Question

To find the sum of \[\left( {42505 + 27807 + 21397} \right)\] to the nearest thousand.

Answer 1

Hint: We have to first find the sum of the numbers and then estimate the sum of the numbers to the nearest thousand. Rounding off a number means to trim a number to make it simple while representing the value as close as possible. To round off a number to the nearest 1000, check the hundredth digit of the number. Round off the number to the previous thousand if the hundredth digit is 0, 1, 2, 3 and 4. Round off the number to the next thousand if the thousandth digit is 5, 6, 7, 8 and 9.

Complete step-by-step solution:
First we have to find the sum of the number, so the sum is
\[
  \begin{array}{*{20}{c}}
  4&2&5&0&5 \\
  2&7&8&0&7 \\
  2&1&3&9&7
\end{array} \\
   - - - - - - - - - \\
  \begin{array}{*{20}{c}}
  9&1&7&0&9
\end{array} \\
 \]
So the estimated value is \[91709\].
Now the value at hundreds place is greater than 5, that is \[709 > 500\]. Hence, we need to maximize the value of the digit at thousands place.
Therefore,
\[
  \begin{array}{*{20}{c}}
  9&1&0&0&0
\end{array} \\
  \begin{array}{*{20}{c}}
  {}&1&0&0&0
\end{array} \\
   - - - - - - - - - \\
  \begin{array}{*{20}{c}}
  9&2&0&0&0
\end{array} \\
 \]

Hence, the number \[91,709\] is rounded off to the nearest thousands as \[92,000\].

Note: Alternatively, it can be solved by rounding off to the nearest hundred and then we can add them.
\[
  \begin{array}{*{20}{c}}
  4&2&5&0&0 \\
  2&7&8&0&0 \\
  2&1&4&0&0
\end{array} \\
   - - - - - - - - - \\
  \begin{array}{*{20}{c}}
  9&1&7&0&0
\end{array} \\
 \]
Now the value at hundreds places is greater than 5, that is \[7 > 5\]. Hence, we need to maximize the value of the digit at thousands place.
Therefore,
\[
  \begin{array}{*{20}{c}}
  9&1&0&0&0
\end{array} \\
  \begin{array}{*{20}{c}}
  {}&1&0&0&0
\end{array} \\
   - - - - - - - - - \\
  \begin{array}{*{20}{c}}
  9&2&0&0&0
\end{array} \\
 \]
Hence, the number \[91,709\] is rounded off to the nearest thousands as \[92,000\].