Question

6 is the mean proportion between two numbers x and y and 48 is the third proportion of X and Y find the numbers​

Answer 1

SOLUTION:-

Given:

•6 is the mean proportion between two numbers x & y.

•48 is the third proportion of x & y.

To find:

The number.

Explanation:

We have,

6 is the mean proportional between x & y.

Therefore,

x, 6, y are continued proportion.

$= >x \ratio 6 \ratio \ratio 6 \ratio y \\ \\ = > \frac{x}{6} = \frac{6}{y} \\ [cross \: multiplication] \\ = > xy = 36 \\ \\ = > x = \frac{36}{y} .............(1)$

&

We have,

48 is the third proportional between x & y.

Therefore,

x, y, 48 are continued proportion.

$= > x \ratio y \ratio \ratio y \ratio 48 \\ \\ = > \frac{x}{y} = \frac{y}{48} \\ [Cross \: multiplication] \\ \\ = > {y}^{2} = 48x..............(2)$

Putting the value of equation (1) in equation (2), we get;

${y}^{2} = 48 \times \frac{36}{y} \\ \\ {y}^{3} = 48 \times 36 \\ \\ {y}^{3} = 1728 \\ \\ {y} = \sqrt[3]{1728} \\ \\ y = 12$

&

Putting the value of y in equation (1), we get;

$x = \frac{36}{12} \\ \\ x = 3$

Thus,

The number of x & y is 3, 12.

Answer 2

Step-by-step explanation:

SOLUTION:-

Given:

•6 is the mean proportion between two numbers x & y.

•48 is the third proportion of x & y.

To find:

The number.

Explanation:

We have,

6 is the mean proportional between x & y.

Therefore,

x, 6, y are continued proportion.

\begin{gathered} = > x \ratio 6 \ratio \ratio 6 \ratio y \\ \\ = > \frac{x}{6} = \frac{6}{y} \\ [cross \: multiplication] \\ = > xy = 36 \\ \\ = > x = \frac{36}{y} .............(1)\end{gathered}

=>x:6::6:y

=>

6

x

=

y

6

[crossmultiplication]

=>xy=36

=>x=

y

36

.............(1)

&

We have,

48 is the third proportional between x & y.

Therefore,

x, y, 48 are continued proportion.

\begin{gathered} = > x \ratio y \ratio \ratio y \ratio 48 \\ \\ = > \frac{x}{y} = \frac{y}{48} \\ [Cross \: multiplication] \\ \\ = > {y}^{2} = 48x..............(2)\end{gathered}

=>x:y::y:48

=>

y

x

=

48

y

[Crossmultiplication]

=>y

2

=48x..............(2)

Putting the value of equation (1) in equation (2), we get;

\begin{gathered} {y}^{2} = 48 \times \frac{36}{y} \\ \\ {y}^{3} = 48 \times 36 \\ \\ {y}^{3} = 1728 \\ \\ {y} = \sqrt[3]{1728} \\ \\ y = 12\end{gathered}

y

2

=48×

y

36

y

3

=48×36

y

3

=1728

y=

3

1728

y=12

&

Putting the value of y in equation (1), we get;

\begin{gathered}x = \frac{36}{12} \\ \\ x = 3\end{gathered}

x=

12

36

x=3

Thus,

The number of x & y is 3, 12.