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Question Number 17610 by Tinkutara last updated on 08/Jul/17

$$\mathrm{A}\mathrm{train},\mathrm{after}\mathrm{travelling}70\mathrm{km}\mathrm{from}\mathrm{a}$$$$\mathrm{station}\mathrm{A}\mathrm{towards}\mathrm{a}\mathrm{station}\mathrm{B},\mathrm{develops}$$$$\mathrm{a}\mathrm{fault}\mathrm{in}\mathrm{the}\mathrm{engine}\mathrm{at}\mathrm{C},\mathrm{and}\mathrm{covers}$$$$\mathrm{the}\mathrm{remaining}\mathrm{journey}\mathrm{to}\mathrm{B}\mathrm{at}\frac{3}{4}\mathrm{of}\mathrm{its}$$$$\mathrm{earlier}\mathrm{speed}\mathrm{and}\mathrm{arrives}\mathrm{at}\mathrm{B}1\mathrm{hour}$$$$\mathrm{and}20\mathrm{minutes}\mathrm{late}.\mathrm{If}\mathrm{the}\mathrm{fault}\mathrm{had}$$$$\mathrm{developed}35\mathrm{km}\mathrm{further}\mathrm{on}\mathrm{at}\mathrm{D},\mathrm{it}$$$$\mathrm{would}\mathrm{have}\mathrm{arrived}20\mathrm{minutes}\mathrm{sooner}.$$$$\mathrm{Find}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{train}\mathrm{and}\mathrm{the}$$$$\mathrm{distance}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{B}.$$

Commented by Tinkutara last updated on 08/Jul/17

Answered by alex041103 last updated on 08/Jul/17

$$$$$$LetSbethedistancefromAtoB$$$$andv-thespeedofthetrain$$$$Then$$$$\frac{S}{v}+\frac{4}{3}=\frac{S-70}{\frac{3}{4}v}+\frac{70}{v}\mid -\frac{70}{v}$$$$\frac{S-70}{v}+\frac{4}{3}=\frac{4}{3}\frac{S-70}{v}$$$$\Rightarrow \frac{S-70}{v}=4orS=4v+70$$$$Alsothefollowingistrue$$$$\frac{S}{v}+1=\frac{S-105}{\frac{3}{4}v}+\frac{105}{v}\mid -\frac{105}{v}$$$$\frac{S-105}{v}+1=\frac{4}{3}\frac{S-105}{v}\mid -\frac{S-105}{v}$$$$1=\frac{1}{3}\frac{S-105}{v}\Rightarrow S=3v+105$$$$\Rightarrow 3v+105=4v+70$$$$\Rightarrow v=35km/handS=210km$$$$Note:$$$$20minutessoonermeans20mins$$$$soonerthan1hourand20minutes$$$$i.e.1hourlaterthanplanned$$

Commented by Tinkutara last updated on 09/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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