Place Value

Estimate a Sum by Rounding

Estimate 384 + 257 by rounding each addend to the nearest ten first: 380 + 260 = 640, which is close to the exact answer of 641.

Example

Round 384 and 257 to estimate their sum.

highlighted = computed this step

Step 1 — Set up

Line up the addends by place value.

xxxx384+257x□□□\begin{array}{rrrr}\phantom{x}&\phantom{x}&\phantom{x}&\phantom{x}\\&3&8&4\\+&2&5&7\\\hline\\\phantom{x}&\square&\square&\square\end{array}
Round 384 and 257 to nearby tens3843803843853905ones 4384 -> 3802572502552572605ones 7257 -> 260round each addend, then addEstimate first by rounding each addend to the nearest ten.

Step 2 — Round first addend

Round 384: ones digit 4, round down to 380.

384→380384\rightarrow \hl{380}
384 has ones digit 4, so round down3843803843853905ones 4384 -> 3802572502552572605ones 7257 -> 260round each addend, then add4 is less than 5, so 384 rounds to 380.

Step 3 — Round second addend

Round 257: ones digit 7, round up to 260.

257→260257\rightarrow \hl{260}
257 has ones digit 7, so round up3843803843853905ones 4384 -> 3802572502552572605ones 7257 -> 260round each addend, then add7 is at least 5, so 257 rounds to 260.

Step 4 — Add rounded numbers

Add the rounded numbers: 380 + 260 = 640.

380+260=640380+260=\hl{640}
Add rounded values: 380 + 260 = 6403843803843853905ones 4384 -> 3802572502552572605ones 7257 -> 260380 + 260 = 640rounded sumThe estimate uses the rounded numbers, not the original ones.

Step 5 — Exact sum

Check the exact sum: 384 + 257 = 641.

xxxx384+257x641,384+257=641\begin{array}{rrrr}\phantom{x}&\phantom{x}&\phantom{x}&\phantom{x}\\&3&8&4\\+&2&5&7\\\hline\\\phantom{x}&6&4&1\end{array},\quad 384+257=641
Estimate 640; exact sum 6413843803843853905ones 4384 -> 3802572502552572605ones 7257 -> 260380 + 260 = 640rounded sumThe estimate 640 is close to the exact sum 641.
estimation Round each number to the nearest ten, then add the rounded values. The result is an estimate — close to the exact answer but easier to compute mentally. Rounding before adding reduces the chance of arithmetic errors.