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by Mclaughlin » Sun Aug 17, 2008 3:46 pm
B i l l a n d S a l l y s e e e a c h o t h e r a c r o s s a f i e l d . T h e y a r e 5 0 0 f e e t a p a r t . I f a t t h e s a m e i n s t a n t o f t i m e t h e y s t a r t r u n n i n g t o w a r d e a c h o t h e r i n a d i r e c t l i n e h o w f a r w i l l B i l l h a v e t r a v e l e d w h e n t h e y m e e t ?

( 1 ) B i l l r a n a t a n a v e r a g e s p e e d t h a t w a s 5 0 % g r e a t e r t h a n S a l l y s a v e r a g e s p e e d .

( 2 ) B i l l r a n a t a n a v e r a g e s p e e d 4 f e e t p e r s e c o n d f a s t e r t h a n S a l l y s a v e r a g e s p e e d .

OA is A
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Source: — Data Sufficiency |

by Suyog » Sun Aug 17, 2008 5:01 pm
Boy! Where did u copy the que from? :o

Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?

(1) Bill ran at an average speed that was 50% greater than Sally’s average speed.

If Sally’s average speed 100m then Bill ran at an average speed that was 150m.

then Bill travel 300m when he meet Sally.

Similarly, If Sally’s average speed 50m then Bill ran at an average speed that was 75m.

again, Bill travel 300m when he meet Sally.

So everytime Bill will travel 300m when they meet.

Suff.

(2) Bill ran at an average speed 4 feet per second faster than Sally’s average speed.

Here if Sally's average speed 50m then Bill's average speed will be 54m.

again, if Sally's average speed 25m then Bill's average speed will be 29m.

So everytime the distance Bill travels will be different.

Insuff.

Choose(a).
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