Linear Equations in One Variable (Word Problems) - Free Printable

Let's solve each problem step by step.

---

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.

#### Solution:
1. Let the present age of Rene's sister be \( x \) years.
2. Then, Rene's present age is \( x + 6 \) years.
3. After 10 years:
- Rene's age will be \( x + 6 + 10 = x + 16 \).
- Sister's age will be \( x + 10 \).
4. The sum of their ages after 10 years is given as 50:
\[
(x + 16) + (x + 10) = 50
\]
5. Simplify the equation:
\[
x + 16 + x + 10 = 50
\]
\[
2x + 26 = 50
\]
6. Solve for \( x \):
\[
2x = 50 - 26
\]
\[
2x = 24
\]
\[
x = 12
\]
7. Therefore:
- Sister's present age is \( x = 12 \) years.
- Rene's present age is \( x + 6 = 12 + 6 = 18 \) years.

Answer:
\[
\boxed{18 \text{ yrs}, 12 \text{ yrs}}
\]

---

Problem 2:

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

#### Solution:
1. Let the breadth of the rectangle be \( b \) meters.
2. Then, the length of the rectangle is \( b + 10 \) meters.
3. The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})
\]
4. Substitute the given perimeter and expressions for length and breadth:
\[
80 = 2 \times ((b + 10) + b)
\]
5. Simplify inside the parentheses:
\[
80 = 2 \times (2b + 10)
\]
6. Distribute the 2:
\[
80 = 4b + 20
\]
7. Solve for \( b \):
\[
4b = 80 - 20
\]
\[
4b = 60
\]
\[
b = 15
\]
8. Therefore:
- Breadth \( b = 15 \) meters.
- Length \( b + 10 = 15 + 10 = 25 \) meters.

Answer:
\[
\boxed{15 \text{ m}, 25 \text{ m}}
\]

---

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 of the plot.

#### Solution:
1. Let the width of the rectangle be \( w \) meters.
2. Then, the length of the rectangle is \( 2w \) meters.
3. The formula for the perimeter of a rectangle is:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
4. Substitute the given perimeter and expressions for length and width:
\[
300 = 2 \times (2w + w)
\]
5. Simplify inside the parentheses:
\[
300 = 2 \times 3w
\]
6. Simplify further:
\[
300 = 6w
\]
7. Solve for \( w \):
\[
w = \frac{300}{6}
\]
\[
w = 50
\]
8. Therefore:
- Width \( w = 50 \) meters.
- Length \( 2w = 2 \times 50 = 100 \) meters.

Answer:
\[
\boxed{50 \text{ m}, 100 \text{ m}}
\]

---

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 my age and my mother’s age.

#### Solution:
1. Let my present age be \( x \) years.
2. Then, my mother's present age is \( 2x + 12 \) years.
3. After 8 years:
- My age will be \( x + 8 \).
- My mother's age will be \( 2x + 12 + 8 = 2x + 20 \).
4. According to the problem, after 8 years, my mother's age will be 20 years less than three times my age:
\[
2x + 20 = 3(x + 8) - 20
\]
5. Simplify the right-hand side:
\[
2x + 20 = 3x + 24 - 20
\]
\[
2x + 20 = 3x + 4
\]
6. Solve for \( x \):
\[
20 - 4 = 3x - 2x
\]
\[
16 = x
\]
7. Therefore:
- My present age is \( x = 16 \) years.
- My mother's present age is \( 2x + 12 = 2(16) + 12 = 32 + 12 = 44 \) years.

Answer:
\[
\boxed{16 \text{ yrs}, 44 \text{ yrs}}
\]

---

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 in the class.

#### Solution:
1. Let the number of girls be \( g \).
2. Then, the number of boys is \( \frac{2}{5}g \).
3. The total number of students is given as 42:
\[
g + \frac{2}{5}g = 42
\]
4. Combine the terms on the left-hand side:
\[
\left(1 + \frac{2}{5}\right)g = 42
\]
\[
\frac{5}{5}g + \frac{2}{5}g = 42
\]
\[
\frac{7}{5}g = 42
\]
5. Solve for \( g \):
\[
g = 42 \times \frac{5}{7}
\]
\[
g = 6 \times 5
\]
\[
g = 30
\]
6. Therefore:
- Number of girls \( g = 30 \).
- Number of boys \( \frac{2}{5}g = \frac{2}{5} \times 30 = 12 \).

Answer:
\[
\boxed{12, 30}
\]

---

Final Answers:

1. \(\boxed{18 \text{ yrs}, 12 \text{ yrs}}\)
2. \(\boxed{15 \text{ m}, 25 \text{ m}}\)
3. \(\boxed{50 \text{ m}, 100 \text{ m}}\)
4. \(\boxed{16 \text{ yrs}, 44 \text{ yrs}}\)
5. \(\boxed{12, 30}\)