Informative line

# Word Problems on Fractions

• To solve the word problems, we need to follow the given steps:

Step 1: Read and understand the given situations.

• In this step, we need to identify the keyword that will help us to determine which operation symbol to be used.

Step 2: Choose the correct operation and solve the problem.

To understand the above steps in a better way, we should understand how problems involve the basic four operations:

1. Addition:  A problem in which two or more quantities are being combined.

Keywords: Sum, in all, combined, altogether, added etc.

For example: Emma bought 2 cupcakes and Carl bought 5. How many cupcakes did they buy in all?

Keyword = 'in all'

2. Subtraction:  A problem in which two quantities are given and one is being removed from another.

Keywords: Left, left over, decreased by, take away, less, less than etc.

For example: Jose takes $$\dfrac{1}{4}$$ minutes less than Marc to reach the staff-room from the corridor. If Marc takes $$\dfrac{3}{4}$$ minutes to reach the staff-room, how many minutes does Jose take?

In this example, 'less than' is the keyword that helps to identify that subtraction is to be done.

3. Multiplication: A problem in which repeated addition needs to be done.

Keywords: Each, every, how much, of etc.

For example: What is $$\dfrac{1}{2}$$ of $$4$$?

In this example, 'of'  is the keyword that helps us to identify that we should use multiplication to solve this problem.

4. Division: A problem in which a single quantity is divided into many equal parts.

Keywords: Split, divide, shared, per etc.

For example: Kelvin teaches 8 students. He divides them equally into two groups. How many students are in each group?

Here, the keyword is 'divided ', which means we need to divide.

#### Lara runs a mile in an average of $$9 \dfrac{3}{6}$$ minutes. She does a cool down jog in an average of $$3 \dfrac{2}{3}$$ minutes. If her daily workout is running a mile and then doing the cool down jog, how long does it take her to do this workout?

A $$\dfrac{54}{5}$$ minutes

B $$13\dfrac{5}{6}$$ minutes

C $$\dfrac{58}{3}$$ minutes

D $$13\dfrac{1}{6}$$ minutes

×

Average time taken by Lara to run a mile  $$=9\dfrac{3}{6}$$ minutes

$$=\dfrac{57}{6}$$ minutes

Average time taken by her for cool down jog $$=3\dfrac{2}{3}$$ minutes

$$=\dfrac{11}{3}$$ minutes

Total time taken for workout

$$=\dfrac{57}{6}+\dfrac{11}{3}$$

$$=\dfrac{57+22}{6} = \dfrac{79}{6}$$

$$=\dfrac{79}{6}$$ minutes

$$=13\dfrac{1}{6}$$ minutes

Thus, the total time taken for workout is $$13\dfrac{1}{6}$$ minutes.

Hence, option (D) is correct.

### Lara runs a mile in an average of $$9 \dfrac{3}{6}$$ minutes. She does a cool down jog in an average of $$3 \dfrac{2}{3}$$ minutes. If her daily workout is running a mile and then doing the cool down jog, how long does it take her to do this workout?

A

$$\dfrac{54}{5}$$ minutes

.

B

$$13\dfrac{5}{6}$$ minutes

C

$$\dfrac{58}{3}$$ minutes

D

$$13\dfrac{1}{6}$$ minutes

Option D is Correct

# Mathematical Modeling on Addition and Subtraction

Many times in real life problems, we need to deal with more than one operations. Only addition or only subtraction would not help. We need to make use of the combination of operations in order to solve the problem.

For example:

A town has a total population of $$20,000$$. There are $$2,000$$ children, $$1,250$$ youths, $$750$$ infants and rest are aged people. What fraction of the total population represents the aged people?

Total population of town $$=20,000$$

Number of children $$=2,000$$

Fraction of children to total population $$=\dfrac {2,000}{20,000}$$

Number of youths $$=1,250$$

Fraction of youths to total population $$=\dfrac {1,250}{20,000}$$

Number of infants $$=750$$

Fraction of infants to total population $$=\dfrac {750}{20,000}$$

Total fraction of children, youths and infants

$$=\dfrac {2,000}{20,000}+\dfrac {1,250}{20,000}+\dfrac {750}{20,000}$$

$$=\dfrac {4,000}{20,000}=\dfrac {1}{5}$$

$$\therefore\;$$ Fraction of aged people  $$=1-\dfrac {1}{5}=\dfrac {4}{5}$$

Thus, $$\dfrac {4}{5}$$ of the total population represents the aged people.

#### In a chutes and ladders game board, there are $$100$$ square blocks, $$9$$ chutes and $$10$$ ladders. What fraction of the total square blocks have the chutes and the ladders? Also find the fraction of the rest of the blocks.(chutes and ladders belong to the blocks where they begin)

A $$\dfrac {19}{100},\dfrac {81}{100}$$

B $$\dfrac {18}{100},\dfrac {82}{100}$$

C $$\dfrac {17}{100},\dfrac {83}{100}$$

D $$\dfrac {16}{100},\dfrac {84}{100}$$

×

Total number of square blocks $$=100$$

Number of chutes $$=9$$

Number of ladders $$=10$$

Fraction of $$9$$ chutes to total square blocks $$=\dfrac {9}{100}$$

Fraction of $$10$$ ladders to total square blocks $$=\dfrac {10}{100}$$

Total fraction $$=\dfrac {9}{100}+\dfrac {10}{100}$$

$$=\dfrac {19}{100}$$

Fraction for remaining blocks $$=1-\dfrac {19}{100}$$

$$=\dfrac {81}{100}$$

Thus, $$\dfrac {19}{100}$$ part of the total square blocks have chutes and ladders while remaining blocks constitute $$\dfrac {81}{100}$$ part of the total square blocks.

Hence, option (A) is correct.

### In a chutes and ladders game board, there are $$100$$ square blocks, $$9$$ chutes and $$10$$ ladders. What fraction of the total square blocks have the chutes and the ladders? Also find the fraction of the rest of the blocks.(chutes and ladders belong to the blocks where they begin)

A

$$\dfrac {19}{100},\dfrac {81}{100}$$

.

B

$$\dfrac {18}{100},\dfrac {82}{100}$$

C

$$\dfrac {17}{100},\dfrac {83}{100}$$

D

$$\dfrac {16}{100},\dfrac {84}{100}$$

Option A is Correct

# Mathematical Modeling on Addition and Division

Here, we will use the combination of addition and division to solve the problem.

For example:

Carl puts $$\dfrac{4}{5}$$ kilogram of fruits in a basket. After a while he adds $$\dfrac{1}{4}$$ kilogram of fruits more to it. He then distributes the total fruits equally among $$3$$ children. What quantity of fruits does each child get?

Quantity of fruits added to the basket by Carl at first $$=\dfrac{4}{5}$$ kg

Quantity of fruits added later $$=\dfrac{1}{4}$$ kg

Total quantity of fruits in the basket $$=\dfrac{4}{5}+\dfrac{1}{4}$$

$$=\dfrac{16+5}{20}$$

$$=\dfrac{21}{20}$$ kg

Carl distributes the total fruits equally among $$3$$ children.

So, the quantity of fruits each child got $$=\dfrac{21}{20}\div3$$

$$=\dfrac{21}{20}\times \dfrac{1}{3}$$

$$=\dfrac{7}{20}$$ kg

Thus, each child got $$\dfrac{7}{20}$$ kilogram of fruits.

#### Sam has $$\dfrac{5}{3}$$ pounds of tomatoes in a basket. He adds $$\dfrac{3}{2}$$ pounds of tomatoes more to it and distributes them equally among $$4$$ kids. What quantity of tomatoes does each child get?

A $$\dfrac{19}{15}$$ pounds

B $$\dfrac{90}{19}$$ pounds

C $$\dfrac{20}{15}$$ pounds

D $$\dfrac{19}{24}$$ pounds

×

Quantity of tomatoes Sam already had in the basket $$=\dfrac{5}{3}$$ pounds

Quantity of tomatoes he added $$=\dfrac{3}{2}$$ pounds

$$\therefore$$ Total quantity of tomatoes in the basket$$=\dfrac{5}{3}+\dfrac{3}{2}$$

$$=\dfrac{10+9}{6}$$

$$=\dfrac{19}{6}$$ pounds

Sam distributed the total quantity of tomatoes equally among $$4$$ kids.

So, the quantity of tomatoes that each kid got $$=\dfrac{19}{6}\div4$$

$$=\dfrac{19}{6}\times\dfrac{1}{4}$$

$$=\dfrac{19}{24}$$ pounds

Thus, each kid got $$\dfrac{19}{24}$$ pounds of tomatoes.

Hence, option (D) is correct.

### Sam has $$\dfrac{5}{3}$$ pounds of tomatoes in a basket. He adds $$\dfrac{3}{2}$$ pounds of tomatoes more to it and distributes them equally among $$4$$ kids. What quantity of tomatoes does each child get?

A

$$\dfrac{19}{15}$$ pounds

.

B

$$\dfrac{90}{19}$$ pounds

C

$$\dfrac{20}{15}$$ pounds

D

$$\dfrac{19}{24}$$ pounds

Option D is Correct

# Mathematical Modeling on Division and Subtraction

Here, we will use the combination of division and subtraction to solve the problem.

For example:

Chris and Sarah together had 4 apples. They divided all the apples equally among themselves. If Chris ate $$\dfrac{1}{2}$$ of the apples he got, how many apples were left with him?

Total number of apples $$=4$$

Number of apples each got $$=4\div 2$$

$$=4\times\dfrac{1}{2}$$

$$=2$$

Number of apples Chris ate $$=\dfrac{1}{2}$$ of the apples he got $$=2\times \dfrac{1}{2}=1$$

Number of apples left with Chris $$=2-1$$

$$=1$$

Thus, 1 apple was left with Chris.

#### Keith and Kara have a $$5$$ liter juice can. They pour $$\dfrac{2}{3}$$ fraction of juice in a bottle and divide it equally among themselves. If Keith drinks $$\dfrac{1}{2}$$ of the juice he gets, how much quantity of juice is left with him?

A $$\dfrac{42}{5}$$ liters

B $$\dfrac{5}{6}$$ liters

C $$\dfrac{31}{2}$$ liters

D $$\dfrac{10}{3}$$ liters

×

Total quantity of juice $$=5$$ liters

Quantity of juice Kara and Keith poured $$=\dfrac{2}{3}$$ of juice

$$=5\times \dfrac{2}{3}$$

$$=\dfrac{10}{3}$$ liters

Quantity of juice each got $$=\dfrac{10}{3}\div2$$

$$=\dfrac{10}{3}\times\dfrac{1}{2}$$

$$=\dfrac{5}{3}$$ liters

Quantity of juice Keith drinks $$=\dfrac{1}{2}$$ of juice he got

$$=\dfrac{1}{2}\times\dfrac{5}{3}=\dfrac{5}{6}$$ liters

$$\therefore$$ Quantity of juice left with him

$$=\dfrac{5}{3}-\dfrac{5}{6}$$

$$=\dfrac{30-15}{18}$$

$$=\dfrac{15}{18} =\dfrac{5}{6}$$ liters

Thus, $$\dfrac{5}{6}$$ liters of juice is left with Keith.

Hence, option (B) is correct.

### Keith and Kara have a $$5$$ liter juice can. They pour $$\dfrac{2}{3}$$ fraction of juice in a bottle and divide it equally among themselves. If Keith drinks $$\dfrac{1}{2}$$ of the juice he gets, how much quantity of juice is left with him?

A

$$\dfrac{42}{5}$$ liters

.

B

$$\dfrac{5}{6}$$ liters

C

$$\dfrac{31}{2}$$ liters

D

$$\dfrac{10}{3}$$ liters

Option B is Correct

#### Which one of the following options represents the mixed number which has the value between $$\dfrac{7}{5} \;and\; \dfrac{8}{5}$$?

A $$2 \dfrac{1}{5}$$

B $$7 \dfrac{1}{3}$$

C $$1 \dfrac{1}{2}$$

D $$1 \dfrac{1}{5}$$

×

Given: $$\dfrac{7}{5}$$ and $$\dfrac{8}{5}$$

Finding the equivalent fractions by multiplying numerator and denominator with the same non-zero number.

$$\dfrac{7}{5} = \dfrac{7 \times2}{5 \times2} = \dfrac{14}{10}$$

$$\dfrac{8}{5} = \dfrac{8 \times2}{5 \times2} = \dfrac{16}{10}$$

The two new fractions are:

$$\dfrac{14}{10}$$ and $$\dfrac{16}{10}$$

Thus, we can say that the fraction between the above two fractions is:

$$\dfrac{15}{10}\;\;\;\;\;\;\;\;\;\Bigg(\because\dfrac{14}{10},\dfrac{15}{10},\dfrac{16}{10}\Bigg)$$

The greatest common factor of $$15$$ and $$10$$ is $$5$$. So, on simplifying the fraction $$\dfrac{15}{10}$$, we get

$$\dfrac{15 \div 5}{10 \div 5}=\dfrac{3}{2}$$

$$3$$ and $$2$$ do not have any common factor other than 1. So, the fraction $$\dfrac{3}{2}$$ is in its simplest form.

On dividing, we get the mixed fraction $$=1\dfrac{1}{2}$$

So, the mixed fraction between  $$\dfrac{7}{5}$$ and $$\dfrac{8}{5}$$  is  $$1\dfrac{1}{2}$$.

Hence, option (C) is correct.

### Which one of the following options represents the mixed number which has the value between $$\dfrac{7}{5} \;and\; \dfrac{8}{5}$$?

A

$$2 \dfrac{1}{5}$$

.

B

$$7 \dfrac{1}{3}$$

C

$$1 \dfrac{1}{2}$$

D

$$1 \dfrac{1}{5}$$

Option C is Correct

# Mathematical Modeling on Addition and Multiplication

Many times in real life problems, we need to deal with more than one operations. Only addition or only multiplication would not help. We need to make use of the combination of operations in order to solve the problem.

For example:

Carl bought $$14$$ cookies at $$\dfrac{2}{7}$$ each and $$20$$ candies at $$\dfrac{4}{9}$$ each. What is the total amount he spent?

Number of cookies Carl bought $$=14$$

Price of $$1$$ cookie $$=\dfrac{2}{7}$$

Price of $$14$$ cookies $$= 14 \times \dfrac{2}{7}$$ $$=4$$

Number of candies he bought $$=20$$

Price of $$1$$ candy $$=\dfrac{4}{9}$$

Price of 20 candies $$=20 \times \dfrac{4}{9}$$ $$=\dfrac{80}{9}$$

$$\therefore$$ Total money he spent $$=4+ \dfrac{80}{9}$$

$$=(\dfrac{36+80}{9})$$

$$=\dfrac{116}{9}$$

#### Sam bought $$13$$ chocolates and ate $$\dfrac{1}{2}$$ of them. Casey bought $$10$$ chocolates and ate $$\dfrac{3}{4}$$ of them. Find the total number of chocolates they both ate in all.

A $$10$$

B $$14$$

C $$13$$

D $$15$$

×

Chocolates bought by Sam $$=13$$

He ate $$\dfrac{1}{2}$$ of the chocolates $$=13\times\dfrac{1}{2}=\dfrac{13}{2}$$

Chocolates bought by Casey $$=10$$

She ate $$\dfrac{3}{4}$$ of the chocolates $$=\dfrac{3}{4}\times10 = \dfrac{15}{2}$$

$$\therefore$$ Total number of chocolates they both ate $$=\dfrac{13}{2}+\dfrac{15}{2}$$

$$=\dfrac{13+15}{2}$$

$$=\dfrac{28}{2}$$

$$=14$$

Thus , the total number of chocolates they both ate equals $$14$$.

Hence, option (B) is correct.

### Sam bought $$13$$ chocolates and ate $$\dfrac{1}{2}$$ of them. Casey bought $$10$$ chocolates and ate $$\dfrac{3}{4}$$ of them. Find the total number of chocolates they both ate in all.

A

$$10$$

.

B

$$14$$

C

$$13$$

D

$$15$$

Option B is Correct

# Mathematical Modeling on Subtraction and Multiplication

Here, we will use the combination of subtraction and multiplication to solve the problem.

For example:

Maria buys $$\dfrac{1}{2}$$ kilogram(kg) of strawberries at the rate of $$10$$ per kg and Jacob buys the same amount of strawberries at the rate of $$12$$ per kg. Find who spends more and by how much.

Rate at which Maria buys strawberries $$=10$$ per kg

Quantity of strawberries she buys $$=\dfrac{1}{2}$$ kg

Total price of $$\dfrac{1}{2}$$ kilogram of strawberries $$=\dfrac{1}{2}\times10$$ $$=5$$

Rate at which Jacob buys strawberries $$=12$$ per kg

Quantity of strawberries he buys $$=\dfrac{1}{2}$$ kg

Total price of $$\dfrac{1}{2}$$ kilogram of strawberries $$=\dfrac{1}{2}\times12$$ $$=6$$

Thus, Jacob spends more.

The amount Jacob spends more  $$= 6-5$$ $$=1$$

Hence, Jacob spends  $$1$$ more than Maria.

#### Kyle had $$3$$ bottles of milk. Each bottle contained $$\dfrac{3}{5}$$ liters of milk. She emptied the milk of each bottle into a big jar, out of which she used  $$\dfrac{2}{5}$$ liters of milk for making coffee. How much quantity of milk was left in the jar?

A $$\dfrac{3}{4}\;liters$$

B $$\dfrac{7}{5}\;liters$$

C $$\dfrac{2}{3}\;liters$$

D $$\dfrac{2}{5}\;liters$$

×

Total number of bottles of milk $$=3$$

Quantity of milk in each bottle $$=\dfrac{3}{5}$$ liters

Total quantity of milk in $$3$$ bottles $$=3\times \dfrac{3}{5}$$

$$=\dfrac{9}{5}$$  [Total quantity of milk in the jar before making coffee]

Quantity of milk she used for coffee $$=\dfrac{2}{5}$$ liters

$$\therefore$$ Quantity of milk left in the jar $$=\dfrac{9}{5}-\dfrac{2}{5}$$

$$=\dfrac{9-2}{5}$$

$$=\dfrac{7}{5}$$ liters

Thus, $$\dfrac{7}{5}$$ liters of milk was left in the jar.

Hence, option (B) is correct.

### Kyle had $$3$$ bottles of milk. Each bottle contained $$\dfrac{3}{5}$$ liters of milk. She emptied the milk of each bottle into a big jar, out of which she used  $$\dfrac{2}{5}$$ liters of milk for making coffee. How much quantity of milk was left in the jar?

A

$$\dfrac{3}{4}\;liters$$

.

B

$$\dfrac{7}{5}\;liters$$

C

$$\dfrac{2}{3}\;liters$$

D

$$\dfrac{2}{5}\;liters$$

Option B is Correct

# Mathematical Modeling on Multiplication and Division

Here, we will use the combination of multiplication and division to solve the problem.

For example:

For a project, Isaac buys $$6$$ packets of pens for $$\dfrac{20}{3}$$ each and decides to share the price among the $$5$$ project team members including him. How much does each member have to pay?

Packets of pens Isaac buys $$=6$$

Price of $$1$$ packet $$=\dfrac{20}{3}$$

Price of $$6$$ packets of pens $$=6\times \dfrac{20}{3}$$

$$=40$$

Isaac shares the price with his $$5$$ project team members $$=40\div 5$$

$$=40\times \dfrac{1}{5}$$

$$=8$$

Thus, each member pays $$8$$.

#### Julie has to clean her house, so she decides to do $$200\,\text{m}^2$$ area of her house today. If she cleans only $$\dfrac{1}{5}$$ area of what she decided and asks her $$4$$ sisters to clean the rest of the area, find the area that each sister has to clean. Assume each sister has to clean equal amount of area.

A $$25\;\text{m}^2$$

B $$40\;\text{m}^2$$

C $$30\;\text{m}^2$$

D $$35\;\text{m}^2$$

×

Total area to be cleaned $$=200\;\text{m}^2$$

Julie cleans only $$\dfrac{1}{5}$$ area of what she decided  $$=200\times\dfrac{1}{5}$$

$$=40\;\text{m}^2$$

Area remained uncleaned by Julie  $$=200\;\text{m}^2$$ $$-40\;\text{m}^2$$

$$=160\;\text{m}^2$$

She divides the rest of her work equally among her $$4$$ sisters

$$=160\div 4=\dfrac{160}{4}$$

$$=40\;\text{m}^2$$

Thus, each sister has to clean  $$40\;\text{m}^2$$ area.

Hence, option (B) is correct.

### Julie has to clean her house, so she decides to do $$200\,\text{m}^2$$ area of her house today. If she cleans only $$\dfrac{1}{5}$$ area of what she decided and asks her $$4$$ sisters to clean the rest of the area, find the area that each sister has to clean. Assume each sister has to clean equal amount of area.

A

$$25\;\text{m}^2$$

.

B

$$40\;\text{m}^2$$

C

$$30\;\text{m}^2$$

D

$$35\;\text{m}^2$$

Option B is Correct