Inequality Word Problems Worksheets with Answer Key - Free Printable

Explanation:
Let’s solve each problem one by one, step by step.

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Problem 1:
*The sum of two consecutive even integers is at most 400. Find the pair of integers with the greatest sum.*

- Let the first even integer be $ x $. Then the next consecutive even integer is $ x + 2 $.
- Their sum: $ x + (x + 2) = 2x + 2 $
- We’re told this sum is at most 400:
$ 2x + 2 \leq 400 $
- Solve:
$ 2x \leq 398 $
$ x \leq 199 $

But $ x $ must be even, so the largest even integer ≤ 199 is 198.

Then the pair is:
$ x = 198 $, $ x + 2 = 200 $
Sum = $ 198 + 200 = 398 $, which is ≤ 400 and the greatest possible.

✔ Check: Next even pair would be 200 and 202 → sum = 402 > 400 → too big.

So answer for #1: 198 and 200

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Problem 2:
*The sum of two consecutive integers is less than 55. Find the pair with the greatest sum.*

- Let the first integer be $ x $, next is $ x + 1 $
- Sum: $ x + (x + 1) = 2x + 1 $
- Inequality: $ 2x + 1 < 55 $
- Solve:
$ 2x < 54 $
$ x < 27 $

So the largest integer $ x $ can be is 26 (since it must be less than 27).

Then pair: $ x = 26 $, $ x+1 = 27 $
Sum = $ 26 + 27 = 53 $, which is < 55.

Next pair: 27 and 28 → sum = 55, but we need less than 55 → not allowed.

So answer for #2: 26 and 27

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Problem 3:
*The sum of three consecutive odd integers is no less than 51. Find the middle integer.*

- Let the middle odd integer be $ x $. Then the three are:
$ x - 2 $, $ x $, $ x + 2 $
- Sum: $ (x - 2) + x + (x + 2) = 3x $
- Given: $ 3x \geq 51 $
- Solve:
$ x \geq 17 $

We want the middle integer, and since the inequality is “no less than”, the smallest possible middle integer is 17 — but the question doesn’t ask for the *smallest* middle integer; it says “Find the middle integer” given the condition. Since there are infinitely many such triples (e.g., 17,19,21 sum to 57; 19,21,23 sum to 63), but likely they want the least possible middle integer satisfying the condition (standard in such problems unless otherwise stated), because otherwise there’s no unique answer.

Let’s verify:
- If middle = 17 → numbers: 15, 17, 19 → sum = 51 ✔ meets “no less than 51”
- If middle = 15 → 13+15+17 = 45 < 51 ✘

So the smallest valid middle integer is 17, and since the problem asks to “find the middle integer” (singular), it's expecting 17.

Answer for #3: 17

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Problem 4:
*The length of a rectangle is 5 cm less than twice its width. If the perimeter is no more than 80 cm, what are the maximum possible dimensions?*

Let width = $ w $ cm
Then length = $ 2w - 5 $ cm

Perimeter of rectangle:
$ P = 2(\text{length} + \text{width}) = 2[(2w - 5) + w] = 2(3w - 5) = 6w - 10 $

Given: $ P \leq 80 $
So:
$ 6w - 10 \leq 80 $
$ 6w \leq 90 $
$ w \leq 15 $

Also, length must be positive:
$ 2w - 5 > 0 $ → $ 2w > 5 $ → $ w > 2.5 $, which is fine since $ w \leq 15 $ and width is positive.

To maximize dimensions, we want the largest possible width (since length increases with width), so take $ w = 15 $ cm.

Then length = $ 2(15) - 5 = 30 - 5 = 25 $ cm
Perimeter = $ 2(15 + 25) = 2(40) = 80 $ cm ✔ exactly the max allowed.

So maximum dimensions: width = 15 cm, length = 25 cm

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Problem 5:
*Find two consecutive integers such that 7 times the smaller is less than 6 times the greater. Which is the greater of such integers?*

Let smaller = $ x $, greater = $ x + 1 $

Inequality:
$ 7x < 6(x + 1) $

Solve:
$ 7x < 6x + 6 $
$ 7x - 6x < 6 $
$ x < 6 $

So any integer less than 6 works. But the question asks: *Which is the greater of such integers?* — again, likely the greatest possible greater integer satisfying the condition.

Since $ x < 6 $, the largest integer $ x $ can be is 5.

Then greater integer = $ x + 1 = 6 $

Check:
7 × 5 = 35
6 × 6 = 36
35 < 36 ✔

If x = 6 → greater = 7 → 7×6 = 42, 6×7 = 42 → 42 < 42? No, false.

So greatest possible greater integer is 6

Answer for #5: 6

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Final Answer:
1) 198 and 200
2) 26 and 27
3) 17
4) width = 15 cm, length = 25 cm
5) 6