Question

for 1-12, find each sum.
1.
$\begin{array}{r}2\frac{10}{12}\\+3\frac{3}{12}\hlineend{array}$
2.
$\begin{array}{r}1\frac{3}{8}\\+3\frac{6}{8}\hlineend{array}$
3.
$\begin{array}{r}5\frac{4}{10}\\+4\frac{2}{10}\hlineend{array}$
4.
$\begin{array}{r}10\frac{2}{6}\\+ \frac{3}{6}\hlineend{array}$

Explanation:

Problem 1:

Step1: Add the whole numbers

$2 + 3 = 5$

Step2: Add the fractions

$\frac{10}{12} + \frac{3}{12} = \frac{13}{12}$

Step3: Convert improper fraction to mixed number and add to whole number

$\frac{13}{12} = 1\frac{1}{12}$, so $5 + 1\frac{1}{12} = 6\frac{1}{12}$

Step1: Add the whole numbers

$1 + 3 = 4$

Step2: Add the fractions

$\frac{3}{8} + \frac{6}{8} = \frac{9}{8}$

Step3: Convert improper fraction to mixed number and add to whole number

$\frac{9}{8} = 1\frac{1}{8}$, so $4 + 1\frac{1}{8} = 5\frac{1}{8}$

Step1: Add the whole numbers

$5 + 4 = 9$

Step2: Add the fractions

$\frac{4}{10} + \frac{2}{10} = \frac{6}{10}$

Step3: Simplify the fraction and add to whole number

$\frac{6}{10} = \frac{3}{5}$, so $9 + \frac{3}{5} = 9\frac{3}{5}$

Answer:

$6\frac{1}{12}$

Problem 2: