Regrouping in calculations

Worked solution

1. 1

   Add the ones and exchange ten of them.

   $7+8=15=10+5$

   Keep five ones and pass one new ten to the tens place.

2. 2

   Add the tens, including the new ten.

   $6+5+1=12$

   Twelve tens become one hundred and two tens.

3. 3

   Add the hundreds, including the new hundred.

   $2+1+1=4$

   The regrouped hundred joins the original hundreds.

4. 4

   Combine the remaining place values.

   $400+20+5=425$

   The result contains four hundreds, two tens and five ones.

Worked solution

1. 1

   Exchange a hundred, then a ten.

   $\begin{array}{rl}403& =300+100+3\\ & =300+90+13\end{array}$

   One hundred supplies ten tens; one of those tens supplies ten extra ones.

2. 2

   Subtract in the ones place.

   $13-6=7$

   There are now enough ones to remove six.

3. 3

   Subtract in the tens and hundreds places.

   $\begin{array}{rl}90-70& =20\\ 300-100& =200\end{array}$

   Use the reduced counts after both exchanges: nine tens and three hundreds.

4. 4

   Combine the places and check.

   $200+20+7=227227+176=403$

   The difference and the removed amount recover the starting number.