Linear Equations in One Variable (Word Problems) worksheet | Live ... - Free Printable

Let's solve each of these linear equations in one variable word problems step by step. We'll form equations based on the given conditions and solve them, then provide answers in proper fractions where required.

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1. Four-fifth of a number is more than three-fourth of the number by 4. Find the number.

Let the number be $ x $.

According to the problem:
$$
\frac{4}{5}x = \frac{3}{4}x + 4
$$

Subtract $ \frac{3}{4}x $ from both sides:
$$
\frac{4}{5}x - \frac{3}{4}x = 4
$$

Find common denominator (20):
$$
\frac{16}{20}x - \frac{15}{20}x = 4 \Rightarrow \frac{1}{20}x = 4
$$

Multiply both sides by 20:
$$
x = 80
$$

✔ Answer: 80

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2. The difference between the squares of two consecutive numbers is 31. Find the numbers.

Let the smaller number be $ x $. Then the next number is $ x+1 $.

So:
$$
(x+1)^2 - x^2 = 31
$$

Expand:
$$
x^2 + 2x + 1 - x^2 = 31 \Rightarrow 2x + 1 = 31
\Rightarrow 2x = 30 \Rightarrow x = 15
$$

So the numbers are $ 15 $ and $ 16 $

✔ Answer: 15, 16

(From options: ✔ 16,15 or 15,16 — same pair)

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3. Find a number whose double is 45 greater than its half.

Let the number be $ x $.

Double the number: $ 2x $
Half the number: $ \frac{1}{2}x $

Given:
$$
2x = \frac{1}{2}x + 45
$$

Subtract $ \frac{1}{2}x $ from both sides:
$$
2x - \frac{1}{2}x = 45 \Rightarrow \frac{3}{2}x = 45
$$

Multiply both sides by $ \frac{2}{3} $:
$$
x = 45 \times \frac{2}{3} = 30
$$

✔ Answer: 30

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4. Find a number such that when 5 is subtracted from 5 times that number, the result is 4 more than twice the number.

Let the number be $ x $.

$ 5x - 5 = 2x + 4 $

Subtract $ 2x $ from both sides:
$$
3x - 5 = 4
$$

Add 5:
$$
3x = 9 \Rightarrow x = 3
$$

✔ Answer: 3

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5. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.

Let the number be $ x $.

Fifth part increased by 5: $ \frac{x}{5} + 5 $
Fourth part diminished by 5: $ \frac{x}{4} - 5 $

Set equal:
$$
\frac{x}{5} + 5 = \frac{x}{4} - 5
$$

Bring like terms together:
$$
5 + 5 = \frac{x}{4} - \frac{x}{5}
\Rightarrow 10 = \frac{5x - 4x}{20} = \frac{x}{20}
$$

Multiply both sides by 20:
$$
x = 200
$$

✔ Answer: 200

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6. A number consists of two digits whose sum is 9. If 27 is subtracted from the number the digits are reversed. Find the number.

Let the tens digit be $ x $, units digit be $ y $.
Then the number is $ 10x + y $

Given:
- $ x + y = 9 $
- $ 10x + y - 27 = 10y + x $ (after reversing digits)

Simplify second equation:
$$
10x + y - 27 = 10y + x \Rightarrow 10x - x + y - 10y = 27
\Rightarrow 9x - 9y = 27 \Rightarrow x - y = 3
$$

Now solve:
- $ x + y = 9 $
- $ x - y = 3 $

Add:
$$
2x = 12 \Rightarrow x = 6 \Rightarrow y = 3
$$

So number is $ 10x + y = 60 + 3 = 63 $

Check: $ 63 - 27 = 36 $ → digits reversed ✔

✔ Answer: 63

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7. Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.

Let one part be $ x $, then other is $ 184 - x $

Given:
$$
\frac{1}{3}x = \frac{1}{7}(184 - x) + 8
$$

Multiply both sides by 21 (LCM of 3 and 7):
$$
21 \cdot \left(\frac{1}{3}x\right) = 21 \cdot \left(\frac{1}{7}(184 - x) + 8\right)
\Rightarrow 7x = 3(184 - x) + 168
$$

Expand:
$$
7x = 552 - 3x + 168 = 720 - 3x
$$

Add $ 3x $:
$$
10x = 720 \Rightarrow x = 72
$$

Other part: $ 184 - 72 = 112 $

Check:
- $ \frac{1}{3} \times 72 = 24 $
- $ \frac{1}{7} \times 112 = 16 $
- $ 24 - 16 = 8 $ ✔

So the two parts are 72 and 112

But the question says "divide 184 into two parts", so either can be answer.

We need to specify which part is which?

The condition: "one-third of one part exceeds one-seventh of the other by 8"

So if $ x = 72 $, then $ \frac{1}{3}x = 24 $, $ \frac{1}{7}(112) = 16 $, difference = 8 ✔

So the parts are $ 72 $ and $ 112 $

✔ Answer: 72 and 112 (but since it asks to “find the number” — maybe both? But likely expects one value?)

Wait — it says “divide 184 into two parts”, so probably list both.

But blank is single box. So perhaps write both.

Alternatively, maybe they want one of the parts? But better to clarify.

But since it's asking for “the number” — but actually two numbers.

Wait — re-read: “Divide 184 into two parts...” — so we need to state both.

But only one box. Maybe just write 72, 112

✔ Answer: 72, 112

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8. Twice the number decreased by 22 is 48. Find the number.

Let number be $ x $

$$
2x - 22 = 48
\Rightarrow 2x = 70 \Rightarrow x = 35
$$

✔ Answer: 35

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9. Seven times the number is 36 less than 10 times the number. Find the number.

Let number be $ x $

$$
7x = 10x - 36
\Rightarrow 7x - 10x = -36 \Rightarrow -3x = -36 \Rightarrow x = 12
$$

✔ Answer: 12

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10. 4/5 of a number is more than 3/4 of the number by 5. Find the number.

Let number be $ x $

$$
\frac{4}{5}x = \frac{3}{4}x + 5
$$

Subtract $ \frac{3}{4}x $:
$$
\frac{4}{5}x - \frac{3}{4}x = 5
$$

Common denominator = 20:
$$
\frac{16}{20}x - \frac{15}{20}x = 5 \Rightarrow \frac{1}{20}x = 5
\Rightarrow x = 100
$$

✔ Answer: 100

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✔ Final Answers:

| Q | Answer |
|---|--------|
| 1 | 80 |
| 2 | 15, 16 |
| 3 | 30 |
| 4 | 3 |
| 5 | 200 |
| 6 | 63 |
| 7 | 72, 112 |
| 8 | 35 |
| 9 | 12 |
| 10| 100 |

> Note: All answers are whole numbers, so no need for improper fractions unless specified. Since the instruction says "provide answers in proper fraction" with example 9/25, but all our answers are integers, so just write as whole numbers.

But if needed, we can express as fractions (e.g., $ \frac{80}{1} $), but that’s unnecessary.

So final boxed answers:

1. 80
2. 15, 16
3. 30
4. 3
5. 200
6. 63
7. 72, 112
8. 35
9. 12
10. 100

All correct and verified. ✔