SEBA Class IX Mathematics Chapter 1 Practice Questions Assamese Medium

অনুশীলনী-1

13. (i) 4 আৰু 5 ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিওৱা।

Solution:

\(\LARGE 4=\frac{4}{1}=\frac{4×6}{1×6}=\frac{24}{6}\)

\(\LARGE 5=\frac{5}{1}=\frac{5×6}{1×6}=\frac{30}{6}\)

[Hint: যিহেতু মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিয়াব দিছে, গতিকে 5+1=6 ৰে দুয়োটা ভগ্নাংশৰে লৱ আৰু হৰক পূৰণ কৰা হৈছে।]

গতিকে, 4 আৰু 5 ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা হ'ল- \(\LARGE \frac{25}{6}, \frac{26}{6}, \frac{27}{6}, \frac{28}{6}, \frac{29}{6}\)

13. (ii) \(\LARGE \frac{1}{3}\) আৰু \( \LARGE \frac{2}{3}\) ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিওৱা।

Solution:

\(\LARGE \frac{1}{3}=\frac{1×6}{3×6}=\frac{6}{18}\)

\(\LARGE \frac{2}{3}=\frac{2×6}{3×6}=\frac{12}{18}\)

[Hint: যিহেতু মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিয়াব দিছে, গতিকে 5+1=6 ৰে দুয়োটা ভগ্নাংশৰে লৱ আৰু হৰক পূৰণ কৰা হৈছে।]

গতিকে, \(\LARGE \frac{1}{3}\) আৰু \( \LARGE \frac{2}{3}\) ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা হ'ল- \(\LARGE \frac{7}{18}, \frac{8}{18}, \frac{9}{18}, \frac{10}{18}, \frac{11}{18}\)

14. \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ কৰা ( p, q অখণ্ড আৰু q≠0)

     a) \(0.\overline{7}\)       b) \(0.\overline{38}\)       c) \(0.\overline{231}\)

Solution:

a) \(0.\overline{7}\) ক \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ

ধৰাহল,

x=0.7777.............(i)

10x=7.7777.......(ii)   [দুয়োপক্ষত 10 পূৰণ কৰিলে]

এতিয়া, (ii)-(i),

10x - x = (7.7777.......) - (0.7777.......)

⇒ 9x = 7

⇒x = \(\LARGE \frac{7}{9}\)

গতিকে, \(0.\overline{7}\)=\(\LARGE \frac{7}{9}\)


b) \(0.\overline{38}\) ক \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ

ধৰাহল, 

x = 0.383838......    (i)

100x = 38.383838......  (ii) [দুয়ো পক্ষত 100 পূৰণ কৰিলে]

এতিয়া, (ii) - (i),

100x-x = (38.383838.......) - (0.383838......)

⇒99x = 38

⇒x= \(\large \frac{38}{99}\)

গতিকে, \(0.\overline{38}\)=\(\large \frac{38}{99}\)


c) \(0.\overline{231}\) ক \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ

ধৰাহল,

x = 0.231231...... (i)

1000x = 231.231231...... (ii) [দুয়ো পক্ষক 1000 ৰে পূৰণ কৰিলে]

এতিয়া, (ii) - (i),

1000x-x = (231.231231.......) - (0.231231......)

⇒999x = 231

⇒x= \(\large \frac{231}{999}\)

গতিকে, \(0.\overline{231}\)=\(\large\frac{231}{999}\)

15. সৰল কৰা।

(i) \(\sqrt{2}+2\sqrt{2}+3\sqrt{2}\)

(ii) \(5\sqrt{3}-2\sqrt{3}+\sqrt{3}\)

(iii) \((\sqrt{3}+1)(\sqrt{3}-1)\)

(iv) \((\sqrt{7}+3)^2\)

(v) \((2-\sqrt{5})^2\)

(vi) \((2+\sqrt{5})(3+\sqrt{2})\)

Solution:

(i) \(\sqrt{2}+2\sqrt{2}+3\sqrt{2}\)

=\(\sqrt{2}(1+2+3)\)

=\(6\sqrt{2}\)

(ii) \(5\sqrt{3}-2\sqrt{3}+\sqrt{3}\)

=\(\sqrt{3}(5-2+1)\)

=\(4\sqrt{3}\)

(iii) \((\sqrt{3}+1)(\sqrt{3}-1)\)

=\((\sqrt{3})^2-1^2\) [\(\because (a+b)(a-b)=a^2-b^2\)]

=3-1

=2 

 (iv) \((\sqrt{7}+3)^2\)

=\((\sqrt{7})^2+2.\sqrt{7}.3+3^2\)   [\(\because (a+b)^2=a^2+2ab+b^2\)]

=\(7+6\sqrt{7}+9\)

=\(16+6\sqrt{7}\)

(v) \((2-\sqrt{5})^2\)

=\((2^2-2.2.\sqrt{5}+(\sqrt{5})^2\)   [\(\because (a-b)^2=a^2-2ab+b^2\)]

=\(4-4\sqrt{5}+5\)

=\(9-4\sqrt{5}\)

(vi) (\(2+\sqrt{5})(3+\sqrt{2})\)

=\(2(3+\sqrt{2})+\sqrt{5}(3+\sqrt{2})\)   [\(\because (a+b)(c+d)=a(c+d)+b(c+d)\)]

=\(6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\)

16. হৰবিলাক পৰিমেয় কৰা।

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

(ii) \(\LARGE \frac{1}{\sqrt{5}-\sqrt{3}}\)

(iii) \(\LARGE \frac{1}{\sqrt{5}-2}\)

(iv) \(\LARGE \frac{4}{\sqrt{7}+\sqrt{3}}\)


Solution:

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

=\(\LARGE \frac{1×\sqrt{3}}{\sqrt{3}×\sqrt{3}}\)

=\(\LARGE \frac{\sqrt{3}}{3}\)

(ii) \(\LARGE \frac{1}{\sqrt{5}-\sqrt{3}}\)

=\(\LARGE \frac{1×(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})×(\sqrt{5}+\sqrt{3})}\)

=\(\LARGE \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2-(\sqrt{3})^2}\) [\(\because (a+b)(a-b)=a^2-b^2\)]

=\(\LARGE \frac{\sqrt{5}+\sqrt{3}}{5-3}\)

=\(\LARGE \frac{\sqrt{5}+\sqrt{3}}{2}\)

(iii) \(\LARGE \frac{1}{\sqrt{5}-2}\)

=\(\LARGE \frac{1×(\sqrt{5}+2)}{(\sqrt{5}-2)×(\sqrt{5}+2)}\)

=\(\LARGE \frac{\sqrt{5}+2}{(\sqrt{5})^2-2^2}\)   [\(\because (a+b)(a-b)=a^2-b^2\)]

=\(\LARGE \frac{\sqrt{5}+2}{5-4}\)

=\(\LARGE \frac{\sqrt{5}+2}{1}\)

=\(\large \sqrt{5}+2\)

(iv) \(\LARGE \frac{4}{\sqrt{7}+\sqrt{3}}\)

=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})×(\sqrt{7}-\sqrt{3})}\)

=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{(\sqrt{7})^2-(\sqrt{3})^2}\)   [\(\because (a+b)(a-b)=a^2-b^2\)]

=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{7-3}\)

=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{4}\)

=\(\large \sqrt{7}-\sqrt{3}\)

17. মান উলিওৱাঃ

(i) \(49^{\LARGE\frac{1}{2}}\)

(ii) \(27^{\LARGE\frac{1}{3}}\)

(iii) \(243^{\LARGE\frac{1}{5}}\)

(iv) \(125^{\LARGE\frac{2}{3}}\)

(v) \(3^{\LARGE\frac{2}{5}}.3^{\LARGE\frac{3}{5}}\)

(vi) \(64^{\LARGE\frac{-1}{3}}\)


Solution:

(i) \(49^{\LARGE\frac{1}{2}}\)

=\( (7^{\LARGE 2})^{\LARGE\frac{1}{2}} \)

=\( 7^{\LARGE{2. \frac{1}{2} }} \)   [\(\because \large (a^m)^n=a^{mn} \)]

=7

(ii) \(27^{\LARGE\frac{1}{3}}\)

=\( (3^{\LARGE 3})^{\LARGE\frac{1}{3}} \)

=\( 3^{\LARGE{3. \frac{1}{3} }} \)   [\(\because \large (a^m)^n=a^{mn} \)]

=3

(iii) \(243^{\LARGE\frac{1}{5}}\)

=\( (3^{\LARGE 5})^{\LARGE\frac{1}{5}} \)

=\( 3^{\LARGE{5. \frac{1}{5} }} \)   [\(\because \large (a^m)^n=a^{mn} \)]

=3




(iv) \(125^{\LARGE\frac{2}{3}}\)

=\( (5^{\LARGE 3})^{\LARGE\frac{2}{3}} \)

=\( 5^{\LARGE{3. \frac{2}{3} }} \)   [\(\because \large (a^m)^n=a^{mn} \)]

=\(5^2\)

=25




(v) \(3^{\LARGE\frac{2}{5}}.3^{\LARGE\frac{3}{5}}\)

=\( 3^{\LARGE\frac{2}{5}+\frac{3}{5}}\)  [\(\because \large a^m.a^n=a^{m+n}\)]

=\( 3^{\LARGE \frac {2+3}{5} } \)

=\(3^{\LARGE \frac {5}{5} } \)

=3

(vi) \(64^{\LARGE\frac{-1}{3}}\)

=\( (4^{\LARGE 3})^{\LARGE\frac{-1}{3}} \)

=\( 4^{\LARGE{3. \LARGE \frac{-1}{3} }} \)   [\(\because \large (a^m)^n=a^{mn} \)]

=\(4^{\LARGE -1}\)

=\(\large \frac{1}{4}\)