A man travelled from town A to B, a distance of 360km. He left A one hour later than he had planned so he decided to drive at 5km/h faster than his normal speed, in order to reach B on schedule. If he arrived B at exactly the scheduled time, find the normal speed.


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Question Number 202079 by necx122 last updated on 19/Dec/23

$A m a n t r a v e l l e d f r o m t o w n A t o B, a$ $d i s t a n c e o f 360 k m . H e l e f t A o n e h o u r$ $l a t e r t h a n h e h a d p l a n n e d s o h e d e c i d e d$ $t o d r i v e a t 5 k m / h f a s t e r t h a n h i s$ $n o r m a l s p e e d, i n o r d e r t o r e a c h B o n$ $s c h e d u l e . I f h e a r r i v e d B a t e x a c t l y t h e$ $s c h e d u l e d t i m e, f i n d t h e n o r m a l s p e e d .$
	Answered by deleteduser1 last updated on 19/Dec/23

	$t = \frac{360}{x + 5} = \frac{360}{x} - 1 \Rightarrow \frac{- 360 \times 5}{x^{2} + 5 x} = - 1 \Rightarrow x^{2} + 5 x - 1800 = 0$ $\Rightarrow x = 40$
	Commented by necx122 last updated on 20/Dec/23
	Sir AST, you've also been very helpful on this platform since you joined. This is very clear. Thank you so much sir.
	Commented by necx122 last updated on 20/Dec/23
	textbook says the answer is 80.
	Commented by deleteduser1 last updated on 20/Dec/23

	$\frac{360}{80} - \frac{360}{85} \neq + 1$
	Commented by necx122 last updated on 20/Dec/23
	true proof. Thanks
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