Long Division with Multi-Digit Divisors — Answer Key
Part A: Fix the Sentence
Each sentence has an error. Rewrite it correctly on the line.
1. Fix the sentence:
To divide 480 by 16 using partial quotients, I subtract 16 once and write 16 as my first partial quotient.
Corrected: To divide 480 by 16 using partial quotients, I subtract a multiple of 16 (like 160) and write the multiplier (10) as my first partial quotient.
Grade 5 partial-quotient work records multipliers (10, 20, 5), then sums them; writing 16 confuses the divisor with the partial quotient.
2. Fix the sentence:
When I add my partial quotient of 30 and 5, I should multiply them together to get the final quotient.
Corrected: When I have partial quotients of 30 and 5, I should add them together to get the final quotient of 35.
Each partial quotient counts a chunk of groups removed, so Grade 5 students add the chunks to find the total quotient.
3. Fix the sentence:
If 17 goes into 850 about 50 times, my first partial quotient should be 17 because that is the divisor.
Corrected: If 17 goes into 850 about 50 times, my first partial quotient should be 50 because that is how many groups of 17 I am removing.
Grade 5 partial-quotient strategy records the multiplier you used, not the divisor itself.
Part B: Fill in the Blank
Write the missing word or number on each line.
1. In partial quotients for 624 ÷ 24, removing 20 groups of 24 leaves 144, so 20 is the first partial quotient.
Grade 5 students label each multiplier they record as a partial quotient before summing.
2. After taking 20 groups and then 6 groups of 24 from 624, the total quotient is the sum of 20 and 6, which is 26.
Adding the partial quotients gives the full quotient in the Grade 5 partial-quotient method.
3. Choosing 10, 20, or 50 as a first partial quotient is called using a friendly multiplier in Grade 5 division.
Friendly multipliers keep the subtraction manageable without using the standard algorithm.
4. When the remaining number is less than the divisor, the partial-quotients work is finished and you write the remainder.
Grade 5 students record any leftover amount as the remainder once no more groups can be removed.
Part C: Short Answer
Answer each question in one or two complete sentences.
1. Use partial quotients to divide 432 ÷ 18 and explain each step in Grade 5 language.
Sample answer: I removed 20 groups of 18 (360) from 432, leaving 72. Then I removed 4 more groups of 18 (72), leaving 0. The partial quotients are 20 and 4, so 432 ÷ 18 = 24 with no remainder.
Recording each chunk and summing them shows the Grade 5 partial-quotients reasoning clearly.
2. Why might a Grade 5 student prefer partial quotients over the standard algorithm for 728 ÷ 14?
Sample answer: Partial quotients let me use friendly numbers like 50 (50 × 14 = 700), then just 2 more groups of 14, instead of guessing one digit at a time. I get 52 with less risk of place-value mistakes.
Partial quotients reduce single-digit guessing errors that often appear in the standard algorithm.