Question

1. what addition problem is represented by the fraction strips?

options:
$\frac{3}{10} + \frac{2}{10} = \frac{5}{10}$
$\frac{3}{10} + \frac{3}{10} = \frac{6}{10}$
$\frac{3}{10} + \frac{4}{10} = \frac{7}{10}$
$\frac{3}{10} + \frac{1}{10} = \frac{4}{10}$

Explanation:

Step1: Analyze the fraction strips

The first part (the vertical red strip with the three small red parts on top) represents $\frac{3}{10}$? Wait, no, looking at the fraction strips: the bottom red block has three $\frac{1}{10}$ parts, so that's $\frac{3}{10}$. The top part: the vertical red strip with the three small red parts (each $\frac{1}{10}$) and then the larger red area? Wait, no, maybe the initial part: the first addend is the number of $\frac{1}{10}$ in the first strip, and the second addend is the number of $\frac{1}{10}$ in the second strip. Wait, the bottom red block has three $\frac{1}{10}$ (so $\frac{3}{10}$), and the top part: the three small red parts (each $\frac{1}{10}$) plus the larger red area? Wait, no, looking at the options, the correct addition should be adding two fractions with denominator 10. Let's check the options:

Looking at the fraction strips, the first strip (the one with the three small red parts on top) – wait, maybe the first addend is $\frac{3}{10}$ (three $\frac{1}{10}$) and the second addend is another $\frac{3}{10}$? Wait, no, the bottom red block has three $\frac{1}{10}$, and the top part: the three small red parts (each $\frac{1}{10}$) and then the larger red area? Wait, no, the options include $\frac{3}{10} + \frac{3}{10} = \frac{6}{10}$. Let's check: if we have two groups of three $\frac{1}{10}$ parts, then adding them gives six $\frac{1}{10}$ parts, which is $\frac{6}{10}$. So the addition problem is $\frac{3}{10} + \frac{3}{10} = \frac{6}{10}$.

Step2: Match with the options

Looking at the options, the third option (from the left) is $\frac{3}{10} + \frac{3}{10} = \frac{6}{10}$, which matches our analysis.

Answer:

$\frac{3}{10} + \frac{3}{10} = \frac{6}{10}$ (corresponding to the option with this equation)