Algebra Word Problems Sheet 2 | PDF | Algebra | Teaching Mathematics - Free Printable

Let's solve each problem step by step and write the algebraic expressions for them.

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Problem 1:

Word Problem:
In a stable, there are \( h \) horses. 6 of them are taken out into the yard to exercise. How many are left in the stable?

Solution:
If there are \( h \) horses initially and 6 are taken out, the number of horses left in the stable is:
\[
h - 6
\]

Algebraic Expression:
\[
h - 6
\]

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Problem 2:

Word Problem:
There are \( c \) cyclists in a cycle race. \( \frac{3}{4} \) of the cyclists finish the race. How many cyclists did not finish?

Solution:
If \( \frac{3}{4} \) of the cyclists finish the race, then the fraction of cyclists who did not finish is:
\[
1 - \frac{3}{4} = \frac{1}{4}
\]
The number of cyclists who did not finish is:
\[
\frac{1}{4} \times c = \frac{c}{4}
\]

Algebraic Expression:
\[
\frac{c}{4}
\]

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Problem 3:

Word Problem:
There are 56 people on a bus. \( t \) people get off at the next stop and 3 more people get on. How many people are on the bus now?

Solution:
Initially, there are 56 people on the bus. After \( t \) people get off, the number of people remaining is:
\[
56 - t
\]
Then, 3 more people get on, so the total number of people on the bus now is:
\[
(56 - t) + 3 = 59 - t
\]

Algebraic Expression:
\[
59 - t
\]

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Problem 4:

Word Problem:
In a class of 30 children, there are \( g \) girls. What fraction of the class are girls?

Solution:
The total number of children in the class is 30, and the number of girls is \( g \). The fraction of the class that are girls is:
\[
\frac{g}{30}
\]

Algebraic Expression:
\[
\frac{g}{30}
\]

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Problem 5:

Word Problem:
In a class of \( c \) children, there are 16 boys. What fraction of the class are boys?

Solution:
The total number of children in the class is \( c \), and the number of boys is 16. The fraction of the class that are boys is:
\[
\frac{16}{c}
\]

Algebraic Expression:
\[
\frac{16}{c}
\]

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Problem 6:

Word Problem:
There are \( b \) people on a bus. At the next stop, 7 people get off and 10 more get on. How many more people are on the bus now?

Solution:
Initially, there are \( b \) people on the bus. After 7 people get off, the number of people remaining is:
\[
b - 7
\]
Then, 10 more people get on, so the total number of people on the bus now is:
\[
(b - 7) + 10 = b + 3
\]
To find how many more people are on the bus now compared to before, we subtract the initial number of people (\( b \)) from the current number of people (\( b + 3 \)):
\[
(b + 3) - b = 3
\]

Algebraic Expression:
\[
3
\]

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Problem 7:

Word Problem:
I cut a long piece of wood into 50cm pieces. I manage to cut \( w \) pieces of wood, and there is 20cm left over. How long was the wood to start with?

Solution:
Each piece of wood is 50cm long, and you cut \( w \) pieces. The total length of the wood used for cutting is:
\[
50w \text{ cm}
\]
Additionally, there is 20cm of wood left over. Therefore, the original length of the wood is:
\[
50w + 20 \text{ cm}
\]

Algebraic Expression:
\[
50w + 20
\]

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Problem 8:

Word Problem:
I have \( c \) chocolates which I share equally between 5 friends. How many do they each get?

Solution:
If you have \( c \) chocolates and you share them equally among 5 friends, each friend gets:
\[
\frac{c}{5}
\]

Algebraic Expression:
\[
\frac{c}{5}
\]

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Problem 9:

Word Problem:
I have 5 pens already. I am given 2 packs of pens. Each pack contains \( t \) pens. How many pens do I have now?

Solution:
You initially have 5 pens. You receive 2 packs, and each pack contains \( t \) pens. The total number of pens you receive is:
\[
2t
\]
Adding this to the 5 pens you already have, the total number of pens is:
\[
5 + 2t
\]

Algebraic Expression:
\[
5 + 2t
\]

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Problem 10:

Word Problem:
There are \( d \) deer and \( p \) pheasants in the woods. How many legs in total?

Solution:
Deer have 4 legs each, so \( d \) deer have:
\[
4d \text{ legs}
\]
Pheasants have 2 legs each, so \( p \) pheasants have:
\[
2p \text{ legs}
\]
The total number of legs is the sum of the legs of the deer and the pheasants:
\[
4d + 2p
\]

Algebraic Expression:
\[
4d + 2p
\]

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Final Answer:

\[
\boxed{
\begin{array}{ll}
1) & h - 6 \\
2) & \frac{c}{4} \\
3) & 59 - t \\
4) & \frac{g}{30} \\
5) & \frac{16}{c} \\
6) & 3 \\
7) & 50w + 20 \\
8) & \frac{c}{5} \\
9) & 5 + 2t \\
10) & 4d + 2p \\
\end{array}
}
\]