a two digit number is a such that the digit at the one place is 4 more than the tens place the ratio of these number and number formed by reversing its digits is 47 find the number

Let's denote the two-digit number as \( N \), and let the tens digit be \( x \) and the ones digit be \( y \).

From the problem statement, we have the following conditions:

1. The digit at the ones place is 4 more than the tens place:
   \[
   y = x + 4
   \]

2. The ratio of the number \( N \) and the number formed by reversing its digits is 47. The number \( N \) can be written as:
   \[
   N = 10x + y
   \]
   The number formed by reversing the digits would be:
   \[
   \text{Reversed number} = 10y + x
   \]
   The given ratio is:
   \[
   \frac{N}{\text{Reversed number}} = 47
   \]
   Substituting the expressions for \( N \) and the reversed number:
   \[
   \frac{10x + y}{10y + x} = 47
   \]

Now, substitute \( y = x + 4 \) into this equation:
\[
\frac{10x + (x + 4)}{10(x + 4) + x} = 47
\]
Simplifying the numerator and denominator:
\[
\frac{10x + x + 4}{10x + 40 + x} = 47
\]
\[
\frac{11x + 4}{11x + 40} = 47
\]

Now, cross-multiply to solve for \( x \):
\[
11x + 4 = 47(11x + 40)
\]
\[
11x + 4 = 517x + 1880
\]
\[
4 - 1880 = 517x - 11x
\]
\[
-1876 = 506x
\]
\[
x = \frac{-1876}{506} = -3.71
\]

It seems there was a miscalculation in some part of the problem. Let me verify this!

A two digit number is such that the digit at the one place is 4 more than the tens place. The ratio of these number and number formed by reversing its digits is 47. Find the number.

Solve a two-digit number problem where the ones place is 4 more than the tens place, and the ratio of the number to the number formed by reversing its digits is 47. This algebraic problem is ideal for students in Grades 7-9.

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