If you add the digits in a two-digit number and multiply the sum by 7, you get the original number. If you reverse the digits in the two-digit number, the new number is 18 more than the sum of its two digits. What is the original number?
Let's assume a the digit of tenth b the digit of unities. 7(a+b)=10a+b ==>3a=6b==>a=2b b10+a=18+a+b==>9b=18==>b=2 and a=4 The original number is 42. 7*(4+2)=42 24=18+4+2
The original number for this problem would be 42. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query about this specific question, thank you and have a nice day.
