Class 8 Maths worksheet on linear equations in one variable with word problems involving ages, geometry, and ratios.

Let's solve each problem step-by-step using linear equations in one variable.

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Problem 1:

> Rene is 6 years older than her younger sister. After 10 years, the sum of their ages will be 50 years. Find their present ages.

#### Step 1: Let’s define variables
Let the present age of the sister = $ x $ years
Then, Rene’s present age = $ x + 6 $ years

#### Step 2: After 10 years
- Sister’s age = $ x + 10 $
- Rene’s age = $ (x + 6) + 10 = x + 16 $

Sum of their ages after 10 years:
$$
(x + 10) + (x + 16) = 50
$$

#### Step 3: Solve the equation
$$
2x + 26 = 50 \\
2x = 24 \\
x = 12
$$

So:
- Sister’s age = $ x = 12 $ years
- Rene’s age = $ x + 6 = 18 $ years

✔ Answer:
Rene: 18 yrs, Sister: 12 yrs

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Problem 2:

> The length of a rectangle is 10 m more than its breadth. If the perimeter is 80 m, find the dimensions.

#### Step 1: Define variable
Let breadth = $ x $ m
Then length = $ x + 10 $ m

#### Step 2: Perimeter formula
Perimeter of rectangle = $ 2(\text{length} + \text{breadth}) $

$$
2((x + 10) + x) = 80 \\
2(2x + 10) = 80 \\
4x + 20 = 80 \\
4x = 60 \\
x = 15
$$

So:
- Breadth = $ x = 15 $ m
- Length = $ x + 10 = 25 $ m

✔ Answer:
Breadth: 15 m, Length: 25 m

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Problem 3:

> A 300 m long wire is used to fence a rectangular plot whose length is twice its width. Find the length and breadth.

#### Step 1: Define variable
Let breadth = $ x $ m
Then length = $ 2x $ m

#### Step 2: Perimeter = 300 m
$$
2(\text{length} + \text{breadth}) = 300 \\
2(2x + x) = 300 \\
2(3x) = 300 \\
6x = 300 \\
x = 50
$$

So:
- Breadth = $ x = 50 $ m
- Length = $ 2x = 100 $ m

✔ Answer:
Breadth: 50 m, Length: 100 m

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Problem 4:

> My mother is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age. Find our ages.

#### Step 1: Define variable
Let my age = $ x $ years
Then mother’s age = $ 2x + 12 $ years

After 8 years:
- My age = $ x + 8 $
- Mother’s age = $ 2x + 12 + 8 = 2x + 20 $

According to the condition:
> Mother’s age after 8 years = 3 times my age after 8 years – 20

$$
2x + 20 = 3(x + 8) - 20
$$

#### Step 2: Solve
$$
2x + 20 = 3x + 24 - 20 \\
2x + 20 = 3x + 4 \\
20 - 4 = 3x - 2x \\
16 = x
$$

So:
- My age = $ x = 16 $ years
- Mother’s age = $ 2(16) + 12 = 32 + 12 = 44 $ years

✔ Answer:
My age: 16 yrs, Mother’s age: 44 yrs

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Problem 5:

> In a class of 42 students, the number of boys is $ \frac{2}{5} $ of the girls. Find the number of boys and girls.

#### Step 1: Define variable
Let number of girls = $ x $
Then number of boys = $ \frac{2}{5}x $

Total students:
$$
x + \frac{2}{5}x = 42
$$

#### Step 2: Solve
$$
\frac{5x + 2x}{5} = 42 \\
\frac{7x}{5} = 42 \\
7x = 210 \\
x = 30
$$

So:
- Girls = $ x = 30 $
- Boys = $ \frac{2}{5} \times 30 = 12 $

✔ Answer:
Boys: 12, Girls: 30

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✔ Final Answers:

1. Rene: 18 yrs, Sister: 12 yrs
2. Breadth: 15 m, Length: 25 m
3. Breadth: 50 m, Length: 100 m
4. My age: 16 yrs, Mother’s age: 44 yrs
5. Boys: 12, Girls: 30

All answers are in whole numbers, which can be written as fractions if needed (e.g., $ \frac{18}{1} $), but since they're integers, they’re acceptable as is.

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