Question

the sum of the digits of a two-digit number is 13 the number obtained by interchanging the digits of the given number exceed that number by 27 find the number

Answer 1

♧♧HERE IS YOUR ANSWER♧♧

Let us consider that the two digits are a and b.

Then the number is (10a + b).

When the digits are interchanged, the new number is (10b + a).

By the given condition :

a + b = 13 .....(i)

and

10b + a = (10a + b) + 27

=> 9b = 9a + 27

=> b = a + 3 .....(ii)

Putting b = a + 3 in (i), we get :

a + a + 3 = 13

=> 2a = 10

=> a = 5

From (ii), putting a = 5, we get :

b = 5 + 3 = 8

Therefore, the number is 58.

♧♧HOPE THIS HELPS YOU♧♧

Answer 2

Hi there !!
Here's your answer

Let the digit in the tens place be x
the sum of digits is 13
So,
the digit in units place = 13 - x
The original Number formed will be
10(x) + 13 - x
= 10x + 13 - x
= 9x + 13 __________(i)

Given,
if the digits are reversed, the number formed exceeds the original number by 27

So,
by interchanging the digits,
we have,
digit in tens place = 13 - x
digit in units place = x
The new number is
10(13 - x) + x
= 130 - 10x + x
= 130 - 9x _________(ii)

So,
the following balanced equation will be formed

130 - 9x - 27 = 9x + 13
103 - 9x = 9x + 13
103 - 13 = 9x + 9x
90 = 18x
x = 90/18
x = 5

Therefore,
digit in tens place = x = 5
digit in units place = 13 - x = 13 - 5 = 8

Thus,
the new number is 58


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Hope it helps !!