AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers InText Questions

**Do This**

Question 1.

Represent \(\frac{-3}{4}\) on a number line. (Page No.3)

Solution:

Step – 1: Divide each unit into four equal parts to the right and left side of zero on the number line.

Step – 2 : Take 3 parts after zero on its left side.

Step -3 : It represents \(\frac{-3}{4}\)

Question 2.

Write 0, 7, 10, -4 in p/q form (Page No.2)

Solution:

Question 3.

Guess my number : Your friend chooses an integer between 0 and 100. You have to find out that number by asking questions, but your friend can only answer ‘Yes’ or ‘No’. What strategy would you use ? (Page No.3)

Solution:

Let my friend choosen 73; then my questions may be like this.

Q : Does it lie in the first 50 numbers ?

A: No .

Q : Does it lie between 50 and 60 ?

A: No

Q : Does it lie between 60 and 70 ?

A: No

Q : Does it lie between 70 and 80 ?

A: Yes

[then my guess would be “it is a number from 70 to 80]

Q : Is it an even number ?

A: No

[my guess : it should be one of the numbers 71, 73, 75, 77 and 79]

Q : Is it a prime number ?

A: Yes

[my guess : it may be 71, 73 or 79]

Q : Is it less than 75 ?

A: Yes

[my guess : it may be either 71 or 73]

Q : Is it less than 72 ?

A: No

Then the number is 73.

Therefore the strategy is

→ Use number properties such as even, odd, composite or prime to determine the number.

**Do This**

Question (i)

Find five rational numbers between 2 and 3 by mean method. (Page No.4)

Solution:

We know that \(\frac{a+b}{2}\) is a rational

between any two numbers a and b.

Let a = 2 and b = 3

Question (ii)

Find 10 rational numbers between \(\frac{-3}{11}\) and \(\frac{8}{11}\) (Page No.4)

Solution:

\(\frac{-3}{11}<\left[\frac{-2}{11}, \frac{-1}{11}, \frac{0}{11}, \frac{1}{11}, \frac{2}{11}, \frac{3}{11}, \frac{4}{11}, \frac{5}{11}, \frac{6}{11}, \frac{7}{11}\right]<\frac{8}{11}\)

**Do This**

Question

Find the decimal form of

(i) \(\frac{1}{17}\) (Page No.5)

Solution:

\(\frac{1}{19}\) = 0.052631

(ii) \(\frac{1}{19}\)(Page No.5)

Solution:

**Try These**

Question 1.

Find the decimal values of the following. (Page No.6)

Solution:

i) \(\frac{1}{2}\) = 0.5

ii) \(\frac{1}{2^{2}}=\frac{1}{4}\) = 0.25

iii) \(\frac{1}{5}\) = 0.2

iv) \(\frac{1}{5 \times 2}=\frac{1}{10}\) = 0.1

v) \(\frac{3}{10}\) = 0.3

vi) \(\frac{27}{25}\)

vii) \(\frac{1}{3}\)

Solution:

viii) \(\frac{7}{6}\)

Solution:

ix) \(\frac{5}{12}\)

Solution:

x) \(\frac{1}{7}\)

Solution:

**Think, Discuss and Write**

Question 1.

Kruthi said √2 can be written as \(\frac{\sqrt{2}}{1}\) which is in form. So √2 is a rational number. Do you agree with her argument ? (Page No.10)

Solution:

No.

Writing √2 = \(\frac{\sqrt{2}}{1}\) is not in the \(\frac { p }{ q }\) form.

Since p and q are integers and √2 is not an integer.

**Try These**

Question 1.

Find the value of √3 upto six decimals.(Page No.10)

Solution:

Step 1 :

Write 3 as 3.00 00 00 00 00 00 00

Step – 2 :

Group the zeros in pairs (i.e.) make periods.

Step – 3 :

Find the square root using long division method.

∴ √3 = 1.732050

**Try These**

Question 1.

Locate √5 and -√5 on the number line [Hint 5 = 2^{2} + 1^{2}](Page No.12)

Solution:

Step – 1 :

At zero draw a rectangle of length 2 units and breadth 1 unit.

Step – 2 :

Step – 3 :

Using compass draw an arc of radius OB with centre ‘O’ which cuts the num-ber line at D and D^{1}.

Step – 4 :

D represents √5 and D^{1} represents – √5 on the number line.

**Activity**

Question

Constructing the ‘Square root spiral’. (Page No. 15)

Solution:

Take a large sheet of paper and construct the ‘Square root spiral’ in the following manner.

Step – 1 : Start with point ‘O’ and draw a line segment \(\overline{\mathrm{OP}}\) of 1 unit length.

Step – 2 : Draw a line segment \(\overline{\mathrm{PQ}}\) perpendicular to \(\overline{\mathrm{OP}}\) of unit length (where OP = PQ = 1) (see Fig.)

Step – 3 : Join 0, Q. (OQ = √2 )

Step – 4 : Draw a line segment OR of unit length perpendicular to \(\overline{\mathrm{OQ}}\)

Step – 5 : Join O, R. (OR = √3 )

Step – 6 : Draw a line segment RS of unit length perpendicular to \(\overline{\mathrm{OR}}\).

Step – 7 : Continue in this manner for some more number of steps, you will create a

beautiful spiral made of line segment \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}, \overline{\mathrm{RS}}, \overline{\mathrm{ST}}, \overline{\mathrm{TU}}\) …………etc. Note that the line segments \(\overline{\mathrm{OQ}}, \overline{\mathrm{OR}}, \overrightarrow{\mathrm{OS}}, \overline{\mathrm{OT}}, \overline{\mathrm{OU}}\) …… etc., denote the lengths √2,√3, √4, √5, √6 respectively.

**Do This**

Question 1.

Find rationalising factors of the denominators of (Page No. 20)

i) \(\frac{1}{2 \sqrt{3}}\)

Solution:

\(\frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{2 \times 3}=\frac{\sqrt{3}}{6}\)

∴ The R.F is √3

ii) \(\frac{1}{\sqrt{5}}\)

Solution:

\(\frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}\)

∴ The R.F is √5

iii) \(\frac{1}{\sqrt{8}}\)

Solution:

**Do This;**

Question

Simplify : (Page No. 23)

i) (16) ^{1/2}

Solution:

(16)^{1/2} = (4 x 4)^{1/2} = (42)^{1/2} = 4^{2/2} = 4

ii) (128)^{1/7}

Solution:

(128)^{1/7} =(2 X 2 X 2 X 2 X 2 X 2 X 2)^{1/7} = (2^{7})^{1/7} = 2

iii) (343)^{1/5}

Solution:

(343)^{1/5}= (3 x 3 x 3 x 3 x 3)^{1/5} = (3^{5})^{1/5}= 3

Question 1.

Write the following surds in exponen-tial form (Page No. 24)

i) √2

Solution:

√2 = 2^{1/2}

ii) 3√9

Solution:

\(\sqrt[3]{9}=\sqrt[3]{3 \times 3}=\sqrt[3]{3^{2}}=3^{\frac{2}{3}}\)

iii) \(\sqrt[5]{20}\)

Solution:

\(\begin{aligned}

\sqrt[5]{20}=\sqrt[5]{2 \times 2 \times 5} &=\sqrt[5]{2^{2} \times 5} \\

&=2^{\frac{2}{5}} \times 5^{\frac{1}{5}}

\end{aligned}\)

iv) 17√19

Solution:

\(\sqrt[17]{19}=19^{\frac{1}{17}}\)

Question 2.

Write the surds in radical form. (Page No. 24)

i) 5^{1/7} = \(\sqrt[7]{5}\)

ii) 17^{1/6} = \(\sqrt[6]{17}\)

iii) 5^{2/5} = \(\sqrt[5]{5^{2}}=\sqrt[5]{5 \times 5}=\sqrt[5]{25}\)

iv) 142^{1/2} = \(\sqrt{142}\)