28 Linear Equations Word Problems for Class 7 Check more at http ... - Free Printable

Let's solve each problem step by step.

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1) The sum of four consecutive odd integers is equal to 320. Set up an equation and find the values for these four odd numbers.

Step 1: Represent the four consecutive odd integers.

Consecutive odd integers differ by 2. Let the first odd integer be:

$$
x
$$

Then the next three are:

$$
x + 2,\quad x + 4,\quad x + 6
$$

Step 2: Set up the equation.

$$
x + (x + 2) + (x + 4) + (x + 6) = 320
$$

Simplify:

$$
4x + 12 = 320
$$

Step 3: Solve for $x$.

$$
4x = 320 - 12 = 308 \\
x = \frac{308}{4} = 77
$$

Step 4: Find the four numbers.

$$
77,\quad 79,\quad 81,\quad 83
$$

✔ Answer: The four consecutive odd integers are 77, 79, 81, and 83.

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2) The cost of making a phone call is $0.40 plus $0.07 for each additional minute. Find the number of minutes you can talk for $2.00 or less.

Step 1: Define variables.

Let $ m $ = number of additional minutes beyond the initial call.

Total cost:
$$
0.40 + 0.07m \leq 2.00
$$

Step 2: Solve the inequality.

$$
0.07m \leq 2.00 - 0.40 = 1.60 \\
m \leq \frac{1.60}{0.07} \approx 22.857
$$

Since $ m $ must be a whole number (minutes), we take the greatest integer less than or equal to 22.857 → 22

But wait: this is the number of additional minutes. So total minutes = 1 (initial) + 22 = 23 minutes?

Wait — actually, the problem says “$0.40 plus $0.07 for each additional minute.” That implies that the $0.40 covers the first minute, and then each extra minute costs $0.07.

So if $ m $ = number of additional minutes, then total minutes = $ 1 + m $

We found:
$$
m \leq 22.857 \Rightarrow m \leq 22
$$

So maximum additional minutes = 22 → total minutes = $ 1 + 22 = 23 $

Check:
Cost = $0.40 + 0.07 × 22 = 0.40 + 1.54 = 1.94 ≤ 2.00$ ✔
If $ m = 23 $: $ 0.40 + 0.07×23 = 0.40 + 1.61 = 2.01 > 2.00 $ ✘

✔ Answer: You can talk for 23 minutes (including the first minute).

> But note: some interpretations might assume $ m $ is total minutes. But since it says "plus $0.07 for each additional minute", the $0.40 likely includes the first minute.

So final answer: You can talk for up to 23 minutes.

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3) A shipping company charges $3.50 plus 55 cents per lb to send packages. What is the largest weight package that can be sent for no more than $25.00?

Step 1: Define variable.

Let $ w $ = weight in pounds.

Cost = $ 3.50 + 0.55w \leq 25.00 $

Step 2: Solve inequality.

$$
0.55w \leq 25.00 - 3.50 = 21.50 \\
w \leq \frac{21.50}{0.55} \approx 39.09
$$

So the largest whole number weight is 39 pounds.

Check:
- $ 3.50 + 0.55×39 = 3.50 + 21.45 = 24.95 \leq 25.00 $ ✔
- $ 3.50 + 0.55×40 = 3.50 + 22.00 = 25.50 > 25.00 $ ✘

✔ Answer: The largest weight is 39 pounds.

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4) Fiona wants to earn at least $30 this week. Her mother said that she can earn $8.00 for organizing the pantry, and $3.50 per hour for babysitting. Set up and solve an inequality to find out how many hours Fiona needs to babysit to earn at least $30.

Step 1: Define variable.

Let $ h $ = number of hours babysitting.

Earnings: $ 8.00 + 3.50h \geq 30 $

Step 2: Solve inequality.

$$
3.50h \geq 30 - 8 = 22 \\
h \geq \frac{22}{3.50} = \frac{2200}{350} = \frac{44}{7} \approx 6.2857
$$

So $ h \geq 6.2857 $ → Since she can't work a fraction of an hour in practice (unless partial hours allowed), but we want at least $30, so she needs at least 6.2857 hours.

But if only full hours are allowed, then she must work 7 hours to reach at least $30.

But let’s check:

- At 6 hours: $ 8 + 3.50×6 = 8 + 21 = 29 < 30 $ ✘
- At 7 hours: $ 8 + 3.50×7 = 8 + 24.50 = 32.50 \geq 30 $ ✔

So if only full hours, 7 hours.

But the question asks: "how many hours... to earn at least $30?" — it may allow fractional hours.

So technically, she needs at least $ \frac{44}{7} $ hours, which is about 6.29 hours.

But typically, such problems expect the smallest number of hours (possibly fractional) needed.

✔ Answer: Fiona needs to babysit for at least $ \frac{44}{7} $ hours, or approximately 6.29 hours.

But if only whole hours, then 7 hours.

Since the problem doesn’t specify, we’ll give the exact answer.

Final Answer: $ h \geq \frac{44}{7} $, so at least $ \frac{44}{7} $ hours (or about 6.29 hours).

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5) Jim has the following test grades this quarter: 85, 74, 82, 89, and 94. What must Jim get on the next test to ensure that his test average will be at least an 86?

Step 1: Let $ x $ = score on next test.

He has 5 tests so far, and one more → total 6 tests.

Average = $ \frac{85 + 74 + 82 + 89 + 94 + x}{6} \geq 86 $

Step 2: Add known scores.

$$
85 + 74 = 159 \\
159 + 82 = 241 \\
241 + 89 = 330 \\
330 + 94 = 424
$$

So:
$$
\frac{424 + x}{6} \geq 86
$$

Step 3: Multiply both sides by 6.

$$
424 + x \geq 516
$$

Step 4: Solve for $x$.

$$
x \geq 516 - 424 = 92
$$

✔ Answer: Jim must score at least 92 on the next test to have an average of at least 86.

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✔ Final Answers Summary:

1) 77, 79, 81, 83
2) Up to 23 minutes
3) 39 pounds
4) At least $ \frac{44}{7} $ hours (≈ 6.29 hours)
5) At least 92

Let me know if you'd like these written neatly for submission!