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Question

A train, after travelling 70 km from a station A towards a station B, develops a fault in the engine at C and covers the remaining journey to B at 34 of its earlier speed and arrives at B 1 hour and 20 minutes late. If the fault had developed 35 km further on at D, it would have arrived 20 minutes sooner. Find the speed (in km / hr) of the train.

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Solution

The correct option is **C** 45

Let, the train's speed = y

∴ Total time =10+xy

Case I :

Total time =70y+x34y=70y+4x3y=210+4x3y

Let, the train's speed = y

∴ Total time =10+xy

Case I :

Total time =70y+x34y=70y+4x3y=210+4x3y

∴210+4x3y=70+xy+1+13

⇒210+4x3y=210+3x+3y+y3y

⇒210+4x=210+3x+4y

⇒x−4y=0 ......... (i)

Case II :

Total time =70+35y+x−3534y

=70+35y+4x−1403y=105×3+4x−1403y

=315−140+4x3y=175+4x3y

∴175+4x3y=70+xy+1

⇒175+4x3y=70+x+yy

⇒175+4x=210+3x+3y

⇒4x−3x=210−175

⇒x−3y=45 ............. (ii)

We have,

(i) ⇒x−4y=0

(i) ⇒x−3y=45

x−4y=0

x−3y=0

____________

−y=−45 ⇒y=45 km/h

∴ Speed of train = 45 km/h

⇒210+4x3y=210+3x+3y+y3y

⇒210+4x=210+3x+4y

⇒x−4y=0 ......... (i)

Case II :

Total time =70+35y+x−3534y

=70+35y+4x−1403y=105×3+4x−1403y

=315−140+4x3y=175+4x3y

∴175+4x3y=70+xy+1

⇒175+4x3y=70+x+yy

⇒175+4x=210+3x+3y

⇒4x−3x=210−175

⇒x−3y=45 ............. (ii)

We have,

(i) ⇒x−4y=0

(i) ⇒x−3y=45

x−4y=0

x−3y=0

____________

−y=−45 ⇒y=45 km/h

∴ Speed of train = 45 km/h

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