$$
∴ Time taken by trains to cross each other
$$\eqalign{
& = \left( {\frac{{{\text{x}} + \frac{{4{\text{x}}}}{3}}}{{2{\text{y}}}}} \right){\text{sec}} \cr
& = \frac{{7{\text{x}}}}{{6{\text{y}}}} \cr
& = \frac{7}{6} \times \frac{{\text{x}}}{{\text{y}}} \cr
& = \left( {\frac{7}{6} \times 120} \right){\text{sec}} \cr
& = 140\,{\text{sec}} \cr} $$
Hence, difference in times taken
= (140 - 120) sec
= 20 sec
Was this answer helpful ?
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