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Subtraction Of Decimals

Subtraction of Decimal Numbers from Whole Numbers

  • Subtraction of decimals can be easily done by dealing with the wholes and the parts of the numbers separately.
  • To subtract decimal numbers from whole numbers, we are going to work with the wholes and parts of the numbers separately.
  • Suppose we want to subtract \(2.5\) from \(4,\) i.e. \(4-2.5\)

\(4\) can be written as \(4.0\)

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c|c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &4&\cdot&0\\ -&2&\cdot&5\\ \hline &&\cdot&\\ \hline \end{array}\)

  • First, subtract the digits of the tenths column. We can not take \(5\) from zero, so we 'borrow' \(10\) from the ones column such that the \(4\) becomes a \(3.\)

Now, we can take \(10\) in place of \(0\) \(\left[\underbrace{\underline{10}}_\text{borrowed}+0=10\right]\)

Now, we subtract \(5\) from \(10\)

\(10-5=5\)

\(5\) is to be written in the tenths column.

\(\begin{array} {c} &\not{4}^3&\cdot&^\underline10\\ -&2&\cdot&\;\;\;5\\ \hline &&&\;\;\;5\\ \hline \end{array}\)

  • Now, subtract the digits of the ones column.

\(3-2=1\)

\(1\) is to be written in the ones column.

\(\begin{array} {c|c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &4&\cdot&0\\ -&2&\cdot&5\\ \hline &1&\cdot&5\\ \hline \end{array}\)

  • Hence, \(1.5\) is the difference of \(4\) and \(2.5\)

Illustration Questions

What is the difference of \(8\) and \(3.6\)?

A \(4.4\)

B \(11.6\)

C \(3.4\)

D \(4.6\)

×

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c|c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &8&\cdot&0\\ -&3&\cdot&6\\ \hline &&\cdot&\\ \hline \end{array}\)

First, subtract the digits of the tenths column. We can not take \(6\) from \(0,\) so we borrow \(10\) from the ones column, such that the \(8\) becomes a \(7.\)

Now, we subtract \(6\) from \(10\) \([0+10\) (borrowed) \(=10]\)

\(4\) is to be written in the tenths column.

\(\begin{array} {c} &\not{8}^7&\cdot&0\\ -&3&\cdot&6\\ \hline &&\cdot&4\\ \hline \end{array}\)

Now, we subtract the digits of the ones column.

\(7-3=4\)

\(4\) is to be written in the ones column.

\(\begin{array} {c|c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &8&\cdot&0\\ -&3&\cdot&6\\ \hline &4&\cdot&4\\ \hline \end{array}\)

So, \(4.4\) is the difference of \(8\) and \(3.6\)

Hence, option (A) is correct.

What is the difference of \(8\) and \(3.6\)?

A

\(4.4\)

.

B

\(11.6\)

C

\(3.4\)

D

\(4.6\)

Option A is Correct

Subtraction of Decimals up to Tenths Place

  • Subtraction of decimals can be easily done by dealing with the wholes and the parts of the numbers separately.
  • Suppose, we want to subtract \(12.4\) from \(25.6\)

i.e. \(25.6-12.4\)

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text{Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &2&5&\cdot&6\\ -&1&2&\cdot&4\\ \hline &&&\cdot&\\ \hline \end{array}\)

  • First, subtract the digits of the tenths column.

\(6-4=2\)

\(2\) is to be written in the tenths column.

\(\begin{array} {c} &2&5&\cdot&6\\ -&1&2&\cdot&4\\ \hline &&&\cdot&2\\ \hline \end{array}\)

  • Now, subtract the digits of the ones column.

\(5-2=3\)

\(3\) is to be written in the ones column.

\(\begin{array} {c} &2&5&\cdot&6\\ -&1&2&\cdot&4\\ \hline &&3&\cdot&2\\ \hline \end{array}\)

  • Subtract the digits of the tens column.

\(2-1=1\)

\(1\) is to be written in the tens column.

\(\begin{array} {c} \hline &\text{Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &2&5&\cdot&6\\ -&1&2&\cdot&4\\ \hline &1&3&\cdot&2\\ \hline \end{array}\)

Hence, \(13.2\) is the difference of \(25.6\) and \(12.4\)

Illustration Questions

Evaluate \(9.4-6.7\)

A \(16.1\)

B \(2.7\)

C \(96.47\)

D \(3.7\)

×

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &9&\cdot&4\\ -&6&\cdot&7\\ \hline &&\cdot&\\ \hline \end{array}\)

First, subtract the digits of the tenths column.

We can not take \(7\) from \(4,\) so we borrow \(10\) from the ones column such that the \(9\) becomes an \(8.\)

We can take \(14\) in place of \(4\) because \(\underbrace{\underline{10}}_{\text{borrowed}}+4=14\)

Now, subtract \(7\) from \(14\) 

\(14-7=7\)

\(7\) is to be written in the tenths column.

\(\begin{array} {c} &\not{9}^8&\cdot&^\underline14\\ -&6&\cdot&\;\;7\\ \hline &&\cdot&7\\ \hline \end{array}\)

Now, subtract the digits of the ones column.

\(8-6=2\)

\(2\) is to be written in the ones column.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}\\ \hline &9&\cdot&4\\ -&6&\cdot&7\\ \hline &2&\cdot&7\\ \hline \end{array}\)

So, \(2.7\) is the difference of \(9.4\) and \(6.7\)

Hence, option (B) is correct.

Evaluate \(9.4-6.7\)

A

\(16.1\)

.

B

\(2.7\)

C

\(96.47\)

D

\(3.7\)

Option B is Correct

Subtraction of Decimals up to Hundredths Place

  • Subtraction of decimals can be easily done by dealing with the wholes and the parts of the numbers separately.
  • Suppose, we want to subtract \(4.25\) from \(8.75\)
  • Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}&\text{Hundredths}\\ \hline &8&\cdot&7&5\\ -&4&\cdot&2&5\\ \hline &&\cdot&\\ \hline \end{array}\)

  • First, subtract the digits of the hundredths column.

\(5-5=0\)

\(0\) is to be written in the hundredths column.

\(\begin{array} {c} &8&\cdot&7&5\\ -&4&\cdot&2&5\\ \hline &&\cdot&&0\\ \hline \end{array}\)

  • Now, we subtract the digits of the tenths column.

\(7-2=5\)

\(5\) is to be written in the tenths column.

\(\begin{array} {c} &8&\cdot&7&5\\ -&4&\cdot&2&5\\ \hline &&\cdot&5&0\\ \hline \end{array}\)

  • Subtract the digits of the ones column.

\(8-4=4\)

\(4\) is to be written in the ones column.

\(\begin{array} {c} &8&\cdot&7&5\\ -&4&\cdot&2&5\\ \hline &4&\cdot&5&0\\ \hline \end{array}\)

  • Hence, \(4.50\) is the difference of \(8.75\) and \(4.25\)

Illustration Questions

Evaluate \(7.50-2.75\)

A \(10.25\)

B \(5.75\)

C \(4.75\)

D \(9.25\)

×

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}&\text{Hundredths}\\ \hline &7&\cdot&5&0\\ -&2&\cdot&7&5\\ \hline &&\cdot&\\ \hline \end{array}\)

First, subtract the digits of the hundredths column.

We can not take \(5\) from zero, so we borrow \(10\) from the tenths column such that the \(5\) becomes a \(4.\)

We can take \(10\) in place of \(0\) because \(\underbrace{\underline{10}}_\text{Borrowed}+0=10\)

Now, subtract \(5\) from \(10\).

\(10-5=5\)

\(5\) is to be written in the hundredths column.

\(\begin{array} {c} &7&\cdot&\not{5}^4&^\underline10\\ -&2&\cdot&7&\;\;\,5\\ \hline &&\cdot&&\;\;\,5\\ \hline \end{array}\)

Now, subtract the digits of the tenths column.

We can not take \(7\) from \(4,\) so we borrow \(10\) from the ones column such that the \(7\) (at one place) becomes a \(6.\) 

We can take \(14\) in place of \(4\) because \(\underbrace{\underline{10}}_{\text{Borrowed}}+4=14\)

Now, we subtract \(7\) from \(14\).

\(14-7=7\)

\(7\) is to be written in the tenths column.

\(\begin{array} {c} &\not{7}^6&\cdot&5&0\\ -&2&\cdot&7&5\\ \hline &&\cdot&7&5\\ \hline \end{array}\)

Now, subtract the digits of the ones column.

\(6-2=4\)

\(4\) is to be written in the ones column.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}&\text{Hundredths}\\ \hline &7&\cdot&5&0\\ -&2&\cdot&7&5\\ \hline &4&\cdot&7&5\\ \hline \end{array}\)

So, \(4.75\) is the difference of \(7.50\) and \(2.75\)

Hence, option (C) is correct.

Evaluate \(7.50-2.75\)

A

\(10.25\)

.

B

\(5.75\)

C

\(4.75\)

D

\(9.25\)

Option C is Correct

Subtraction of Decimals up to Thousandths Place

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

Suppose, we want to subtract \(6.125\) from \(7.375\)

i.e.

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &7&\cdot&3&7&5\\ -&6&\cdot&1&2&5\\ \hline &&\cdot&\\ \hline \end{array}\)

First, subtract the digits of the thousandths column.

\(5-5=0\)

\(0\) is to be written in the thousandths column.

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

Subtract the digits of the hundredths column.

\(7-2=5\)

\(5\) is to be written in the hundredths column.

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

Subtract the digits of the tenths column.

\(3-1=2\)

\(2\) is to be written in the tenths column.

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

Now, subtract the digits of the ones column.

\(7-6=1\)

\(1\) is to be written in the ones column.

\(\begin{array} {c} \hline &\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &7&\cdot&3&7&5\\ -&6&\cdot&1&2&5\\ \hline &1&\cdot&2&5&0\\ \hline \end{array}\)

\(1.250\) can be written as \(1.25\)

Hence, \(1.25\) is the difference of \(7.375\) and \(6.125\)

Illustration Questions

What is the difference of \(20.735\) and \(11.475\)?

A \(32.1210\)

B \(9.026\)

C \(9.360\)

D \(9.26\)

×

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text {Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &2&0&\cdot&7&3&5\\ -&1&1&\cdot&4&7&5\\ \hline &&&\cdot&\\ \hline \end{array}\)

First, we subtract the digits of the thousandths column.

\(5-5=0\)

\(0\) is to be written in the thousandths column.

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

Now, we subtract the digits of the hundredths column.

We can not take \(7\) from \(3\), so we borrow \(10\) from the tenths column such that the \(7\) (at tenths place) becomes a \(6\).

We can take \(13\) in place of \(3\) because \([10\;(\text {borrowed})+3=13]\)

Now, subtract \(7\) from \(13\)

\(13-7=6\)

\(6\) is to be written in the hundredths column.

\(\begin{array} {c} &2&0&\cdot&^6\not7&^13&5\\ -&1&1&\cdot&\;\;4&7&5\\ \hline &&&\cdot&&6&0\\ \hline \end{array}\)

Now, we subtract the digits of the tenths column.

\(6-4=2\)

\(2\) is to be written in the tenths column.

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

Now, subtract the digits of the ones column.

We can not take \(1\) from \(0\), so we borrow \(10\) from the tens column that makes the \(2\) into a \(1.\)

Subtract \(1\) from \(10\)

\(10-1=9\)

\(9\) is to be written in the ones column.

\(\begin{array} {c} &\not2^1&^{\underline{1}}0&\cdot&7&3&5\\ -&1&1&\cdot&4&7&5\\ \hline &&9&\cdot&2&6&0\\ \hline \end{array}\)

Now, we subtract the digits of the tens column.

\(1-1=0\)

\(0\) is to be written in the tens column.

\(\begin{array} {c} \hline &\text {Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &2&0&\cdot&7&3&5\\ -&1&1&\cdot&4&7&5\\ \hline &0&9&\cdot&2&6&0\\ \hline \end{array}\)

\(09.260\) can be written as \(9.26\)

So, \(9.26\) is the difference of \(20.735\) and \(11.475\)

Hence, option (D) is correct.

What is the difference of \(20.735\) and \(11.475\)?

A

\(32.1210\)

.

B

\(9.026\)

C

\(9.360\)

D

\(9.26\)

Option D is Correct

Subtraction of Mixed Decimals (Place value) 

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

Suppose, we want to subtract \(5.125\) from \(25.25\)

i.e. \(25.25-5.125\)

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

The given decimal numbers can be written as:

\(\begin{array} {c} &2&5&\cdot&2&5&0\\ -&0&5&\cdot&1&2&5\\ \hline &&&\cdot&&&\\ \hline \end{array}\)

First, subtract the digits of the thousandths column.

\(0-5\)

We can not take \(5\) from zero, so we borrow \(10\) from the hundredths column that makes the \(5\) (at hundredths place) into a \(4\).

Now, subtract \(5\) from \(10\).

\(10-5=5\)

\(5\) is to be written in the thousandths column.

\(\begin{array} {c} &2&5&\cdot&2&\not5^4&^{\underline1}0\\ -&0&5&\cdot&1&2&5\\ \hline &&&\cdot&&&5\\ \hline \end{array}\)

Now, we subtract the digits of the hundredths column.

\(4-2=2\)

\(2\) is to be written in the hundredths column.

\(\begin{array} {c} &2&5&\cdot&2&5&0\\ -&0&5&\cdot&1&2&5\\ \hline &&&\cdot&&2&5\\ \hline \end{array}\)

Now, subtract the digits of the tenths column.

\(2-1=1\)

\(1\) is to be written in the tenths column.

\(\begin{array} {c} &2&5&\cdot&2&5&0\\ -&0&5&\cdot&1&2&5\\ \hline &&&\cdot&1&2&5\\ \hline \end{array}\)

Now, subtract the digits of the ones column.

\(5-5=0\)

\(0\) is to be written in the ones column.

\(\begin{array} {c} &2&5&\cdot&2&5&0\\ -&0&5&\cdot&1&2&5\\ \hline &&0&\cdot&1&2&5\\ \hline \end{array}\)

Now, subtract the digits of the tens column.

\(2-0=2\)

\(2\) is to be written in the tens column.

\(\begin{array} {c} \hline &\text {Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &2&5&\cdot&2&5&0\\ -&0&5&\cdot&1&2&5\\ \hline &2&0&\cdot&1&2&5\\ \hline \end{array}\)

Hence, \(20.125\) is the difference of \(25.25\) and \(5.125\)

Illustration Questions

Evaluate \(79.558-45.01\) 

A \(127.873\)

B \(31.000\)

C \(127.500\)

D \(34.548\)

×

Write one number just below the other, so that the bottom decimal point is directly below and lined up with the top decimal point.

\(\begin{array} {c} \hline &\text {Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &7&9&\cdot&5&5&8\\ -&4&5&\cdot&0&1&0\\ \hline &&&\cdot&\\ \hline \end{array}\)

First, subtract the digits of the thousandths column.

\(8-0=8\)

\(8\) is to be written in the thousandths column.

\(\begin{array} {c} &7&9&\cdot&5&5&8\\ -&4&5&\cdot&0&1&0\\ \hline &&&\cdot&&&8\\ \hline \end{array}\)

Now, subtract the digits of the hundredths column.

\(5-1=4\)

\(4\) is to be written in the hundredths column.

\(\begin{array} {c} &7&9&\cdot&5&5&8\\ -&4&5&\cdot&0&1&0\\ \hline &&&\cdot&&4&8\\ \hline \end{array}\)

Subtract the digits of tenths column. 

\(5-0=5\)

\(5\) is to be written in the tenths column.

\(\begin{array} {c} &7&9&\cdot&5&5&8\\ -&4&5&\cdot&0&1&0\\ \hline &&&\cdot&5&4&8\\ \hline \end{array}\)

Now, subtract the digits of the ones column. 

\(9-5=4\)

\(4\) is to be written in the ones column.

\(\begin{array} {c} &7&9&\cdot&5&5&8\\ -&4&5&\cdot&0&1&0\\ \hline &&4&\cdot&5&4&8\\ \hline \end{array}\)

Now, we subtract the digits of the tens column. 

\(7-4=3\)

\(3\) is to be written in the tens column.

\(\begin{array} {c} \hline &\text {Tens}&\text{Ones}&\text{Decimal point}&\text{Tenths}& \text {Hundredths} & \text {Thousandths}\\ \hline &7&9&\cdot&5&5&8\\ -&4&5&\cdot&0&1&0\\ \hline &3&4&\cdot&5&4&8\\ \hline \end{array}\)

So, \(34.548\) is the difference of \(79.558\) and \(45.01\)

Hence, option (D) is correct.

Evaluate \(79.558-45.01\) 

A

\(127.873\)

.

B

\(31.000\)

C

\(127.500\)

D

\(34.548\)

Option D is Correct

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