Question

the sum of two number is 62 and their product is 960. find the sum of their reciprocal​

Answer 1

$\large\green{\underline{\underline{\bf{\green{Given}}}}}$

• Sum of two numbers = 62

• Product of two numbers = 960

$\large\green{\underline{\underline{\bf{\green{To Find}}}}}$

• Sum of reciprocal of two numbers

$\large\green{\underline{\underline{\bf{\green{Solution}}}}}$

As per given in question,

➨ Let the first no be 'x'

Then,

➡ Second Number = ( 62 - x)

Now,

➡ Product of two numbers = 960

So,

➭ x ( 62 - x) = 960

➭ 62x - x² = 960

➭ x² - 62x + 960 = 0

➭ x² - 32x - 30x + 960 = 0

➭ x ( x - 32) - 30( x - 32) = 0

➭ ( x - 30) ( x - 32) = 0

➭ x = 30 or x = 32

➨ Here we get two different values of x

So,

➨ Both of them satisfy the conditions of question.

Hence,

➡ First number = x = 30 or 32

➡ Second Number = ( 62 - x)

Put x = 32 or 30

We get,

➦ Second Number = (62 - x) = 62 - 30 = 32

Or

➦ Second Number = (62 - x) = (62 - 32) = 30

Hence,

Either First no is 32 or 30

Or

Second no is 32 or 30

➦ Reciprocal of First number = 1/30 or 1/32

➦ Reciprocal of Second Number = 1/30 or 1/32

So,

➢ Sum of their Reciprocal = 1/32 + 1/30 = 62/960 = 31/480

Answer 2

Let the 2 numbers be x and y respectively

Given

Sum of 2 numbers = 62

$\implies x+y=62\\\implies x = 62-y--(1)$

Product of the numbers = 960

$\implies xy = 960 --(2)$

From (1) and (2), we get,

$\implies (62-y)(y)=960$

$\implies -y^{2} +62y=960$

$\implies y^{2}-62y+960=0$

$\implies y^{2} -30y-32y+960=0$

$\implies y(y-30)-32(y-30)=0$

$\implies (y-30)(y-32)=0$

Now, if x = 30, then y = 32, or vice versa

Sum of reciprocals

$\implies \dfrac{1}{30} +\dfrac{1}{32}$

$\implies \dfrac{62}{960}$

$\implies \boxed{\boxed{\bold{\frac{31}{480} }}}}}$