Question

The sum of two numbers is 16. The sum of their reciprocal is 1/3. Find the numbers.

Answer 1

AnSwER

${\tt{\red{\underline{\large{Given}}}}}$

• a world problem
• in which sum of two number is 16
• sum of their reciprocal is 1/3

${\sf{\green{\underline{\large{To\:find}}}}}$

• Number's

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Let the no be x and y

Then

sum of no is 16

=>x+y =16____(1)

According to the question

• sum of their reciprocal

$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$

$\implies\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$

$\implies\dfrac{x+y}{xy}=\frac{1}{3}$

Putting values

$\tt\frac{16}{xy}=\frac{1}{3}$

$\implies\tt\frac{1}{xy}=\frac{1}{48}$

=>xy=48

Now x-y

(x-y)²=(x+y)²-4xy

=>(x-y)²=(16)²-192

=>(x-y)²=256-192

=>x-y=√64

=>x-y=8________(2)

From equation (1) and (2)

x+y =16

x-y=8

_____

=>2x=24

=>x=12

For y

=>12+y =16

=>y=4

Answer 2

Question

The sum of two numbers is 16. The sum of their reciprocal is 1/3. Find the numbers.

Answer

Let the two numbers be " x " and " y "

The sum of two numbers is 16

⇒ x + y = 16 ... (1)

The sum of their reciprocal is 1/3

$\implies \bold{\frac{1}{x}+\frac{1}{y}=\frac{1}{3} }\\\\\implies \bold{\frac{x+y}{xy}=\frac{1}{3} }\\\\\implies \bold{\frac{16}{xy}=\frac{1}{3} }\\\\\implies \bold{xy=48}$

Now (x-y)² = (x+y)² - 4xy

$\implies \bold{(x-y)^2=16^2-4(48)}\\\\\implies \bold{(x-y)^2=256-192}\\\\\implies \bold{x-y=\sqrt{64} }\\\\\implies \bold{x-y=8...(2)}$

Now adding (1) & (2) , we get ,

⇒ 2x = 24

⇒ x = 12

sub. x value in (1) , we get ,

⇒ y = 16 - 12

⇒ y = 4

So the two numbers are 12 & 4 .