Multiplication Word Problems and Finding Mistakes | Grade 5 Math | LevelKite

You already know how to multiply large numbers. Today you put that skill to work in two ways. First, you solve word problems. Then you play detective: you look at finished work, find the mistake, and fix it.

How do you know that a word problem calls for multiplying? Look for equal groups: the same amount again and again. When you know the number of groups and the size of each group, you multiply to find the total. The answer to a multiplication is called the product.

With large numbers, it is easy to make a slip. So before you multiply, make an estimate: a number that is close to the exact answer and quick to find.

To estimate a product, round each number. That means you change it to a nearby number that is easy to use. A two-digit number goes to the nearest ten. A three-digit number goes to the nearest hundred. Then multiply the rounded numbers. Rounding changes each number only a little, so the estimate lands near the exact product.

Take $62 \times 29$ . Round $62$ to $60$ and $29$ to $30$ . Since $60 \times 30 = 1{,}800$ , the exact product should be near $1{,}800$ . It is: $62 \times 29 = 1{,}798$ . If your work gave $682$ instead, you would know to look for a mistake, because $682$ is far from $1{,}800$ .

Key concept

Estimate, multiply, then compare

An estimate tells you how big a product should be. Make one before you multiply. Then compare your product with it.

If your product is far from the estimate, a digit is probably in the wrong place. Look at your placeholder zeros and at how your partial products are lined up.
If your product is close to the estimate, the size is right. A small mistake can still hide, so check each partial product once more.

Worked example

Pencils in boxes

A school orders $24$ boxes of pencils. Each box holds $36$ pencils. How many pencils does the school order in all?

Find the equal groups. There are 24 boxes, and each box holds 36 pencils. That is 24 groups of 36, so you multiply to find the total.
$24 \times 36$
Estimate first. Round 24 down to 20 and round 36 up to 40, because those are the nearest tens. The exact answer should be near 800.
$20 \times 40 = 800$
Split 24 into 2 tens and 4 ones, and start with the 4 ones. 4 times 6 ones is 24 ones, so write 4 ones and regroup 2 tens. 4 times 3 tens is 12 tens, and the 2 regrouped tens make 14 tens. This partial product is 144, the number of pencils in 4 boxes.
$4 \times 36 = 144$
Now use the 2 tens. Write a placeholder zero in the ones place, because 2 tens times 36 gives a number of tens. Then 2 times 36 is 72, so this partial product is 72 tens, or 720. That is the number of pencils in 20 boxes.
$20 \times 36 = 720$
Add the two partial products, because 4 boxes and 20 boxes make all 24 boxes. Then compare with the estimate. 864 is close to 800, so the size is right.
$144 + 720 = 864$

Answer

The school orders $864$ pencils in all. This is close to the estimate of $800$ , so the answer makes sense.

Worked example

Pages at the print shop

A print shop prints $214$ copies of a book. Each copy has $487$ pages. How many pages does the shop print in all?

There are 214 copies with 487 pages each. That is 214 equal groups of 487, so the problem is 214 times 487. Estimate first: round 214 down to 200 and 487 up to 500, the nearest hundreds. The exact answer should be near 100,000.
$200 \times 500 = 100{,}000$
Split 214 into 2 hundreds, 1 ten and 4 ones. Multiply 487 by the 4 ones first. 4 times 7 ones is 28 ones: write 8, regroup 2 tens. 4 times 8 tens is 32 tens, plus the 2 regrouped tens is 34 tens: write 4, regroup 3 hundreds. 4 times 4 hundreds is 16 hundreds, plus the 3 regrouped hundreds is 19 hundreds. That is the number of pages in 4 copies.
$4 \times 487 = 1{,}948$
Multiply 487 by the 1 ten. Write one placeholder zero, because the answer is a number of tens. 1 times 487 is 487, so this partial product is 487 tens. That is the number of pages in 10 copies.
$10 \times 487 = 4{,}870$
Multiply 487 by the 2 hundreds. Write two placeholder zeros, because the answer is a number of hundreds. 2 times 487 is 974, so this partial product is 974 hundreds. That is the number of pages in 200 copies.
$200 \times 487 = 97{,}400$
Add the three partial products, lined up by place value, because 4 copies, 10 copies and 200 copies make all 214 copies. Then compare with the estimate. 104,218 is close to 100,000, so the size is right.
$1{,}948 + 4{,}870 + 97{,}400 = 104{,}218$

Answer

The shop prints $104{,}218$ pages in all. This is close to the estimate of $100{,}000$ , so the answer makes sense.

Now for the detective work. Start with a multiplication that is done correctly: $63 \times 47$ .

The $47$ is $4$ tens and $7$ ones, so there are two partial products. The first row is $63 \times 7 = 441$ . The second row is $63 \times 40 = 2{,}520$ , and it ends in a placeholder zero. The rows are lined up by place value: ones under ones, tens under tens. They add up to $2{,}961$ .

An estimate agrees: $60 \times 50 = 3{,}000$ , and $2{,}961$ is close to $3{,}000$ .

Here is the correct work. After it, a table shows three ways the same problem can go wrong.

\begin{array}{rl} 63 & \\ \times\ 47 & \\ \hline 441 & \leftarrow 63 \times 7 \\ +\ 2{,}520 & \leftarrow 63 \times 40 \\ \hline 2{,}961 & \leftarrow 63 \times 47 \end{array}

Three common mistakes when multiplying 63 by 47

Mistake	What the work shows	Answer it gives	How to fix it
Missing placeholder zero	The second row is $252$ , not $2{,}520$	$441 + 252 = 693$	Write a $0$ in the ones place first, so the second row is $2{,}520$
Regrouped digit not added	The first row is $421$ , not $441$	$421 + 2{,}520 = 2{,}941$	Add the $2$ regrouped tens: $42$ tens and $2$ tens make $44$ tens, so the first row is $441$
Rows lined up in the wrong places	The left edges line up, so the $4$ hundreds of $441$ sit in the thousands place and it counts as $4{,}410$	$4{,}410 + 2{,}520 = 6{,}930$	Line up ones under ones and tens under tens, then add again

Compare each wrong answer with the estimate of $3{,}000$ .

$693$ is far too small. The estimate catches it.
$6{,}930$ is more than double the estimate. The estimate catches it too.
$2{,}941$ is close to $3{,}000$ . The estimate does not catch it.

So an estimate finds mistakes that change the size of the answer. A regrouping mistake often changes the answer only a little. To find it, you check each row.

Worked example

Find and fix the mistake

Leo multiplied $386 \times 64$ . His first row was $1{,}524$ . His second row was $23{,}160$ . He added the rows and got $24{,}684$ . Find Leo's mistake and fix it.

Check the size first. Round 386 up to 400 and 64 down to 60. Leo's answer of 24,684 is close to 24,000, so the size is right. There is no big place-value mistake. A small mistake can still hide, so check each row.
$400 \times 60 = 24{,}000$
Check the first row. It should be 386 times the 4 ones. Since 386 is 300 and 80 and 6, multiply each part by 4 to get 1,200 and 320 and 24. Then add. The row should be 1,544, but Leo wrote 1,524. His first row is 20 too small.
$1{,}200 + 320 + 24 = 1{,}544$
Find out why. 4 times 6 ones is 24 ones, so you write 4 ones and regroup 2 tens. 4 times 8 tens is 32 tens, and the 2 regrouped tens make 34 tens. Leo has a 2 in the tens place where a 4 belongs. He did not add the 2 regrouped tens. That is the missing 20.
$32 \text{ tens} + 2 \text{ tens} = 34 \text{ tens}$
Check the second row. It should be 386 times the 6 tens. The row ends in a placeholder zero, as it should. Multiply each part of 386 by 6 to get 1,800 and 480 and 36, which make 2,316. So the row is 2,316 tens. Leo's second row is correct.
$386 \times 60 = 23{,}160$
Fix the work. The product is the sum of the two rows, so add the corrected first row to the second row. The new product is still close to the estimate of 24,000, as it should be.
$1{,}544 + 23{,}160 = 24{,}704$

Answer

Leo did not add the $2$ regrouped tens in his first row. That row should be $1{,}544$ , so the correct product is $386 \times 64 = 24{,}704$ .

The big idea: when a word problem gives you equal groups, you multiply. Then you give your product two checks.

Compare it with an estimate. A product far from the estimate points to a place-value mistake, such as a missing placeholder zero or rows lined up in the wrong places.
Check each partial product. A regrouped digit that was not added may change the answer only a little, so the estimate can miss it.

When both checks pass, you can feel sure about your answer.