# AP 9th Class Maths Important Questions Chapter 1 Real Numbers

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.
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
= (52)3/2 × (54)-1/4
= $$5^{2 \times \frac{3}{2}} \times 5^{4 \times\frac{-1}{4}}$$ = 53 × 5-1 = 53-1 = 52 – 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 x2 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
x2 = 4k2 4k + 1 = 2(2k2 + 2k) + 1 = 2l + 1
x2 = 2l + 1 {l = 2k2 + 2k}
∴ x2 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 a3 + b3
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
∴ a3 + b3 = (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 x2 + y2.
Solution:
Given that x = $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$
R.F. of √3 – √2 is √3 + √2

x2 + y2 = (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