269is it 315...????
In a forest, ratio of no. of deers, bear and fox is 3 : 7 : 5. If the difference between no. of deers and bears is a multiple of 3 as well as 7, what is min. no. of animals in the park?
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3 Answers
36Let the constant be x.
No. of deer,bears and foxes= 3x,7x,5x
Therefore,
(7x-3x)=7*3*y(where y is any constant)
or, 4x=21y
or, x is divisible by 21, since 4 is not divisible by 21
Therefore, min. value of x=21.
or, Min. no of animals= (7x+3x+5x)=15*21=315(Ans.)
206Show how you arrived at this solution........
- Astha Gupta is it wrong sir.....?????Upvote·0· Reply ·2013-06-27 10:12:59