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Operations On Fractions

Subtraction of Fractions

  • Subtraction of fractions can be done by following the given steps:

Step 1: Make sure that the numbers are in fraction form. If they are not, then convert them into fraction form as follows: 

Case 1. If the number is a whole number, convert it into a fraction by putting it over \(1\).

For example: \(2-\dfrac{1}{3}\)

Here, \(2=\dfrac{2}{1}\)

Thus, \(\dfrac{2}{1}-\dfrac{1}{3}\)

Case 2. If the number is a mixed number, convert it into an improper fraction.

For example: \(1\dfrac{1}{3}-\dfrac{1}{2}\)

Here, \(1\dfrac{1}{3}=\dfrac{4}{3}\)

Thus,  \(\dfrac{4}{3}-\dfrac{1}{2}\)

Step 2: Make sure that the denominators are equal. If they are not, convert them into like fractions.

For example: \(\dfrac{3}{4}-\dfrac{2}{7}\)

LCM of \(4\) and \(7=28\)

Now, LCM becomes the least common denominator.

So, \(\dfrac{3}{4}×\dfrac{7}{7}=\dfrac{21}{28}\)

\(\dfrac{2}{7}×\dfrac{4}{4}=\dfrac{8}{28}\)

Thus, \(\dfrac{21}{28}\) and \(\dfrac{8}{28}\) are like fractions.

Step 3: Subtract the like fractions.

In the above example,

\(\dfrac{21}{28}-\dfrac{8}{28}=\dfrac{21-8}{28}\)

\(=\dfrac{13}{28}\)

Step 4: Simplify the result.

In the above example, \(13\) and \(28\) do not have any common factor other than \(1.\)

\(\therefore\) \(\dfrac{13}{28}\) is in its simplest form.

Illustration Questions

Aron's weight is \(176\dfrac{1}{5}\) pounds. Through workouts, he reduces \(21\dfrac{4}{5}\) pounds in a month. What is the weight of Aron after a month?

A \(\dfrac{771}{5}\) pounds

B \(\dfrac{621}{5}\) pounds

C \(145\dfrac{7}{5}\) pounds

D \(154\dfrac{2}{5}\) pounds

×

Given: 

Aron's weight before \(1\) month \(=176\dfrac{1}{5}\) pounds

Weight reduced by Aron in \(1\) month \(=21\dfrac{4}{5}\) pounds

So, Aron's weight after \(1\) month

\(\text{= Weight of Aron before 1 month}- \text{ Weight reduced by Aron in 1 month }\)

So, we have to subtract \(21\dfrac{4}{5}\) from \(176\dfrac{1}{5}\).

\(\Rightarrow 176\dfrac{1}{5}-21\dfrac{4}{5}\)

First, convert the mixed numbers into the improper fractions.

Here, \(176\dfrac{1}{5}=\dfrac{881}{5}\)

\(21\dfrac{4}{5}=\dfrac{109}{5}\)

Both the numbers have same denominators.

So, subtract \(\dfrac{109}{5}\) from \(\dfrac{881}{5}\) 

\(\Rightarrow \dfrac{881}{5}-\dfrac{109}{5}=\dfrac{772}{5}\)

Simplifying the result. 

\(772 \) and \(5\) do not have any common factor other than \(1\).

\(\therefore \;\;\dfrac{772}{5}\) is in its simplest from.

We can write \(\dfrac{772}{5}\) as: \(154\dfrac{2}{5}\)  

Thus, the weight of Aron after a month is \(154\dfrac{2}{5}\) pounds.

Hence, option (D) is correct.

Aron's weight is \(176\dfrac{1}{5}\) pounds. Through workouts, he reduces \(21\dfrac{4}{5}\) pounds in a month. What is the weight of Aron after a month?

A

\(\dfrac{771}{5}\) pounds

.

B

\(\dfrac{621}{5}\) pounds

C

\(145\dfrac{7}{5}\) pounds

D

\(154\dfrac{2}{5}\) pounds

Option D is Correct

Multiplication of Fractions

  • Multiplication of fractions can be done by following the given steps:

Step 1: Make sure that the numbers are in fraction form. If they are not, convert them into fraction form as follows:

Case 1: If the number is a whole number, convert it into a fraction by putting it over \(1\).

For example: \(5×\dfrac{1}{6}\)

Here \(5=\dfrac{5}{1}\)

Thus, \(\dfrac{5}{1}×\dfrac{1}{6}\)

Case 2: If the number is a mixed number, convert it into an improper fraction.

For example: \(\dfrac{3}{7}×2\dfrac{1}{5}\)

Here \(2\dfrac{1}{5}=\dfrac{11}{5}\)

Thus, \(\dfrac{3}{7}×\dfrac{11}{5}\)

Step 2: Multiply the numerator by numerator and denominator by denominator.

For example: \(\dfrac{5}{9}×\dfrac{3}{8}\)

\(\dfrac{5×3}{9×8}=\dfrac{15}{72}\)

Step 3. Simplify the result.

In the above example: \(\dfrac{15}{72}\)

The Greatest Common Factor (GCF) of \(15\) and \(72\) is \(3\).

\(\therefore \;\;\;\;\dfrac{15\div3}{72\div3}=\dfrac{5}{24}\)

Thus, \(\dfrac{5}{24}\) is in its simplest form.

Illustration Questions

In a box, there are  \(50\) chocolates. The cost of one chocolate is \($1\dfrac{2}{25}\). If Kara buys \(2\) boxes of chocolates, how much price will she pay?

A \($108\)

B \($109.5\)

C \($120.2\)

D \($102\)

×

Given:

Number of chocolates in one box \(=50\) 

Cost of one chocolate \(=$1\dfrac{2}{25}\)

To find the total amount paid by Kara, we first find out the cost of \(50\) chocolates.

\(\text{Cost of 50 chocolates = Number of chocolates in one box × Cost of one chocolate}\)

\(=50×1\dfrac{2}{25}\)

 

To solve this, first we convert the whole number into a fraction by putting it over \(1\).

Here,  \(50=\dfrac{50}{1}\)

Now, we convert the mixed number into an improper fraction.

\(1\dfrac{2}{25}=\dfrac{27}{25}\)

Now, the problem becomes \(\dfrac{50}{1}×\dfrac{27}{25}\)

Multiply the numerator by numerator and denominator by denominator.

\(=\dfrac{50×27}{1×25}\)

\(=\dfrac{1350}{25}\)

Simplify the fraction obtained.

The Greatest Common Factor (GCF) of \(1350\) and \(25\) is \(25\).

\(\therefore\;\;\;\dfrac{1350\div25}{25\div25}=\dfrac{54}{1}\)

Thus, the total cost of \(50\) chocolates is \($54\).

One box contains \(50\) chocolates that cost \($54\) and Kara buys \(2\) boxes, so the total cost of two boxes is :

Cost of \(2\) boxes \(=2×54\)

\(=$108\)

Thus, Kara pays \($108\) for the  \(2\) boxes of chocolates.

Hence, option (A) is correct.

In a box, there are  \(50\) chocolates. The cost of one chocolate is \($1\dfrac{2}{25}\). If Kara buys \(2\) boxes of chocolates, how much price will she pay?

A

\($108\)

.

B

\($109.5\)

C

\($120.2\)

D

\($102\)

Option A is Correct

Addition of Fractions

  • Addition of fractions can be done by following the given steps:

Step 1: Make sure that the numbers are in fraction from. If they are not, convert them into fraction form as follows:

Case 1: If the number is a whole number, then convert it into a fraction by putting it over \(1\).

For example: \(3+\dfrac{1}{2}\)

\(3=\dfrac{3}{1}\)

Thus, \(\dfrac{3}{1}+\dfrac{1}{2}\)

Case 2: If the number is a mixed number, convert it into an improper fraction.

For example: \(3\dfrac{2}{3}+\dfrac{4}{5}\)

\(3\dfrac{2}{3}=\dfrac{11}{3}\)

Thus, \(\dfrac{11}{3}+\dfrac{4}{5}\)

Step 2: Make sure that the denominators are equal, if they are not, convert them into like fraction.

For example: \(\dfrac{2}{3}+\dfrac{4}{5}\)

LCM of \(3\) and \(5\;=\;15\)

Now, LCM becomes the least common denominator (LCD).

So, \(\dfrac{2}{3}×\dfrac{5}{5}=\dfrac{10}{15}\) and

\(\dfrac{4}{5}×\dfrac{3}{3}=\dfrac{12}{15}\)

Thus, \(\dfrac{10}{15}\, \text{and }\dfrac{12}{15}\) are like fractions.

Step 3. Add the like fractions.

In the above example,

\(\dfrac{10}{15}+\dfrac{12}{15}=\dfrac{10+12}{15}=\dfrac{22}{15}\)

Step 4. Simplify the result.

In the above example,

\(22\) and \(15\) do not have any common factor other than \(1\).

\(\therefore\)  \(\dfrac{22}{15}\) is in its simplest form.

Illustration Questions

Alex, Kevin, and Jose are solving the same problem individually, given by their teacher. The problem is: \(\dfrac{2}{21}+1\dfrac{1}{3}+\dfrac{6}{7}\) Alex calculates the result as \(\dfrac{42}{5}\), Kevin's result is \(2\dfrac{2}{7}\) while Jose's answer is \(\dfrac{16}{7}\). Which option is correct?

A \(\text{Alex is right}.\)

B \(\text{Kevin is right}.\)

C \(\text{Jose is right}.\)

D \(\text{Both Kevin and Jose are right}.\)

×

Given problem: \(\dfrac{2}{21}+1\dfrac{1}{3}+\dfrac{6}{7}\)

To solve this, first we convert the mixed number into a fraction.

\(1\dfrac{1}{3}=\dfrac{3×1+1}{3}=\dfrac{4}{3}\)

Now, rewrite the problem \(=\dfrac{2}{21}+\dfrac{4}{3}+\dfrac{6}{7}\)

Here, denominators are not equal. So, we convert the fractions into like fractions.

LCM of \(21,\;3\;\text{and 7 = 21}\)

\(\therefore\) Take \(21\) as the least common denominator.

Thus, \(\dfrac{4}{3}×\dfrac{7}{7}=\dfrac{28}{21}\)

\(\dfrac{6}{7}×\dfrac{3}{3}=\dfrac{18}{21}\)

Add the like fractions.

\(\dfrac{2}{21}+\dfrac{28}{21}+\dfrac{18}{21}\)

\(=\dfrac{2+28+18}{21}\)

\(=\dfrac{48}{21}\)

Simplifying the result.

Greatest Common Factor (GCF) of \(48\) and \(21=3\)

\(\therefore\) \(\dfrac{48\div3}{21\div3}=\dfrac{16}{7}\)

We can write \(\dfrac{16}{7}\) as \(2\dfrac{2}{7}\).

Thus, the answer is \(\dfrac{16}{7}\) or \(2\dfrac{2}{7}\).

So, both Kevin and Jose calculate the correct value.

Hence, option (D) is correct.

Alex, Kevin, and Jose are solving the same problem individually, given by their teacher. The problem is: \(\dfrac{2}{21}+1\dfrac{1}{3}+\dfrac{6}{7}\) Alex calculates the result as \(\dfrac{42}{5}\), Kevin's result is \(2\dfrac{2}{7}\) while Jose's answer is \(\dfrac{16}{7}\). Which option is correct?

A

\(\text{Alex is right}.\)

.

B

\(\text{Kevin is right}.\)

C

\(\text{Jose is right}.\)

D

\(\text{Both Kevin and Jose are right}.\)

Option D is Correct

Division of Fractions

  • Dividing a fraction by another fraction is same as multiplying by its reciprocal.
  • To understand it, consider an example.

\(\dfrac{5}{4}\div\dfrac{2}{3}\)

Step 1: Find the reciprocal of the divisor.

Here, divisor \(=\dfrac{2}{3}\) and dividend \(=\dfrac{5}{4}\) 

So, reciprocal of \(\dfrac{2}{3}\) is \(\dfrac{3}{2}\).

Step 2: Multiply the dividend by the reciprocal of divisor.

\(\dfrac{5}{4}×\dfrac{3}{2}\)

\(=\dfrac{5×3}{4×2}\)

\(=\dfrac{15}{8}\)

Step 3: Simplify the result.

\(15\) and \(8\) do not have any common factor other than \(1\).

\(\therefore\) \(\dfrac{15}{8}\) is in its simplest form.

Thus, \(\dfrac{5}{4}\div\dfrac{2}{3}=\dfrac{15}{8}\)

Note: Convert the mixed number into an improper fraction.

For example: \(2\dfrac{5}{7}=\dfrac{19}{7}\)

Illustration Questions

Kyle wants to decorate the room for a party. She buys a roll of ribbon which is \(46\dfrac{1}{5}\) inches long. To decorate the room, she needs to cut the ribbon into equal lengths of \(\dfrac{7}{5}\) inches. How many such pieces can be cut?

A \(40\)

B \(33\)

C \(28\)

D \(38\)

×

Given: 

Total length of the ribbon \(=46\dfrac{1}{5}\) inches

Length of each small piece \(=\dfrac{7}{5}\) inches

Total number of small pieces of ribbon \(=\) Total length of ribbon \(\div\) Length of each small piece

\(=46\dfrac{1}{5}\div\dfrac{7}{5}\)                            \(\bigg\{\;\because\;46\dfrac{1}{5}=\dfrac{46\times5+1}{5}=\dfrac{231}{5}\;\bigg\}\)

\(=\dfrac{231}{5}\div\dfrac{7}{5}\)

Here, dividend \(=\dfrac{231}{5}\) and divisor \(=\dfrac{7}{5}\)

Reciprocal of \(\dfrac{7}{5}\) is \(\dfrac{5}{7}\).

Multiply the dividend with the reciprocal of divisor.

\(\dfrac{231}{5}×\dfrac{5}{7}\)

\(=\dfrac{231×5}{5×7}\)                  \(\bigg\{\;\because\;\dfrac{5}{5}=1\;\bigg\}\)

\(=\dfrac{231}{7}\)

\(=\;33\)

Thus, Kyle can cut the ribbon into \(33\) equal small pieces.

Hence, option (B) is correct.

Kyle wants to decorate the room for a party. She buys a roll of ribbon which is \(46\dfrac{1}{5}\) inches long. To decorate the room, she needs to cut the ribbon into equal lengths of \(\dfrac{7}{5}\) inches. How many such pieces can be cut?

A

\(40\)

.

B

\(33\)

C

\(28\)

D

\(38\)

Option B is Correct

Mix Operations

A solution of a problem can be achieved by following steps:

1. Identify the keywords 

2. Apply appropriate operation to solve the problem.

Example: A bank charges an overdraft fee on a standard current account when the balance falls below zero at any point of time. Mr. Watson had a balance of \($4500\) in his account. He withdrew \(\dfrac{6}{5}\) of the account balance for business purposes. The bank charged \(\dfrac{1}{20}\) of the overdrawn balance. Next day, he deposited \($3000\) into the bank account. Calculate the current balance of his bank account.

Solution: Original balance in bank account \(=$4500\)

Fraction of amount withdrew \(=\dfrac{6}{5}\) of the account balance

\(=\dfrac{6}{5}\) of \(4500\)

\(=\dfrac{6}{5}×4500\)

\(=\dfrac{6×4500}{5×1}=5400\)

Now, the overdrawn amount  \(=\text{Total withdrawal }-\text{Original deposit}\)

\(=5400-4500\)

\(=$900\)

Thus, the overdrawn amount is \($900\).

Now, the overdrawn fees  \(=\dfrac{1}{20}\text{ of Total overdrawn amount}\)

\(=\dfrac{1}{20}\text{ of }(900)\)

\(=\dfrac{1}{20}×(900)\)

\(=\dfrac{1×(900)}{20×1}=$45\)

Overdrawn amount and fee is taken as negative amount as bank takes away this amount.

So, the total amount taken away by bank

\(=900+45\\=$945\)

Now, the total amount deposited by Mr. Watson \(=$3000\)

So, the current balance of the bank account 

\(\text{=Total amount deposited after withdrawal - Total amount taken away by bank}\\=$3000 \;-\;$945\\=$2055\)

Illustration Questions

Terrence decided to walk \(7\) miles a day. Today, he walked \(2\dfrac{1}{2}\) miles to school, \(1\dfrac{1}{2}\) miles to market and \(2\dfrac{1}{4}\) miles to his friend's house. How many miles does he have to cover?

A \(\dfrac{2}{3}\) miles

B \(\dfrac{3}{4}\) miles

C \(\dfrac{4}{5}\) miles

D \(\dfrac{5}{6}\) miles

×

Given: 

Terrence to walk per day \(= 7\) miles

Terrence walked:

to school \(=2\dfrac{1}{2}\) miles

to market \(=1\dfrac{1}{2}\) miles

to his friend's house \(=2\dfrac{1}{4}\) miles

Converting mixed numbers into improper fractions.

\(2\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)

\(1\dfrac{1}{2}=1+\dfrac{1}{2}=\dfrac{3}{2}\)

\(2\dfrac{1}{4}=2+\dfrac{1}{4}=\dfrac{9}{4}\)

First, we find out the total distance covered by Terrence, till now.

\(=\dfrac{5}{2}+\dfrac{3}{2}+\dfrac{9}{4}\)

\(=\dfrac{10}{4}+\dfrac{6}{4}+\dfrac{9}{4}\)  {make denominators same}

\(=\dfrac{25}{4}\)

Now, to calculate the miles left for the day.

Here, 'left' shows subtraction.

Total miles to be covered in a day \(=7\) miles

Miles left for the day \(=\) Total miles to be walked in a day \(-\) Total miles walked by Terrence

\(=\dfrac{7}{1}-\dfrac{25}{4}\)    {writing a whole number as afraction}}

\(=\dfrac{28}{4}-\dfrac{25}{4}\)   {making denominators same}

\(=\dfrac{3}{4}\)

Simplify the result.

\(3\) and \(4\) do not have any common factor other than \(1\).

\(\therefore\) \(\dfrac{3}{4}\) is in its simplest form.

Thus, Terrence needs to walk \(\dfrac{3}{4}\) miles more to complete the target.

Hence, option (B) is correct.

Terrence decided to walk \(7\) miles a day. Today, he walked \(2\dfrac{1}{2}\) miles to school, \(1\dfrac{1}{2}\) miles to market and \(2\dfrac{1}{4}\) miles to his friend's house. How many miles does he have to cover?

A

\(\dfrac{2}{3}\) miles

.

B

\(\dfrac{3}{4}\) miles

C

\(\dfrac{4}{5}\) miles

D

\(\dfrac{5}{6}\) miles

Option B is Correct

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