CBSE Class 9 Mental Maths Number System Worksheet - Free Printable

Let's solve each problem one by one with detailed explanations.

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32. Express $\frac{16}{\sqrt{41} - 5}$ with a rational denominator.

To rationalize the denominator, multiply numerator and denominator by the conjugate of the denominator:

$$
\frac{16}{\sqrt{41} - 5} \cdot \frac{\sqrt{41} + 5}{\sqrt{41} + 5}
= \frac{16(\sqrt{41} + 5)}{(\sqrt{41})^2 - (5)^2}
= \frac{16(\sqrt{41} + 5)}{41 - 25}
= \frac{16(\sqrt{41} + 5)}{16}
= \sqrt{41} + 5
$$

✔ Answer: $\boxed{\sqrt{41} + 5}$

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33. Find the value of $(512)^{-\frac{2}{9}}$

First, express 512 as a power of 2:
$$
512 = 2^9
$$

So:
$$
(512)^{-\frac{2}{9}} = (2^9)^{-\frac{2}{9}} = 2^{9 \cdot (-\frac{2}{9})} = 2^{-2} = \frac{1}{4}
$$

✔ Answer: $\boxed{\frac{1}{4}}$

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34. Find the value of $(125)^{\frac{2}{9}}$

Express 125 as a power of 5:
$$
125 = 5^3
$$

So:
$$
(125)^{\frac{2}{9}} = (5^3)^{\frac{2}{9}} = 5^{3 \cdot \frac{2}{9}} = 5^{\frac{6}{9}} = 5^{\frac{2}{3}}
$$

This is the cube root of $5^2 = 25$, so:
$$
5^{\frac{2}{3}} = \sqrt[3]{25}
$$

✔ Answer: $\boxed{\sqrt[3]{25}}$

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35. If $x = \sqrt{2} + 1$, find the value of $\left(x - \frac{1}{x}\right)^2$

We are given $x = \sqrt{2} + 1$. First compute $\frac{1}{x}$:

$$
\frac{1}{x} = \frac{1}{\sqrt{2} + 1}
$$

Rationalize:
$$
\frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1
$$

Now:
$$
x - \frac{1}{x} = (\sqrt{2} + 1) - (\sqrt{2} - 1) = \sqrt{2} + 1 - \sqrt{2} + 1 = 2
$$

Then:
$$
\left(x - \frac{1}{x}\right)^2 = 2^2 = 4
$$

✔ Answer: $\boxed{4}$

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36. Which of the following is rational or irrational?

#### (i) $(2 + \sqrt{3})^2$

Expand:
$$
(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}
$$

Since $\sqrt{3}$ is irrational, this is irrational.

#### (ii) $(3 + \sqrt{4})^2$

Note: $\sqrt{4} = 2$, which is rational.

So:
$$
(3 + 2)^2 = 5^2 = 25
$$

This is rational.

✔ Answer:
- (i) Irrational
- (ii) Rational

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37. If $P = 3 - 2\sqrt{2}$, what is the value of $P^2 + \frac{1}{P^2}$?

Let’s first compute $P^2$:

$$
P = 3 - 2\sqrt{2}
$$
$$
P^2 = (3 - 2\sqrt{2})^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2}
$$

Now compute $\frac{1}{P}$. Rationalize:

$$
\frac{1}{P} = \frac{1}{3 - 2\sqrt{2}} \cdot \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{9 - 8} = 3 + 2\sqrt{2}
$$

Then:
$$
\frac{1}{P^2} = (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2}
$$

Now:
$$
P^2 + \frac{1}{P^2} = (17 - 12\sqrt{2}) + (17 + 12\sqrt{2}) = 34
$$

✔ Answer: $\boxed{34}$

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38. Express as an irrational number: $2xy \sqrt{xy}$

This expression is already in terms of $x$ and $y$. But we need to interpret it.

Assuming $x$ and $y$ are positive real numbers, then $2xy\sqrt{xy}$ is generally irrational unless $xy$ is a perfect square.

But since no values are given, we can say that if $xy$ is not a perfect square, then $\sqrt{xy}$ is irrational, and thus the whole expression is irrational.

But if we must express it as an irrational number, perhaps we are meant to write it in simplified radical form?

Wait — the question says "express as an irrational number", but without specific values for $x$ and $y$, it's ambiguous.

Alternatively, maybe it's asking to simplify or rewrite it in another form.

Let’s suppose $x$ and $y$ are variables. Then:

$$
2xy \sqrt{xy} = 2xy \cdot (xy)^{1/2} = 2(xy)^{3/2}
$$

So:
$$
2(xy)^{3/2} = 2 \sqrt{(xy)^3}
$$

This is irrational unless $xy$ is a perfect square.

But since the problem asks to "express as an irrational number", perhaps we just leave it in this form.

But likely, it's a typo or misphrasing. Probably means "simplify" or "write in radical form".

So:
$$
2xy \sqrt{xy} = 2 \sqrt{x^3 y^3} = 2 \sqrt{(xy)^3}
$$

But again, unless $xy$ is known, we can't assign a numeric value.

Perhaps the intent is to recognize it as an irrational expression.

✔ Answer: $\boxed{2(xy)^{3/2}}$ or $\boxed{2\sqrt{(xy)^3}}$ — this is generally an irrational number unless $xy$ is a perfect square.

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39. Express as an irrational number: $\frac{1}{a^2} \sqrt[3]{ab^2}$

We are to express this as an irrational number.

Assume $a, b > 0$, and not such that the expression becomes rational.

Let’s simplify:
$$
\frac{1}{a^2} \cdot (ab^2)^{1/3} = a^{-2} \cdot a^{1/3} b^{2/3} = a^{-2 + 1/3} b^{2/3} = a^{-5/3} b^{2/3}
$$

So:
$$
= \frac{b^{2/3}}{a^{5/3}} = \frac{\sqrt[3]{b^2}}{\sqrt[3]{a^5}} = \sqrt[3]{\frac{b^2}{a^5}}
$$

This is irrational unless $\frac{b^2}{a^5}$ is a perfect cube.

So in general, this is an irrational number.

✔ Answer: $\boxed{\sqrt[3]{\frac{b^2}{a^5}}}$ — an irrational number (in general)

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40. Find the value of $4\sqrt{12} \times 7\sqrt{6}$

Simplify:

First, $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$

So:
$$
4 \cdot 2\sqrt{3} = 8\sqrt{3}
$$

Now:
$$
8\sqrt{3} \times 7\sqrt{6} = 56 \cdot \sqrt{3} \cdot \sqrt{6} = 56 \cdot \sqrt{18}
$$

Now $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$

So:
$$
56 \cdot 3\sqrt{2} = 168\sqrt{2}
$$

✔ Answer: $\boxed{168\sqrt{2}}$

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41. Simplify $\sqrt{\frac{27}{80}}$

Break into prime factors:

- $27 = 3^3$
- $80 = 16 \cdot 5 = 2^4 \cdot 5$

So:
$$
\sqrt{\frac{27}{80}} = \frac{\sqrt{27}}{\sqrt{80}} = \frac{\sqrt{9 \cdot 3}}{\sqrt{16 \cdot 5}} = \frac{3\sqrt{3}}{4\sqrt{5}}
$$

Now rationalize:
$$
\frac{3\sqrt{3}}{4\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{15}}{20}
$$

✔ Answer: $\boxed{\frac{3\sqrt{15}}{20}}$

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42. Find the value of $\frac{1}{\sqrt{10}}$ when $\sqrt{10} = 3.162$

$$
\frac{1}{\sqrt{10}} = \frac{1}{3.162} \approx ?
$$

Divide:
$$
1 \div 3.162 \approx 0.3162
$$

✔ Answer: $\boxed{0.3162}$

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43. If $x = 7 + 4\sqrt{3}$, what is the value of $\sqrt{x} + \frac{1}{\sqrt{x}}$?

Let $y = \sqrt{x} + \frac{1}{\sqrt{x}}$

We want to find $y$. Let’s square both sides:

$$
y^2 = \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = x + 2 + \frac{1}{x}
$$

So:
$$
y^2 = x + \frac{1}{x} + 2
$$

Given $x = 7 + 4\sqrt{3}$, compute $\frac{1}{x}$:

$$
\frac{1}{x} = \frac{1}{7 + 4\sqrt{3}} \cdot \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{7 - 4\sqrt{3}}{49 - 48} = 7 - 4\sqrt{3}
$$

So:
$$
x + \frac{1}{x} = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14
$$

Then:
$$
y^2 = 14 + 2 = 16 \Rightarrow y = \sqrt{16} = 4
$$

(Since $\sqrt{x} > 0$, $y > 0$)

✔ Answer: $\boxed{4}$

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44. Find three rational numbers between 0 and 1.

Any fractions between 0 and 1 work.

Examples: $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

✔ Answer: $\boxed{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}}$ (or any other three)

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45. What is the value of $\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$?

Rationalize the denominator:

Multiply numerator and denominator by the conjugate $2 + \sqrt{3}$:

$$
\frac{2 + \sqrt{3}}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}
= \frac{4 + 4\sqrt{3} + 3}{4 - 3} = \frac{7 + 4\sqrt{3}}{1} = 7 + 4\sqrt{3}
$$

✔ Answer: $\boxed{7 + 4\sqrt{3}}$

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✔ Final Answers Summary:

| Q | Answer |
|---|--------|
| 32 | $\sqrt{41} + 5$ |
| 33 | $\frac{1}{4}$ |
| 34 | $\sqrt[3]{25}$ |
| 35 | $4$ |
| 36 | (i) Irrational, (ii) Rational |
| 37 | $34$ |
| 38 | $2(xy)^{3/2}$ or $2\sqrt{(xy)^3}$ |
| 39 | $\sqrt[3]{\frac{b^2}{a^5}}$ |
| 40 | $168\sqrt{2}$ |
| 41 | $\frac{3\sqrt{15}}{20}$ |
| 42 | $0.3162$ |
| 43 | $4$ |
| 44 | $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ |
| 45 | $7 + 4\sqrt{3}$ |

Let me know if you'd like these formatted differently or explained further!