Class 8 Maths Worksheet: Linear Equations in One Variable - Word Problems

Let's solve each problem step by step.

---

Problem 1:

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

#### Solution:
Let the number be \( x \).

- Four-fifth of the number: \( \frac{4}{5}x \)
- Three-fourth of the number: \( \frac{3}{4}x \)

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

To eliminate the fractions, find the least common denominator (LCD) of 5 and 4, which is 20. Multiply both sides by 20:
\[
20 \cdot \frac{4}{5}x = 20 \cdot \left( \frac{3}{4}x + 4 \right)
\]
\[
8x = 15x + 80
\]

Rearrange the equation to isolate \( x \):
\[
8x - 15x = 80
\]
\[
-7x = 80
\]
\[
x = -\frac{80}{7}
\]

Thus, the number is:
\[
\boxed{-\frac{80}{7}}
\]

---

Problem 2:

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

#### Solution:
Let the two consecutive numbers be \( n \) and \( n+1 \).

The difference between their squares is:
\[
(n+1)^2 - n^2 = 31
\]

Expand and simplify:
\[
(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1
\]
\[
2n + 1 = 31
\]

Solve for \( n \):
\[
2n = 30
\]
\[
n = 15
\]

The two consecutive numbers are \( n = 15 \) and \( n+1 = 16 \).

Thus, the numbers are:
\[
\boxed{16, 15}
\]

---

Problem 3:

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

#### Solution:
Let the number be \( x \).

- Double of the number: \( 2x \)
- Half of the number: \( \frac{x}{2} \)

According to the problem:
\[
2x = \frac{x}{2} + 45
\]

To eliminate the fraction, multiply both sides by 2:
\[
2 \cdot 2x = 2 \cdot \left( \frac{x}{2} + 45 \right)
\]
\[
4x = x + 90
\]

Rearrange the equation to isolate \( x \):
\[
4x - x = 90
\]
\[
3x = 90
\]
\[
x = 30
\]

Thus, the number is:
\[
\boxed{30}
\]

---

Problem 4:

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

#### Solution:
Let the number be \( x \).

- Five times the number: \( 5x \)
- Subtract 5: \( 5x - 5 \)
- Twice the number: \( 2x \)
- Four more than twice the number: \( 2x + 4 \)

According to the problem:
\[
5x - 5 = 2x + 4
\]

Rearrange the equation to isolate \( x \):
\[
5x - 2x = 4 + 5
\]
\[
3x = 9
\]
\[
x = 3
\]

Thus, the number is:
\[
\boxed{3}
\]

---

Problem 5:

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

#### Solution:
Let the number be \( x \).

- Fifth part of the number: \( \frac{x}{5} \)
- Increased by 5: \( \frac{x}{5} + 5 \)
- Fourth part of the number: \( \frac{x}{4} \)
- Diminished by 5: \( \frac{x}{4} - 5 \)

According to the problem:
\[
\frac{x}{5} + 5 = \frac{x}{4} - 5
\]

To eliminate the fractions, find the least common denominator (LCD) of 5 and 4, which is 20. Multiply both sides by 20:
\[
20 \cdot \left( \frac{x}{5} + 5 \right) = 20 \cdot \left( \frac{x}{4} - 5 \right)
\]
\[
4x + 100 = 5x - 100
\]

Rearrange the equation to isolate \( x \):
\[
4x - 5x = -100 - 100
\]
\[
-x = -200
\]
\[
x = 200
\]

Thus, the number is:
\[
\boxed{200}
\]

---

Problem 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.

#### Solution:
Let the two-digit number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the units digit.

From the problem:
1. The sum of the digits is 9:
\[
a + b = 9
\]

2. If 27 is subtracted from the number, the digits are reversed:
\[
10a + b - 27 = 10b + a
\]

Simplify the second equation:
\[
10a + b - 27 = 10b + a
\]
\[
10a - a + b - 10b = 27
\]
\[
9a - 9b = 27
\]
\[
a - b = 3
\]

Now we have a system of linear equations:
\[
a + b = 9
\]
\[
a - b = 3
\]

Add the two equations:
\[
(a + b) + (a - b) = 9 + 3
\]
\[
2a = 12
\]
\[
a = 6
\]

Substitute \( a = 6 \) into \( a + b = 9 \):
\[
6 + b = 9
\]
\[
b = 3
\]

Thus, the number is:
\[
10a + b = 10 \cdot 6 + 3 = 63
\]

Thus, the number is:
\[
\boxed{63}
\]

---

Problem 7:

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

#### Solution:
Let the two parts be \( x \) and \( 184 - x \).

According to the problem:
\[
\frac{1}{3}x = \frac{1}{7}(184 - x) + 8
\]

To eliminate the fractions, find the least common denominator (LCD) of 3 and 7, which is 21. Multiply both sides by 21:
\[
21 \cdot \frac{1}{3}x = 21 \cdot \left( \frac{1}{7}(184 - x) + 8 \right)
\]
\[
7x = 3(184 - x) + 168
\]
\[
7x = 552 - 3x + 168
\]
\[
7x = 720 - 3x
\]

Rearrange the equation to isolate \( x \):
\[
7x + 3x = 720
\]
\[
10x = 720
\]
\[
x = 72
\]

The other part is:
\[
184 - x = 184 - 72 = 112
\]

Thus, the two parts are:
\[
\boxed{72, 112}
\]

---

Problem 8:

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

#### Solution:
Let the number be \( x \).

According to the problem:
\[
2x - 22 = 48
\]

Solve for \( x \):
\[
2x = 48 + 22
\]
\[
2x = 70
\]
\[
x = 35
\]

Thus, the number is:
\[
\boxed{35}
\]

---

Problem 9:

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

#### Solution:
Let the number be \( x \).

According to the problem:
\[
7x = 10x - 36
\]

Rearrange the equation to isolate \( x \):
\[
7x - 10x = -36
\]
\[
-3x = -36
\]
\[
x = 12
\]

Thus, the number is:
\[
\boxed{12}
\]

---

Problem 10:

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

#### Solution:
Let the number be \( x \).

- Four-fifth of the number: \( \frac{4}{5}x \)
- Three-fourth of the number: \( \frac{3}{4}x \)

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

To eliminate the fractions, find the least common denominator (LCD) of 5 and 4, which is 20. Multiply both sides by 20:
\[
20 \cdot \frac{4}{5}x = 20 \cdot \left( \frac{3}{4}x + 5 \right)
\]
\[
16x = 15x + 100
\]

Rearrange the equation to isolate \( x \):
\[
16x - 15x = 100
\]
\[
x = 100
\]

Thus, the number is:
\[
\boxed{100}
\]

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Final Answers:

1. \(\boxed{-\frac{80}{7}}\)
2. \(\boxed{16, 15}\)
3. \(\boxed{30}\)
4. \(\boxed{3}\)
5. \(\boxed{200}\)
6. \(\boxed{63}\)
7. \(\boxed{72, 112}\)
8. \(\boxed{35}\)
9. \(\boxed{12}\)
10. \(\boxed{100}\)