the tens digit of a two - digit number exceeds the units digit by 5. if the digits are reversed, the new number is less by 45. if the sum of their digits is 9, find the numbers.
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Let the units digit be y.
and the tens digit be x.
The number will be 10x+y. --------- (1)
The reversed number will be 10y + x. --------- (2)
Given, x + y = 9 -------------- (3)
Solve (1) and (2) with the given number, we get
(10x + y) - (10y + x) = 45
10x + y - 10y - x = 45
9x - 9y = 45
9(x - y)=45
x- y = 45 ------------- (4)
on solving (3) and (4) we get,
2x = 14
x = 7
Substitute x = 7 in (3)
y = 2.
Hence the original number will be 72.
and the tens digit be x.
The number will be 10x+y. --------- (1)
The reversed number will be 10y + x. --------- (2)
Given, x + y = 9 -------------- (3)
Solve (1) and (2) with the given number, we get
(10x + y) - (10y + x) = 45
10x + y - 10y - x = 45
9x - 9y = 45
9(x - y)=45
x- y = 45 ------------- (4)
on solving (3) and (4) we get,
2x = 14
x = 7
Substitute x = 7 in (3)
y = 2.
Hence the original number will be 72.
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