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The digit at the tens place of a 2-digit number is three times the digit at the ones place. if the number and the number formed by reversing the digits is

Let the digit at the ones place be x Then the digit at the tens place will be = 3x Therefore the number is 10(3x) +x = 30x+x = 31x By reversing the digits we get 10(x)+3x = 10x+3x = 13x Their sum = 88 ⇒31x + 13x = 88 ⇒ 44x = 88 ⇒ x = 88/44 ∴ x = 2 Then the digits are x, 3x = 2 and 3(2) = 2 and 6 The number = 31x = 31(2) = 62 ∴ The number is 62

Let the digit in one's place be x. the digit in ten's place= 3x the original number= 10(3x) + x (we multiplied by 10 because we have to find the value in ten's place.) when the digits are interchanged the number = 10x + 3x a/q 10x+3x + (10(3x)+x) = 88 = 13x+30x+x = 88 = 44x = 88 = x = 88/44 ∴ x= 2 digit in unit's place= 2 digit in ten's place = 3x= 3*2= 6 the original number= 6*10 + 2 = 60+2 = 62 so the required number is 62.