Question

7 wholes and 1 forth minus 3 wholes and 2 thirds

Answer 1

To solve the problem, we need to first convert the mixed numbers into improper fractions.

1. Convert 7 whole and 1 fourth to an improper fraction:
\[
7 \frac{1}{4} = \frac{(7 \times 4) + 1}{4} = \frac{28 + 1}{4} = \frac{29}{4}
\]

2. Convert 3 whole and 2 thirds to an improper fraction:
\[
3 \frac{2}{3} = \frac{(3 \times 3) + 2}{3} = \frac{9 + 2}{3} = \frac{11}{3}
\]

Now, we need to perform the subtraction:
\[
\frac{29}{4} - \frac{11}{3}
\]

To subtract these fractions, we need a common denominator. The least common multiple of 4 and 3 is 12.

Convert each fraction to have the denominator of 12:

1. Convert \(\frac{29}{4}\) to a denominator of 12:
\[
\frac{29}{4} = \frac{29 \times 3}{4 \times 3} = \frac{87}{12}
\]

2. Convert \(\frac{11}{3}\) to a denominator of 12:
\[
\frac{11}{3} = \frac{11 \times 4}{3 \times 4} = \frac{44}{12}
\]

Now we can subtract:
\[
\frac{87}{12} - \frac{44}{12} = \frac{87 - 44}{12} = \frac{43}{12}
\]

Finally, we can convert \(\frac{43}{12}\) back into a mixed number:
\[
\frac{43}{12} = 3 \frac{7}{12} \quad (\text{since } 43 \div 12 = 3 \text{ R } 7)
\]

So, the result of \(7\) wholes and \(1\) fourth minus \(3\) wholes and \(2\) thirds is:
\[
3 \frac{7}{12}
\]