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Multiplication Of Decimals

Multiplication of Decimals up to Tenths Place

Factors are referred to as the numbers which are being multiplied.

Product refers to the result of a multiplication problem.

  • In the multiplication of decimals up to tenths place, there is one digit after the decimal point in the first factor and one digit after the decimal point in the second factor also. Both the numbers have collectively two decimal digits. Therefore, the product will also have two decimal digits.

Steps:

  1. Ignore the decimal point and just multiply as whole numbers.
  2. Put the decimal point in the product by counting the correct number of decimal places in the original problem.

Suppose, we want to multiply \(2.4\) by \(6.2\).

Write one decimal number just below the other decimal number.

Here, \(6.2\) is the multiplier because it is being multiplied and \(2.4\) is the multiplicand.

First, we multiply each digit of the multiplicand by \(2\).

We multiply \(2\) by \(4\).

\(2×4=8\)

and \(2\) by \(2\) 

\(2×2=4\)

Now, we multiply each digit of the multiplicand by \(6\).

\(6×4=24\)

\(4\) is written after placing \(0\) in the second row as we are multiplying by a number in the tens place and \(20\) is carried to \(2\).

Now, \(6×2=12\) and \(2+12=14\)

Here, we are adding \(2\) and not \(20\), because we are already dealing with tens column.

Now, add the products of the given numbers.

In the given example, both decimal numbers have one decimal digit each, both the numbers together have two decimal digits. Therefore, the product will also have two decimal digits.

So, \(14.88\) is the product of \(2.4\) and \(6.2\) .

Illustration Questions

Evaluate \(4.8×3.5\)

A \(8.3\)

B \(16.8\)

C \(1.3\)

D \(12.85\)

×

Write one decimal number below the other decimal number.

image

First, we multiply \(48\) by \(5\).

\(48×5=240\)

Now, we multiply \(48\) by \(3\).

\(48×3=144\)

Now we add the products of both the numbers.

 

image

In the given problem, both decimal numbers have one decimal digit each, both the numbers together have two decimal digits. Therefore, the product will also have two decimal digits.

Product = \(16.80\)

So, \(16.8\) is the product of \(4.8\) and \(3.5\)

 

Hence, option (B) is correct.

Evaluate \(4.8×3.5\)

A

\(8.3\)

.

B

\(16.8\)

C

\(1.3\)

D

\(12.85\)

Option B is Correct

Multiplication of Decimals up to Hundredths Place

Factors are referred to as the numbers which are being multiplied.

Product refers to the result of a multiplication problem.

In the multiplication of decimals up to hundredths place, there are two digits after the decimal point in both the factors. Thus, both the numbers together have four decimal digits. Therefore, the product will also have four decimal digits.

Steps:

  1. Ignore the decimal point and just multiply as whole numbers.
  2. Put the decimal point in the product by counting the correct number of decimal places in the original problem. 

For example: We want to multiply \(7.35\) by \(2.50\).

Write one decimal number below the other decimal number.

\(\begin{array} {r l} 7.35 & \rightarrow\text{Multiplicand} \\ \times2.50 & \rightarrow\text{Multiplier} \\ \hline & \\ \hline \end{array}\)

Multiply each digit of the multiplier by each digit of the multiplicand, one at a time.

First, we multiply \(735\) by \(0\).

\(735×0=000\)

Multiply \(735\) by \(5\).

\(735×5=3675\)

Now, multiply \(735\) by \(2\).

\(735×2=1470\)

Now, we add the products.

\(\begin{array} {r l} 7.35 & \rightarrow\text{Multiplicand} \\ \times2.50 & \rightarrow\text{Multiplier} \\ \hline 000 \\ 36750 & \rightarrow\text{We add in a zero as we are multiplying by a number at the tens place} \\ +147000 &\rightarrow\text{We add in two zeros as we are multiplying by a number at the hundreds place} \\ \hline 183750 \\ \hline \end{array}\)

In the given example, both the decimal numbers have two decimal digits. Thus, there are altogether four decimal digits. Therefore, the product will also have four decimal digits.

\(18.3750\)

Hence, \(18.375\) is the product of \(7.35\) and \(2.50\)

Illustration Questions

Evaluate \(5.13×3.18\)

A \(16.3134\)

B \(53.1318\)

C \(8.31\)

D \(1.85\)

×

Write one decimal number below the other decimal number.

\(\begin{array} {r l} 5.13 & \rightarrow\text{Multiplicand} \\ \times 3.18 & \rightarrow\text{Multiplier} \\ \hline & \\ \hline \end{array}\)

First, we multiply \(513\) by \(8\).

\(513×8=4104\)

Multiply \(513\) by \(1\).

\(513×1=513\)

Now, multiply \(513\) by \(3\).

\(513×3=1539\)

Now, we add the products.

\(\begin{array} {r l} 5.13 & \rightarrow\text{Multiplicand} \\ \times3.18 & \rightarrow\text{Multiplier} \\ \hline ^\underline{\color{red}1}4104 \\{\color{red}1} 5130 & \rightarrow\text{We add in a zero as we are multiplying by a number at the tens place} \\ +153900 &\rightarrow\text{We add in two zeros as we are multiplying by a number at the hundreds place} \\ \hline 163134 \\ \hline \end{array}\)

In the given problem, both decimal numbers have two decimal digits each. Thus, there are altogether four decimal digits. Therefore, the product will also have four decimal digits.

\(16.3134\)

So, \(16.3134\) is the product of \(5.13\) and \(3.18\)

 

Hence, option (A) is correct.

Evaluate \(5.13×3.18\)

A

\(16.3134\)

.

B

\(53.1318\)

C

\(8.31\)

D

\(1.85\)

Option A is Correct

Miscellaneous

Factors are referred to as the numbers which are being multiplied.

Product refers to the result of a multiplication problem.

In the multiplication of mixed decimal digits, if there are two digits after the decimal point in the one factor and one digit in the other factor, then all the digits together have three decimal digits. Therefore, the product will also have three decimal digits.

Steps:

  1. Ignore the decimal point and just multiply as whole numbers.
  2. Put the decimal point in the product by counting the correct number of decimal places in the original problem. 

Suppose, we want to multiply \(2.7\) by \(0.15\)

Write one decimal number below the other decimal number.

\(\begin{array} {r l} 2.7 & \rightarrow\text{Multiplicand} \\ \times 0.15 & \rightarrow\text{Multiplier} \\ \hline & \\ \hline \end{array}\)

Multiply each digit of the multiplicand by each digit of the multiplier, one at a time.

First, multiply \(27\) by \(5\).

\(27×5=135\)

Multiply \(27\) by \(1\).

\(27×1=27\)

Now, multiply \(27\) by \(0\).

\(27×0=00\)

Now, we add all the products.

\(\begin{array} {r l} 27 & \rightarrow\text{Multiplicand} \\ \times 015 & \rightarrow\text{Multiplier} \\ \hline ^\underline{\color{red}1}135 \\ 270 & \rightarrow\text{We add in a zero as we are multiplying by a number at the tens place} \\ +0000 &\rightarrow\text{We add in two zeros as we are multiplying by a number at the hundreds place} \\ \hline 0405 \\ \hline \end{array}\)

In the given example, the first factor has one decimal digit and the second factor has two decimal digits. Thus, all the numbers together have three decimal digits. Therefore, the product will also have three decimal digits.

\(0.405\)

Hence, \(0.405\) is the product of \(2.7\) and \(0.15\)

Illustration Questions

Evaluate \(15.3×0.85\)

A \(15.88\)

B \(14.45\)

C \(13.005\)

D \(13.500\)

×

Write one decimal number below the other decimal number.

\(\begin{array} {r l} 15.3 & \rightarrow\text{Multiplicand} \\ \times 0.85 & \rightarrow\text{Multiplier} \\ \hline & \\ \hline \end{array}\)

First, we multiply \(153\) by \(5\).

\(153×5=765\)

Multiply \(153\) by \(8\).

\(153×8=1224\)

Now, multiply  \(153\) by \(0\).

\(153×0=000\)

Now, add all the products.

\(\begin{array} {r l} 153 & \rightarrow\text{Multiplicand} \\ \times 085 & \rightarrow\text{Multiplier} \\ \hline \underline{\color{red}1}\;\;\;\\\underline{\color{red}1}765 \\ 12240 & \rightarrow\text{We add in a zero as we are multiplying by a number at the tens place} \\ +00000 &\rightarrow\text{We add in two zeros as we are multiplying by a number at the hundreds place} \\ \hline 13005\\ \hline \end{array}\)

In the given problem, the first factor has one decimal digit and the second factor has two decimal digits. Thus, altogether there are three decimal digits. Therefore, the product will also have three decimal digits.

\(13.005\)

So, \(13.005\) is the product of \(15.3×0.85\) 

 

Hence, option (C) is correct.

Evaluate \(15.3×0.85\)

A

\(15.88\)

.

B

\(14.45\)

C

\(13.005\)

D

\(13.500\)

Option C is Correct

Multiplying Decimals Using Area Model

  • An area model is used to show multiplication and division.
  • The numbers that are being multiplied are known as factors.
  • For multiplication of decimal, shade squares vertically for first factor.
  • Shade squares horizontally for second factor.
  • The common shaded area created by the overlapping of the grids is the product.
  • For example: We want to multiply \(0.7\) by \(0.5\) by using hundreds grids.
  • Since \(0.7=0.70\), shade \(70\) squares vertically.

  • Since \(0.5=0.50\), shade \(50\) squares horizontally.

  • Combine the shaded squares in the same grid.

Shaded area is \(0.35\) or \(35\) hundredths.

  • The common shaded area created by the overlapping of the grids is the product.

Here, the yellow shaded squares represent the product of \(0.7\) and \(0.5\)

Thus, \(0.7×0.5=0.35\)

  • We can also multiply large decimal numbers using the area model.
  • For example: Multiplying \(2.7\) by \(3.2\) by using hundreds grid.
  • Representing one decimal number vertically and other horizontally.

Since, \(2.7=2.70=270\) hundredths \(=(100+100+70)\)

So, shade two whole hundred grids and \(70\) squares out of \(100\).

  • Since, \(3.2=3.20=320\) hundredths \(=(100+100+100+20)\)

So, shade three whole hundred grids and \(20\) squares out of \(100\).

  • Now, combine both the rectangles.

  • Form a larger grid and shade the squares to show the complete overlapping.

  • From the figure, calculate the areas of all the rectangles of different colors.

  • Now, add all the areas.

\(6+2.1+0.4+0.14=8.64\)

  • Hence, the product of \(2.7\) and \(3.2\) is \(8.64\)

Illustration Questions

Which area model represents the product of \(0.6\) and \(0.4\)?

A

B

C

D

×

To represent the product of \(0.6\) and \(0.4\), first shade \(60\) squares of grid vertically to show \(0.6[0.6=0.60=60\;\text{hundredths}=60\;\text{out of}\;100]\)

image

Now, shade \(40\) squares of grid horizontally to show \(0.4[0.4=0.40=40\;\text{hundredths}=40\;\text{out of}\;100]\)

image

Combine the shaded squares in the same grid.

image

The common shaded area created by the overlapping of the grids is the product.

\(\therefore\) The overlapping squares (yellow in color) represent the product of \(0.6\) and \(0.4\)

Thus, \(0.6×0.4=0.24\)

Hence, option (B) is correct.

Which area model represents the product of \(0.6\) and \(0.4\)?

A image
B image
C image
D image

Option B is Correct

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