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