Informative line

Operations On Decimals

Decimals and their Place Values

  • Decimal is a part of whole.
  • Decimal numbers have wholes, a decimal point and parts in it.

Example: \(1.50\)

Each digit in a decimal number has a place value.

Example\(15.75\)

Hundreds Tens Ones

Decimal Point

Tenths Hundredths Thousandths Ten Thousandths

Hundred

Thousandths

Millionths
  1 5 . 7 5      

 

 

  • We can also write decimal numbers in words, by naming the place value of the last digit.

Examples: 

  1. Seventy-two and one-tenths \(=72.1\)
  2. One and two hundred fifteen thousandths \(=1.215\)
  3. \(25.002=\) Twenty-five and two-thousandths
  4. \(2.5=\) Two and five-tenths

Illustration Questions

Which one of the following options represents forty-nine and fifty-eight thousandths, in numbers?

A \(49.58\)

B \(49.058\)

C \(58.49\)

D \(58.049\)

×

In forty-nine and fifty-eight thousandths, forty-nine is the whole and fifty-eight thousandths is the parts.

Forty-nine \(=49\)

Fifty-eight thousandths \(=0.058\)

\(\therefore\) The number is \(49.058\).

Hence, option (B) is correct.

Which one of the following options represents forty-nine and fifty-eight thousandths, in numbers?

A

\(49.58\)

.

B

\(49.058\)

C

\(58.49\)

D

\(58.049\)

Option B is Correct

Subtraction of Decimals

  • Subtraction of decimals can be easily done by dealing with the wholes and the parts of the numbers separately.

Example: \(5.674-2.5\)

  • Here, both decimals have different place values, so we can add zeros to make them same.

\(5.674-2.500\)

  • Now, both decimal numbers have same place values.

Next, we write the decimal numbers vertically, lining up the decimal points and each digit according to its place value.

\(\begin{array}\\ &5&.&6&7&4\\ -&2&.&5&0&0\\ \hline\\ \hline \end{array}\)

  • Now, subtracting the columns vertically.

\(\begin{array}\\ &5&.&6&7&4\\ -&2&.&5&0&0 \\ \hline &3&.&1&7&4 \\\hline \end{array}\)

  • Difference of \(5.674\) and \(2.5\) is \(3.174\)

Illustration Questions

Find:​ \(10.1-2.85\)

A \(8.15\)

B \(12.95\)

C \(7.25\)

D \(12.86\)

×

Since both numbers have different place values, so we add zero to make them same.

\(10.10-2.85\)

Now, we write the decimals vertically, lining up the decimal points and each digit according to its place value.

\(\begin{array}\\ &1&0&.&1&0\\ -&&2&.&8&5\\ \hline\\ \hline \end{array}\)

On subtracting the columns vertically, we get

\(\begin{array}\\ &1&0&.&1&0\\ -&&2&.&8&5\\ \hline&&7&.&2&5\\ \hline \end{array}\)

Hence, option (C) is correct.

Find:​ \(10.1-2.85\)

A

\(8.15\)

.

B

\(12.95\)

C

\(7.25\)

D

\(12.86\)

Option C is Correct

Multiplication of Decimals

  • Factors are referred to as the numbers which are being multiplied.
  • Product refers to the result of a multiplication problem.
  • To multiply two decimal numbers, first line them up in right alignment.
  • Then, multiply each digit of multiplier by each digit of multiplicand.
  • Count the number of digits after the decimal point in multiplier and multiplicand.
  • The product will have same number, of total digits, after the decimal point.

Example: \(5.23×2.1\)

  • First, line up the decimals in right alignment.

\(\begin{array}\\ &5&.&2&3\\ ×&&2&.&1\\ \hline\\ \hline \end{array}\)

  • Now, multiply each digit of the multiplier by each digit of the multiplicand, just like whole numbers, by ignoring the decimal point.

\(\begin{array}\\ &&5&.&2&3\ \rightarrow2 \text{ digits}\\ ×&&&2&.&1\ \rightarrow1 \text{ digit}\\ \hline &&&5&2&3\\ +&1&0&4&6&0&\\ \hline &1&0&9&8&3&\\ \end{array}\)

  • There are \(2\) digits after the decimal point in multiplicand and \(1\) digit in multiplier.
  • So, the product will have \(3\) digits after the decimal point.

\(10.983\)

  • Thus, \(10.983\) is the product of \(5.23\) and \(2.1\)

Note:

When we multiply decimals by power of \(10\), then we can simply move the decimal point to the right, by the number of places per multiple of \(10\).

Example: \(6.45\times10=64.5\)

Here, we move decimal by one place value.

\(7.234×1000=7234\)

Here, we move decimal by three place values.

Illustration Questions

Solve: \(6.35×5.9\)

A \(37.465\)

B \(35.465\)

C \(39.465\)

D \(38.469\)

×

First, we will line up the decimal numbers in right alignment.

\(\begin{array}\\ &6&.&3&5\\ ×&&5&.&9\\ \hline\\ \hline \end{array}\)

Now, we multiply each digit of the multiplier by each digit of the multiplicand, just like whole numbers, by ignoring the decimal point.

\(\begin{array}\\ &&6&.&3&5\rightarrow2 \text{ digits}\\ ×&&&5&.&9\rightarrow1 \text{ digit}\\\hline\ &&5&7&1&5\\ +&3&1&7&5&0\\ \hline &3&7&4&6&5\\ \end{array}\)

  • There are \(2\) digits after the decimal point in the multiplicand and \(1\) digit in the multiplier.
  • So, the product will have \(3\) digits after the decimal point.

\(37.465\)

Hence, option (A) is correct.

Solve: \(6.35×5.9\)

A

\(37.465\)

.

B

\(35.465\)

C

\(39.465\)

D

\(38.469\)

Option A is Correct

Representation of Decimals on Number Line

  • Decimals can be represented on the number line.
  • To find the decimal number on the number line, we move in the forward direction from zero.
  • First, find the wholes and then the parts on the number line.

For example: We want to represent \(3.7\) on the number line.

  • To find the decimal number \(3.7\) on the number line, find the whole \(3\).
  • To find \(3\) on the number line, move in the forward direction and mark the whole at \(3\).

Now, we have to find the part \(0.7\) on the number line. To find \(0.7\) on the number line, divide the interval between \(3\) and \(4\) into \(10\) equal sections, each having a scale of \(0.1\).

  • Now, start from \(3\) and count \(7\) slashes by moving forward. Mark the \(7\)th slash.

  • We have reached the whole \(3\) and the parts \(0.7\) on the number line.
  • So, by combining these, we get the decimal number \(3.7\) on the number line.

Illustration Questions

Which one of the following points is closest to ​2.65 on the given number line?

A Point A

B Point B

C Point C

D Point D

×

First, we need to find intervals between whole numbers.

image

There are four equal intervals between whole numbers, this means that each interval has a scale of \(.25\).

image

So, the given number \(2.65\) should be placed between \(2.50\) and \(2.75\).

Among \(2.50\) and \(2.75,\;2.65\) is closer to \(2.75\).

Thus, point \(D\) is closest to \(2.65\).

image

Hence, option (D) is correct.

Which one of the following points is closest to ​2.65 on the given number line?

image
A

Point A

.

B

Point B

C

Point C

D

Point D

Option D is Correct

Addition of Decimals 

  • Addition of decimals can be easily done by dealing with the wholes and the parts of the numbers separately.

Example: Add \(3.4\) and \(7.347\)

  • Both decimal numbers have different place values, so we can add zeros to make their place value same.

\(3.400+7.347\)

  • Now, both the numbers have same place values.

  • Write the decimal numbers vertically, lining up the decimal points and each digit according to its place value

\(\begin{array}\\ &3&.&4&0&0\\ +&7&.&3&4&7\\ \hline\\ \hline \end{array}\)

  • Add the columns vertically-

\(\begin{array}\\ &3&.&4&0&0\\ +&7&.&3&4&7\\ \hline 1&0&.&7&4&7\\ \hline \end{array}\)

  • Sum of \(3.4 \) and \(7.347\) is \(10.747\)

Illustration Questions

Find:​ \(20.6+1.225+2.75\)

A \(23.575\)

B \(24.575\)

C \(24.306\)

D \(23.306\)

×

All decimal numbers have different place values, so we can add zeros to make them same.

\(20.600+1.225+2.750\)

Now, write the numbers vertically, lining up the decimal points and each digit according to its place value.

\(\begin{array}\\ &2&0&.&6&0&0\\ +&&1&.&2&2&5\\ &&2&.&7&5&0\\ \hline \\ \hline \end{array}\)

On adding the columns vertically, we get

\(\begin{array}\\ &2&0&.&6&0&0\\ +&&1&.&2&2&5\\ &&2&.&7&5&0\\ \hline &2&4&.&5&7&5\\ \hline \end{array}\)

Hence, option (B) is correct.

Find:​ \(20.6+1.225+2.75\)

A

\(23.575\)

.

B

\(24.575\)

C

\(24.306\)

D

\(23.306\)

Option B is Correct

Division of Decimals

  • Dividing decimals by other decimals is almost same as dividing whole numbers.
  • To make divisor into a whole number, multiply the divisor by a power of ten.
  • If we multiply the divisor by a power of ten, then multiply the dividend also by the same power of ten.

Example: Evaluate \(7.2\div3.6\)

  • First, make the divisor into a whole number.
  • Multiply \(3.6\) by \(10\).

\(3.6×10=36\)

  • Now, we also need to multiply dividend by \(10\).

\(7.2×10=72\)

  • Now, divide as whole numbers.
  • So, quotient of \(7.2\) by \(3.6\) is \(2\).

Note: 

While dividing a decimal number by another decimal number, if a part of the quotient is repeating, then it is called repeating or recurring decimal.

Example: 

We divide \(0.1\) by \(0.3\).

  • We get \(3\), which is repeating again and again, it is known as repeating or recurring decimal.
  • It is written as \(0.\overline{3}\)
  • Thus, \(0.1\div0.3=0.\overline{3}\)

Note:

When we divide decimals by power of \(10\), then we can simply move the decimal point to the left, by the number of places per multiple of \(10\).

Example 1: \(17.25\div1000\)

Here, we will move decimal by three place values.

Example 2: \(57.6\div10\)

Here, we will move decimal point by one place value.

Illustration Questions

Evaluate:

A \(1.35\)

B \(0.14\)

C \(1.14\)

D \(14\)

×

First, we make the divisor a whole number by multiplying \(.09\) by \(100\).

\(.09×100=9\)

Now, we also need to multiply dividend by \(100\).

\(1.26×100=126\)

Now, we divide them as whole numbers.

 

image

Thus, \(1.26\div0.09=14\)

Hence, option (D) is correct.

Evaluate:

image
A

\(1.35\)

.

B

\(0.14\)

C

\(1.14\)

D

\(14\)

Option D is Correct

Ordering Decimals

  • We can arrange decimals from least to greatest or greatest to least.
  • To do so, first line up the decimal point and each digit in the place value chart.
  • Then add zeros to make the same number of digits in the decimal numbers.
  • Now, compare from left to right.

Example: Compare \(13.2,\;13.299,\;13.213,\;13.226\)

  • Lining up each decimal point and each digit in the place value chart, we get 
Tens  Ones   .  Tenths  Hundredths  Thousandths 
1 3 . 2 0 0
1 3 . 2 9 9
1 3 . 2 1 3
1 3 . 2 2 6
  • Now we start comparing from tens column (left most column).
  • Tens, ones and tenths columns have same digits, i.e. \(1,\,3,\,2\), respectively.
  • So, we compare hundredths column.
  • In the hundredths column,

\(0<1<2<9\)

  • Hence, \(13.200<13.213<13.226<13.299\)

Illustration Questions

Arrange \(6.531,\;6.446,\;6.921,\;6.189\) from least to greatest.

A \(6.921,\;6.531,\;6.466,\;6.189\)

B \(6.921,\;6.189,\;6.446,\;6.531\)

C \(6.189,\;6.921,\;6.446,\;6.531\)

D \(6.189,\;6.446,\;6.531,\;6.921\)

×

Line up the decimal point and each digit in the place value chart.

Ones   .  Tenths  Hundredths  Thousandths
6 . 5 3 1
6 . 4 4 6
6 . 9 2 1
6 . 1 8 9

Start comparing from ones column.

Ones column has same number i.e. \(6\).

So, we compare tenths column.

In the tenths column,

\(1<4<5<9\)

Thus, \(6.189,\;6.446,\;6.531,\;6.921\)

Hence, option (D) is correct.

Arrange \(6.531,\;6.446,\;6.921,\;6.189\) from least to greatest.

A

\(6.921,\;6.531,\;6.466,\;6.189\)

.

B

\(6.921,\;6.189,\;6.446,\;6.531\)

C

\(6.189,\;6.921,\;6.446,\;6.531\)

D

\(6.189,\;6.446,\;6.531,\;6.921\)

Option D is Correct

Word Problems (Two or More Operations)

  • Decimal numbers are used in real world problems too.
  • There are countless examples which show that 'How to deal with decimals in real life?'
  • Consider an example to understand real world problems.
  • Alex needs \(6\) paint brushes and some colors for drawing. He has \($12.5\) to spend and takes \($10\) more from his father. If the cost of one paint brush is \($1.2\), then how much money he is left with to buy colors?
  • We can solve these type of problems by going step by step, using suitable operations.
  • Cost of one paint brush is \($1.2\) and there are \(6\) brushes.

\(\therefore\) We need to multiply \(1.2\) by \(6\) to find out the cost of \(6\) brushes.

\(=1.2×6\)

\(=7.2\)

So, cost of \(6\) brushes \(=$7.2\)

Alex has \($12.5\) and he takes \($10\) more from his father.

\(\therefore\) Total money Alex has \(=12.5+10\)

                                           \(=$22.5\)

  • Since he spent \($7.2\) for brushes, so we need to subtract \($7.2\) from the total money, i.e. \($22.5\)

\(=$22.5-$7.2\)

\(=$15.3\)

Thus, he is left with \($15.3\) to buy colors.

Illustration Questions

Ms. Wendy has \(0.8\) liters of energy drink, she adds twice of the quantity of energy drink that she already had. If she distributes total quantity of energy drink to her \(9\) players, then find the quantity of energy drink that each player gets?

A \(\text{0.26 liters}\)

B \(\text{0.8 liters}\)

C \(\text{2.4 liters}\)

D \(\text{1.8 liters}\)

×

Given: 

Energy drink that Ms. Wendy has \(=0.8\) liters

Twice of the energy drink that she had 

\(=2×0.8\)

\(=1.6\) liters

Total energy drink Ms. Wendy has, after adding twice of the energy drink that she had 

\(=0.8+1.6\)

\(=2.4\) liters

Now according to the situation, she distributes energy drink among her \(9\) players.

\(=2.4\div9\)

\(=0.26\) liters

Thus, each player gets \(0.26\) liters of energy drink.

Hence, option (A) is correct.

Ms. Wendy has \(0.8\) liters of energy drink, she adds twice of the quantity of energy drink that she already had. If she distributes total quantity of energy drink to her \(9\) players, then find the quantity of energy drink that each player gets?

A

\(\text{0.26 liters}\)

.

B

\(\text{0.8 liters}\)

C

\(\text{2.4 liters}\)

D

\(\text{1.8 liters}\)

Option A is Correct

Practice Now