Grade 9 Rational and Irrational Numbers worksheet with questions on fractions, number lines, and rational number properties.

Let's solve each question step by step from the worksheet titled "Rational and Irrational Numbers" for Grade 9.

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(1) If $\frac{-7}{x} = \frac{x}{7}$, then $x$ is _______.

We are given:
$$
\frac{-7}{x} = \frac{x}{7}
$$

Cross-multiply:
$$
(-7) \cdot 7 = x \cdot x \Rightarrow -49 = x^2
$$

So,
$$
x^2 = -49
$$

But $x^2 = -49$ has no real solution, because a square of a real number cannot be negative.

However, if we consider complex numbers, $x = \pm 7i$, but since this is a Grade 9 math problem focusing on rational/irrational numbers, we assume $x$ is real.

Thus, there is no real number $x$ satisfying this equation.

But wait — let’s double-check. Is there any real value of $x$ that satisfies it?

Try plugging in values:
- Try $x = 7$: LHS = $-7/7 = -1$, RHS = $7/7 = 1$ → Not equal.
- Try $x = -7$: LHS = $-7/-7 = 1$, RHS = $-7/7 = -1$ → Not equal.

So no real solution exists.

But perhaps the question assumes $x$ is such that the equation holds, so maybe we need to interpret differently.

Wait — suppose we solve algebraically:

From:
$$
\frac{-7}{x} = \frac{x}{7}
\Rightarrow x^2 = -49
\Rightarrow x = \pm 7i
$$

This is not a real number, hence not rational, not natural, not whole, not prime.

So the correct choice is:

> a. not a rational number

✔ Answer: (a)

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(2) If $\frac{p}{q}$ lies to the left of $\frac{r}{s}$ on the number line, then

On the number line, left means smaller.

So:
$$
\frac{p}{q} < \frac{r}{s}
$$

So option (c) says: $\frac{p}{q} < \frac{r}{s}$ → True

Option (a): $\frac{p}{q} = \frac{r}{s}$ → False
Option (b): $\frac{p}{q} > \frac{r}{s}$ → False
Option (d): Both (c) and (b) → False

✔ Answer: (c)

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(3) Which of the following statements is false?

Let’s evaluate each:

(a) Every prime number is a whole number.
→ Prime numbers are positive integers greater than 1 (e.g., 2, 3, 5...), which are all whole numbers. ✔ True

(b) Every prime number is a rational number.
→ Any integer can be written as $\frac{p}{1}$, so yes, primes are rational. ✔ True

(c) Any rational number, when multiplied by 9, is a rational number.
→ Rational × rational = rational. 9 is rational, so product is rational. ✔ True

(d) Every rational number is a natural number.
→ No! For example, $\frac{1}{2}$ is rational but not a natural number. ✘ False

✔ Answer: (d)

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(4) The rational number lying between $\frac{1}{7}$ and $\frac{1}{5}$ is:

First, approximate:
- $\frac{1}{7} \approx 0.1428$
- $\frac{1}{5} = 0.2$

We need a rational number between them.

Check options:

(a) $\frac{6}{35} = ?$
$6 ÷ 35 ≈ 0.1714$ → Between 0.1428 and 0.2 → ✔ Yes

(b) $\frac{1}{9} ≈ 0.111$ → Less than $\frac{1}{7}$ → ✘ No

(c) $\frac{1}{2} = 0.5$ → Too big → ✘ No

(d) $\frac{1}{11} ≈ 0.0909$ → Too small → ✘ No

Only (a) lies between them.

✔ Answer: (a)

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(5) Which of the following statements is true for a rational number $\frac{a}{b}$?

Recall: A rational number is of the form $\frac{a}{b}$, where $a$ and $b$ are integers, and $b \ne 0$.

Now check:

(a) The numerator $a$ cannot be a prime number.
→ False. E.g., $\frac{2}{3}$, numerator is prime. So this is false.

(b) The denominator $b$ cannot be 0.
→ This is true by definition. Division by zero is undefined.

(c) The denominator $b$ cannot be a prime number.
→ False. E.g., $\frac{1}{2}$, denominator is prime.

(d) The denominator $b$ cannot be 1.
→ False. $\frac{5}{1} = 5$ is rational.

✔ Answer: (b)

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(6) Fill in the blanks:

If $A = 3^x$, $B = 9^y$, $C = 27^z$, where $x = 0.\overline{5}$, $y = 0.\overline{8}$, $z = 0.\overline{2}$, then find $A \times B \times C$.

First, convert repeating decimals to fractions.

#### Step 1: Convert decimals

(i) $x = 0.\overline{5} = \frac{5}{9}$

(ii) $y = 0.\overline{8} = \frac{8}{9}$

(iii) $z = 0.\overline{2} = \frac{2}{9}$

Now compute:

- $A = 3^x = 3^{5/9}$
- $B = 9^y = (3^2)^{8/9} = 3^{2 \cdot 8/9} = 3^{16/9}$
- $C = 27^z = (3^3)^{2/9} = 3^{3 \cdot 2/9} = 3^{6/9} = 3^{2/3}$

Now multiply:
$$
A \times B \times C = 3^{5/9} \times 3^{16/9} \times 3^{2/3}
$$

Add exponents:
- First two: $5/9 + 16/9 = 21/9 = 7/3$
- Then: $7/3 + 2/3 = 9/3 = 3$

So:
$$
A \times B \times C = 3^3 = 27
$$

✔ Answer: 27

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(7) Examine if the variable 'x' here represents a rational or an irrational number.

Wait — the problem says "Examine if the variable 'x' here..." but there is no context provided for this question in the image. It seems like part of the worksheet was cut off or missing.

Possibly, this refers back to Question (1), where $x$ was used in $\frac{-7}{x} = \frac{x}{7}$, but we saw that no real $x$ satisfies it, so $x$ would have to be imaginary.

But in real numbers, no solution exists, so $x$ does not represent a real number at all.

But if we assume the question is asking about whether $x$ could be rational or irrational based on some context, but without more info, we can’t determine.

Alternatively, perhaps the intended question is to analyze whether $x$ in Question (1) is rational.

But from earlier:
$$
x^2 = -49 \Rightarrow x = \pm 7i
$$

These are imaginary numbers, not real, so neither rational nor irrational (since both are subsets of real numbers).

So, $x$ is not a real number, hence neither rational nor irrational.

But since the question asks to examine whether $x$ is rational or irrational, and it's not real, the answer is:

> $x$ is not a real number, so it is neither rational nor irrational.

But if we must choose, we might say it is not rational.

But again, due to lack of context, this question is incomplete.

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## ✔ Final Answers Summary:

| Question | Answer |
|--------|--------|
| (1) | a. not a rational number |
| (2) | c. $\frac{p}{q} < \frac{r}{s}$ |
| (3) | d. Every rational number is a natural number |
| (4) | a. $\frac{6}{35}$ |
| (5) | b. The denominator $b$ cannot be 0 |
| (6) | 27 |
| (7) | Not enough information; likely $x$ is not real, so neither rational nor irrational |

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