Free printable worksheet helping students practice translating real-world scenarios into mathematical inequalities.

Here are the step-by-step solutions for each word problem on the worksheet.

1) The Farmer's Fencing

Step 1: Identify the total amount of fencing.
The problem says the farmer has between 40 and 60 yards, inclusive. "Inclusive" means we use $\le$ (less than or equal to) signs.
Total fencing range: $40 \le \text{Perimeter} \le 60$.

Step 2: Relate perimeter to side length ($s$).
A square has 4 equal sides. So, the Perimeter = $4 \times s$.

Step 3: Write the inequality.
Substitute $4s$ for the Perimeter in the range from Step 1.
$40 \le 4s \le 60$

*(Note: You can also simplify this by dividing everything by 4 to get $10 \le s \le 15$, but the unsimplified version shows the direct translation of the words.)*

2) Alice's Run

Step 1: Find the time for one mile.
Alice takes between 10 and 12 minutes per mile (inclusive).
$10 \le \text{time per mile} \le 12$

Step 2: Calculate the time for a half marathon.
A half marathon is 13 miles. To find the total time ($t$), we multiply the time per mile by 13.
Lower limit: $10 \times 13 = 130$ minutes.
Upper limit: $12 \times 13 = 156$ minutes.

Step 3: Write the inequality.
$130 \le t \le 156$

3) Flame's Shopping

Step 1: Determine how much was spent.
Flame starts with $\$240$.
She spends more than half: $\frac{1}{2}$ of $240 = 120$. So she spends $> 120$.
She spends less than three-quarters: $\frac{3}{4}$ of $240 = 180$. So she spends $< 180$.
Amount Spent: $120 < \text{Spent} < 180$.

Step 2: Determine how much is left ($m$).
Money Left = Total Money - Amount Spent.
If she spends the *most* ($180$), she has the *least* left: $240 - 180 = 60$.
If she spends the *least* ($120$), she has the *most* left: $240 - 120 = 120$.

Step 3: Write the inequality.
Since the spending limits were exclusive (strictly more/less than), the money left limits are also exclusive.
$60 < m < 120$

4) Newton's Swim

Step 1: Identify the time for one length.
It takes between 25 and 32 seconds, exclusive. Exclusive means we use $<$ and $>$ signs (not equal to).
$25 < \text{time for 1 length} < 32$

Step 2: Calculate the time for 20 lengths.
Let $t$ be the total time. We multiply the single length time by 20.
Lower limit: $25 \times 20 = 500$.
Upper limit: $32 \times 20 = 640$.

Step 3: Write the inequality.
$500 < t < 640$

5) Sally's Fimo Models

Step 1: Identify the daily production rate.
Sally makes 6 to 10 models a day. Let $x$ be the number of days.
Minimum models made in $x$ days: $6x$.
Maximum models made in $x$ days: $10x$.

Step 2: Set the goal.
She needs to make 120 models. This means her total production must be greater than or equal to 120.

Step 3: Write the compound inequality.
We know that in $x$ days, she makes at least $6x$ models and at most $10x$ models. For her to finish the job, even her slowest pace ($6x$) must eventually reach 120, and her fastest pace ($10x$) represents the upper bound of what she could have made in that time frame if she went super fast.
However, usually, these problems ask for the range of days required.
If she works at max speed (10/day): $120 / 10 = 12$ days.
If she works at min speed (6/day): $120 / 6 = 20$ days.
So, the number of days $x$ will be between 12 and 20.
$12 \le x \le 20$

*(Alternative interpretation based strictly on "models made": $6x \le 120 \le 10x$ is technically incorrect logic for "how long it takes". The standard answer for "how long" is the range of days derived above.)*

6) Bert's Savings

Step 1: Calculate the annual savings goal range.
Salary: $\$26,000$.
Lower savings goal (5%): $0.05 \times 26,000 = 1,300$.
Upper savings goal (10%): $0.10 \times 26,000 = 2,600$.
Target savings range: $1,300$ to $2,600$.

Step 2: Account for money already saved.
Bert has already saved $\$500$.
Let $s$ be the *additional* money he needs to save.
Total Saved = $500 + s$.

Step 3: Write the inequality.
The Total Saved must be between 1,300 and 2,600.
$1,300 \le 500 + s \le 2,600$

*(You can simplify this by subtracting 500 from all parts: $800 \le s \le 2,100$)*

7) Captain's Journey

Step 1: Determine the remaining distance.
Total journey: 360 miles.
Traveled so far: 60 miles.
Remaining distance: $360 - 60 = 300$ miles.

Step 2: Determine the time range based on speed.
Speed range: 45 to 60 mph (inclusive).
Time = Distance / Speed.
Fastest speed (60 mph) results in shortest time: $300 / 60 = 5$ hours.
Slowest speed (45 mph) results in longest time: $300 / 45 = 6.66...$ hours (or $20/3$ hours).

Step 3: Write the inequality.
Let $r$ be the time in hours.
$5 \le r \le \frac{20}{3}$ (or approx $6.67$)

8) Chris's Age

Step 1: Find Bill's age.
Alice is 9 years old.
Bill is 2 years younger than Anna? Wait, the text says "Bill is 2 years younger than Anna" but then compares Chris to Bill and gives Alice's age. Looking closely at the text: "Bill is 2 years younger than Anna... If Alice is 9 years old..."
There seems to be a typo in the question names (Anna vs Alice). Assuming "Anna" and "Alice" are the same person (a common typo in worksheets):
Alice/Anna = 9.
Bill = $9 - 2 = 7$ years old.

Step 2: Determine Chris's age range relative to Bill.
Chris is more than twice as old as Bill: $2 \times 7 = 14$. So Chris $> 14$.
Chris is less than 3 times as old as Bill: $3 \times 7 = 21$. So Chris $< 21$.

Step 3: Write the inequality.
Let $c$ be Chris's age.
$14 < c < 21$

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Final Answer:

1) $40 \le 4s \le 60$ (or simplified: $10 \le s \le 15$)
2) $130 \le t \le 156$
3) $60 < m < 120$
4) $500 < t < 640$
5) $12 \le x \le 20$
6) $1300 \le 500 + s \le 2600$ (or simplified: $800 \le s \le 2100$)
7) $5 \le r \le \frac{20}{3}$ (or $5 \le r \le 6.67$)
8) $14 < c < 21$