3-Digit Addition with Regrouping Practice Quiz

When adding 134 and 256, the ones digit sums to 10, so you write down 0 and carry over 1. Adding the tens and hundreds digits with the carried value gives the correct total of 390.

In this subtraction, borrowing is required because 5 is less than 7 in the ones place. After borrowing and subtracting each digit correctly, the final answer is 118.

Adding the ones digits gives 8 + 9 = 17, so a 7 is written down and 1 is carried over. Continuing with the tens and hundreds places leads to the sum of 567.

The subtraction requires borrowing in the ones column since 2 is less than 6, and then again in the tens column. Following these steps accurately gives the correct difference of 386.

This addition problem involves carrying over from the ones to the tens place, and then from the tens to the hundreds place. The careful regrouping leads to a final sum of 424.

Adding 487 and 256 involves carrying over from the ones digit (7 + 6 = 13) into the tens column. After a second carry from the tens column, the final sum is correctly calculated as 743.

Direct subtraction in the ones and tens columns isn't possible without borrowing. After borrowing correctly from higher place values, the difference is found to be 236.

The ones digits add up to 15, prompting a carry to the tens digit, and the process is repeated for the tens and hundreds digits. This results in a total sum of 825.

Borrowing from the higher place values is necessary because some digits in the minuend are smaller than those in the subtrahend. After proper regrouping, the subtraction yields a correct difference of 347.

Starting with the ones place, 9 + 8 equals 17, which results in a carry to the tens column. Continuing methodically through the tens and hundreds places gives the sum of 937.

The ones digit subtraction requires borrowing since 0 is less than 6, affecting the tens place. With careful regrouping, the final answer is determined to be 544.

Beginning with the ones digit, 5 + 9 equals 14, leading to a carry over to the tens. After adding tens and hundreds with the carries, the total sum comes out to be 1204.

Because the ones digit in 592 is smaller than in 367, borrowing is needed. Once the regrouping is done correctly across the digits, the answer is found to be 225.

The ones, tens, and hundreds places each require carrying over when their sums exceed 9. This systematic addition results in a final sum of 1010.

Borrowing from the tens digit makes the subtraction in the ones place possible, yielding 17 - 8 = 9, and subsequent subtraction gives the final difference of 319. This problem emphasizes the importance of proper regrouping.

This addition requires regrouping in both the ones and tens places. After carrying over from 4 + 9 and then from the tens digit, the final sum is correctly determined to be 1623.

Multiple borrowings are required in this subtraction, first in the ones column and then in the tens column. Careful regrouping yields the correct result of 443.

Subtraction here requires borrowing twice: first from the tens column and then from the hundreds. This careful process results in the correct difference of 259.

By adding from the ones digit upward, 4 + 7 gives 11, which means you carry over 1, and similar regrouping in the tens produces the final sum. The complete addition results in a total of 1011.

The subtraction requires borrowing in the ones place because 2 is smaller than 4, and then adjusting the tens digit accordingly. Following these careful regrouping steps results in the correct answer of 218.