Question

The digits of a 2-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, we get99. Find the original number.

Answer 1

Hey there!
Let the one's digit be a and the ten's digit be b
Original no. = 10b + a
Reversed no. = 10 a+ b

Case 1 : ( When a > b)
a - b = 5
a = 5+b

(10a + b) + (10b + a ) = 99
10a + a + 10b + b = 99
11a + 11b = 99
11( a + b ) = 99
a + b = 9

Putting the value of a in the equation:
5 + b + b = 9
5 + 2b = 9
2b = 4
b = 2

a = 5+ b
= 5 + 2
= 7

Therefore, the no. will be (10*2) + 7 i.e 27.

Case 2 : ( when b > a )
The digits of the no. will be reversed i.e the no. will be 72.

Hope it helps!

Answer 2

Answer:

Let the one's digit be a and the ten's digit be b

Original no. = 10b + a

Reversed no. = 10 a+ b

Case 1 : ( When a > b)

a - b = 5

a = 5+b

(10a + b) + (10b + a ) = 99

10a + a + 10b + b = 99

11a + 11b = 99

11( a + b ) = 99

a + b = 9

Putting the value of a in the equation:

5 + b + b = 9

5 + 2b = 9

2b = 4

b = 2

a = 5+ b

= 5 + 2

= 7

Therefore, the no. will be (10*2) + 7 i.e 27.

Step-by-step explanation: