Question

the sum of the digits in a two digit no. is 8. if 18 is added to the number the digits are reversed find the number

Answer 1

Let the tens place digit be a

And Unit place digit be b

According to first condition,

Sum of digits = a + b

=> a + b = 8 -------(1)

Now,

According to second condition,

10a+b + 18 = 10b+a

=> 10a + b - 10b - a + 18 = 0

=> 9a - 9b = - 18

=> a - b = - 2 -------(2)


On adding equation 1 and 2, we get

2a = 6

=> a = 3

Now,

On Substituting the value of a in equation 1, we get

3 + b = 8

=> b = 5

Required number = 10 × 3 + 5 = 35

Answer 2

Let the number be 10x+y
10x because x is in tens place..

Sum of the digits is 8
x+y = 8. ;. x = 8-y

10x+y+18 = 10y+x { Number is reversed }
10(8-y)+y+18 = 10y+8-y
80-10y+y+18 = 9y+8
80-9y+18 = 9y+8
80+18-8 = 9y+9y
90 = 18y
y = 5

x = 8-y
x = 8-5
x = 3

Hence the number is 35..

VERIFICATION:-

35 reverse is equal to 53

35+18 = 53


Hope this helps!!
cheers!! (:

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