Set of 15 multiple-choice math problems focusing on algebraic equations, linear expressions, and number concepts.

Let's solve each problem step by step.

---

Problem 1:

The solution of which of the following equations is neither a fraction nor an integer.
(a) \(3x + 2 = 5x + 2\)
(b) \(4x - 18 = 2\)
(c) \(4x + 7 = x + 2\)
(d) \(5x - 8 = x + 4\)

#### Solution:
- Option (a): \(3x + 2 = 5x + 2\)
\[
3x + 2 = 5x + 2 \implies 3x = 5x \implies 0 = 2x \implies x = 0
\]
The solution is \(x = 0\), which is an integer.

- Option (b): \(4x - 18 = 2\)
\[
4x - 18 = 2 \implies 4x = 20 \implies x = 5
\]
The solution is \(x = 5\), which is an integer.

- Option (c): \(4x + 7 = x + 2\)
\[
4x + 7 = x + 2 \implies 4x - x = 2 - 7 \implies 3x = -5 \implies x = -\frac{5}{3}
\]
The solution is \(x = -\frac{5}{3}\), which is a fraction.

- Option (d): \(5x - 8 = x + 4\)
\[
5x - 8 = x + 4 \implies 5x - x = 4 + 8 \implies 4x = 12 \implies x = 3
\]
The solution is \(x = 3\), which is an integer.

None of the options have a solution that is neither a fraction nor an integer. However, if we interpret the question as asking for a solution that is not an integer, then the correct answer is:
\[
\boxed{c}
\]

---

Problem 2:

The solution of the equation \(ax + b = 0\) is
(a) \(x = \frac{a}{b}\)
(b) \(x = -b\)
(c) \(x = -\frac{b}{a}\)
(d) \(x = \frac{b}{a}\)

#### Solution:
Solve the equation \(ax + b = 0\):
\[
ax + b = 0 \implies ax = -b \implies x = -\frac{b}{a}
\]
The correct answer is:
\[
\boxed{c}
\]

---

Problem 3:

If \(8x - 3 = 25 + 17x\), then \(x\) is
(a) a fraction
(b) an integer
(c) a rational number
(d) cannot be solved

#### Solution:
Solve the equation \(8x - 3 = 25 + 17x\):
\[
8x - 3 = 25 + 17x \implies 8x - 17x = 25 + 3 \implies -9x = 28 \implies x = -\frac{28}{9}
\]
The solution is \(x = -\frac{28}{9}\), which is a fraction and a rational number. Since the question asks for the type of number, the most specific answer is:
\[
\boxed{a}
\]

---

Problem 4:

The shifting of a number from one side of an equation to the other is called
(a) Transposition
(b) Distributivity
(c) Commutativity
(d) Associativity

#### Solution:
Shifting a number from one side of an equation to the other is called transposition. This is a fundamental technique in solving linear equations.
\[
\boxed{a}
\]

---

Problem 5:

If \(\frac{5x}{3} - 4 = \frac{2x}{5}\), then the numerical value of \(2x - 7\) is
(a) \(\frac{19}{13}\)
(b) \(-\frac{13}{19}\)
(c) \(0\)
(d) \(\frac{13}{19}\)

#### Solution:
Solve the equation \(\frac{5x}{3} - 4 = \frac{2x}{5}\):
\[
\frac{5x}{3} - 4 = \frac{2x}{5} \implies \frac{5x}{3} - \frac{2x}{5} = 4
\]
Find a common denominator (15):
\[
\frac{25x}{15} - \frac{6x}{15} = 4 \implies \frac{19x}{15} = 4 \implies 19x = 60 \implies x = \frac{60}{19}
\]
Now, find the value of \(2x - 7\):
\[
2x - 7 = 2 \left(\frac{60}{19}\right) - 7 = \frac{120}{19} - 7 = \frac{120}{19} - \frac{133}{19} = \frac{120 - 133}{19} = \frac{-13}{19}
\]
The correct answer is:
\[
\boxed{b}
\]

---

Problem 6:

The value of \(x\) for which the expressions \(3x - 4\) and \(2x + 1\) become equal is
(a) \(-3\)
(b) \(0\)
(c) \(5\)
(d) \(1\)

#### Solution:
Set the expressions equal to each other:
\[
3x - 4 = 2x + 1 \implies 3x - 2x = 1 + 4 \implies x = 5
\]
The correct answer is:
\[
\boxed{c}
\]

---

Problem 7:

If \(a\) and \(b\) are positive integers, then the solution of the equation \(ax = b\) has to be always
(a) positive
(b) negative
(c) one
(d) zero

#### Solution:
Solve the equation \(ax = b\):
\[
x = \frac{b}{a}
\]
Since \(a\) and \(b\) are positive integers, \(\frac{b}{a}\) is a positive number. Therefore, the solution \(x\) is always positive.
\[
\boxed{a}
\]

---

Problem 8:

Linear equation in one variable has
(a) only one variable with any power.
(b) only one term with a variable.
(c) only one variable with power 1.
(d) only constant term.

#### Solution:
A linear equation in one variable has only one variable, and the power of the variable must be 1. For example, \(ax + b = 0\) where \(a \neq 0\).
\[
\boxed{c}
\]

---

Problem 9:

Which of the following is a linear expression:
(a) \(x^2 + 1\)
(b) \(y + y^2\)
(c) \(4\)
(d) \(1 + z\)

#### Solution:
A linear expression contains variables with power 1 and no higher powers.
- Option (a): \(x^2 + 1\) is not linear because it contains \(x^2\).
- Option (b): \(y + y^2\) is not linear because it contains \(y^2\).
- Option (c): \(4\) is a constant and not a linear expression.
- Option (d): \(1 + z\) is linear because \(z\) has power 1.

The correct answer is:
\[
\boxed{d}
\]

---

Problem 10:

A linear equation in one variable has
(a) Only one solution
(b) Two solutions
(c) More than two solutions
(d) No solution

#### Solution:
A linear equation in one variable generally has exactly one solution, unless it is an identity (infinitely many solutions) or a contradiction (no solution). However, the standard case is one solution.
\[
\boxed{a}
\]

---

Problem 11:

Value of \(S\) in \(\frac{1}{3} + S = \frac{2}{5}\)
(a) \(\frac{4}{5}\)
(b) \(\frac{1}{15}\)
(c) \(10\)
(d) \(0\)

#### Solution:
Solve for \(S\):
\[
\frac{1}{3} + S = \frac{2}{5} \implies S = \frac{2}{5} - \frac{1}{3}
\]
Find a common denominator (15):
\[
S = \frac{6}{15} - \frac{5}{15} = \frac{1}{15}
\]
The correct answer is:
\[
\boxed{b}
\]

---

Problem 12:

\(\frac{-4}{3}y = -\frac{3}{4}\), then \(y =\)
(a) \(-\left(\frac{3}{4}\right)^2\)
(b) \(-\left(\frac{4}{3}\right)^2\)
(c) \(\left(\frac{3}{4}\right)^2\)
(d) \(\left(\frac{4}{3}\right)^2\)

#### Solution:
Solve for \(y\):
\[
\frac{-4}{3}y = -\frac{3}{4} \implies y = \frac{-\frac{3}{4}}{-\frac{4}{3}} = \frac{3}{4} \cdot \frac{3}{4} = \left(\frac{3}{4}\right)^2
\]
The correct answer is:
\[
\boxed{c}
\]

---

Problem 13:

The digit in the tens place of a two-digit number is 3 more than the digit in the units place. Let the digit at the units place be \(b\). Then the number is
(a) \(11b + 30\)
(b) \(10b + 30\)
(c) \(11b + 3\)
(d) \(10b + 3\)

#### Solution:
Let the digit in the units place be \(b\). Then the digit in the tens place is \(b + 3\). The number can be expressed as:
\[
10 \times (\text{tens digit}) + \text{units digit} = 10(b + 3) + b = 10b + 30 + b = 11b + 30
\]
The correct answer is:
\[
\boxed{a}
\]

---

Problem 14:

Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was \(x\). Then Arpita’s present age is
(a) \(3(x - 3)\)
(b) \(3x + 3\)
(c) \(3x - 9\)
(d) \(3(x + 3)\)

#### Solution:
If Shilpa’s age three years ago was \(x\), then Shilpa’s present age is \(x + 3\). Arpita’s present age is thrice Shilpa’s present age:
\[
\text{Arpita’s age} = 3 \times (\text{Shilpa’s present age}) = 3(x + 3)
\]
The correct answer is:
\[
\boxed{d}
\]

---

Problem 15:

The sum of three consecutive multiples of 7 is 357. Find the smallest multiple.
(a) 112
(b) 126
(c) 119
(d) 116

#### Solution:
Let the three consecutive multiples of 7 be \(7n\), \(7(n+1)\), and \(7(n+2)\). Their sum is:
\[
7n + 7(n+1) + 7(n+2) = 357
\]
Factor out 7:
\[
7(n + n + 1 + n + 2) = 357 \implies 7(3n + 3) = 357 \implies 3n + 3 = 51 \implies 3n = 48 \implies n = 16
\]
The smallest multiple is:
\[
7n = 7 \times 16 = 112
\]
The correct answer is:
\[
\boxed{a}
\]

---

Final Answers:

1. \(\boxed{c}\)
2. \(\boxed{c}\)
3. \(\boxed{a}\)
4. \(\boxed{a}\)
5. \(\boxed{b}\)
6. \(\boxed{c}\)
7. \(\boxed{a}\)
8. \(\boxed{c}\)
9. \(\boxed{d}\)
10. \(\boxed{a}\)
11. \(\boxed{b}\)
12. \(\boxed{c}\)
13. \(\boxed{a}\)
14. \(\boxed{d}\)
15. \(\boxed{a}\)