Class IX Mathematics worksheet on Number Systems with questions on simplifying expressions, identifying rational and irrational numbers, and evaluating algebraic functions.

Here are the step-by-step solutions for the problems in the image.

SECTION A: (1 MARK)

1. Given that $\sqrt{10} = 3.162$, find the value of $\frac{1}{\sqrt{10}}$.
* Step 1: To find $\frac{1}{\sqrt{10}}$, we can rationalize the denominator by multiplying the top and bottom by $\sqrt{10}$.
$$ \frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{10}}{10} $$
* Step 2: Substitute the given value of $\sqrt{10} = 3.162$.
$$ \frac{3.162}{10} = 0.3162 $$

2. Simplify: $\frac{28}{\sqrt{98} - \sqrt{7}}$
* Step 1: Simplify $\sqrt{98}$. Since $98 = 49 \times 2$, then $\sqrt{98} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2}$.
* Step 2: Rewrite the expression:
$$ \frac{28}{7\sqrt{2} - \sqrt{7}} $$
* Step 3: Rationalize the denominator by multiplying numerator and denominator by the conjugate $(7\sqrt{2} + \sqrt{7})$.
$$ \text{Denominator: } (7\sqrt{2} - \sqrt{7})(7\sqrt{2} + \sqrt{7}) = (7\sqrt{2})^2 - (\sqrt{7})^2 = (49 \times 2) - 7 = 98 - 7 = 91 $$
$$ \text{Numerator: } 28(7\sqrt{2} + \sqrt{7}) = 196\sqrt{2} + 28\sqrt{7} $$
* Step 4: Put it together and simplify the fraction by dividing by 7.
$$ \frac{196\sqrt{2} + 28\sqrt{7}}{91} = \frac{28\sqrt{2} + 4\sqrt{7}}{13} $$
*(Note: Often in these tests, if the answer key says $4\sqrt{7}$, there might be a typo in the question like $\sqrt{98}-\sqrt{8}$, but based strictly on the text provided, this is the simplified form. However, checking standard simplifications, if the question was $\frac{28}{\sqrt{98}-\sqrt{8}}$, the answer would be $4\sqrt{7}$. Given the visual ambiguity, $4\sqrt{7}$ is a likely intended "clean" answer if we assume $\sqrt{98}-\sqrt{8}$).*
*Let's re-evaluate assuming a common textbook pattern where terms cancel out nicely.*
If we assume the question meant $\sqrt{98} - \sqrt{8}$:
$\sqrt{98} = 7\sqrt{2}$ and $\sqrt{8} = 2\sqrt{2}$.
Denominator becomes $5\sqrt{2}$.
$\frac{28}{5\sqrt{2}} = \frac{28\sqrt{2}}{10} = \frac{14\sqrt{2}}{5}$. This doesn't match $4\sqrt{7}$.

Let's look at the answer key hint in brackets: `[4√7]`.
How do we get $4\sqrt{7}$?
If the expression was $\frac{28}{\sqrt{7}}$, the answer is $4\sqrt{7}$.
Is $\sqrt{98} - \sqrt{7}$ equal to $\sqrt{7}$? No.
Is it possible the question is $\frac{28}{\sqrt{98}/\sqrt{2}}$? No.

Let's stick to the most rigorous math for the written text:
Answer: $\frac{28\sqrt{2} + 4\sqrt{7}}{13}$ or approx $7.56$.
*However, looking at the bracketed answer key `[4√7]`, it is highly probable the question contains a typo and should have been simply $\frac{28}{\sqrt{7}}$ or similar. But solving exactly what is written:*

Actually, let's look closer at the image. It says $\sqrt{98} - \sqrt{7}$.
Let's try one more path. Maybe simplify $\sqrt{98}$ as $7\sqrt{2}$.
$\frac{28}{7\sqrt{2}-\sqrt{7}}$.
There is no simple integer cancellation.

*Self-Correction for Student Help:* I will provide the exact calculation steps. If the student needs to match the key `4√7`, they should check if the second term was $\sqrt{0}$ or if the first term was different. But I will provide the mathematically correct derivation for the visible text.

Wait, let's look at Question 2 again. Is it $\sqrt{98} \cdot \sqrt{7}$? No, it's a minus sign.
Let's assume the standard simplification for school level often involves perfect squares.
Let's just provide the calculated value.

Revised Step for Q2:
The provided answer key in the image suggests $4\sqrt{7}$. This result is obtained if the denominator is simply $\sqrt{7}$. It is likely the question intended $\frac{28}{\sqrt{7}}$ or there is a specific context not visible. However, solving $\frac{28}{\sqrt{98}-\sqrt{7}}$ yields $\frac{28\sqrt{2}+4\sqrt{7}}{13}$.

*Let's move to the next questions which are clearer.*

3. Simplify: $16^{-\frac{3}{4}} \times \sqrt[3]{16}$
* Step 1: Convert everything to base 2. $16 = 2^4$.
* Step 2: First part: $16^{-\frac{3}{4}} = (2^4)^{-\frac{3}{4}} = 2^{4 \times -\frac{3}{4}} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}$.
* Step 3: Second part: $\sqrt[3]{16} = \sqrt[3]{2^4} = 2^{\frac{4}{3}}$.
* Step 4: Multiply them: $\frac{1}{8} \times 2^{\frac{4}{3}} = 2^{-3} \times 2^{\frac{4}{3}} = 2^{-3 + \frac{4}{3}} = 2^{-\frac{9}{3} + \frac{4}{3}} = 2^{-\frac{5}{3}}$.
* Step 5: Convert back to radical form: $\frac{1}{\sqrt[3]{2^5}} = \frac{1}{\sqrt[3]{32}} = \frac{1}{2\sqrt[3]{4}}$.

4. If $x = 3 + 2\sqrt{2}$, then find whether $x + \frac{1}{x}$ is rational or irrational.
* Step 1: Find $\frac{1}{x}$.
$$ \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2} $$
* Step 2: Add $x + \frac{1}{x}$.
$$ (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 $$
* Step 3: 6 is an integer, so it is Rational.

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SECTION B: (2 MARKS)

5. Simplify: $5\sqrt{75} + 8\sqrt{108} - \frac{1}{4}\sqrt{48}$
* Step 1: Simplify each radical.
* $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \rightarrow 5(5\sqrt{3}) = 25\sqrt{3}$
* $\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3} \rightarrow 8(6\sqrt{3}) = 48\sqrt{3}$
* $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \rightarrow \frac{1}{4}(4\sqrt{3}) = 1\sqrt{3}$
* Step 2: Combine the terms.
$$ 25\sqrt{3} + 48\sqrt{3} - 1\sqrt{3} = (25 + 48 - 1)\sqrt{3} = 72\sqrt{3} $$

6. Evaluate:
(i) $\left(\frac{125}{216}\right)^{\frac{2}{3}}$
* Step 1: Take the cube root first. $\sqrt[3]{125} = 5$ and $\sqrt[3]{216} = 6$. So, $\left(\frac{5}{6}\right)^2$.
* Step 2: Square the result. $\frac{5^2}{6^2} = \frac{25}{36}$.

(ii) $\left(\frac{25}{49}\right)^{-\frac{1}{2}} \div \left(\frac{8}{27}\right)^{-\frac{1}{3}}$
* Step 1: Handle negative exponents by flipping the fractions.
$\left(\frac{49}{25}\right)^{\frac{1}{2}} \div \left(\frac{27}{8}\right)^{\frac{1}{3}}$
* Step 2: Apply the roots.
$\sqrt{\frac{49}{25}} = \frac{7}{5}$
$\sqrt[3]{\frac{27}{8}} = \frac{3}{2}$
* Step 3: Divide.
$$ \frac{7}{5} \div \frac{3}{2} = \frac{7}{5} \times \frac{2}{3} = \frac{14}{15} $$

7. Express in the form p/q:
(i) $1.\overline{419}$ (Assuming the bar is over 419 based on standard notation, though the image shows dots over 1 and 9? Let's assume standard repeating decimal $1.419419...$)
* Let $x = 1.419419...$
* $1000x = 1419.419419...$
* Subtract $x$: $999x = 1418$
* $x = \frac{1418}{999}$

(ii) $0.\overline{001}$
* Let $x = 0.001001...$
* $1000x = 1.001001...$
* Subtract $x$: $999x = 1$
* $x = \frac{1}{999}$

8. Find a rational number and also an irrational number between $1.101100110001...$ and $2.1011001110001...$
* Rational Number: Any terminating or repeating decimal between them. For example, 1.5 or 2.0.
* Irrational Number: A non-terminating, non-repeating decimal between them. For example, 1.101100110001111... (creating a new pattern) or simply $\sqrt{2} \approx 1.414$ which lies between 1.1 and 2.1.

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SECTION C: (3 MARKS)

9. If $x = 2 + \sqrt{3}$, find the value of:
(i) $x^2 + \frac{1}{x^2}$
* Step 1: Find $\frac{1}{x}$.
$$ \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3} $$
* Step 2: Find $x + \frac{1}{x}$.
$$ (2+\sqrt{3}) + (2-\sqrt{3}) = 4 $$
* Step 3: Square both sides to get $x^2 + \frac{1}{x^2} + 2$.
$$ (x + \frac{1}{x})^2 = 4^2 = 16 $$
$$ x^2 + \frac{1}{x^2} + 2 = 16 $$
$$ x^2 + \frac{1}{x^2} = 14 $$
*(Note: The answer key in the image says 18. Let me re-read the question. Ah, usually these questions ask for $x^2 + \frac{1}{x^2}$ given $x+\frac{1}{x}$. If $x=2+\sqrt{3}$, then $x^2 = 7+4\sqrt{3}$ and $1/x^2 = 7-4\sqrt{3}$. Sum = 14. The key [18] might correspond to a different input like $x=2+\sqrt{5}$? If $x=2+\sqrt{5}$, sum is $4+2\sqrt{5}+4-2\sqrt{5} + ...$ wait. $(2+\sqrt{5}) + (2-\sqrt{5}) = 4$. Same result.
Let's check if the question is $x = \sqrt{3} + \sqrt{2}$? No.
Let's check calculation again.
$x = 2+\sqrt{3}$. $x^2 = 4 + 3 + 4\sqrt{3} = 7+4\sqrt{3}$.
$1/x = 2-\sqrt{3}$. $1/x^2 = 7-4\sqrt{3}$.
Sum = 14.
The image key says 18. This implies the question might have been $x = 2 + \sqrt{2}$?
If $x=2+\sqrt{2}$, $1/x = 2-\sqrt{2}$. Sum $x+1/x = 4$. Square is 16. $x^2+1/x^2 = 14$.
What if $x = 3 + \sqrt{8}$?
Let's trust the math: For $x=2+\sqrt{3}$, the answer is 14.)*

(ii) $x^3 - \frac{1}{x^3}$
* Step 1: We know $x - \frac{1}{x} = (2+\sqrt{3}) - (2-\sqrt{3}) = 2\sqrt{3}$.
* Step 2: Use formula $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$.
$$ (x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x)(\frac{1}{x})(x - \frac{1}{x}) $$
$$ (2\sqrt{3})^3 = x^3 - \frac{1}{x^3} - 3(1)(2\sqrt{3}) $$
$$ 8 \times 3\sqrt{3} = x^3 - \frac{1}{x^3} - 6\sqrt{3} $$
$$ 24\sqrt{3} + 6\sqrt{3} = x^3 - \frac{1}{x^3} $$
$$ 30\sqrt{3} $$
*(Image key says $8\sqrt{3}$. This confirms there is likely a mismatch between the printed question $x=2+\sqrt{3}$ and the answer key provided in the document source. Based on strict calculation for $x=2+\sqrt{3}$, the answers are 14 and $30\sqrt{3}$.)*

10. Simplify by rationalizing the denominator:
(i) $\frac{1}{\sqrt{11} + \sqrt{13}}$
* Multiply by conjugate $\sqrt{13} - \sqrt{11}$ (putting larger first to keep denominator positive).
$$ \frac{\sqrt{13} - \sqrt{11}}{13 - 11} = \frac{\sqrt{13} - \sqrt{11}}{2} $$

(ii) $\frac{3 - 2\sqrt{3}}{2 + \sqrt{3}}$
* Multiply by conjugate $2 - \sqrt{3}$.
$$ \text{Denominator: } 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 $$
$$ \text{Numerator: } (3 - 2\sqrt{3})(2 - \sqrt{3}) = 6 - 3\sqrt{3} - 4\sqrt{3} + 2(3) = 6 - 7\sqrt{3} + 6 = 12 - 7\sqrt{3} $$

(iii) $\frac{3 - \sqrt{5}}{3 + 2\sqrt{5}}$
* Multiply by conjugate $3 - 2\sqrt{5}$.
$$ \text{Denominator: } 3^2 - (2\sqrt{5})^2 = 9 - (4 \times 5) = 9 - 20 = -11 $$
$$ \text{Numerator: } (3 - \sqrt{5})(3 - 2\sqrt{5}) = 9 - 6\sqrt{5} - 3\sqrt{5} + 2(5) = 9 - 9\sqrt{5} + 10 = 19 - 9\sqrt{5} $$
$$ \text{Result: } \frac{19 - 9\sqrt{5}}{-11} = \frac{9\sqrt{5} - 19}{11} $$

11. If $(5)^{p-3} \times (3)^{2p-8} = 225$, then find the value of p.
* Step 1: Express 225 in prime factors. $225 = 25 \times 9 = 5^2 \times 3^2$.
* Step 2: Equate the powers of same bases.
$$ 5^{p-3} \times 3^{2p-8} = 5^2 \times 3^2 $$
* Step 3: Compare exponents for base 5:
$$ p - 3 = 2 \implies p = 5 $$
* Step 4: Verify with base 3:
$$ 2p - 8 = 2 \implies 2p = 10 \implies p = 5 $$
* Both match, so p = 5.

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Final Answer:

Section A
1. 0.3162
2. $\frac{28\sqrt{2} + 4\sqrt{7}}{13}$ (Note: If following the likely intended typo for answer $4\sqrt{7}$, the question was likely $\frac{28}{\sqrt{7}}$)
3. $\frac{1}{2\sqrt[3]{4}}$ or $2^{-5/3}$
4. Rational (Value is 6)

Section B
5. $72\sqrt{3}$
6. (i) $\frac{25}{36}$
(ii) $\frac{14}{15}$
7. (i) $\frac{1418}{999}$
(ii) $\frac{1}{999}$
8. Rational: 1.5; Irrational: 1.101100110001111... (or $\sqrt{2}$)

Section C
9. (i) 14 (Calculated value; Image key says 18)
(ii) $30\sqrt{3}$ (Calculated value; Image key says $8\sqrt{3}$)
10. (i) $\frac{\sqrt{13} - \sqrt{11}}{2}$
(ii) $12 - 7\sqrt{3}$
(iii) $\frac{9\sqrt{5} - 19}{11}$
11. 5