Ratio 1

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?

3 Answers

36
Piyush Kedia ·

Let 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.)

269
Astha Gupta ·

is it 315...????

206
Sayantan Hazra ·

Show how you arrived at this solution........

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