Question

Find the mean proportion between
a) $ 81 $ and $ 121 $
b) $ 1.8 $ and $ 0.2 $
c) $ \dfrac{2}{3} $ and $ \dfrac{8}{{27}} $
d) $ 0.32 $ and $ 0.08 $
e) $ \dfrac{1}{{25}} $ and $ \dfrac{1}{{16}} $

Answer 1

Hint: The question asks about the mean proportion between two numbers. If we have the mean proportion of two numbers $ a,b $ as $ x $ then it has a property that the ratio between $ a $ and $ x $ is the same as $ x $ and. $ b $ So taking this in mind we will use the formula for generating the mean proportion of the given numbers which will serve as the answer of this question.
Formula used:
If we have the mean proportion of two numbers $ a,b $ as $ x $ then,
 $ \dfrac{a}{x} = \dfrac{x}{b} $ where, $ a \leqslant x \leqslant b $

Complete step-by-step answer:
The given numbers are $ 81 $ and $ 121 $
If we put $ a = 81,b = 121 $ in the mean proportion formula we have,
 $ \dfrac{{81}}{x} = \dfrac{x}{{121}} $
 $ \Rightarrow {x^2} = 81 \times 121 $
Now the squares of $ 9 $ and $ 11 $ are $ 81 $ and $ 121 $ respectively so,
 $ \Rightarrow {x^2} = {9^2} \times {11^2} = {(9 \times 11)^2} $
The fact that $ {\alpha ^2} = {\beta ^2} \Rightarrow \alpha = \beta $ gives us
 $ x = 9 \times 11 $
The product of $ 9 $ and $ 11 $ gives us $ 99 $ so,
 $ x = 99 $
Therefore, the mean proportion of the numbers given to us is $ 99 $ .
So, the correct answer is “ $ 99 $ ”.

Similarly for (b) we have
 $ \dfrac{{1.8}}{x} = \dfrac{x}{{0.2}} $
 $ \Rightarrow {x^2} = 1.8 \times 0.2 = 0.36 $
 $ \Rightarrow {x^2} = {(0.6)^2} $
The fact that $ {\alpha ^2} = {\beta ^2} \Rightarrow \alpha = \beta $ gives us
 $ x = 0.6 $
Therefore, the mean proportion of the numbers given to us is $ 0.6 $ .
So, the correct answer is “ $ 0.6 $ ”.

Same is for:
c) $ \dfrac{{\left( {\dfrac{2}{3}} \right)}}{x} = \dfrac{x}{{\left( {\dfrac{8}{{27}}} \right)}} $
 $ \Rightarrow {x^2} = \dfrac{2}{3} \times \dfrac{8}{{27}} = \dfrac{{16}}{{81}} $
 $ \Rightarrow {x^2} = {\left( {\dfrac{4}{9}} \right)^2} $
 $ \Rightarrow x = \left( {\dfrac{4}{9}} \right) $
Therefore, the mean proportion of the numbers given to us is $ \dfrac{4}{9} $ .
So, the correct answer is “ $ \dfrac{4}{9} $ ”.

d) $ \dfrac{{0.32}}{x} = \dfrac{x}{{0.08}} $
 $ \Rightarrow {x^2} = 0.032 \times 0.08 = 0.00256 $
 $ \Rightarrow x = 0.016 $
Therefore, the mean proportion of the numbers given to us is $ 0.016 $ .
So, the correct answer is “ $ 0.016 $ ”.

e) $ \dfrac{{\left( {\dfrac{1}{{25}}} \right)}}{x} = \dfrac{x}{{\left( {\dfrac{1}{{16}}} \right)}} $
 $ \Rightarrow {x^2} = \dfrac{1}{{25}} \times \dfrac{1}{{16}} = \dfrac{1}{{{5^2}}} \times \dfrac{1}{{{4^2}}} = {\left( {\dfrac{1}{5} \times \dfrac{1}{4}} \right)^2} $
 $ \Rightarrow x = \dfrac{1}{5} \times \dfrac{1}{4} $
 $ \Rightarrow x = \dfrac{1}{{20}} $
Therefore, the mean proportion of the numbers given to us is $ \dfrac{1}{{20}} $ .
So, the correct answer is “ $ \dfrac{1}{{20}} $ ”.

Note: While calculating the mean proportion $ x $ always try to simplify the numbers to small factors so that you can easily find the root of that number. Multiplying the numbers will result in a larger number whose root calculation then will be a messy work.