The ratio between a two-digit number and the sum of the digits of that number is 4:1. If the digit in the unit place is 3 more than the digit in the ten's place, what is the number?
Question
The ratio between a two-digit number and the sum of the digits of that number is 4:1. If the digit in the unit place is 3 more than the digit in the ten's place, what is the number?
Solution
Let's solve this step by step:
Step 1: Let's denote the digit in the tens place as 'x' and the digit in the units place as 'y'. According to the problem, y = x + 3.
Step 2: The two-digit number can be expressed as 10x + y.
Step 3: The sum of the digits of the number is x + y.
Step 4: According to the problem, the ratio between the two-digit number and the sum of the digits is 4:1. So, we can write the equation as (10x + y) / (x + y) = 4/1.
Step 5: Simplifying the equation, we get 10x + y = 4x + 4y.
Step 6: Rearranging the terms, we get 6x = 3y.
Step 7: Substituting y = x + 3 into the equation, we get 6x = 3(x + 3).
Step 8: Solving for x, we get x = 3.
Step 9: Substituting x = 3 into y = x + 3, we get y = 6.
Step 10: Therefore, the two-digit number is 36.
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