Question 68

What is the ratio of the mean proportional between 14.4 and 3.6 and the third proportional of 5 and 4?

Solution

Let the mean proportional between 14.4 and 3.6 = $$x$$

$$=$$> $$x^2=14.4\times3.6$$

$$=$$> $$x^2=\frac{144}{10}\times\frac{36}{10}$$

$$=$$> $$x=\frac{72}{10}$$

Let the third proportional of 5 and 4 = $$y$$

$$=$$> $$4^2=5\times y$$

$$=$$> $$y=\frac{16}{5}$$

$$\therefore\ $$Required ratio$$=\frac{72}{10}:\frac{16}{5}=\frac{72}{10}:\frac{32}{10}=9:4$$

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