Molding costs dollars a meter and a customer wants meters. How do you multiply when both numbers have a decimal point?

Math 6 · 6.NS.B.3

Multiplying multi-digit decimals with the standard algorithm

Molding costs 4.85 dollars a meter and a customer wants 3.6 meters. How do you multiply when both numbers have a decimal point?

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Multiplying decimalsGrade 5 Multiplying and dividing by ten, one hundred and one thousandGrade 4

Where you are 8 in Decimal arithmetic

Number9

Arithmetic5

Decimal arithmetic8

▴ 1 above Multiplying decimals Dividing decimals

Multiplying multi-digit decimals with the standard algorithm

Where the point goes in a product Decimal problems in words ▾ Show all 8 (2 more)

When you reach for it

Ximena works the counter at a woodworking shop that sells molding and skirting board by the meter. When a customer wants a whole number of meters, the bill is a decimal times a whole number, which Multiplying decimals already handles: 3 meters at 4.85 dollars a meter is 4.85 \times 3. But walls are rarely a whole number of meters long. A wall that needs 3.6 meters of molding at 4.85 dollars a meter asks for 4.85 \times 3.6, and now both factors have a decimal point.

The move in this lesson works for any two positive decimals, however many digits they have: multiply as if the points were not there, using the standard multiplication algorithm for whole numbers, then put the point back in the right place. It is taught over two days. Day 1 is the algorithm and placing the point by counting decimal places. Day 2 checks the answer with an estimate, deals with zeros in the product, and shows why counting places gives the right answer.

The procedure

Day 1: multiply the whole numbers, then count places

Take 4.85 \times 3.6. Multiplying by 100 moves every digit of 4.85 two places to the left, so 4.85 \times 100 = 485; in the same way 3.6 \times 10 = 36 (Multiplying and dividing by ten, one hundred and one thousand). These are whole numbers, so the ordinary algorithm multiplies them, one partial product for each digit of the lower factor: 485 \times 6 = 2910 and 485 \times 30 = 14550, which add to 485 \times 36 = 17460.

That is not yet the answer. The first factor was made 100 times larger and the second 10 times larger, so 17460 is 100 \times 10 = 1000 times larger than 4.85 \times 3.6. Dividing by 1000 moves the digits three places back to the right:

4.85 \times 3.6 = 17460 \div 1000 = 17.460 = 17.46

The customer pays 17.46 dollars.

Look at where the three came from. The number of digits after a decimal point is its number of decimal places. 4.85 has two decimal places and needed \times 100; 3.6 has one and needed \times 10. Together the factors were scaled by 1000, a one followed by 2 + 1 = 3 zeros, so the point goes back three places. The same reasoning works for any factors, which gives the rule of Where the point goes in a product: the product, before any zero at its end is dropped, has as many decimal places as the two factors have together.

So in practice the points are simply ignored until the end. Write the factors with their last digits lined up on the right, not their points. This is unlike adding decimals, where tenths must sit over tenths because tenths are being combined with tenths; in a product no place is combined with another until the point is placed. Multiply, count the places in both factors, and count that many digits in from the right of the product:

\begin{array}{r} 4.85 \\ \times\ \ 3.6 \\ \hline 2910 \\ 14550 \\ \hline 17.460 \end{array}

A zero at the end, after the point is placed, can be dropped: 17.460 = 17.46, because 0 thousandths adds nothing. Drop it only after the point is placed; Day 2 shows why the order matters.

Day 2: estimate, zeros, and why the count works

Estimate to check. Counting places is exact, but a slip in the count is easy to miss unless you know roughly what to expect. Round each factor to a number that is easy to multiply in your head (Estimating by rounding). 4.85 \times 3.6 is near 5 \times 4 = 20, or more closely 5 \times 3.6 = 18. The answer 17.46 fits; 1.746 and 174.6 do not. The estimate does not replace the count. What it catches is a point in the wrong place, which makes the answer wrong by a factor of 10 or more, far more than rounding can explain.

Zeros in the product. Two kinds of zero need care. If the whole-number product has fewer digits than the places you must count, write zeros in front to make up the count. 0.3 \times 0.2 gives 3 \times 2 = 6 and needs 1 + 1 = 2 places, so the product is 0.06, with a zero holding the tenths place. And if the whole-number product ends in zeros, count the places first and drop the zeros afterwards. 2.5 \times 0.4 gives 25 \times 4 = 100 and needs 2 places, so it is 1.00 = 1. Dropping the zeros of 100 first would leave 1, and counting two places in would give 0.01, which is wrong: the zeros of 100 are digits of the product and take part in the count.

Why counting places works. A decimal that stops is a fraction whose denominator is a power of ten: \displaystyle 4.85 = \frac{485}{100} and \displaystyle 3.6 = \frac{36}{10}. Multiplying fractions multiplies the numerators and the denominators:

\begin{aligned} 4.85 \times 3.6 &= \frac{485}{100} \times \frac{36}{10} \\ &= \frac{485 \times 36}{100 \times 10} = \frac{17460}{1000} \end{aligned}

The numerator is the whole-number product. The denominator is a product of powers of ten, and their zeros add: two zeros and one zero make the three zeros of 1000. Thousandths need three decimal places, so \displaystyle \frac{17460}{1000} = 17.460. That is the counting rule. It also explains the zeros in front: \displaystyle 0.3 \times 0.2 = \frac{3 \times 2}{10 \times 10} = \frac{6}{100}, six hundredths, which is 0.06 whether or not 6 has two digits.

The figure shows the smallest case. A square of side 1 cut into a 10 by 10 grid has 100 small squares, each \displaystyle \frac{1}{100} of the whole. A rectangle 0.3 wide and 0.4 tall covers 3 \times 4 = 12 of them, so 0.3 \times 0.4 = 0.12: tenths times tenths are hundredths.

Figure 10.3 \times 0.4 is the shaded rectangle, 12 small squares out of 100, so it is 0.12.

Worked examples

The lesson is taught in two parts. Examples 1 to 3 belong to Day 1: multiplying and placing the point. Examples 4 to 6 belong to Day 2: estimating, zeros, and the reason behind the count.

Example 1

Tenths times tenths

Kaito sells pine skirting board at 2.6 dollars a meter. A customer needs 3.4 meters. What does it cost?

Solution

The cost is the price of one meter times the number of meters, 2.6 \times 3.4. Ignore the points and multiply 26 \times 34: the partial products are 26 \times 4 = 104 and 26 \times 30 = 780, which add to 884.

Each factor has one decimal place, so the product has 1 + 1 = 2. Counting two digits in from the right of 884 puts the point after the 8: 2.6 \times 3.4 = 8.84. The board costs 8.84 dollars, a little under 3 \times 3 = 9 dollars, as it should be.

Example 2

Hundredths times tenths

Lorenzo’s oak molding weighs 1.35 kilograms per meter. How much does a 4.7 meter length weigh?

Solution

The weight is the weight of one meter times the length, 1.35 \times 4.7. Multiply 135 \times 47 with the points ignored:

\begin{array}{r} 1.35 \\ \times\ \ 4.7 \\ \hline 945 \\ 5400 \\ \hline 6.345 \end{array}

The rows are 135 \times 7 = 945 and 135 \times 40 = 5400, which add to 6345. The factors have 2 and 1 decimal places, so the product has 3, and 6345 becomes 6.345. The length weighs 6.345 kilograms.

The product has more decimal places than either factor. That is to be expected: hundredths of a kilogram per meter, times tenths of a meter, make thousandths of a kilogram.

Example 3

Three rows of partial products

Yara prices 3.85 meters of hardwood skirting at 14.25 dollars a meter. What is the price, to the nearest cent?

Solution

Multiply 1425 \times 385. The lower factor has three digits, so there are three partial products:

\begin{array}{r} 1425 \\ \times\ \ 385 \\ \hline 7125 \\ 114000 \\ 427500 \\ \hline 548625 \end{array}

They are 1425 \times 5, 1425 \times 80 and 1425 \times 300. Both factors have 2 decimal places, so the product has 2 + 2 = 4: 14.25 \times 3.85 = 54.8625.

A price is charged in whole cents, which are hundredths of a dollar. The digit after the hundredths is 2, less than 5, so the price rounds down to 54.86 dollars (Rounding decimals).

Example 4

An estimate catches a misplaced point

Finn sells 23.8 meters of thin beading at 0.46 dollars a meter and writes the bill as 109.48 dollars. Check the bill with an estimate, then find the right amount.

Solution

Round 23.8 to 24 and 0.46 to 0.5. Half of 24 is 12, so the bill should be about 12 dollars. A bill of 109.48 dollars is roughly ten times too large, which points to a misplaced decimal point rather than a wrong digit.

Multiply again. 238 \times 46 has partial products 238 \times 6 = 1428 and 238 \times 40 = 9520, which add to 10948. The places are 1 + 2 = 3, so 23.8 \times 0.46 = 10.948, close to the estimate of 12. Finn’s digits were right; he counted two places instead of three. To the nearest cent the bill is 10.95 dollars, since the digit after the hundredths is 8, which rounds the hundredths up.

Example 5

Zeros in front and zeros at the end

Abena has two jobs to price. (a) A 0.2 meter offcut of beading at 0.35 dollars a meter. (b) 1.6 meters of skirting at 4.25 dollars a meter.

Solution

(a) 35 \times 2 = 70. The factors have 2 + 1 = 3 decimal places, but 70 has only two digits. A zero written in front makes three, 070, and the point goes in front of all three: 0.070. The final zero is 0 thousandths and can now be dropped, so 0.35 \times 0.2 = 0.07. The offcut costs 0.07 dollars, seven cents. The estimate 0.4 \times 0.2 = 0.08 agrees.

(b) 425 \times 16 has partial products 425 \times 6 = 2550 and 425 \times 10 = 4250, which add to 6800. The places are 2 + 1 = 3, so count three digits in from the right of 6800: 6.800. Only now drop the zeros at the end: 4.25 \times 1.6 = 6.8, and the price is 6.80 dollars. The estimate 4 \times 1.6 = 6.4 is close. Dropping the zeros of 6800 first would have left 68 and a product of 0.068, which the estimate rejects at once.

Example 6

One multiplication, many products

Lorenzo knows that 268 \times 15 = 4020. Without multiplying again, find 2.68 \times 1.5 and 26.8 \times 0.015, and use fractions to explain the second.

Solution

Every product made from the digits 268 and 15 has the digits of 4020; only the point moves.

For 2.68 \times 1.5 the places are 2 + 1 = 3, so the product is 4.020 = 4.02. The estimate 3 \times 1.5 = 4.5 is close.

For 26.8 \times 0.015 the places are 1 + 3 = 4, so the product is 0.4020 = 0.402. The estimate 30 \times 0.015 = 0.45 is close. As fractions, \displaystyle 26.8 = \frac{268}{10} and \displaystyle 0.015 = \frac{15}{1000}, so

\begin{aligned} 26.8 \times 0.015 &= \frac{268 \times 15}{10 \times 1000} \\ &= \frac{4020}{10000} = 0.402 \end{aligned}

The denominator 10 \times 1000 = 10000 has 1 + 3 = 4 zeros. Ten-thousandths need four decimal places, which is the count of four.

Boundaries

The count needs decimals that stop. A repeating decimal such as 0.333\ldots has no last digit to line up and no number of decimal places to count; write it as a fraction first (Turning a repeating decimal into a fraction).

The count gives the exact product, which often has more decimal places than a price or a measurement can use. Rounding to cents or to a sensible unit is a separate step taken after the exact product is known, as in Example 3.

A whole number has 0 decimal places, so multiplying a decimal by a whole number is the special case where only one factor adds to the count.

Every factor here is positive. A product with a negative factor has the same digits and the same point, with its sign settled by the rule in Multiplying and dividing rational numbers.

Ignoring the points is a move for multiplication only. Adding and subtracting need the points lined up, and dividing needs the divisor scaled to a whole number first.

Where this goes next

Division undoes this lesson. Dividing multi-digit decimals with the standard algorithm scales the divisor and the dividend by the same power of ten, the same trick as Day 1 run in reverse, and then lets long division do the work. Decimal problems in words then puts all four operations with decimals into stories about money, measures and rates.

Exercises

Exercises 1 to 11 go with Day 1 (Examples 1 to 3), Exercises 12 to 20 with Day 2 (Examples 4 to 6), and the reasoning group draws on both.

Day 1: multiply and place the point

0.7 \times 0.9
3.2 \times 1.4
5.7 \times 0.8
2.15 \times 3.3
12.4 \times 2.6
0.48 \times 7.3
6.07 \times 4.9
1.25 \times 2.25

Day 1: in the shop

Kaito sells cedar skirting board at 4.6 dollars a meter. What do 3.8 meters cost?
Abena’s ogee molding weighs 0.85 kilograms per meter. How much does a 6.3 meter length weigh?
Ximena prices 2.7 meters of cove molding at 8.36 dollars a meter. What is the price, to the nearest cent?

Day 2: estimate first, then multiply

Estimate 9.8 \times 3.1, then find it exactly.
Estimate 0.62 \times 48.5, then find it exactly.

Day 2: zeros in the product

0.4 \times 0.2
0.05 \times 0.4
1.25 \times 0.08
7.5 \times 1.6
Given that 46 \times 23 = 1058, find (a) 4.6 \times 2.3, (b) 0.46 \times 0.23 and (c) 46 \times 0.023.

Day 2: in the shop

Finn prices a 0.75 meter offcut of skirting at 2.4 dollars a meter and writes 18 dollars. Use an estimate to show this cannot be right, then find the correct price.
Lorenzo cuts 3.25 meters of skirting priced at 10.4 dollars a meter. Estimate the price, then find it exactly.

Reasoning

Use fractions to explain why 0.4 \times 0.06 has three decimal places, and give the product.
Yara works out 3.5 \times 1.2 and Kaito works out 3.50 \times 1.2. They count different numbers of decimal places. Explain why they still get the same answer.
Without multiplying, explain why 6.8 \times 0.9 is less than 6.8, while 6.8 \times 1.2 is more than 6.8.
Can two decimals, each with exactly one decimal place and a last digit that is not 0, have a whole-number product? Give an example or explain why not.

Answers

0.63. 7 \times 9 = 63, and 1 + 1 = 2 places.
4.48. 32 \times 14 = 448, and 1 + 1 = 2 places.
4.56. 57 \times 8 = 456, and 1 + 1 = 2 places.
7.095. 215 \times 33 = 6450 + 645 = 7095, and 2 + 1 = 3 places.
32.24. 124 \times 26 = 2480 + 744 = 3224, and 1 + 1 = 2 places.
3.504. 48 \times 73 = 3360 + 144 = 3504, and 2 + 1 = 3 places.
29.743. 607 \times 49 = 24280 + 5463 = 29743, and 2 + 1 = 3 places.
2.8125. 125 \times 225 = 28125, and 2 + 2 = 4 places.
17.48 dollars. 46 \times 38 = 1380 + 368 = 1748, and 1 + 1 = 2 places.
5.355 kilograms. 85 \times 63 = 5100 + 255 = 5355, and 2 + 1 = 3 places.
22.57 dollars. 836 \times 27 = 16720 + 5852 = 22572, and 2 + 1 = 3 places give 22.572; the digit after the hundredths is 2, so round down.
About 10 \times 3 = 30; exactly 30.38. 98 \times 31 = 2940 + 98 = 3038, and 2 places.
About 0.6 \times 50 = 30; exactly 30.07. 62 \times 485 = 29100 + 970 = 30070, and 2 + 1 = 3 places give 30.070 = 30.07.
0.08. 4 \times 2 = 8 needs 2 places, so a zero goes in front: 0.08.
0.02. 5 \times 4 = 20 needs 2 + 1 = 3 places: 0.020 = 0.02.
0.1. 125 \times 8 = 1000 needs 2 + 2 = 4 places: 0.1000 = 0.1.
12. 75 \times 16 = 1200 needs 2 places: 12.00 = 12.
(a) 10.58, 2 places. (b) 0.1058, 4 places. (c) 1.058, 0 + 3 = 3 places.
Less than a meter must cost less than 2.4 dollars, so 18 dollars is far too much (about 0.75 \times 2 = 1.5). Exactly: 75 \times 24 = 1800, and 2 + 1 = 3 places give 1.800 = 1.8, so the price is 1.80 dollars.
About 3 \times 10 = 30 dollars. Exactly: 325 \times 104 = 32500 + 1300 = 33800, and 2 + 1 = 3 places give 33.800 = 33.8, so the price is 33.80 dollars.
0.024. \displaystyle 0.4 = \frac{4}{10} and \displaystyle 0.06 = \frac{6}{100}, so the product is \displaystyle \frac{4 \times 6}{10 \times 100} = \frac{24}{1000}. The denominator has 1 + 2 = 3 zeros, so the answer is in thousandths and has three decimal places.
Both get 4.2. Yara: 35 \times 12 = 420 with 2 places, 4.20. Kaito: 350 \times 12 = 4200 with 3 places, 4.200. The extra 0 in 3.50 adds one zero to the end of the digit product and one place to the count, and those cancel, because 3.50 and 3.5 are the same number.
0.9 is less than 1, so 6.8 \times 0.9 is only part of 6.8, nine tenths of it. And 1.2 = 1 + 0.2, so 6.8 \times 1.2 is 6.8 plus two tenths of 6.8 more (the distributive property). Both comparisons use that 6.8 is positive: part of a positive number is less than the number, and adding a positive amount makes it larger. For any positive number, multiplying by a factor less than 1 makes it smaller and by a factor more than 1 makes it larger.
Yes. For example 2.5 \times 1.2 = 3: 25 \times 12 = 300 needs 2 places, giving 3.00 = 3. It works whenever the digit product ends in at least two zeros.