- 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'

So, we need to add.

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.

A \(\dfrac{54}{5}\) minutes

B \(13\dfrac{5}{6}\) minutes

C \(\dfrac{58}{3}\) minutes

D \(13\dfrac{1}{6}\) minutes

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.

A \(\dfrac {19}{100},\dfrac {81}{100}\)

B \(\dfrac {18}{100},\dfrac {82}{100}\)

C \(\dfrac {17}{100},\dfrac {83}{100}\)

D \(\dfrac {16}{100},\dfrac {84}{100}\)

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.

A \(\dfrac{19}{15}\) pounds

B \(\dfrac{90}{19}\) pounds

C \(\dfrac{20}{15}\) pounds

D \(\dfrac{19}{24}\) pounds

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.

A \(\dfrac{42}{5}\) liters

B \(\dfrac{5}{6}\) liters

C \(\dfrac{31}{2}\) liters

D \(\dfrac{10}{3}\) liters

A \(2 \dfrac{1}{5}\)

B \(7 \dfrac{1}{3}\)

C \(1 \dfrac{1}{2}\)

D \(1 \dfrac{1}{5}\)

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 \(=$4+ $\dfrac{80}{9}\)

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

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

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.

A \(\dfrac{3}{4}\;liters\)

B \(\dfrac{7}{5}\;liters\)

C \(\dfrac{2}{3}\;liters\)

D \(\dfrac{2}{5}\;liters\)

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 \($8\).

A \(25\;\text{m}^2\)

B \(40\;\text{m}^2\)

C \(30\;\text{m}^2\)

D \(35\;\text{m}^2\)