the sum of two natural numbers is 9 and sum of their reciprocals is 9÷20 find the number
Answers
Answer:
4,5
Step-by-step explanation:
that is the answereuehsjsjsjjs
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 number ?
Solution:-
Let the two natural numbers be X and Y
The sum of two natural numbers is 9
X+Y = 9 -------------------(1)
Reciprocal of X = 1/X
Reciprocal of Y = 1/Y
Sum of their reciprocals is 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 applying cross multiplication then
=> XY×9 = 9×20
=>9 XY = 180
=> XY = 180/9 = 20
XY = 20 ---------------(2)
we know that
(a-b)^2=(a+b)^2-4ab
(X-Y)^2 = (X+Y)^2-4XY
From (1)&(2)
=> (X-Y)^2 = 9^2-4(20)
=>(X-Y)^2 = 81-80
=> (X-Y)^2 = 1
=> (X-Y)=±√1
=> (X-Y)=±1
X-Y=1 ------------------(3)
(on taking positive value since given numbers are natural numbers so -1 is not a natural number)
On adding (1)&(3) then
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
The number = 54
Answer:-
The required number for the given problem is 54
Check:-
Sum of the digits = 5+4=9
Reciprocal of 5 = 1/5
Reciprocal of 4 = 1/4
Their sum = (1/5)+(1/4)
LCM of 5 and 4 = 20
=> (4+5)/20
=> 9/20
Verified the given relations.
Used formula:-
- (a-b)^2=(a+b)^2-4ab