Informative line

# 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.

#### 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.

#### 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 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.

#### 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, $$\dfrac{4}{3}×\dfrac{7}{7}=\dfrac{28}{21}$$

$$\dfrac{6}{7}×\dfrac{3}{3}=\dfrac{18}{21}$$

$$\dfrac{2}{21}+\dfrac{28}{21}+\dfrac{18}{21}$$

$$=\dfrac{2+28+18}{21}$$

$$=\dfrac{48}{21}$$

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: $$2\dfrac{5}{7}=\dfrac{19}{7}$$

#### 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

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

#### 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