Free Math Worksheets

- What whole number is nearest to \(4 \frac{2}{3}\)? - Rewrite each mixed number as an improper fraction before finding the exact product. - Look for common factors to cancel before multiplying.

1. For a), \(4 \frac{2}{3}\approx5\) and \(1 \frac{1}{2}\approx2\), so the estimate is \(5\times2=10\). 2. For b), rewrite the mixed numbers: \(4 \frac{2}{3}=\frac{14}{3}\) and \(1 \frac{1}{2}=\frac{3}{2}\). 3. Multiply and simplify: \(\frac{14}{3}\times\frac{3}{2}=\frac{14}{2}=7\).

a) Estimate: \(10\) b) Exact product: \(7\)

- Rewrite each mixed number as an improper fraction. - Look for common factors between numerators and denominators before multiplying. - Simplify the result completely.

1. For a), \(1 \frac{1}{4}=\frac{5}{4}\). Then \(\frac{5}{4}\times\frac{8}{15}=\frac{2}{3}\) after canceling common factors. 2. For b), \(2 \frac{2}{3}=\frac{8}{3}\) and \(1 \frac{1}{8}=\frac{9}{8}\). Then \(\frac{8}{3}\times\frac{9}{8}=3\).

a) \(\frac{2}{3}\) b) \(3\)

- Imagine splitting the rectangle into four smaller rectangles by separating each side length into a whole-number part and a fractional part. - How do you rewrite a mixed number as an improper fraction? - Look for common factors to cancel before multiplying.

1. For a), Jordan counted only two of the four partial areas. Splitting the side lengths into whole-number and fractional parts creates two additional positive products, \(2\times\frac{1}{5}\) and \(\frac{1}{2}\times3\), so the claimed area is too small. 2. For b), rewrite the side lengths: \(2 \frac{1}{2}=\frac{5}{2}\) and \(3 \frac{1}{5}=\frac{16}{5}\). 3. Multiply and simplify: \(\frac{5}{2}\times\frac{16}{5}=\frac{16}{2}=8\). 4. The actual area is \(8\,\text{cm}^2\).

a) Jordan omitted the two partial areas formed by multiplying a whole-number part by a fractional part, so the result is too small. b) \(8\,\text{cm}^2\)

- Write a mixed number as a sum of a whole number and a fraction. - Apply the distributive property when one factor is a whole number. - For two mixed numbers, expand \(\left(2 + \frac{1}{3}\right)\left(1 + \frac{1}{2}\right)\) and count the partial products.

1. Multiplying by a whole number: The method works because of the distributive property. For example, \(4 \times \left(2 + \frac{1}{3}\right) = 4 \times 2 + 4 \times \frac{1}{3} = 8 + \frac{4}{3} = 9\frac{1}{3}\). 2. Check by converting the mixed number: \(\frac{7}{3} \times 4 = \frac{28}{3} = 9\frac{1}{3}\). 3. Multiplying two mixed numbers: Multiplying only the whole-number parts and only the fractional parts does not work because it omits two partial products. 4. For example, the incorrect shortcut gives \(2 \times 1 + \frac{1}{3} \times \frac{1}{2} = 2\frac{1}{6}\) for \(2\frac{1}{3} \times 1\frac{1}{2}\). 5. The correct product is \(\frac{7}{3} \times \frac{3}{2} = \frac{21}{6} = 3\frac{1}{2}\). Since \(2\frac{1}{6} \ne 3\frac{1}{2}\), the shortcut is invalid for two mixed numbers.

The claim is true when one factor is a whole number. It is false for two mixed numbers because multiplying only the matching parts omits the cross-products. For example, \(2\frac{1}{3} \times 1\frac{1}{2} = 3\frac{1}{2}\), not \(2\frac{1}{6}\).