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Expanded Fractions and Decimals


Alignments to Content Standards: 4.NF.C.5 4.NF.C.6

Task

Complete the table.

Fraction Expanded Fraction Form Expanded Decimal Form Decimal
$43 \frac{65}{100}$ $40 + 3 + \frac{6}{10} + \frac{5}{100} $ $40 + 3 + 0.6 + 0.05$ $43.65$
$21 \frac{90}{100}$
$40 \frac{76}{100}$
$7 \frac{82}{100}$
$18 \frac{3}{100}$

IM Commentary

The purpose of this task is for students to show they understand the connection between fraction and decimal notation by writing the same numbers both ways. Comparing and contrasting the two solutions shown below shows why decimal notation can be confusing. The first solution shows the briefest way to represent each number, and the second solution makes all the zeros explicit.

Solutions

Solution: Supressing unnecessary zeros

Fraction Expanded Fraction Form Expanded Decimal Form Decimal
$43 \frac{65}{100}$ $40 + 3 + \frac{6}{10} + \frac{5}{100} $ $40 + 3 + 0.6 + 0.05$ $43.65$
$21 \frac{90}{100}$ $20 + 1 + \frac{9}{10} $ $20 + 1 + 0.9 $ $21.9$
$40 \frac{76}{100}$ $40 + \frac{7}{10} + \frac{6}{100} $ $40 + 0.7 + 0.06$ $40.76$
$7 \frac{82}{100}$ $7 + \frac{8}{10} + \frac{2}{100} $ $7 + 0.8 + 0.02$ $\hskip 6pt 7.82$
$18 \frac{3}{100}$ $10 + 8 + \frac{3}{100} $ $10 + 8 + 0.03$ $18.03$

Solution: Keeping all zeros

Fraction Expanded Fraction Form Expanded Decimal Form Decimal
$43 \frac{65}{100}$ $40 + 3 + \frac{6}{10} + \frac{5}{100} $ $40 + 3 + 0.6 + 0.05$ $43.65$
$21 \frac{90}{100}$ $20 + 1 + \frac{9}{10} + \frac{0}{100} $ $20 + 1 + 0.9 + 0.00$ $21.90$
$40 \frac{76}{100}$ $40 + 0 + \frac{7}{10} + \frac{6}{100} $ $40 + 0 + 0.7 + 0.06$ $40.76$
$7 \frac{82}{100}$ $\hskip 6pt 0 + 7 + \frac{8}{10} + \frac{2}{100} $ $\hskip 6pt 0 + 7 + 0.8 + 0.02$ $\hskip 6pt 7.82$
$18 \frac{3}{100}$ $10 + 8 + \frac{0}{10} + \frac{3}{100} $ $10 + 8 + 0.0 + 0.03$ $18.03$

tsalerno@cox.net says:

over 4 years

What if a student writes the answer to 21 90/100 and 20 + 1 + 9/10 and 20 + 1 + 0.9? This is correct mathematically. Will it get full credit?

Kristin says:

over 4 years

If I understand your question, the answer to this is yes (that is how it is shown in the first solution). The idea is that students might do it either way, so two (correct) solution approaches are shown.