Rs Aggarwal 2017 for Class 9 Math Chapter 1 - Real Numbers

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Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 9:

Question 1:

What are rational numbers? Give ten examples of rational numbers.

Answer:

The numbers that can be written in the $\frac{p}{q}$ form, where p and q are integers and q ≠ 0 are known as rational numbers.

Examples of rational numbers:
(1) $\frac{4}{5}$ (2) $\frac{1}{5}$ (3) $\frac{5}{4}$ (4) $\frac{4}{1}$ = 4 (5) $\frac{5}{2}$ (6) $\frac{1}{7}$ (7) $\frac{0}{1}$ = 0 (8) $\frac{9}{5}$ (9) $\frac{5}{5}$ = 1 (10) $\frac{4}{9}$

Page No 9:

Question 2:

Represent each of the following rational numbers on the number line:
(i) 5
(ii) −3
(iii) $\frac{5}{7}$
(iv) $\frac{8}{3}$
(v) 1.3
(vi) −2.4
(vii) $\frac{23}{6}$

Answer:

(i)

(ii)

(iii)

(iv) $\frac{8}{3} = 2 \frac{2}{3}$

(v) $1.3 = \frac{13}{10} = 1 \frac{3}{10}$

(vi) $- 2.4 = \frac{- 24}{10} = \frac{- 12}{5} = - 2 \frac{2}{5}$

(vii) $\frac{23}{6} = 3 \frac{5}{6}$

Question 32:

An irrational number between $\sqrt{2} and \sqrt{3}$ is
(a) $(\sqrt{2} + \sqrt{3})$
(b) $\sqrt{2} \times \sqrt{3}$
(c) 5^1/4
(d) 6^1/4

Answer:

(d) 6^1/4
An irrational number between $\sqrt{2} and \sqrt{3} : \sqrt{\sqrt{2} \times \sqrt{3}} = 6^{\frac{1}{4}}$

Page No 36:

Question 33:

An irrational number between $\frac{1}{7} and \frac{2}{7}$ is
(a) $\frac{1}{2} (\frac{1}{7} + \frac{2}{7})$
(b) $(\frac{1}{7} \times \frac{2}{7})$
(c) $\sqrt{\frac{1}{7} \times \frac{2}{7}}$
(d) none of these

Answer:

Answer:

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $π$ is an irrational number.

(c) $\frac{1}{7} = . 142857142857$ ...
Hence, its period is 6.

(d)
$x^{2} + \frac{1}{x^{2}} = {(2 - \sqrt{3})}^{2} + \frac{1}{{(2 - \sqrt{3})}^{2}} = (2^{2} + {(\sqrt{3})}^{3} - 2 \times 2 \times \sqrt{3}) + \frac{1}{(2^{2} + {(\sqrt{3})}^{3} - 2 \times 2 \times \sqrt{3})} = (4 + 3 - 4 \times \sqrt{3}) + \frac{1}{(4 + 3 - 4 \times \sqrt{3})} = (7 - 4 \times \sqrt{3}) + \frac{1}{(7 - 4 \times \sqrt{3})} = \frac{{(7 - 4 \times \sqrt{3})}^{2}}{(7 - 4 \times \sqrt{3})} + \frac{1}{(7 - 4 \times \sqrt{3})} = \frac{7^{2} + {(4 \sqrt{3})}^{2} - 2 \times 7 \times 4 \sqrt{3} + 1}{(7 - 4 \times \sqrt{3})} = \frac{49 + 48 - 56 \sqrt{3} + 1}{(7 - 4 \times \sqrt{3})} = \frac{98 - 56 \sqrt{3}}{(7 - 4 \times \sqrt{3})} = 14 \times \frac{((7 - 4 \times \sqrt{3}))}{(7 - 4 \times \sqrt{3})} = 14$

Page No 37:

Question 39:

Match the following columns:

Column I		Column II
(a)	$\sqrt[4]{{(81)}^{- 2}} = . . . . . . .$	(p)	4
(b)	If ${(\frac{a}{b})}^{x - 2} = {(\frac{b}{a})}^{x - 4}$ , then x = ........ .	(q)	$\frac{2}{9}$
(c)	If $x = (9 + 4 \sqrt{5})$ , then $(\sqrt{x} - \frac{1}{\sqrt{x}})$ = ...... .	(r)	$\frac{1}{9}$
(d)	${(\frac{81}{16})}^{- 3 / 4} \times {(\frac{64}{27})}^{- 1 / 3} = ?$	(s)	3

Between two rational numbers
(a) there is no rational number
(b) there is exactly one rational number
(c) there are infinitely many irrational numbers
(d) there is no irrational number

Answer:

Answer:

${(\frac{81}{16})}^{\frac{- 3}{4}} \times \{{(\frac{25}{9})}^{\frac{- 3}{2}} \div {(\frac{5}{2})}^{- 3}\} = {(\frac{16}{81})}^{\frac{3}{4}} \times \{{(\frac{9}{25})}^{\frac{3}{2}} \times {(\frac{5}{2})}^{- (- 3)}\} = {(\frac{2}{3})}^{4 \times \frac{3}{4}} \times {(\frac{3}{5})}^{2 \times \frac{3}{2}} \times {(\frac{5}{2})}^{3} = {(\frac{2}{3})}^{3} \times {(\frac{3}{5})}^{3} \times {(\frac{5}{2})}^{3} = \frac{2^{3} \times 3^{3} \times 5^{3}}{3^{3} \times 5^{3} \times 2^{3}} = 1$

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