Question

sum of two number is 24 and product is 108. find the sum of their reciprocal​

Answer 1

Question:

Some of two numbers is 24 and product is 108. Find the sum of their reciprocals.

Answer:

Required sum is 2/9.

Note:

• (A+B)^2 = A^2 + B^2 + 2•A•B

• (A-B)^2 = A^2 + B^2 - 2•A•B

• (A+B)(A-B) = A^2 - B^2

Solution:

Let the two numbers be x and y .

It is given that;

The sum of the two numbers is 24.

Thus,

x + y = 24 --------(1)

Also,

The product of the numbers is 108.

Thus,

x•y = 108 ----------(2)

Thus,

The sum of the reciprocals of the two numbers

= 1/x + 1/y

= (y+x)/x•y

= (x+y)/x•y

= 24/108 {using eq-(1) and (2)}

= 2/9

Hence,

The required sum of the reciprocals of the two numbers is 2/9 .

Answer 2

$\Large\underline{\underline{\sf{Given}:}}$

• sum of two numbers = 24
• Product of these two numbers = 108 .

$\Large\underline\mathfrak{Question}$

• sum of reciprocal of these two numbers ??

$\large\star{\underline{\tt{\red{Answer}}}}\star$

$\textbf{let the 2 numbers be x and y.}$

A/q,

x + y = 24 ---------------- Equation (1)

x * y = 108 -------------- Equation (2)

we have to Find ,,

1/x + 1/y .= ?

Taking LCM we get,

→ (y+x)/xy = ?

Putting values from Equation (1) & Eqn (2) , we get,

→ 24/108

→ 12×2/12×19

→ 2/9 ...

Hence,,,,

$\red{\large\boxed{\bold{ \frac{1}{x} + \frac{1}{y} = \frac{2}{9} }}}$

$\large\underline\textbf{Hope it Helps You.}$