Math worksheet featuring 16 algebraic problems, including linear equations, consecutive numbers, and word problems, designed for practice and learning.

Let's solve each problem step by step:

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Problem 16

In a linear equation, the _______ power of the variable appearing in the equation is one.

- Solution: In a linear equation, the highest power of the variable is one.
- Answer: The blank should be filled with "first."

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Problem 17

The solution of the equation \(3x - 4 = 1 - 2x\) is _______.

- Solution:
1. Start with the equation: \(3x - 4 = 1 - 2x\).
2. Add \(2x\) to both sides: \(3x + 2x - 4 = 1\).
\[
5x - 4 = 1
\]
3. Add 4 to both sides: \(5x = 5\).
4. Divide by 5: \(x = 1\).

- Answer: \(x = 1\).

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Problem 18

The solution of the equation \(2y = 5y - \frac{18}{5}\) is _______.

- Solution:
1. Start with the equation: \(2y = 5y - \frac{18}{5}\).
2. Subtract \(5y\) from both sides: \(2y - 5y = -\frac{18}{5}\).
\[
-3y = -\frac{18}{5}
\]
3. Divide by \(-3\): \(y = \frac{-\frac{18}{5}}{-3}\).
\[
y = \frac{18}{5} \cdot \frac{1}{3} = \frac{18}{15} = \frac{6}{5}
\]

- Answer: \(y = \frac{6}{5}\).

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Problem 19

Any value of the variable which makes both sides of an equation equal is known as a _______ of the equation.

- Solution: Such a value is called a solution or root of the equation.
- Answer: The blank should be filled with "solution" or "root."

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Problem 20

\(9x - \_\_\_\_\_\_ = -21\) has the solution \((-2)\).

- Solution:
1. Let the missing term be \(k\). The equation becomes: \(9x - k = -21\).
2. Substitute \(x = -2\): \(9(-2) - k = -21\).
\[
-18 - k = -21
\]
3. Add 18 to both sides: \(-k = -3\).
4. Multiply by \(-1\): \(k = 3\).

- Answer: The missing term is \(3\).

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Problem 21

Three consecutive numbers whose sum is 12 are _______, _______, and _______.

- Solution:
1. Let the three consecutive numbers be \(n\), \(n+1\), and \(n+2\).
2. Their sum is given as 12: \(n + (n+1) + (n+2) = 12\).
\[
3n + 3 = 12
\]
3. Subtract 3 from both sides: \(3n = 9\).
4. Divide by 3: \(n = 3\).
5. The three consecutive numbers are \(3\), \(4\), and \(5\).

- Answer: \(3\), \(4\), and \(5\).

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Problem 22

The share of A when Rs 25 are divided between A and B so that A gets Rs 8 more than B is _______.

- Solution:
1. Let B's share be \(x\). Then A's share is \(x + 8\).
2. The total amount is Rs 25: \(x + (x + 8) = 25\).
\[
2x + 8 = 25
\]
3. Subtract 8 from both sides: \(2x = 17\).
4. Divide by 2: \(x = 8.5\).
5. A's share is \(x + 8 = 8.5 + 8 = 16.5\).

- Answer: Rs 16.5.

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Problem 23

A term of an equation can be transposed to the other side by changing its _______.

- Solution: When a term is moved to the other side of the equation, its sign changes (e.g., positive becomes negative, and vice versa).
- Answer: The blank should be filled with "sign."

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Problem 24

On subtracting 8 from \(x\), the result is 2. The value of \(x\) is _______.

- Solution:
1. The equation is: \(x - 8 = 2\).
2. Add 8 to both sides: \(x = 10\).

- Answer: \(x = 10\).

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Problem 25

\(\frac{x}{5} + 30 = 18\) has the solution as _______.

- Solution:
1. Start with the equation: \(\frac{x}{5} + 30 = 18\).
2. Subtract 30 from both sides: \(\frac{x}{5} = -12\).
3. Multiply by 5: \(x = -60\).

- Answer: \(x = -60\).

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Problem 26

When a number is divided by 8, the result is \(-3\). The number is _______.

- Solution:
1. Let the number be \(x\). The equation is: \(\frac{x}{8} = -3\).
2. Multiply by 8: \(x = -24\).

- Answer: \(-24\).

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Problem 27

9 is subtracted from the product of \(p\) and 4, the result is 11. The value of \(p\) is _______.

- Solution:
1. The equation is: \(4p - 9 = 11\).
2. Add 9 to both sides: \(4p = 20\).
3. Divide by 4: \(p = 5\).

- Answer: \(p = 5\).

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Problem 28

If \(\frac{2}{5}x - 2 = 5 - \frac{3}{5}x\), then \(x =\) _______.

- Solution:
1. Start with the equation: \(\frac{2}{5}x - 2 = 5 - \frac{3}{5}x\).
2. Add \(\frac{3}{5}x\) to both sides: \(\frac{2}{5}x + \frac{3}{5}x - 2 = 5\).
\[
x - 2 = 5
\]
3. Add 2 to both sides: \(x = 7\).

- Answer: \(x = 7\).

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Problem 29

After 18 years, Swarnim will be 4 times as old as he is now. His present age is _______.

- Solution:
1. Let Swarnim's present age be \(x\).
2. After 18 years, his age will be \(x + 18\).
3. According to the problem: \(x + 18 = 4x\).
4. Subtract \(x\) from both sides: \(18 = 3x\).
5. Divide by 3: \(x = 6\).

- Answer: \(6\).

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Problem 30

Convert the statement "Adding 15 to 4 times \(x\) is 39" into an equation.

- Solution:
1. "4 times \(x\)" is \(4x\).
2. "Adding 15 to 4 times \(x\)" is \(4x + 15\).
3. This equals 39: \(4x + 15 = 39\).

- Answer: \(4x + 15 = 39\).

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Problem 31

The denominator of a rational number is greater than the numerator by 10. If the numerator is increased by 1 and the denominator is decreased by 1, then the new denominator is _______.

- Solution:
1. Let the numerator be \(n\) and the denominator be \(d\).
2. According to the problem: \(d = n + 10\).
3. If the numerator is increased by 1, the new numerator is \(n + 1\).
4. If the denominator is decreased by 1, the new denominator is \(d - 1\).
5. Substitute \(d = n + 10\): \(d - 1 = (n + 10) - 1 = n + 9\).

- Answer: \(n + 9\).

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Problem 32

The sum of two consecutive multiples of 10 is 210. The smaller multiple is _______.

- Solution:
1. Let the two consecutive multiples of 10 be \(10n\) and \(10(n+1)\).
2. Their sum is 210: \(10n + 10(n+1) = 210\).
\[
10n + 10n + 10 = 210
\]
3. Simplify: \(20n + 10 = 210\).
4. Subtract 10 from both sides: \(20n = 200\).
5. Divide by 20: \(n = 10\).
6. The smaller multiple is \(10n = 10 \times 10 = 100\).

- Answer: \(100\).

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Final Answers

\[
\boxed{
\begin{array}{l}
16. \text{first} \\
17. x = 1 \\
18. y = \frac{6}{5} \\
19. \text{solution} \\
20. 3 \\
21. 3, 4, 5 \\
22. 16.5 \\
23. \text{sign} \\
24. 10 \\
25. x = -60 \\
26. -24 \\
27. p = 5 \\
28. x = 7 \\
29. 6 \\
30. 4x + 15 = 39 \\
31. n + 9 \\
32. 100 \\
\end{array}
}
\]