Mr. Crone started from his house at 8.00 a.m. and walked t to his office at an average speed of 4 km/h. Mettle started from Mr. Crones at 8.30 a.m. and travelled by cycle in the same direction as Mr. Crone at average speed of of 6 km/h. If the two arrive arrived at the office at the same time .(i) Find the time of arrival (ii) Find the distance between Mr. Crones house and the office


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Question Number 35472 by Issifu1 last updated on 19/May/18

$M r . C r o n e s t a r t e d f r o m h i s$ $h o u s e a t 8.00 a . m . a n d w a l k e d t$ $t o h i s o f f i c e a t a n a v e r a g e s p e e d$ $o f 4 k m / h . M e t t l e s t a r t e d f r o m$ $M r . C r o n e s a t 8.30 a . m . a n d t r a v e l l e d$ $b y c y c l e i n t h e s a m e d i r e c t i o n$ $a s M r . C r o n e a t a v e r a g e s p e e d$ $o f o f 6 k m / h . I f t h e t w o a r r i v e$ $a r r i v e d a t t h e o f f i c e a t t h e s a m e t i m e$ $. (i) F i n d t h e t i m e o f a r r i v a l$ $(i i) F i n d t h e d i s t a n c e b e t w e e n M r .$ $C r o n e s h o u s e a n d t h e o f f i c e$
	Answered by Rasheed.Sindhi last updated on 19/May/18

	$Let Mr Crone takes t hours and$ $covers s km in arriving office .$ $∴ Mr Mettle takes t - \frac{1}{2} hours in$ $covering s km (same distance .$ $In case of Mr Crone :$ $s = 4 t$ $In case of Mr Mettle :$ $s = 6 (t - \frac{1}{2}) = 6 \times \frac{2 t - 1}{2} = 6 t - 3$ $Comparing in both cases$ $s = 4 t = 6 t - 3 \Rightarrow t = \frac{3}{2} hours = 1 \frac{1}{2} hours$ $(i) So Mr Crone 8 am + 1 \frac{1}{2} hours = 9 : 30 am$ $(Mr Mettle starts at 8 : 30 am and he$ $takes time 1 \frac{1}{2} - \frac{1}{2} = 1 hour and arrives$ $at same time 9 : 30 am)$ $(ii) Distence between Mr Crone and his office :$ $s = 4 t = 4 \times \frac{3}{2} = 6 km$
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