a number consist of two digits whose sum is 5 when the digits are reversed the number becomes greater by 9. find the number
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let the two digits of the number be a and b
(a+b)= 5
Original number= 10a+b
New number= 10b+a
So, we have 10a+b= (10b+a)+9
9a-9b= 9
9(a-b)= 9
(a-b)= 9/9= 1
(a+b)+(a-b)= 5+1
2a= 6
a= 6/2= 3
By substituting the value of a
(3+b)= 5
b= 2
The original number is 32
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(a+b)= 5
Original number= 10a+b
New number= 10b+a
So, we have 10a+b= (10b+a)+9
9a-9b= 9
9(a-b)= 9
(a-b)= 9/9= 1
(a+b)+(a-b)= 5+1
2a= 6
a= 6/2= 3
By substituting the value of a
(3+b)= 5
b= 2
The original number is 32
HOPE IT HELPS
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