Question

The sum of two natural numbers is 9 and sum of their reciprocals is 9/20 Find the numbers​

Answer 1

Answer:

Let the numbers be a and b.

» a + b = 9 ... (1)

» 1/a + 1/b = 9/20

» 20 (a + b) = 9ab ... (2)

» ab = 20 from (1)

On solving we get,

Therefore, numbers are 4 and 5.

Answer 2

Step-by-step explanation:

Given:-

The sum of two natural numbers is 9 and sum of their reciprocals is 9/20

To find:-

Find the numbers

Solution:-

Let the required numbers be X and Y

Their sum =9

X +Y = 9----------(1)

Reciprocal of X = 1/X

Reciprocal of Y = 1/Y

Sum of their reciprocals = 9/20

=>(1/ X ) + (1/Y) = 9/20

=>(Y+X)/XY = 9/20

=>(X+Y)/XY = 9/20

=>9/XY = 9/20

From (1)

On Cancelling 9 both sides

=>1/XY = 1/20

=>XY = 20

We know that

(a-b)^2 = (a+b)^2 -4ab

=>(X-Y)^2 = (X+Y)^2 -4XY

=>(X-Y)^2 = 9^2-4(20)

=>(X-Y)^2 = 81-80

=>(X-Y)^2 = 1

=>X-Y = ±√1

X-Y = 1-----(3) (on taking positive value)

From (1)&(3)

X+Y = 9

X-Y = 1

(+)

________

2X+0 = 10

_________

=>2X = 10

=>X = 10/2

=>X = 5

from (1)

5+Y = 9

Y = 9-5

Y = 4

X = 5 and Y = 4

Answer:-

The two required Numbers are 5 and 4

Check:-

The numbers = 5 and 4

Their sum = 5+4=9

Reciprocal of 5 = 1/5

Reciprocal of 4 = 1/4

Their sum = (1/5)+(1/4)

=>(4+5)/20

=>9/20

Verified the given relations