# A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Answer

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Hint : If the number is ab it can also be written as 10a+b since a is at tens place and b is at ones place.

Let the digits at the tens place be x and ones place be y.

Hence the number is 10x + y ……(i)

By reversing the number it becomes 10y + x ……(ii)

According to the question we have

x + y = 5 ……(iii)

Also when 9 is added to the number the digits get interchanged.

$

10x + y + 9 = 10y + x \\

9x - 9y + 9 = 0\, \\

$

$y - x = 1\,\,\,$ ……(iv)

Adding (iii) &(iv) we get,

$

2y = 6 \\

y = 3 \\

\\

$ (v)

Then x can be calculated from (iii)or(iv) we get,

x = 2 (vi)

From (i),(v),(vi) the number is

10(2) + 3 = 23

Hence the number is 23.

Note :- In these types of problems of finding numbers we have to assume variables and assign them a place it may be ones, tens, hundredth etc. Then according to the question make the equation and solve it to get the number.

Let the digits at the tens place be x and ones place be y.

Hence the number is 10x + y ……(i)

By reversing the number it becomes 10y + x ……(ii)

According to the question we have

x + y = 5 ……(iii)

Also when 9 is added to the number the digits get interchanged.

$

10x + y + 9 = 10y + x \\

9x - 9y + 9 = 0\, \\

$

$y - x = 1\,\,\,$ ……(iv)

Adding (iii) &(iv) we get,

$

2y = 6 \\

y = 3 \\

\\

$ (v)

Then x can be calculated from (iii)or(iv) we get,

x = 2 (vi)

From (i),(v),(vi) the number is

10(2) + 3 = 23

Hence the number is 23.

Note :- In these types of problems of finding numbers we have to assume variables and assign them a place it may be ones, tens, hundredth etc. Then according to the question make the equation and solve it to get the number.

Last updated date: 20th Sep 2023

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