CBSE Class 7 Mental Maths Simple Equations Worksheet - Free Printable

Problem Analysis:

The image contains a list of algebraic equations and word problems. Below, I will solve a few representative problems from the list to demonstrate the solution approach. If you need solutions for specific problems, please let me know.

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Solutions to Selected Problems:

#### Problem 6:
\[
\frac{P}{4} + 7 = 5
\]

Solution:
1. Subtract 7 from both sides:
\[
\frac{P}{4} = 5 - 7
\]
\[
\frac{P}{4} = -2
\]

2. Multiply both sides by 4 to solve for \( P \):
\[
P = -2 \times 4
\]
\[
P = -8
\]

Answer:
\[
\boxed{-8}
\]

---

#### Problem 10:
\[
5x - \frac{7}{3} = 3x - \frac{2}{3}
\]

Solution:
1. Subtract \( 3x \) from both sides:
\[
5x - 3x - \frac{7}{3} = -\frac{2}{3}
\]
\[
2x - \frac{7}{3} = -\frac{2}{3}
\]

2. Add \( \frac{7}{3} \) to both sides:
\[
2x = -\frac{2}{3} + \frac{7}{3}
\]
\[
2x = \frac{5}{3}
\]

3. Divide both sides by 2:
\[
x = \frac{5}{3} \div 2
\]
\[
x = \frac{5}{3} \times \frac{1}{2}
\]
\[
x = \frac{5}{6}
\]

Answer:
\[
\boxed{\frac{5}{6}}
\]

---

#### Problem 18:
\[
\frac{3x}{8} = 27
\]

Solution:
1. Multiply both sides by 8 to eliminate the denominator:
\[
3x = 27 \times 8
\]
\[
3x = 216
\]

2. Divide both sides by 3:
\[
x = \frac{216}{3}
\]
\[
x = 72
\]

Answer:
\[
\boxed{72}
\]

---

#### Problem 26:
If \( x = y + 2 \), find the value of \( y \) from the equation:
\[
y - \frac{x - 2}{2} = \frac{2}{3}
\]

Solution:
1. Substitute \( x = y + 2 \) into the equation:
\[
y - \frac{(y + 2) - 2}{2} = \frac{2}{3}
\]
Simplify the numerator inside the fraction:
\[
y - \frac{y + 2 - 2}{2} = \frac{2}{3}
\]
\[
y - \frac{y}{2} = \frac{2}{3}
\]

2. Combine the terms on the left-hand side:
\[
y - \frac{y}{2} = \frac{2y}{2} - \frac{y}{2} = \frac{y}{2}
\]
So the equation becomes:
\[
\frac{y}{2} = \frac{2}{3}
\]

3. Multiply both sides by 2:
\[
y = \frac{2}{3} \times 2
\]
\[
y = \frac{4}{3}
\]

Answer:
\[
\boxed{\frac{4}{3}}
\]

---

#### Problem 29:
Find 3 consecutive numbers whose sum is 45.

Solution:
1. Let the three consecutive numbers be \( n \), \( n+1 \), and \( n+2 \).
2. Their sum is given as 45:
\[
n + (n+1) + (n+2) = 45
\]
Simplify the left-hand side:
\[
3n + 3 = 45
\]

3. Subtract 3 from both sides:
\[
3n = 42
\]

4. Divide both sides by 3:
\[
n = 14
\]

5. The three consecutive numbers are:
\[
n = 14, \quad n+1 = 15, \quad n+2 = 16
\]

Answer:
\[
\boxed{14, 15, 16}
\]

---

#### Problem 35:
A number when added to its half gives 72. Find the number.

Solution:
1. Let the number be \( x \).
2. Its half is \( \frac{x}{2} \).
3. According to the problem:
\[
x + \frac{x}{2} = 72
\]

4. Combine the terms on the left-hand side:
\[
x + \frac{x}{2} = \frac{2x}{2} + \frac{x}{2} = \frac{3x}{2}
\]
So the equation becomes:
\[
\frac{3x}{2} = 72
\]

5. Multiply both sides by 2:
\[
3x = 144
\]

6. Divide both sides by 3:
\[
x = 48
\]

Answer:
\[
\boxed{48}
\]

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Summary of Solutions:

- Problem 6: \(\boxed{-8}\)
- Problem 10: \(\boxed{\frac{5}{6}}\)
- Problem 18: \(\boxed{72}\)
- Problem 26: \(\boxed{\frac{4}{3}}\)
- Problem 29: \(\boxed{14, 15, 16}\)
- Problem 35: \(\boxed{48}\)

If you need solutions for other problems, please specify!