To find the number, let us denote it by [tex]\( x \)[/tex].
According to the problem, [tex]\(\frac{1}{4}\)[/tex] of [tex]\( x \)[/tex] is added to [tex]\( 4 \frac{1}{3} \)[/tex]. The expression for this part is:
[tex]\[
\frac{1}{4}x + 4 \frac{1}{3}
\][/tex]
We can convert the mixed number [tex]\( 4 \frac{1}{3} \)[/tex] into an improper fraction:
[tex]\[
4 \frac{1}{3} = 4 + \frac{1}{3} = \frac{12}{3} + \frac{1}{3} = \frac{13}{3}
\][/tex]
Now, substituting this back in, the left-hand side of the equation is:
[tex]\[
\frac{1}{4}x + \frac{13}{3}
\][/tex]
The problem also states that this sum is the same as when [tex]\(\frac{1}{3}\)[/tex] of [tex]\( x \)[/tex] is subtracted from [tex]\( 20 \frac{2}{3} \)[/tex]. The expression for this part is:
[tex]\[
20 \frac{2}{3} - \frac{1}{3}x
\][/tex]
Similarly, convert the mixed number [tex]\( 20 \frac{2}{3} \)[/tex] into an improper fraction:
[tex]\[
20 \frac{2}{3} = 20 + \frac{2}{3} = \frac{60}{3} + \frac{2}{3} = \frac{62}{3}
\][/tex]
Now, substituting this back in, the right-hand side of the equation is:
[tex]\[
\frac{62}{3} - \frac{1}{3}x
\][/tex]
We set up the equation as given in the problem:
[tex]\[
\frac{1}{4}x + \frac{13}{3} = \frac{62}{3} - \frac{1}{3}x
\][/tex]
To solve this equation, first clear the fractions by finding a common denominator for all terms. The common denominator for 4 and 3 is 12. Multiply every term by 12 to clear the fractions:
[tex]\[
12 \left( \frac{1}{4}x + \frac{13}{3} \right) = 12 \left( \frac{62}{3} - \frac{1}{3}x \right)
\][/tex]
Applying the multiplication:
[tex]\[
3x + 4 \cdot 13 = 4 \cdot 62 - 4x
\][/tex]
Simplify each term:
[tex]\[
3x + 52 = 248 - 4x
\][/tex]
Now, combine like terms:
[tex]\[
3x + 4x + 52 = 248
\][/tex]
[tex]\[
7x + 52 = 248
\][/tex]
Next, isolate [tex]\( x \)[/tex] by subtracting 52 from both sides:
[tex]\[
7x = 196
\][/tex]
Finally, divide by 7:
[tex]\[
x = 28
\][/tex]
Thus, the number is:
[tex]\[
\boxed{28}
\][/tex]
