Question

4 minus 2 wholes and 7 twelths

Answer 1

To subtract \(2\) wholes and \(7\) twelfths from \(4\), we can express the whole number \(4\) as a fraction with a denominator of \(12\).

First, convert \(4\) into twelfths:

\[
4 = \frac{4 \times 12}{1 \times 12} = \frac{48}{12}
\]

Next, we want to subtract \(2\) wholes and \(7\) twelfths. In fraction form, \(2\) wholes is:

\[
2 = \frac{2 \times 12}{1 \times 12} = \frac{24}{12}
\]

Now, \(2\) wholes and \(7\) twelfths can be written as:

\[
2 + \frac{7}{12} = \frac{24}{12} + \frac{7}{12} = \frac{31}{12}
\]

Now we can perform the subtraction:

\[
4 - (2 + \frac{7}{12}) = \frac{48}{12} - \frac{31}{12}
\]

Subtract the fractions:

\[
\frac{48 - 31}{12} = \frac{17}{12}
\]

So, \(4\) minus \(2\) wholes and \(7\) twelfths equals:

\[
\frac{17}{12}
\]

This can also be expressed as:

\[
1 \frac{5}{12}
\]

Thus, \(4 - 2 \text{ wholes and } 7 \text{ twelfths} = \frac{17}{12}\) or \(1 \frac{5}{12}\).