Question: A farmer placed an order for 15\frac{3}{8} tons of seed. He calculated that his first project…

A farmer placed an order for 15\frac{3}{8} tons of seed. He calculated that his first project would require 11\frac{3}{4} tons of seed. How much seed will he have left over for his remaining projects?

Solution

The problem involves finding how much seed is left after the farmer uses some for his project. First, convert the mixed numbers to improper fractions. For \(15 \frac{3}{8}\): \[ 15 \frac{3}{8} = \frac{15 \times 8 + 3}{8} = \frac{123}{8} \] For \(11 \frac{3}{4}\): \[ 11 \frac{3}{4} = \frac{11 \times 4 + 3}{4} = \frac{47}{4} \] Next, subtract the two fractions. First, find a common denominator: The least common denominator of 8 and 4 is 8. Convert \(\frac{47}{4}\) to the common denominator: \[ \frac{47}{4} = \frac{47 \times 2}{4 \times 2} = \frac{94}{8} \] Now subtract: \[ \frac{123}{8} - \frac{94}{8} = \frac{123 - 94}{8} = \frac{29}{8} \] Convert \(\frac{29}{8}\) back to a mixed number: \[ \frac{29}{8} = 3 \frac{5}{8} \] The farmer will have \(3 \frac{5}{8}\) tons of seed left for his remaining projects.