sum of digits of a two digit no in 12 new no is formed by reversing the digit is greater than the original no. by 54 . Find the original no.
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Given sum of the digits is 12 Let the digits in ones place be x Hence the digit in tens place is (12 – x) The original number = 10(12 – x) + x = 120 – 9x Number formed by reversing the digits = 10x + (12 – x) = 9x + 12 Given that number formed by reversing the digits is 54 greater than the original number. ⇒ 9x + 12 = (120 – 9x) + 54 = 174 – 9x ⇒ 18x = 174 – 12 = 162 ∴ x = 9 The original number = 120 – 9x = 120 – 9(9) = 39
Given sum of the digits is 12 Let the digits in ones place be x Hence the digit in tens place is (12 – x) The original number = 10(12 – x) + x = 120 – 9x Number formed by reversing the digits = 10x + (12 – x) = 9x + 12 Given that number formed by reversing the digits is 54 greater than the original number. ⇒ 9x + 12 = (120 – 9x) + 54 = 174 – 9x ⇒ 18x = 174 – 12 = 162 ∴ x = 9 The original number = 120 – 9x = 120 – 9(9) = 39
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