Question

6 is the mean proportion between two numbers x and y and 48 is third prportion to x and y . Find the numbers.

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Answer 1

${\underline{\large{\sf{\pink{\bigstar{ \:Given:}}}}}}$

• 6 is the mean proportion between two numbers x and y.
• 48 is third prportion to x and y.

${ \underline{\large{\sf{\pink{\bigstar{ \:To \: find:}}}}}}$

• Find the numbers.

${ \underline{\large{\sf{\red{\bigstar{ \:Solution:}}}}}}$

We know that 6 is the mean proportion between two numbers x and y.

So,

$\: \: \: \: \: \: \: \: {\sf{\large{\bull{ \: \: \: \: 6 = \sqrt{xy} }}}}$

So, here x is :-

${\sf{ \implies{ \: \: \: \: 6 = \sqrt{xy} }}}$

By squaring of both side we will get:-

${\sf{\implies{36 = xy}}}$

${\sf{\implies{x = \frac{36}{y} .............(1)}}}$

It is given that 48 is third proportional to x & y. So it will be :-

$\: \: \: \: \: \: \: \: {\sf{\large{\bull{ \: \: \: \: {y}^{2} = 48 \: x \: \: \: .......(2)}}}}$

From equation 1 & 2 ,we get :-

${\sf{\implies{ {y}^{2} = 48 \: x}}}$

Putting the value of x :-

${\sf{\implies{ {y}^{2} = 48 \: \times ( \frac{36}{y} )}}}$

${\sf{\implies{ {y}^{2} \times y= 48 \: \times 36}}}$

${\sf{\implies{ {y}^{3} =1728}}}$

${\sf{\implies{ y = \sqrt[3]{1728} }}}$

${ \boxed{ \large{ \color{green}{\sf{\implies{ y =12}}}}}}$

So, value of x is 12.

Value of y is :-

$\\ {\sf{\implies{x = \frac{36}{12}}}}$

${ \boxed{ \large{ \color{green}{\sf{\implies{ x = 3}}}}}}$

So, final answer is :-

The numbers are 3 and 12.

Answer 2

Answer:

refer to the attachment