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Question Number 195661 by otchereabdullai@gmail.com last updated on 06/Aug/23

Commented by otchereabdullai@gmail.com last updated on 07/Aug/23

$t h a n k s p r o f$
Commented by mr W last updated on 07/Aug/23

$e v e n w h e n t h e w h o l e r o a d i s g o o d, h e$ $n e e d s \frac{23}{52} \times 60 = 26.5 m i n u t e s . t h a t$ $m e a n s h e c a n n e v e r d r i v e 23 k m i n$ $o n l y 25 m i n u t e s .$ $\Rightarrow q u e s t i o n i s w r o n g!$
	Answered by som(math1967) last updated on 07/Aug/23

	$l e t g o o d r o a d = x k m$ $∴ b a d r o a d = 23 - x$ $\frac{x}{52} + \frac{(23 - x)}{28} = \frac{25}{60}$ $\Rightarrow \frac{7 x + 13 (23 - x)}{4 \times 13 \times 7} = \frac{5}{12}$ $\Rightarrow - 6 x + 13 \times 23 = \frac{5 \times 13 \times 7}{3}$ $\Rightarrow 18 x = 13 \times 23 \times 3 - 35 \times 13$ $\Rightarrow x = \frac{13 (69 - 35)}{18} = \frac{13 \times 17}{9} = 24.5 k m$ $24.5 > 23 ?$ $l e n g t h o f g o o d r o a d > t o t a l l e n g t h$ $?$
	Commented by otchereabdullai@gmail.com last updated on 07/Aug/23

	$t h a n k s s i r$
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