For $$6,\square 48$$ to round down to $$6,000$$, the hundreds digit must make the number less than $$6,500$$. The largest digit that works is $$4$$, giving $$6,448$$, which rounds down to $$6,000$$. Choice A gives a smaller valid digit but not the largest. Choice C gives $$6,548$$, which rounds up to $$7,000$$. Choice D gives $$6,648$$, which also rounds up to $$7,000$$.

First find the total: $$5,847 + 6,239 + 5,156 + 6,891 = 24,133$$. Then round to the nearest ten thousand: since $$4,133 < 5,000$$, the number $$24,133$$ rounds down to $$20,000$$. Choice B rounds to the nearest thousand instead of ten thousand. Choice C is the exact sum without rounding. Choice D incorrectly rounds up to the next ten thousand.

When you see a rounding problem that asks for differences between rounded values, you need to carefully round to each specified place value first, then subtract to find the difference.

Let's start with $$8,429$$ and round it to the nearest thousand. Look at the hundreds digit (4) to decide whether to round up or down. Since 4 is less than 5, you round down to $$8,000$$.

Next, round $$8,429$$ to the nearest hundred. Look at the tens digit (2) to make your decision. Since 2 is less than 5, you round down to $$8,400$$.

Now find the difference: $$8,400 - 8,000 = 400$$.

Let's examine why the other answers are wrong. Answer A ($$371$$) likely comes from subtracting $$8,429 - 8,000 = 429$$, then making an arithmetic error. Answer B ($$571$$) doesn't correspond to any logical rounding calculation with these numbers. Answer C ($$429$$) is the difference between the original number and the number rounded to the nearest thousand ($$8,429 - 8,000$$), but the question asks for the difference between the two rounded numbers, not between a rounded and unrounded number.

The key strategy here is to always round first, then perform any operations on the rounded numbers. Don't mix rounded and unrounded values in your final calculation. Also, remember that when rounding, you only look at the digit immediately to the right of the place value you're rounding to.

When you encounter rounding problems with two different place values, you need to find a number that satisfies both rounding conditions simultaneously. Think about the range of numbers that would round to each given value.

For a number to round to $$3,490$$ when rounded to the nearest ten, it must be between $$3,485$$ and $$3,494$$ (since $$3,485$$ rounds up to $$3,490$$, while $$3,495$$ would round up to $$3,500$$). For the same number to round to $$3,500$$ when rounded to the nearest hundred, it must be between $$3,450$$ and $$3,549$$.

The mystery number must fall in both ranges, so it's between $$3,485$$ and $$3,494$$. Only choice D) $$3,493$$ fits this range. Let's verify: $$3,493$$ rounds to $$3,490$$ (nearest ten) and $$3,500$$ (nearest hundred) ✓

Choice A) $$3,485$$ rounds to $$3,490$$ for the nearest ten, but rounds to $$3,500$$ for the nearest hundred, which seems right at first glance. However, $$3,485$$ is exactly at the boundary and would round to $$3,490$$ for tens, but the hundreds rounding works.

Choice B) $$3,504$$ rounds to $$3,500$$ for both tens and hundreds, not $$3,490$$ for tens.

Choice C) $$3,496$$ rounds to $$3,500$$ for tens (not $$3,490$$) and $$3,500$$ for hundreds.

Study tip: When solving two-condition rounding problems, find the overlap between the two ranges. Draw number lines if it helps visualize which numbers satisfy both conditions.

For $$9,\square 67$$ to round to $$9,600$$ (nearest hundred), the hundreds digit must be $$5$$ or $$6$$. If it's $$5$$: $$9,567$$ rounds to $$9,600$$ because $$67 \geq 50$$. If it's $$6$$: $$9,667$$ rounds to $$9,700$$ because $$67 \geq 50$$. So the hundreds digit must be $$5$$. For $$9,567$$ to round to $$10,000$$ (nearest thousand), we need $$567 \geq 500$$, which is true. Therefore, the hundreds digit is $$5$$.

The number must be between $$2,250$$ and $$2,349$$ to round to $$2,300$$ (nearest hundred), and between $$2,345$$ and $$2,354$$ to round to $$2,350$$ (nearest ten). Only $$2,349$$ satisfies both conditions. Choice A rounds to $$2,340$$ (nearest ten). Choice C rounds to $$2,400$$ (nearest hundred). Choice D rounds to both $$2,400$$ (nearest hundred) and $$2,360$$ (nearest ten).

This question tests 4th grade ability to use place value understanding to round multi-digit whole numbers to any place (CCSS.4.NBT.3). Rounding simplifies a number by changing it to the nearest value at a specified place. The process: (1) Identify the rounding place, (2) Look at the digit immediately to the right, (3) If that digit is 0-4, round DOWN (keep the rounding digit the same), if it's 5-9, round UP (increase the rounding digit by 1), (4) All digits to the right of the rounding place become 0. To round 45,678 to the nearest thousand, we look at the 5 in the thousands place and check the 6 in the hundreds place to determine whether to round up. Choice B is correct because the digit to the right of the thousands is 6, which is ≥ 5 so we round UP, changing the thousands from 5 to 6 and making all digits to the right 0, giving us 46,000. This demonstrates correct application of rounding rules and place value understanding. Choice A represents rounding to the nearest hundred instead, which happens when students don't identify the correct rounding place. To help students, use place value charts to identify the rounding place and decision digit clearly, and teach the 'neighbor rule'—look at the neighbor to the right (0-4 = stay, 5-9 = go up). Use number lines to show which benchmark the number is closer to, and emphasize that all digits to the right become 0.

This question tests 4th grade ability to use place value understanding to round multi-digit whole numbers to any place (CCSS.4.NBT.3). Rounding simplifies a number by changing it to the nearest value at a specified place. The process: (1) Identify the rounding place, (2) Look at the digit immediately to the right, (3) If that digit is 0-4, round DOWN (keep the rounding digit the same), if it's 5-9, round UP (increase the rounding digit by 1), (4) All digits to the right of the rounding place become 0. To round 8,456 to the nearest hundred, we look at the 4 in the hundreds place and check the 5 in the tens place to determine whether to round up. Choice C is correct because the digit to the right of the hundreds place is 5, which is ≥5 so we round up, changing the hundreds from 4 to 5 and making all digits to the right 0, giving us 8,500. This demonstrates correct application of rounding rules and place value understanding. Choice B represents rounding to the nearest ten instead of hundred, which happens when students don't identify the correct rounding place. To help students: Use place value charts to identify rounding place and decision digit clearly. Teach the 'neighbor rule'—look at the neighbor to the right (0-4 = stay, 5-9 = go up). Use number lines to show which benchmark (8,400 or 8,500) a number is closer to. Emphasize that ALL digits to the right become 0 (8,456 rounded to nearest hundred is 8,500, not 8,456). Practice with benchmark numbers: 8,450 is exactly halfway between 8,400 and 8,500—by convention we round up to 8,500. Remind students to check their answer makes sense (should be close to original). Watch for: rounding to wrong place, rounding wrong direction (up when should be down), truncating (chopping off) instead of rounding, and forgetting to add all the necessary zeros.

This question tests 4th grade ability to use place value understanding to round multi-digit whole numbers to any place (CCSS.4.NBT.3). Rounding simplifies a number by changing it to the nearest value at a specified place. The process: (1) Identify the rounding place, (2) Look at the digit immediately to the right, (3) If that digit is 0-4, round DOWN (keep the rounding digit the same), if it's 5-9, round UP (increase the rounding digit by 1), (4) All digits to the right of the rounding place become 0. To round 28,345 to the nearest ten thousand, we look at the 2 in the ten thousands place and check the 8 in the thousands place to determine whether to round up. Choice C is correct because the digit to the right of the ten thousands place is 8, which is ≥5 so we round up, changing the ten thousands from 2 to 3 and making all digits to the right 0, giving us 30,000. This demonstrates correct application of rounding rules and place value understanding. Choice B represents rounding to the nearest thousand instead of ten thousand, which happens when students don't identify the correct rounding place. To help students: Use place value charts to identify rounding place and decision digit clearly. Teach the 'neighbor rule'—look at the neighbor to the right (0-4 = stay, 5-9 = go up). Use number lines to show which benchmark (20,000 or 30,000) a number is closer to. Emphasize that ALL digits to the right become 0 (28,345 rounded to nearest ten thousand is 30,000, not 28,345). Practice with benchmark numbers: 25,000 is exactly halfway between 20,000 and 30,000—by convention we round up to 30,000. Remind students to check their answer makes sense (should be close to original). Watch for: rounding to wrong place, rounding wrong direction (up when should be down), truncating (chopping off) instead of rounding, and forgetting to add all the necessary zeros.

This question tests 4th grade ability to use place value understanding to round multi-digit whole numbers to any place (CCSS.4.NBT.3). Rounding simplifies a number by changing it to the nearest value at a specified place. The process: (1) Identify the rounding place, (2) Look at the digit immediately to the right, (3) If that digit is 0-4, round DOWN (keep the rounding digit the same), if it's 5-9, round UP (increase the rounding digit by 1), (4) All digits to the right of the rounding place become 0. To round 24,678 to the nearest thousand, we look at the 4 in the thousands place and check the 6 in the hundreds place to determine whether to round up. Choice C is correct because the digit to the right of the thousands place is 6, which is ≥5 so we round up, changing the thousands from 4 to 5 and making all digits to the right 0, giving us 25,000. This demonstrates correct application of rounding rules and place value understanding. Choice B represents rounding down incorrectly, which happens when students misapply the 0-4 down, 5-9 up rule. To help students: Use place value charts to identify rounding place and decision digit clearly. Teach the 'neighbor rule'—look at the neighbor to the right (0-4 = stay, 5-9 = go up). Use number lines to show which benchmark (24,000 or 25,000) a number is closer to. Emphasize that ALL digits to the right become 0 (24,678 rounded to nearest thousand is 25,000, not 25,678). Practice with benchmark numbers: 24,500 is exactly halfway between 24,000 and 25,000—by convention we round up to 25,000. Remind students to check their answer makes sense (should be close to original). Watch for: rounding to wrong place, rounding wrong direction (up when should be down), truncating (chopping off) instead of rounding, and forgetting to add all the necessary zeros.