The digit in the units place of a two digit number is 4 times the digit in the 10th place the number obtained by reversing the digits exceeds the given number by 54 find the given number.?

Question

Answer

Understanding the Problem
To solve the problem, let's denote the two-digit number as $10a + b$, where $a$ is the digit in the tens place and $b$ is the digit in the units place.

Given Conditions
- The units digit $b$ is 4 times the tens digit $a$:

$b = 4a$

- The number obtained by reversing the digits exceeds the original number by 54:

$10b + a = (10a + b) + 54$

Setting Up Equations
1. From the first condition $b = 4a$.
2. From the second condition, we can simplify the equation:

$10b + a - 10a - b = 54$
$9b - 9a = 54$
$b - a = 6$

Substituting Values
Now, substitute $b = 4a$ into $b - a = 6$:

$4a - a = 6$
$3a = 6$
$a = 2$

Now, substitute $a$ back to find $b$:

$b = 4 \times 2 = 8$

Finding the Final Number
Thus, the two-digit number is:

$10a + b = 10 \times 2 + 8 = 28$

Conclusion
The given two-digit number is **28**.

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