One Variable Inequalities Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve each of the 6 word problems step by step, writing an inequality, solving it, and then describing how to graph the solution. Since we can’t draw graphs here, I’ll describe them clearly.

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Problem 1:

> A boat can hold at most 1,000 pounds. If there is 300 pounds of equipment plus 25-pound boxes, what is the maximum number of boxes the ship can carry?

Step 1: Define variable
Let \( x \) = number of 25-pound boxes.

Step 2: Write inequality
Total weight = equipment + boxes ≤ 1000
→ \( 300 + 25x \leq 1000 \)

Step 3: Solve
Subtract 300 from both sides:
→ \( 25x \leq 700 \)

Divide by 25:
→ \( x \leq 28 \)

Answer: The boat can carry at most 28 boxes.

Graph description:
On a number line, place a closed circle at 28 (since ≤ includes 28), and shade everything to the left (all numbers less than or equal to 28).

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Problem 2:

> A hotel costs $100 per night plus a $35 hotel fee. If George has $500, what is the maximum number of nights he could stay?

Step 1: Define variable
Let \( x \) = number of nights.

Step 2: Write inequality
Total cost = nightly rate × nights + fee ≤ 500
→ \( 100x + 35 \leq 500 \)

Step 3: Solve
Subtract 35:
→ \( 100x \leq 465 \)

Divide by 100:
→ \( x \leq 4.65 \)

Since George can't stay a fraction of a night (assuming whole nights only), we take the greatest whole number less than or equal to 4.65, which is 4.

Answer: George can stay at most 4 nights.

Graph description:
On a number line, closed circle at 4, shade to the left. (Note: In real context, only integer values make sense, so discrete points 0,1,2,3,4 are valid.)

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Problem 3:

> The school paper says that no less than a third of the juniors went to prom. If 37 juniors went, what’s the minimum amount of juniors at the school?

Step 1: Define variable
Let \( x \) = total number of juniors.

Step 2: Write inequality
“no less than a third” means ≥ one-third
→ \( \frac{1}{3}x \leq 37 \) ✘ Wait — let’s read carefully.

Actually: “no less than a third went to prom” → meaning the number who went is at least one-third of total juniors.

So: \( 37 \geq \frac{1}{3}x \)

Or rewritten: \( \frac{1}{3}x \leq 37 \) → same thing.

But actually, if 37 juniors went, and that is *at least* one-third of all juniors, then:

→ \( 37 \geq \frac{1}{3}x \)

Multiply both sides by 3:
→ \( 111 \geq x \) → \( x \leq 111 \) ✘ That would mean total juniors ≤ 111, but if 37 is at least 1/3, then total must be *at most* 111? That doesn’t make sense.

Wait — let’s think again.

If 37 juniors went, and that is no less than a third of the total, that means:

> 37 ≥ (1/3) × total juniors

So:

\( 37 \geq \frac{1}{3}x \)

Multiply both sides by 3:

→ \( 111 \geq x \) → \( x \leq 111 \)

But this implies total juniors is ≤ 111. But if total juniors were 111, then 1/3 is 37 — so 37 is exactly 1/3. If total were more than 111, say 120, then 1/3 is 40, and 37 < 40 — which violates “no less than a third”.

So yes, total juniors must be ≤ 111.

But the question asks for the minimum amount of juniors at the school.

Wait — if 37 is *at least* one-third, then the smallest possible total occurs when 37 is *exactly* one-third — because if total were smaller, say 100, then 1/3 is ~33.3, and 37 > 33.3 — still satisfies “no less than a third”. So actually, total juniors can be as low as... well, theoretically, any number where 37 ≥ (1/3)x.

So solving:

\( 37 \geq \frac{1}{3}x \)

→ \( x \leq 111 \)

But that gives an upper bound, not lower.

We need to find the minimum total number of juniors such that 37 is at least 1/3 of them.

That means: 37 ≥ (1/3)x → x ≤ 111

But if x is smaller, say x=100, then 1/3 of 100 is ~33.3, and 37 > 33.3 — still satisfies.

In fact, x can be as small as... well, as long as 37 ≥ (1/3)x, x can be very small? No — if x is too small, say x=30, then 1/3 is 10, and 37 > 10 — still true.

Actually, the condition is always satisfied if x ≤ 111, but for x > 111, it fails.

But the question is: what’s the minimum amount of juniors at the school?

This is tricky. Let’s re-read:

> “no less than a third of the juniors went to prom. If 37 juniors went, what’s the minimum amount of juniors at the school?”

The phrase “no less than a third went” means the number who went (37) is ≥ (1/3) of total.

So: 37 ≥ (1/3)x → x ≤ 111

But this gives an upper bound on x. There is no lower bound — in theory, if there were only 1 junior, 37 couldn’t have gone. So we must assume that the 37 who went are part of the total.

Ah! Important: The 37 juniors who went are from the total juniors. So total juniors must be at least 37.

And also, 37 ≥ (1/3)x → x ≤ 111

So combining: 37 ≤ x ≤ 111

The minimum total juniors is 37.

Check: If total juniors = 37, then 1/3 of 37 ≈ 12.33, and 37 ≥ 12.33 — true.

Is 37 the minimum? Yes, because you can’t have fewer than 37 juniors if 37 went to prom.

Answer: Minimum number of juniors is 37.

Graph description:
On a number line, closed circle at 37, shade to the right up to 111 (but since question asks for minimum, we focus on 37 as the lower bound). Actually, the solution set is 37 ≤ x ≤ 111, so graph from 37 to 111 inclusive.

But since the question asks for the minimum, the answer is 37.

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Problem 4:

> Kyle’s doctor told him that in order to lose weight, he needs to eat less than 2,000 calories a day. He eats 875 calories for breakfast and then 3 protein smoothies. How many calories can the protein smoothies have to reach his goal?

Step 1: Define variable
Let \( x \) = calories per smoothie.

Step 2: Write inequality
Total calories = breakfast + 3 smoothies < 2000
→ \( 875 + 3x < 2000 \)

Step 3: Solve
Subtract 875:
→ \( 3x < 1125 \)

Divide by 3:
→ \( x < 375 \)

Answer: Each smoothie must have less than 375 calories.

Graph description:
Open circle at 375 (since <, not ≤), shade to the left.

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Problem 5:

> 2 less than 3 times a number is greater than 10.

Step 1: Define variable
Let \( x \) = the number.

Step 2: Write inequality
“2 less than 3 times a number” → \( 3x - 2 \)
“is greater than 10” → \( 3x - 2 > 10 \)

Step 3: Solve
Add 2 to both sides:
→ \( 3x > 12 \)

Divide by 3:
→ \( x > 4 \)

Answer: The number is greater than 4.

Graph description:
Open circle at 4, shade to the right.

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Problem 6:

> Ben is twice his brother’s age. If the sum of their ages is below 81, how old could Ben’s brother be?

Step 1: Define variable
Let \( x \) = brother’s age.
Then Ben’s age = \( 2x \)

Step 2: Write inequality
Sum of ages < 81
→ \( x + 2x < 81 \)
→ \( 3x < 81 \)

Step 3: Solve
Divide by 3:
→ \( x < 27 \)

Answer: Ben’s brother could be any age less than 27 (and presumably positive, so 0 < x < 27, but since age, likely x > 0).

Graph description:
Open circle at 27, shade to the left. (If considering only positive ages, start shading from 0 to 27, open at 27.)

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✔ Final Answers Summary:

1. 28 boxes
2. 4 nights
3. 37 juniors (minimum)
4. Each smoothie < 375 calories
5. x > 4
6. Brother’s age < 27

Let me know if you’d like me to sketch the number line graphs for any of these!