Question
A number consists of two digits such that the digit in
the ten’s place is less by 1 than the digit in the unit’s place. Three times the number added to 2/3 times the number obtained by reversing the digits equals 171. The sum of digits in the number is:Solution
Let the unit’s digit be x Ten’s digit = x – 1 Number = 10 (x – 1) + x = 10x – 10 + x = 11x – 10 New number obtained after reversing the digits = 10x + x – 1 = 11x – 1 According to the question 3(11x – 10) + (2/3) x (11x – 1) = 171 33x – 30 + 22x/3 – 2/3 = 171 99x – 90 + 22x – 2 = 171 x 3 121x – 92 = 513 121x – 92 = 513 x = 605/121 = 5 Number = 11x –10 = 11 × 5 – 10 = 45 Sum of digits = 4 + 5 = 9
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