CBSE Class 7 Mathematics Simple Equations Worksheet Set 04

Question 1. Solve the following equations:
1. \( 3x - 14 = 16 \)
2. \( 4(2x + 1) = 36 \)
3. \( 3(2x - 1) + 5 = 14 \)
4. \( 2y - (7 - 5y) - 21 = 0 \)
5. \( 2(f - 2) + 3(4f - 1) = 7 \)
6. \( 3m + 2(m + 2) = 20 - (2m - 5) \)
7. \( 3a - 2(2a - 5) = 2(a + 3) - 8 \)
8. \( 5(3x - 3) - 3(3x - 7) = 12 \)
9. \( 2(5y - 3) - 4 = 20 \)
10. \( 3(x - \frac{1}{3}) = 2(2x - 6) \)
11. \( \frac{a + 9}{2} = 10 \)
12. \( \frac{y + 1}{2} = 1 \)
13. \( \frac{2x + 5}{3} = 3x - 10 \)
14. \( \frac{3(d - 2)}{9} + 5 = 7 \)
15. \( \frac{3x + 2}{11} = \frac{x + 5}{8} \)
16. \( 15y - \frac{4}{5} = \frac{1}{5} \)
17. \( \frac{2d + 6}{9} + 10 = 14 \)
18. \( \frac{d - 1}{4} - \frac{d - 2}{3} = 1 \)
19. \( \frac{x}{2} - 1 = \frac{5x + 10}{6} \)
20. \( \frac{4m + 2}{6m + 10} = \frac{1}{2} \)
Answer:
1. \( 3x - 14 = 16 \)
Add 14 to both sides:
\( \implies 3x = 30 \)
Divide both sides by 3:
\( \implies x = 10 \)

2. \( 4(2x + 1) = 36 \)
Divide both sides by 4:
\( \implies 2x + 1 = 9 \)
Subtract 1 from both sides:
\( \implies 2x = 8 \)
Divide both sides by 2:
\( \implies x = 4 \)

3. \( 3(2x - 1) + 5 = 14 \)
Subtract 5 from both sides:
\( \implies 3(2x - 1) = 9 \)
Divide both sides by 3:
\( \implies 2x - 1 = 3 \)
Add 1 to both sides:
\( \implies 2x = 4 \)
Divide both sides by 2:
\( \implies x = 2 \)

4. \( 2y - (7 - 5y) - 21 = 0 \)
Open the brackets:
\( \implies 2y - 7 + 5y - 21 = 0 \)
Combine like terms:
\( \implies 7y - 28 = 0 \)
Add 28 to both sides:
\( \implies 7y = 28 \)
Divide both sides by 7:
\( \implies y = 4 \)

5. \( 2(f - 2) + 3(4f - 1) = 7 \)
Multiply through:
\( \implies 2f - 4 + 12f - 3 = 7 \)
Combine terms:
\( \implies 14f - 7 = 7 \)
Add 7 to both sides:
\( \implies 14f = 14 \)
Divide both sides by 14:
\( \implies f = 1 \)

6. \( 3m + 2(m + 2) = 20 - (2m - 5) \)
Multiply through:
\( \implies 3m + 2m + 4 = 20 - 2m + 5 \)
Combine terms:
\( \implies 5m + 4 = 25 - 2m \)
Add \( 2m \) to both sides:
\( \implies 7m + 4 = 25 \)
Subtract 4 from both sides:
\( \implies 7m = 21 \)
Divide both sides by 7:
\( \implies m = 3 \)

7. \( 3a - 2(2a - 5) = 2(a + 3) - 8 \)
Multiply through:
\( \implies 3a - 4a + 10 = 2a + 6 - 8 \)
Simplify:
\( \implies -a + 10 = 2a - 2 \)
Add \( a \) to both sides:
\( \implies 10 = 3a - 2 \)
Add 2 to both sides:
\( \implies 12 = 3a \)
Divide both sides by 3:
\( \implies a = 4 \)

8. \( 5(3x - 3) - 3(3x - 7) = 12 \)
Multiply through:
\( \implies 15x - 15 - 9x + 21 = 12 \)
Combine terms:
\( \implies 6x + 6 = 12 \)
Subtract 6 from both sides:
\( \implies 6x = 6 \)
Divide both sides by 6:
\( \implies x = 1 \)

9. \( 2(5y - 3) - 4 = 20 \)
Add 4 to both sides:
\( \implies 2(5y - 3) = 24 \)
Divide both sides by 2:
\( \implies 5y - 3 = 12 \)
Add 3 to both sides:
\( \implies 5y = 15 \)
Divide both sides by 5:
\( \implies y = 3 \)

10. \( 3(x - \frac{1}{3}) = 2(2x - 6) \)
Multiply through:
\( \implies 3x - 1 = 4x - 12 \)
Subtract \( 3x \) from both sides:
\( \implies -1 = x - 12 \)
Add 12 to both sides:
\( \implies x = 11 \)

11. \( \frac{a + 9}{2} = 10 \)
Multiply both sides by 2:
\( \implies a + 9 = 20 \)
Subtract 9 from both sides:
\( \implies a = 11 \)

12. \( \frac{y + 1}{2} = 1 \)
Multiply both sides by 2:
\( \implies y + 1 = 2 \)
Subtract 1 from both sides:
\( \implies y = 1 \)

13. \( \frac{2x + 5}{3} = 3x - 10 \)
Multiply both sides by 3:
\( \implies 2x + 5 = 3(3x - 10) \)
\( \implies 2x + 5 = 9x - 30 \)
Subtract \( 2x \) from both sides:
\( \implies 5 = 7x - 30 \)
Add 30 to both sides:
\( \implies 35 = 7x \)
Divide both sides by 7:
\( \implies x = 5 \)

14. \( \frac{3(d - 2)}{9} + 5 = 7 \)
Simplify the fraction:
\( \implies \frac{d - 2}{3} + 5 = 7 \)
Subtract 5 from both sides:
\( \implies \frac{d - 2}{3} = 2 \)
Multiply both sides by 3:
\( \implies d - 2 = 6 \)
Add 2 to both sides:
\( \implies d = 8 \)

15. \( \frac{3x + 2}{11} = \frac{x + 5}{8} \)
Cross-multiply:
\( \implies 8(3x + 2) = 11(x + 5) \)
\( \implies 24x + 16 = 11x + 55 \)
Subtract \( 11x \) from both sides:
\( \implies 13x + 16 = 55 \)
Subtract 16 from both sides:
\( \implies 13x = 39 \)
Divide both sides by 13:
\( \implies x = 3 \)

16. \( 15y - \frac{4}{5} = \frac{1}{5} \)
Add \( \frac{4}{5} \) to both sides:
\( \implies 15y = \frac{1}{5} + \frac{4}{5} \)
\( \implies 15y = 1 \)
Divide both sides by 15:
\( \implies y = \frac{1}{15} \)

17. \( \frac{2d + 6}{9} + 10 = 14 \)
Subtract 10 from both sides:
\( \implies \frac{2d + 6}{9} = 4 \)
Multiply both sides by 9:
\( \implies 2d + 6 = 36 \)
Subtract 6 from both sides:
\( \implies 2d = 30 \)
Divide both sides by 2:
\( \implies d = 15 \)

18. \( \frac{d - 1}{4} - \frac{d - 2}{3} = 1 \)
Multiply both sides by 12 (the LCM of 4 and 3):
\( \implies 3(d - 1) - 4(d - 2) = 12 \)
\( \implies 3d - 3 - 4d + 8 = 12 \)
Combine terms:
\( \implies -d + 5 = 12 \)
Subtract 5 from both sides:
\( \implies -d = 7 \)
Multiply by -1:
\( \implies d = -7 \)

19. \( \frac{x}{2} - 1 = \frac{5x + 10}{6} \)
Multiply both sides by 6 (the LCM of 2 and 6):
\( \implies 3x - 6 = 5x + 10 \)
Subtract \( 3x \) from both sides:
\( \implies -6 = 2x + 10 \)
Subtract 10 from both sides:
\( \implies -16 = 2x \)
Divide both sides by 2:
\( \implies x = -8 \)

20. \( \frac{4m + 2}{6m + 10} = \frac{1}{2} \)
Cross-multiply:
\( \implies 2(4m + 2) = 1(6m + 10) \)
\( \implies 8m + 4 = 6m + 10 \)
Subtract \( 6m \) from both sides:
\( \implies 2m + 4 = 10 \)
Subtract 4 from both sides:
\( \implies 2m = 6 \)
Divide both sides by 2:
\( \implies m = 3 \)
In simple words: To solve these math sentences, we do the same action to both sides until our letter is left all alone on one side.

Exam Tip: Always make sure to write down each step clearly. Doing math steps on both sides prevents errors.

Question 2. Write equations for the following statements and solve:
1. the sum of a number and 12 is 27
2. The product of a number and 7 is 91
3. Half of a number is 19.
4. Eight times a number is 120
5. Three times a number plus 10 is 97
6. Twenty less than a number is 54.
7. Twice a number plus 13 is 47.
8. One-fourth of the sum of a number and 6 is 12.
9. Half the product of a number and 20 is 50.
10. Ten less than twice a number is 86.
11. Twice the sum of x and 5 is 12.
12. 5 added to thrice the number y is 17.
13. Adding 1 to one-third of m is 5.
14. The sum of three times x and 9 is 15.
15. 8 less than three times a number is 85.
Answer:
Let us use the letter \( n \) to represent the unknown number.

1. Equation: \( n + 12 = 27 \)
Subtract 12 from both sides:
\( \implies n = 15 \)

2. Equation: \( 7n = 91 \)
Divide both sides by 7:
\( \implies n = 13 \)

3. Equation: \( \frac{n}{2} = 19 \)
Multiply both sides by 2:
\( \implies n = 38 \)

4. Equation: \( 8n = 120 \)
Divide both sides by 8:
\( \implies n = 15 \)

5. Equation: \( 3n + 10 = 97 \)
Subtract 10 from both sides:
\( \implies 3n = 87 \)
Divide both sides by 3:
\( \implies n = 29 \)

6. Equation: \( n - 20 = 54 \)
Add 20 to both sides:
\( \implies n = 74 \)

7. Equation: \( 2n + 13 = 47 \)
Subtract 13 from both sides:
\( \implies 2n = 34 \)
Divide both sides by 2:
\( \implies n = 17 \)

8. Equation: \( \frac{n + 6}{4} = 12 \)
Multiply both sides by 4:
\( \implies n + 6 = 48 \)
Subtract 6 from both sides:
\( \implies n = 42 \)

9. Equation: \( \frac{20n}{2} = 50 \) which is \( 10n = 50 \)
Divide both sides by 10:
\( \implies n = 5 \)

10. Equation: \( 2n - 10 = 86 \)
Add 10 to both sides:
\( \implies 2n = 96 \)
Divide both sides by 2:
\( \implies n = 48 \)

11. Equation: \( 2(x + 5) = 12 \)
Divide both sides by 2:
\( \implies x + 5 = 6 \)
Subtract 5 from both sides:
\( \implies x = 1 \)

12. Equation: \( 3y + 5 = 17 \)
Subtract 5 from both sides:
\( \implies 3y = 12 \)
Divide both sides by 3:
\( \implies y = 4 \)

13. Equation: \( \frac{m}{3} + 1 = 5 \)
Subtract 1 from both sides:
\( \implies \frac{m}{3} = 4 \)
Multiply both sides by 3:
\( \implies m = 12 \)

14. Equation: \( 3x + 9 = 15 \)
Subtract 9 from both sides:
\( \implies 3x = 6 \)
Divide both sides by 3:
\( \implies x = 2 \)

15. Equation: \( 3n - 8 = 85 \)
Add 8 to both sides:
\( \implies 3n = 93 \)
Divide both sides by 3:
\( \implies n = 31 \)
In simple words: First we turn the English sentences into math equations. Then we find the mystery number by undoing the operations.

Exam Tip: Be careful with phrases like "less than". For example, "20 less than a number" means \( n - 20 \), not \( 20 - n \).

Question 3. Solve the following:
1. Arun's father is three times as old as Arun. If the sum of their ages is 56 years, what are their ages?
2. In an isosceles triangle, the vertex angle is twice the each base angle. Find the vertex angle.
3. the sum of three consecutive numbers is 21. Find the numbers.
4. the length of a rectangle is twice its breadth. If its perimeter is 18cm, find its length and breadth.
Answer:
1. Let Arun's age be \( x \) years.
So, his father's age is \( 3x \) years.
The sum of their ages is 56:
\( \implies x + 3x = 56 \)
\( \implies 4x = 56 \)
Divide by 4:
\( \implies x = 14 \)
Arun's age is 14 years. His father's age is \( 3 \times 14 = \) 42 years.

2. In an isosceles triangle, let the two equal base angles be \( b \) each.
The vertex angle is twice each base angle, which means the vertex angle is \( 2b \).
The sum of angles in a triangle is \( 180^\circ \):
\( \implies b + b + 2b = 180^\circ \)
\( \implies 4b = 180^\circ \)
Divide by 4:
\( \implies b = 45^\circ \)
So, the vertex angle is \( 2 \times 45^\circ = \) 90°.

3. Let the three consecutive numbers be \( n \), \( n + 1 \), and \( n + 2 \).
Their sum is 21:
\( \implies n + (n + 1) + (n + 2) = 21 \)
Combine terms:
\( \implies 3n + 3 = 21 \)
Subtract 3 from both sides:
\( \implies 3n = 18 \)
Divide by 3:
\( \implies n = 6 \)
The numbers are 6, 7, and 8.

4. Let the breadth of the rectangle be \( y \) cm.
Its length is twice the breadth, which is \( 2y \) cm.
The perimeter is 18 cm:
\( \implies 2(\text{length} + \text{breadth}) = 18 \)
\( \implies 2(2y + y) = 18 \)
\( \implies 2(3y) = 18 \)
\( \implies 6y = 18 \)
Divide by 6:
\( \implies y = 3\text{ cm} \)
The breadth is 3 cm. The length is \( 2 \times 3 = \) 6 cm.
In simple words: We can solve real-life puzzles by setting up equations. Just use a letter for the unknown values and solve them step by step.

Exam Tip: Always state what your variable represents at the start of your word problem solutions.

Question 4. Construct equations starting with the given values:
1. Construct 2 equations starting with x = 3
2. Construct 2 equations starting with x = -5
Answer:
1. Starting with \( x = 3 \):
- Equation A: Multiply both sides by 5:
\( \implies 5x = 15 \)
Add 4 to both sides:
\( \implies 5x + 4 = 19 \)
- Equation B: Subtract 2 from both sides:
\( \implies x - 2 = 1 \)
Multiply both sides by 3:
\( \implies 3(x - 2) = 3 \)

2. Starting with \( x = -5 \):
- Equation A: Multiply both sides by 2:
\( \implies 2x = -10 \)
Add 7 to both sides:
\( \implies 2x + 7 = -3 \)
- Equation B: Add 5 to both sides:
\( \implies x + 5 = 0 \)
Divide both sides by 3:
\( \implies \frac{x + 5}{3} = 0 \)
In simple words: Building equations is like starting with a basic fact and doing the same math step to both sides to make a bigger equation.

Exam Tip: You can create endless equations from a single starting point. Just perform the exact same operations on both sides.