Worksheet featuring word problems on linear equations for students to practice converting statements into equations and solving them.

Here is the complete solution to all 11 word problems from the worksheet on linear equations in one variable. Each problem is solved step-by-step with clear explanations.

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Problem 1: Convert statements into equations

(a) 5 added to a number is 9.
Let the number be \( x \).
Equation: \( x + 5 = 9 \)

(b) 3 subtracted from a number is equal to 12.
Let the number be \( x \).
Equation: \( x - 3 = 12 \)

(c) 5 times a number decreased by 2 is 4.
“5 times a number” = \( 5x \), “decreased by 2” = \( 5x - 2 \)
Equation: \( 5x - 2 = 4 \)

(d) 2 times the sum of the number \( x \) and 7 is 13.
Sum of \( x \) and 7 = \( x + 7 \), then multiplied by 2 → \( 2(x + 7) \)
Equation: \( 2(x + 7) = 13 \)

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Problem 2: A number is 12 more than the other. Find the numbers if their sum is 48.

Let the smaller number be \( x \).
Then the larger number is \( x + 12 \).

Their sum is 48:
\[
x + (x + 12) = 48
\]
\[
2x + 12 = 48
\]
\[
2x = 36
\]
\[
x = 18
\]

So, the numbers are 18 and 30.

✔ Check: 18 + 30 = 48 ✔️, 30 - 18 = 12 ✔️

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Problem 3: Twice the number decreased by 22 is 48. Find the number.

Let the number be \( x \).
Twice the number: \( 2x \)
Decreased by 22: \( 2x - 22 \)
Set equal to 48:

\[
2x - 22 = 48
\]
\[
2x = 70
\]
\[
x = 35
\]

✔ Check: 2×35 = 70; 70 - 22 = 48 ✔️

Answer: 35

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Problem 4: Seven times the number is 36 less than 10 times the number. Find the number.

Let the number be \( x \).
Seven times: \( 7x \)
Ten times: \( 10x \)
“36 less than 10 times” → \( 10x - 36 \)

Equation:
\[
7x = 10x - 36
\]

Subtract \( 7x \) from both sides:
\[
0 = 3x - 36
\]
\[
3x = 36
\]
\[
x = 12
\]

✔ Check: 7×12 = 84; 10×12 = 120; 120 - 36 = 84 ✔️

Answer: 12

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

Let the number be \( x \).
\( \frac{4}{5}x \) is 5 more than \( \frac{3}{4}x \):

\[
\frac{4}{5}x = \frac{3}{4}x + 5
\]

Multiply both sides by 20 (LCM of 5 and 4) to eliminate denominators:

\[
20 \cdot \frac{4}{5}x = 20 \cdot \frac{3}{4}x + 20 \cdot 5
\]
\[
16x = 15x + 100
\]
\[
16x - 15x = 100
\]
\[
x = 100
\]

✔ Check:
\( \frac{4}{5} \times 100 = 80 \)
\( \frac{3}{4} \times 100 = 75 \)
80 - 75 = 5 ✔️

Answer: 100

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Problem 6: The sum of two consecutive even numbers is 38. Find the numbers.

Consecutive even numbers differ by 2.
Let first number = \( x \), second = \( x + 2 \)

\[
x + (x + 2) = 38
\]
\[
2x + 2 = 38
\]
\[
2x = 36
\]
\[
x = 18
\]

Numbers: 18 and 20

✔ Check: 18 + 20 = 38 ✔️

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Problem 7: The sum of three consecutive odd numbers is 51. Find the numbers.

Consecutive odd numbers differ by 2.
Let middle number = \( x \), then others are \( x - 2 \) and \( x + 2 \)

\[
(x - 2) + x + (x + 2) = 51
\]
\[
3x = 51
\]
\[
x = 17
\]

Numbers: 15, 17, 19

✔ Check: 15 + 17 + 19 = 51 ✔️

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Problem 8: 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.

Let sister’s present age = \( x \)
Rene’s present age = \( x + 6 \)

After 10 years:
Sister: \( x + 10 \)
Rene: \( x + 6 + 10 = x + 16 \)

Sum after 10 years = 50:
\[
(x + 10) + (x + 16) = 50
\]
\[
2x + 26 = 50
\]
\[
2x = 24
\]
\[
x = 12
\]

Sister: 12 years, Rene: 18 years

✔ Check:
Now: 12 + 18 = 30
In 10 years: 22 + 28 = 50 ✔️

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Problem 9: The length of a rectangle is 10 m more than its breadth. If the perimeter is 80 m, find dimensions.

Let breadth = \( x \) m
Length = \( x + 10 \) m

Perimeter of rectangle = \( 2(\text{length} + \text{breadth}) \)
\[
2((x + 10) + x) = 80
\]
\[
2(2x + 10) = 80
\]
\[
4x + 20 = 80
\]
\[
4x = 60
\]
\[
x = 15
\]

Breadth = 15 m, Length = 25 m

✔ Check: Perimeter = 2(25 + 15) = 2×40 = 80 ✔️

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Problem 10: A 300 m long wire is used to fence a rectangular plot whose length is twice its width. Find length and breadth.

Let width = \( x \) m
Length = \( 2x \) m

Perimeter = \( 2(\text{length} + \text{width}) = 2(2x + x) = 2(3x) = 6x \)

Given perimeter = 300 m:
\[
6x = 300
\]
\[
x = 50
\]

Width = 50 m, Length = 100 m

✔ Check: Perimeter = 2(100 + 50) = 2×150 = 300 ✔️

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Problem 11: The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and denominator is decreased by 1, the number obtained is 3/2. Find the fraction.

Let numerator = \( x \)
Denominator = \( x + 8 \) (since it's 8 more)

New fraction after changes:
Numerator: \( x + 17 \)
Denominator: \( (x + 8) - 1 = x + 7 \)

New fraction = \( \frac{3}{2} \):
\[
\frac{x + 17}{x + 7} = \frac{3}{2}
\]

Cross-multiply:
\[
2(x + 17) = 3(x + 7)
\]
\[
2x + 34 = 3x + 21
\]
\[
34 - 21 = 3x - 2x
\]
\[
13 = x
\]

So, original fraction: Numerator = 13, Denominator = 13 + 8 = 21
→ \( \frac{13}{21} \)

✔ Check:
New numerator: 13 + 17 = 30
New denominator: 21 - 1 = 20
New fraction: \( \frac{30}{20} = \frac{3}{2} \) ✔️

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## ✔ Final Answers Summary:

1. Equations:
(a) \( x + 5 = 9 \)
(b) \( x - 3 = 12 \)
(c) \( 5x - 2 = 4 \)
(d) \( 2(x + 7) = 13 \)

2. Numbers: 18 and 30

3. Number: 35

4. Number: 12

5. Number: 100

6. Consecutive even numbers: 18 and 20

7. Consecutive odd numbers: 15, 17, 19

8. Present ages: Rene = 18, Sister = 12

9. Rectangle: Length = 25 m, Breadth = 15 m

10. Rectangular plot: Length = 100 m, Width = 50 m

11. Fraction: \( \frac{13}{21} \)

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All problems solved using linear equations in one variable. Let me know if you’d like a printable version or further explanation!