// Numbas version: finer_feedback_settings {"name": "Decimals: Addition", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Decimals: Addition", "tags": [], "metadata": {"description": "

Decimals addition algorithm. 2 and 3 digit numbers. Carrying.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Write the following question down on paper and evaluate it without using a calculator.

\n

If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

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$\\var{threedigit1}+\\var{threedigit2} = $ [[0]]

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Generally, we set up $\\var{threedigit1}+\\var{threedigit2}$ with the decimal points lined up vertically so that the columns with the same place value are also lined up vertically:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$0$.$\\var{cdigs[2]}$$\\var{cdigs[1]}$$+$
$0$.$\\var{cdigs[5]}$$\\var{cdigs[4]}$$\\var{cdigs[3]}$
$\\phantom{0}$
\n

\n

Note that we can pad out the decimal with zeros if we prefer:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$0$.$\\var{cdigs[2]}$$\\var{cdigs[1]}$$\\color{red}{\\var{cdigs[0]}}$$+$
$0$.$\\var{cdigs[5]}$$\\var{cdigs[4]}$$\\var{cdigs[3]}$
$\\phantom{0}$
\n

\n

Now we add the digits in the column to the far right (in this case, the thousandths column).

\n

This results in $\\var{cunitsum}$ and so we place $\\var{cunitsumlastdigit}$ under the line in this column.

\n

This results in $\\var{cunitsum}$ and so we place $\\var{cunitsumlastdigit}$ under the line in this column and carry the $1$ into the next column to the left. 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\overset{\\phantom{1}}{0}$.$\\overset{\\phantom{1}}{\\var{cdigs[2]}}$$\\overset{\\color{red}1}{\\var{cdigs[1]}}$ $\\overset{\\phantom{0}}{\\var{cdigs[1]}}$$\\color{green}{\\overset{\\phantom{1}}{\\var{cdigs[0]}}}$$+$
$0$.$\\var{cdigs[5]}$$\\var{cdigs[4]}$$\\color{green}{\\var{cdigs[3]}}$
$\\color{red}{\\var{cunitSumLastDigit}}$
\n

\n

Now we add the digits in the next column to the left (in this case, the hundredths column).

\n

This results in $\\var{ctenSum}$ and so we place $\\var{ctenSumlastdigit}$ under the line in this column.

\n

This results in $\\var{ctenSum}$ and so we place $\\var{ctenSumlastdigit}$ under the line in this column and carry the $1$ into the next column to the left.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\overset{\\phantom{1}}{0}$.$\\overset{\\color{red}{1}}{\\var{cdigs[2]}}$ $\\overset{\\phantom{1}}{\\var{cdigs[2]}}$$\\color{green}{\\overset{1}{\\var{cdigs[1]}}}$ $\\color{green}{\\overset{\\phantom{0}}{\\var{cdigs[1]}}}$$\\overset{\\phantom{1}}{\\var{cdigs[0]}}$$+$
$0$.$\\var{cdigs[5]}$$\\color{green}{\\var{cdigs[4]}}$$\\var{cdigs[3]}$
$\\color{red}{\\var{ctenSumlastdigit}}$${\\var{cunitSumLastDigit}}$
\n

\n

\n

Now we add the digits in the next column to the left (in this case, the tenths column).

\n

This is $\\var{chunsum}$ so we place $\\var{chunsum}$ under the line in this column.

\n

This is $\\var{chunsum}$ so we place $\\var{chunsumlastdigit}$ under the line in this column and carry $\\var{chuncarry}$ into the next column to the left (which in this case is the ones column on the other side of the decimal point).

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\overset{\\color{red}1}{0}$ $\\overset{\\phantom{1}}{0}$.$\\color{green}{\\overset{1}{\\var{cdigs[2]}}}$ $\\color{green}{\\overset{\\phantom{1}}{\\var{cdigs[2]}}}$$\\overset{1}{\\var{cdigs[1]}}$ $\\overset{\\phantom{0}}{\\var{cdigs[1]}}$$\\overset{\\phantom{1}}{\\var{cdigs[0]}}$$+$
$0$.$\\color{green}{\\var{cdigs[5]}}$$\\var{cdigs[4]}$$\\var{cdigs[3]}$
.$\\color{red}{\\var{chunsumlastdigit}}$$\\var{ctenSumlastdigit}$${\\var{cunitSumLastDigit}}$
\n

\n

Now we add the digits in the next column to the left (in this case, the ones column).

\n

This is just $0$ so we place $0$ under the line in this column.

\n

This is just $\\var{chuncarry}$ so we place $\\var{chuncarry}$ under the line in this column.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\color{green}{\\overset{1}{0}}$ $\\color{green}{\\overset{\\phantom{1}}{0}}$.$\\overset{1}{\\var{cdigs[2]}}$ $\\overset{\\phantom{1}}{\\var{cdigs[2]}}$$\\overset{1}{\\var{cdigs[1]}}$ $\\overset{\\phantom{0}}{\\var{cdigs[1]}}$$\\overset{\\phantom{1}}{\\var{cdigs[0]}}$$+$
$\\color{green}{0}$.$\\var{cdigs[5]}$$\\var{cdigs[4]}$$\\var{cdigs[3]}$
$\\color{red}{\\var{chuncarry}}$.$\\var{chunsumlastdigit}$$\\var{ctenSumlastdigit}$${\\var{cunitSumLastDigit}}$
\n

The answer is therefore $\\var{cans}$.

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