Question

In a 3 digit number, sum of the digit is 9 and the unit digit is twice the the the tens digit. on adding 99 to the number, the digit are interchanged. Find the number.​

Answer 1

1. a+b+c=9

2.c= 2b

3. 100c+10b+ a = 100a+10b+c+99

Next substitute given 2 into given 3.

100(2b) + 10b+a =100a+10b+c+99

210b-12b+a = 100a+12b-12b+99

198b+a-a=100a-a+99

198b-99= 99a+99-99

(198b-99)/99=99a/99

a=2b-1

Now substitute the value for a, and for c into given 1.

(2b-1) + b + (2b) = 9

5b-1+1=9+1

5b/5=10/5

b = 2

a=2b -1 so a = 3

c = 2b, so c = 4

So the number is 234.

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