These AP 9th Class Maths Important Questions 1st Lesson Real Numbers will help students prepare well for the exams.

## AP State Syllabus 9th Class Maths 1st Lesson Important Questions and Answers Real Numbers

Question 1.

Represent \(\frac{-13}{5}\) on number line.

Solution:

Draw number line, having, -3, -2, -1,0, 1 integers on it.

→ Divide each unit as 5 equal parts on either side of zero.

→ From these count 13 parts on left side of the zero, which indicates \(\frac{-13}{5}\)

Question 2.

Find any two irrational numbers between 0.5 and 0.55.

Answer:

Two irrational numbers between 0.5 and 0.55 are (0.52515345 ……………….. and

0.541656475 …………….. ).

Question 3.

Simplify (25)^{3/2} × (625)^{-1/4}

Solution:

Given that (25)^{3/2} × (625)^{-1/4}

= (5^{2})^{3/2} × (5^{4})^{-1/4}

= \(5^{2 \times \frac{3}{2}} \times 5^{4 \times\frac{-1}{4}}\) = 5^{3} × 5^{-1} = 5^{3-1 } = 5^{2} – 25.

Question 4.

Find two rational numbers between 0.2 and 0.3.

Solution:

The rational numbers between 0.2 and 0.3 is 0.25 and 0.28.

It means \(\frac{25}{100}, \frac{28}{100}=\frac{1}{4}, \frac{7}{25}\)

Question 5.

Convert \(\sqrt[3]{5}\) from the radical form to the exponential form.

Solution:

As \(\sqrt[n]{a^{m}}=(a)^{\frac{m}{n}}\) = \(\sqrt[n]{a}=(a)^{\frac{1}{n}}\)

\(\sqrt[3]{5}=(5)^{\frac{1}{3}}\)

Question 6.

Find the value of \(\text { (625) } \frac{-1}{4} \text { . }\) .

Solution:

Question 7.

Simplify: \(\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{3}-\sqrt{2}}\)

Solution:

Question 8.

Prove that if x is odd, then x^{2} is also odd.

Solution:

x is an odd number, x = 2k + 1

{definition of odd}

x = 2k + 1

squaring on both sides

(x)^{2} = (2k + 1)^{2}

x^{2} = 4k^{2} 4k + 1 = 2(2k^{2} + 2k) + 1 = 2l + 1

x^{2} = 2l + 1 {l = 2k^{2} + 2k}

∴ x^{2} is an odd number.

Question 9.

Express \(5 . \overline{25}\) in from where q ≠ 0, p and q are co primes.

Solution:

Let x = \(5 . \overline{25}\)

x – 5.25 25 25 ……………… (1)

Multiply by 100 to (1)

100 x = 525.25 25 25 ………….. (2)

Subtract (1) from (2)

Question 10.

Express as a decimal number.

Solution:

For the correct process of division.

For expressing = 0.5222222 …………

= \(0.5 \overline{2}\)

Question 11.

If = a + b√3, then find the value of a^{3} + b^{3}

The Rationalising Factor of 7 – 4√3 and √3 – 2 is 7 + 4√3 and √3 + 2

= 7 + 4√3 – √3 – 2

= 5 + 3√3 = a + b√3

Here a = 5,b = 3

∴ a^{3} + b^{3} = (5)^{3} + (3)^{3} = 125 + 27 = 152

Question 12.

Visualize the value of √5 upto 3 decimals on a number line, using successive magnification.

Solution:

Question 13.

If x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) and y = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\), then find the value of x^{2} + y^{2}.

Solution:

Given that x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)

R.F. of √3 – √2 is √3 + √2

x^{2} + y^{2} = (5 + 2√6 )^{2} + (5 – 2√6 )^{2}

= 25 + 20√6 + 24

+ 25 – 20√6 + 24

= 98

Question 14.

If √2 = 1.414, √3= 1.732, then find then value of \(\frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}\).

Solution:

Given that √2 = 1.414 and √3 = 1.732

Question 15.

If \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}\) = a + b√3, then find the values of a and b.

Solution:

Given that \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}\)

Now rationalise the denominator, (Rational Factor of 7+4√3 is 7 – 4√3 )

= \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}\)

= \(\frac{35-20 \sqrt{3}+14 \sqrt{3}-8 \times 3}{(7)^{2}-(4 \sqrt{3})^{2}}\)

= \(\frac{11-6 \sqrt{3}}{49-48}\) = 11 – 6√3

a + b√3 – 11 – 6√3 =11 + (-6)√3

.’. a = 11, b = -6.

Question 16.

Visualise \(4.6 \overline{7}\) on number line upto 3 decimal places, using successive magnification.

Solution:

Question 17.

Represent -√3 and √3 on number line.

Solution:

Question 18.

Find the value of ‘a’ and ‘b’, if \(\frac{(\sqrt{5}-\sqrt{3})}{2 \sqrt{5}+3 \sqrt{3}}\) = a + b\(\sqrt{15}\)

Solution:

The Rationalising Factor of \(\frac{(\sqrt{5}-\sqrt{3})}{2 \sqrt{5}+3 \sqrt{3}}\)

is 2√5 – 3√3 , multiplying the numerator and denominator with 2√5 – 3√3.

Question 19.

Represent 2.019 on number line using the method of successive magnification.

Solution:

Question 20.

Simplify:

\(15 \sqrt[5]{32}+\sqrt{225}-8 \sqrt[3]{343}+\sqrt[4]{81}\)

Solution:

= \(15 \sqrt[5]{32}+\sqrt{225}-8 \sqrt[3]{343}+\sqrt[4]{81}\)

= \(15 \sqrt[5]{2^{5}}+\sqrt{15^{2}}-8 \sqrt[3]{7^{3}}+\sqrt[4]{3^{4}}\)

= (15 × 2) + 15 – (8 × 7) + 3

= 30 + 15 – 56 + 3 = – 8.

Question 21.

Express \(5.12 \overline{3}\) in \(\frac{\mathbf{p}}{\mathbf{q}}\) form where p and q are co-primes.

Solution:

Let x = \(5.12 \overline{3}\)

=> x = 5.12333 ……………. (1)

Here periodicity is 1.

So, we have to multiply (1) with 10.on both sides.

=> 10 × x = 10 × 5.12333 ……………….

10 x = 51.2333. …………… (2)

(2) – (1)

x = \(\frac{46.11}{9}=\frac{46.11}{9}\) × \(\frac{100}{100}=\frac{4611}{900}\)

∴ x = \(\frac{4611}{900}\)

Question 22.

The length of a rectangular field is \(\sqrt[3]{5}+\sqrt[3]{2}\) Find the measure of its breadth such that the area of the rectangle is a rational number.

Solution:

Given that the length of a rectangular

field (l) = \(\sqrt[3]{5}+\sqrt[3]{2}\).

If we want to get area of rectangle as a rational number, then we should take the breadth as Rationalising Factor of length.

Question 23.

Simplify:

\(\sqrt[4]{81}-8 \cdot \sqrt[5]{243}+15 \cdot \sqrt[5]{32}+\sqrt{225}\)

Solution:

\(\sqrt[4]{81}-8 \cdot \sqrt[5]{243}+15 \cdot \sqrt[5]{32}+\sqrt{225}\)

= \(\sqrt[4]{3^{4}}-8 \cdot \sqrt[5]{3^{5}}+15 \cdot \sqrt[5]{2^{5}}+\sqrt{15^{2}}\)

= \(\left(3^{4}\right)^{\frac{1}{4}}-8 \cdot\left(3^{5}\right)^{\frac{1}{5}}+15\left(2^{5}\right)^{\frac{1}{5}}+\left(15^{2}\right)^{\frac{1}{2}}\)

= 3 – 8.3 + 15.2 + 15

= 3 – 24 + 30+ 15

= 48 – 24 = 24