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-

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