Question

7 is the mean proportion between two numbers x and y and 56 is third proportion to x and y. Find the numbers.
$\left[3\right]$

Answer 1

Solution:
Since, 7 is mean proportional between x and y.

$\Rightarrow x:7=7:y\Rightarrow xy=49\cdots \cdots \left(1\right)$

and 56 is third proportional to x and y

$\Rightarrow x:y=y:56\Rightarrow y^2=56x\cdots \cdots \left(2\right)$

From eq (1)

$xy=49\Rightarrow x=\frac{49}{y}$ [1 Mark]

Substituting $x=\frac{49}{y}$ in eq (2)

$y^2=56\times \frac{49}{y}\Rightarrow y^3=56\times 49$ [1 Mark]

$\Rightarrow y^3=(7^3\times 2^3)$

$y=7\times 2=14$

$\therefore x=\frac{49}{14}=3.5$

$\therefore$ The required numbers are 3.5 and 14. [1 Mark]