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RRB ALP 10th Aug 2018 Shift-2 Question 6

Question 6

Find the mean proportional between 2 and 98?

Solution

Let,

The mean proportional of 2 and 98 is the number ‘n’ in the equation,

                                          $$\frac{2}{n}=\frac{n}{98}$$

 Solving for n by cross-multiplication , get that              

           or, $$\left(n\right)\left(n\right)=\left(98\right)\left(2\right)$$

           or, $$n^2=\left(7\right)\left(7\right)\left(2\right)\left(2\right)$$

           or, $$n^2=7^2\times 2^2$$

           or, $$\sqrt{n^2}=\sqrt{(7^2\times 2^2)}$$   (by taking square root on the both side)

           or, $$n=\left(7\right)\times\left(2\right)$$

           or, $$n=14$$

So, the mean proportion of 2 and 98 is 14 .



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