Math Class 9 Question Bank

Question 1 of 37

1. Question
1 point(s)
Are there two irrational number whose sum and product both are rationals ? Justify
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Hint

Yes, $(3+\sqrt2)$ and $(3-\sqrt2)$ are two irrational numbers
Sum $=3+\sqrt2+3-\sqrt2=6$
Product $=(3+\sqrt2)\times(3-\sqrt2)$
$=(3)^2-(\sqrt2)^2=9-2=7$
Question 2 of 37

2. Question
1 point(s)
Prove that the negative of an irrational number is an irrational number.
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Hint

Let negative of an irrational number be rational number
Let P be an irrational number by the assumption -P is a rational number. Since negative of a rational number is rational -(-P) = P is a rational number but we assumed P as an irrational number. So our assumption is wrong, so negative of an irrational number is irrational.
Question 3 of 37

3. Question
1 point(s)
Find 4 rational numbers between $\dfrac{1}{3} $ and $\dfrac{4}{5} $.
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Hint

$\dfrac{1\times5}{3\times5}=\dfrac{5}{15},\dfrac{4\times3}{5\times3}=\dfrac{12}{15}\\ \\ \dfrac{6}{15},\dfrac{7}{15},\dfrac{8}{15},\dfrac{9}{15}$ are 4 rational numbers
Question 4 of 37

4. Question
1 point(s)
Express $1.\bar{323} $ in the form $\dfrac{p}{q} $, where $p $ and $q $ are integers $q\ne0 $.
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Hint

Let $x=1.\bar{323}$
$1000x=1323.\bar{323}\\ \\ 1000x-x=1323-1\\ \\ \Rightarrow 999x=1322\\ \\ x=\dfrac{1322}{999}$
Question 5 of 37

5. Question
1 point(s)
Simplify : $\left(\dfrac{81}{16}\right)^{\frac{-3}{4}}\times\left\{\left(\dfrac{25}{9}\right)^{\frac{-3}{2}}\div\left(\dfrac{5}{2}\right)^{-3}\right\} $
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Hint

$\left[\left(\dfrac{3}{2}\right)^4\right]^{\frac{-3}{4}}\times\left\{\left[\left(\dfrac{5}{3}\right)^2\right]^{\frac{-3}{2}}\div\left(\dfrac{5}{2}\right)\right\}\\ \\ =\left(\dfrac{3}{2}\right)^{-3}\times\left[\left(\dfrac{5}{3}\right)^{-3}\div\left(\dfrac{5}{2}\right)^{-3}\right]\\ \\ =\left(\dfrac{2}{3}\right)^3\times\left[\left(\dfrac{3}{5}\right)^3\div\left(\dfrac{2}{5}\right)^3\right]\\ \\ =\dfrac{2^3}{3^3}\times\left[\dfrac{3^3}{5^3}\times\dfrac{5^3}{2^3}\right]\\ \\ =\dfrac{2^3}{3^3}\times\dfrac{3^3}{2^3}=1$
Question 6 of 37

6. Question
1 point(s)
Show that $\dfrac{x^{a(b-c)}}{x^{b(a-c)}}\div\left(\dfrac{x^b}{x^a}\right)^c=1 $.
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Hint

$\dfrac{x^{ab-ac}}{x^{ba-bc}}\div\dfrac{x^{bc}}{x^{ac}}\\ \\ \dfrac{x^{ab-ac}}{x^{ba-bc}}\times\dfrac{x^{ac}}{x^{bc}}\\ \\ =\dfrac{x^{ab-ac+ac}}{x^{ba-bc+bc}}=\dfrac{x^{ab}}{x^{ba}}\\ \\ =x^{ab-ba}=x^0=1$
Question 7 of 37

7. Question
1 point(s)
Locate $\sqrt{13} $ on the number line.
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Hint

$AC=\sqrt{9}+\sqrt4\\ \\ =\sqrt{(35+(2)^2}$
Question 8 of 37

8. Question
1 point(s)
Represent $\sqrt{5.6} $ on the number line.
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Hint

Presentation of $\sqrt{5.6}$ on number line:
Mark the distance 5.6 units from a fixed points A on a given line to obtain a point B such that AB=5.6 units. From B, mark a distance of 1 unit and mark the new points as C. Find the mid-point of AC and mark the points as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then $BD=\sqrt{5.6}$ .
Now, draw an arc with centre B and radius BD, which intersects the number line in E.
Thus, E represent $\sqrt{5.6}$
Question 9 of 37

9. Question
1 point(s)
If $a+b\sqrt3=\dfrac{\sqrt3-1}{\sqrt3+1} $, find the values of $a $ and $b $.
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Hint

$\dfrac{\sqrt3-1}{\sqrt3+1}\times\dfrac{\sqrt3-1}{\sqrt3-1}\\ \\ =\dfrac{(\sqrt3-1)^2}{(\sqrt3)^2-(1)^2}=\dfrac{3+1-2\sqrt3}{2}\\ \\ =\dfrac{4-2\sqrt3}{2}\\ \\ =\dfrac{2(2-\sqrt3)}{2}=2-\sqrt3\\ \\ a+b\sqrt3=2-\sqrt3\\ \\ a=2,b=-1$
Question 10 of 37

10. Question
1 point(s)
Express $\dfrac{y^2}{\sqrt{x^2+y^2}+x} $ with rational denominator
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Hint

$\dfrac{y^2}{\sqrt{x^2+y^2}+x}\\ \\ =\dfrac{y^2}{\sqrt{x^2+y^2}+x}\times\dfrac{\sqrt{x^2+y^2}-x}{\sqrt{x^2+y^2}-x}\\ \\ =\dfrac{y^2(\sqrt{x^2+y^2}-x)}{(\sqrt{x^2+y^2})^2-x^2}\\ \\ =\dfrac{y^2(\sqrt{x^2+y^2})-x}{x^2+y^2-x^2}=\dfrac{y^2(\sqrt{x^2+y^2})-x}{y^2}\\ \\ =\sqrt{x^2+y^2}-x$
Question 11 of 37

11. Question
1 point(s)
Find the values of $x $ and $y $ if $\mathrm{(A)\ \dfrac{7+\sqrt5}{7-\sqrt5}}-\dfrac{7-\sqrt5}{7+\sqrt5}=x+\dfrac{7}{11}\sqrt5y $.
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Hint

$\dfrac{7+\sqrt5}{7-\sqrt5}-\dfrac{7-\sqrt5}{7+\sqrt5}=x+\dfrac{7\sqrt5y}{11}\\ \\ =\dfrac{(7+\sqrt5)^2}{49-5}-\dfrac{(7-\sqrt5)^2}{49-5}\\ \\ =\dfrac{49+14\sqrt5+5}{44}-\left(\dfrac{49-14\sqrt5+5}{44}\right)\\ \\ =\dfrac{54+14\sqrt5-54+14\sqrt5}{44}\\ \\ =\dfrac{28\sqrt5}{44}=x+\dfrac{7}{11}\sqrt5\ y\\ \\ a=0,\ b=\dfrac{4}{4}=1$
Question 12 of 37

12. Question
1 point(s)
If $x=\dfrac{\sqrt3-1}{\sqrt3+1} $ and $y=\dfrac{3+2\sqrt2}{3-2\sqrt2} $ then find the value of $ x+y$.
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Hint

$x=\dfrac{\sqrt3-1}{\sqrt3+1},y=\dfrac{3+2\sqrt2}{3-2\sqrt2}\\ \\ x=\dfrac{(\sqrt3-1)^2}{(\sqrt3)^2-1^2}=\dfrac{4-2\sqrt3}{2}=2-\sqrt3\\ \\ y=\dfrac{(3+2\sqrt2)^2}{9-8}=17+12\sqrt2\\ \\ x+y=2-\sqrt3+17+12\sqrt2\\ \\ =19+12\sqrt2-\sqrt3$
Question 13 of 37

13. Question
1 point(s)
Express $3.42\bar5 $ in the form $\dfrac{p}{q} $, where $p $ and $q $ are integers, $q\ne0 $
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Hint

$x=3.42\bar5\\ \\ 10x=34.2\bar5\\ \\ 100x=342.\bar5\\ \\ 1000x=3425.\bar5\\ \\ 1000x-100x=3425.\bar5-342.\bar5\\ \\ \Rightarrow 900x=3083\\ \\ \Rightarrow x=\dfrac{3083}{900}$
Question 14 of 37

14. Question
1 point(s)
If $\dfrac{\sqrt3-1}{\sqrt3+1}=a+b\sqrt3 $. find the values of a and b.
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Hint

$\dfrac{\sqrt3-1}{\sqrt3+1}\times\dfrac{\sqrt3-1}{\sqrt3-1}\\ \\ =\dfrac{(\sqrt3-1)^2}{(\sqrt3)^2-1^2}=\dfrac{3+1-2\sqrt3}{2}\\ \\ =\dfrac{4-2\sqrt3}{2}=2-\sqrt3\\ \\ a=2,\ b=-1$
Question 15 of 37

15. Question
1 point(s)
If $a=\dfrac{3-\sqrt5}{3+\sqrt5} $ and $b=\dfrac{3+\sqrt5}{3-\sqrt5} $, find $ a^2-b^2$.
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Hint

$a=\dfrac{3-\sqrt5}{3+\sqrt5},\ b=\dfrac{3+\sqrt5}{3-\sqrt5}\\ \\ a=\dfrac{(3-\sqrt5)^2}{9-5}=\dfrac{14-6\sqrt5}{4}\\ \\ =\dfrac{7-3\sqrt5}{2}\\ \\ b=\dfrac{3+\sqrt5}{3-\sqrt5}\times\dfrac{3+\sqrt5}{3+\sqrt5}\\ \\ =\dfrac{(3+\sqrt5)^2}{9-5}=\dfrac{7+3\sqrt5}{2}\\ \\ a^2-b^2=\left(\dfrac{7-3\sqrt5}{2}\right)^2-\left(\dfrac{7+3\sqrt5}{2}\right)^2\\ \\ =\dfrac{49-45}{4}-\dfrac{49+45}{4}\\ \\ =\dfrac{49-45-49-45}{4}=\dfrac{0}{4}=0$
Question 16 of 37

16. Question
1 point(s)
If $\dfrac{\sqrt2-\sqrt5}{\sqrt2+\sqrt5}=a+b\sqrt{10} $, find the values of a and b.
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Hint

$\dfrac{\sqrt2-\sqrt5}{\sqrt2+\sqrt5}\times\dfrac{\sqrt2-\sqrt5}{\sqrt2-\sqrt5}\\ \\ =\dfrac{(\sqrt2-\sqrt5)^2}{(\sqrt2)^2-(\sqrt5)^2}=\dfrac{2+5-2\sqrt{10}}{-3}\\ \\ =\dfrac{7-2\sqrt{10}}{-3}\\ \\ =\dfrac{2\sqrt{10}-7}{3}=a+b\sqrt{10}\\ \\ a=\dfrac{-7}{3}\ ,\ b=\dfrac{2}{3}$
Question 17 of 37

17. Question
1 point(s)
If $\dfrac{3+\sqrt7}{3-\sqrt7}+\dfrac{3-\sqrt7}{3+\sqrt7}=a+b\sqrt7 $, find the values of a and b.
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Hint

$\dfrac{3+\sqrt7}{3-\sqrt7}+\dfrac{3-\sqrt7}{3+\sqrt7}=a+b\sqrt7\\ \\ \left(\dfrac{3+\sqrt7}{3-\sqrt7}\times\dfrac{3+\sqrt7}{3+\sqrt7}\right)+\dfrac{(3-\sqrt7)(3-\sqrt7)}{(3+\sqrt7)(3-\sqrt7)}\\ \\ \dfrac{(3+\sqrt7)^2}{2}+\dfrac{(3-\sqrt7)^2}{2}\\ \\ =\dfrac{9+7+6\sqrt7+9+7-6\sqrt7}{2}\\ \\ =\dfrac{16+16}{2}=16\\ \\ a=16,\ b=0$
Question 18 of 37

18. Question
1 point(s)
Find the value of $a $, if $\dfrac{3-\sqrt5}{3+2\sqrt5}=a\sqrt5-\dfrac{19}{11} $
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Hint

$\dfrac{3-\sqrt5}{3+2\sqrt5}\times\dfrac{3-2\sqrt5}{3-2\sqrt5}\\ \\ =\dfrac{9-6\sqrt5-3\sqrt5+10}{9-20}=\dfrac{19-9\sqrt5}{-11}\\ \\ =\dfrac{9\sqrt5-19}{11}\\ \\ \dfrac{9\sqrt5-19}{11}=\dfrac{a\sqrt5-19}{11}\\ \\ a=9$
Question 19 of 37

19. Question
1 point(s)
If $x=2+\sqrt5 $ find the value of $\left(x^2-\dfrac{1}{x^2}\right) $
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Hint

$x=2+\sqrt5\\ \\ x^2=(2+\sqrt5)^2=4+5+4\sqrt5\\ \\ =9+4\sqrt5\\ \\ \dfrac{1}{x^2}=\dfrac{1}{9+4\sqrt5}\times\dfrac{9-4\sqrt5}{9-4\sqrt5}\\ \\ =\dfrac{9-4\sqrt5}{81-80}=9-4\sqrt5\\ \\ (9+4\sqrt5)-(9-4\sqrt5)\\ \\ =9+4\sqrt5-9+4\sqrt5=8\sqrt5$
Question 20 of 37

20. Question
1 point(s)
If $x=\dfrac{\sqrt5-\sqrt2}{\sqrt5+\sqrt2} $ and $y=\dfrac{\sqrt5+\sqrt2}{\sqrt5-\sqrt2} $, find the value of $x^2+xy+y^2 $
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Hint

$x=\dfrac{\sqrt5-\sqrt2}{\sqrt+\sqrt2}\times\dfrac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}\\ \\ =\dfrac{(\sqrt5-\sqrt2)^2}{5-2}=\dfrac{5+2-2\sqrt{10}}{3}\\ \\ =\dfrac{7-2\sqrt{10}}{3}\\ \\ y=\dfrac{\sqrt5+\sqrt2}{\sqrt5-\sqrt2}\times\dfrac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}\\ \\ =\dfrac{(\sqrt5+\sqrt2)^2}{3}=\dfrac{5+2+2\sqrt{10}}{3}\\ \\ =\dfrac{7+2\sqrt{10}}{3}\\ \\ x^2+xy+y^2\\ \\ =\left(\dfrac{7-2\sqrt{10}}{3}\right)^2+\left(\dfrac{7-2\sqrt{10}}{3}\right)\left(\dfrac{7+2\sqrt{10}}{3}\right)+\left(\dfrac{7+2\sqrt{10}}{3}\right)^2\\ \\ =\dfrac{89-28\sqrt{10}}{9}+\dfrac{49-40}{9}+\dfrac{89+28\sqrt{10}}{9}\\ \\ =\dfrac{187}{9}$
Question 21 of 37

21. Question
1 point(s)
Simplify : $\dfrac{\sqrt6}{\sqrt2+\sqrt3}+\dfrac{3\sqrt2}{\sqrt6+\sqrt3}-\dfrac{4\sqrt3}{\sqrt6+\sqrt2} $.
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Hint

$\dfrac{\sqrt6}{\sqrt2+\sqrt3}+\dfrac{3\sqrt2}{\sqrt6+\sqrt3}-\dfrac{4\sqrt3}{\sqrt6+\sqrt2}\\ \\ =\dfrac{\sqrt6\times(\sqrt2-\sqrt3)}{(\sqrt2)^2-(\sqrt3)^2}+\dfrac{3\sqrt2}{\sqrt6+\sqrt3}\times\dfrac{\sqrt6-\sqrt3}{\sqrt6-\sqrt3}\\ \\ \dfrac{-4\sqrt3}{\sqrt6+\sqrt2}\times\dfrac{\sqrt6-\sqrt2}{\sqrt6-\sqrt2}\\ \\ =\dfrac{\sqrt{12}-\sqrt{18}}{-1}+\dfrac{3\sqrt{12}-3\sqrt3}{3}\\ \\ -\left(\dfrac{4\sqrt{18}+4\sqrt6}{6-2}\right)\\ \\ =-\sqrt{12}+\sqrt{18}+\sqrt{12}-\sqrt{3}-\sqrt{18}-\sqrt6\\ \\ =-\sqrt3-\sqrt6$
Question 22 of 37

22. Question
1 point(s)
Express $\dfrac{3}{\sqrt3-\sqrt2+\sqrt5} $ with rational denominator
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Hint

$\dfrac{3}{\sqrt3-\sqrt2+\sqrt5}\\ \\ =\dfrac{3}{\sqrt3+\sqrt5-\sqrt2}=\dfrac{3}{(\sqrt3+\sqrt5)-\sqrt2}\times\dfrac{(\sqrt3+\sqrt5)+\sqrt2}{(\sqrt3+\sqrt5)+\sqrt2}\\ \\ =\dfrac{3(\sqrt2+\sqrt3+\sqrt5)}{(\sqrt3+\sqrt5)^2-2}\\ \\ =\dfrac{3(\sqrt2+\sqrt3+\sqrt5)}{3+5+2\sqrt{15}-2}\\ \\ =\dfrac{3(\sqrt2+\sqrt3+\sqrt5)}{6+2\sqrt{15}}\\ \\ =\dfrac{3(\sqrt2+\sqrt3+\sqrt5)}{2(3+\sqrt{15})}\\ \\ =\dfrac{3}{2}\times\dfrac{\sqrt2+\sqrt3+\sqrt5}{3+\sqrt{15}}\times\dfrac{3-\sqrt{15}}{3-\sqrt{15}}\\ \\ =\dfrac{3}{2}\dfrac{(\sqrt2+\sqrt3+\sqrt5)\ (3-\sqrt{15})}{-6}\\ \\ =\dfrac{-1}{4}(\sqrt2+\sqrt3+\sqrt5)\ (3-\sqrt{15})$
Question 23 of 37

23. Question
1 point(s)
If $x=7+4\sqrt3 $ find the value of $\left(x^3+\dfrac{1}{x^3}\right) $.
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Hint

$x=7+4\sqrt3\\ \\ \dfrac{1}{x}=\dfrac{1}{7+4\sqrt3}\times\dfrac{7-4\sqrt3}{7-4\sqrt3}\\ \\ =\dfrac{7-4\sqrt3}{49-48}=7-4\sqrt3\\ \\ x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\\ \\ x+\dfrac{1}{x}=7+4\sqrt3+7-4\sqrt3=14\\ \\ (14)^3-3(14)=2702$
Question 24 of 37

24. Question
1 point(s)
If $x=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2} $ and $y=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2} $, then find the value of $x^2+y^2-10xy $.
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Hint

$x=\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},\ y=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\\ \\ x=\dfrac{(\sqrt3+\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\\ \\ =\dfrac{3+2+2\sqrt6}{1}=5+2\sqrt6\\ \\ y=\dfrac{(\sqrt3-\sqrt2)^2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\\ \\ x^2=(5+2\sqrt6)^2=49+20\sqrt6\\ \\ y^2=49-20\sqrt6\\ \\ x^2+y^2-10xy\\ \\ xy=\left(\dfrac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\right)\ \left(\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\right)\\ \\ =\dfrac{3-2}{3-2}=1\\ \\ 49+20\sqrt6+49-20\sqrt6-10=88$
Question 25 of 37

25. Question
1 point(s)
Find the square root of 4.2 geometrically.
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Hint

Draw AC = 4.2
Take mid point as O
now draw a circle of OA = OC radius from centre O
Now take a point B from O at OB $=\dfrac{4.2-1}{2}=\dfrac{3.2}{2}=1.6\$ distance and make perpendicular D
Now BD is $\sqrt{4.2}$ measure it
Question 26 of 37

26. Question
1 point(s)
Show that $\dfrac{1}{1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}=-1+\sqrt3 $.
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Hint

$\dfrac{1}{1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}=-1+\sqrt3\\ \\ \dfrac{1}{1+\sqrt2}\times\dfrac{1-\sqrt2}{1-\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}\times\dfrac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}\\ \\ =\dfrac{1-\sqrt2}{-1}+\dfrac{\sqrt2-\sqrt3}{-1}\\ \\ =\sqrt2-1+\sqrt3-\sqrt2=-1+\sqrt3$
Question 27 of 37

27. Question
1 point(s)
If $x=3+2\sqrt2 $, find the value of $\left(\sqrt x-\dfrac{1}{\sqrt x}\right) $
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Hint

$x=3+2\sqrt2\\ \\ =1+2\sqrt2+2\\ \\ =(1+\sqrt2)^2\\ \\ \sqrt x=1+\sqrt2\\ \\ \sqrt x-\dfrac{1}{\sqrt x}=(1+\sqrt2)-\dfrac{1}{1+\sqrt2}\\ \\ =1+\sqrt2-\dfrac{1}{1+\sqrt2}\times\dfrac{1-\sqrt2}{1-\sqrt2}\\ \\ =1+\sqrt2-\left(\dfrac{1-\sqrt2}{-1}\right)\\ \\ =1+\sqrt2-(\sqrt2-1)\\ \\ =1+\sqrt2-\sqrt2+1=2$
Question 28 of 37

28. Question
1 point(s)
Find the square root of 4.5 geometrically.
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Hint

Step 1: Draw a number line. Mark O as the zero on the number line.
Step 2: Mark a point A as -4.5 on the number line.
Step 3: Mark a point C as 1 on the number line.
Step 4: Mark a point B as the mid-point of AC.
Step 5: With point B as the centre and radius as AB draw a semicircle.
Step 6: From O draw a perpendicular line to the number line that intersects the previous drawn semi-circle at D. Here OD = square root of 4.5.
Step 7: With O as centre and radius as OD, draw an arc that intersects the number line at point E. Here E is the point square root of 4.5 on the number line.
Question 29 of 37

29. Question
1 point(s)
If $a = 2, b = 3$, then find the values of the following
(A) $(a^b+b^a)^{-1} $ (B) $(a^a+b^b)^{-1} $
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Hint

$a=2,b=3\\ \\ (2^3+3^2)^{-1}\\ \\ \text{a} \ =(8+9)^{-1}=17^{-1}=\dfrac{1}{17}\ \text{b}\ (a^a+b^b)^{-1}\\ \\ =(2^2+3^3)^{-1}=(4+27)^{-1}=(31)^{-1}=\dfrac{1}{31}$
Question 30 of 37

30. Question
1 point(s)
Prove that $\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}} =1$.
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Hint

$\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}\\ \\ =\dfrac{1}{1+\dfrac{x^a}{x^b}}+\dfrac{1}{1+\dfrac{x^b}{x^a}}\\ \\ =\dfrac{x^b}{x^b+x^a}+\dfrac{x^a}{x^a+x^b}=\dfrac{x^a+x^b}{x^a+x^b}=1$
Question 31 of 37

31. Question
1 point(s)
Express $2.02\bar5 $ in the form of $\dfrac{p}{q} $, where $p $ and $q $ are integers and $q\ne0 $
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Hint

$x=2.02\bar5\\ \\ 100x=202.\bar5\\ \\ 1000x=2025.\bar5\\ \\ 1000x-100x=2025.\bar5-202.\bar5\\ \\ 900x=1823\\ \\ x=\dfrac{1823}{900}$
Question 32 of 37

32. Question
1 point(s)
Express $0.6+0.\bar 7+0.4\bar7 $ in the form $\dfrac{p}{q} $, where $p,q\in Z,q\ne0 $.
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Hint

$0.6+0.\bar7+0.4\bar7$
Let $x=0.777$
$10x=7.77\\ \\ 10x-x=7\\ \\ 9x=7\Rightarrow x=\dfrac{7}{9}\\ \\ y=0.4777\\ \\ 10y=4.777\\ \\ 100y=47.77\\ \\ =100y-10y=47-4\\ \\ 90y=43\\ \\ y=\dfrac{43}{90}$
Required sum $=\dfrac{6}{10}+\dfrac{43}{90}+\dfrac{7}{9}=\dfrac{167}{90}$
Question 33 of 37

33. Question
1 point(s)
If $a=5+2\sqrt6 $ and $b=\dfrac{1}{a} $, then find the value of $a^2+b^2 $.
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Hint

$a=5+2\sqrt6\\ \\ b=\dfrac{1}{a}=\dfrac{1}{5+2\sqrt6}\times\dfrac{5-2\sqrt6}{5-2\sqrt6}\\ \\ =\dfrac{5-2\sqrt6}{25-24}=5-2\sqrt6 \\ \\ a^2+b^2=(5+2\sqrt6)^2+(5-2\sqrt6)^2\\ \\ =25+24+20\sqrt6+25+24-20\sqrt6\\ \\ =50+48=98$
Question 34 of 37

34. Question
1 point(s)
Find the value of : $\dfrac{1}{1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}+\dfrac{1}{\sqrt3+\sqrt4}+…..\dfrac{1}{\sqrt8+\sqrt9} $
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Hint

$\dfrac{1}{1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}+…+\dfrac{1}{\sqrt8+\sqrt9}\\ \\ =\dfrac{1}{1+\sqrt2}\times\dfrac{1-\sqrt2}{1-\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}\times\dfrac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}+….+\dfrac{1}{\sqrt8+\sqrt9}\times\dfrac{\sqrt8-\sqrt9}{\sqrt8-\sqrt9}\\ \\ =\dfrac{1-\sqrt2}{-1}+\dfrac{\sqrt2-\sqrt3}{-1}+\dfrac{\sqrt3-\sqrt4}{-1}+\dfrac{\sqrt4-\sqrt5}{-1}+\dfrac{\sqrt5-\sqrt6}{-1}+\dfrac{\sqrt6-\sqrt7}{-1}+\dfrac{\sqrt7-\sqrt8}{-1}+\dfrac{\sqrt8-\sqrt9}{-1}\\ \\ -1+\sqrt2-\sqrt2+\sqrt3-\sqrt3+\sqrt4-\sqrt4+\sqrt5-\sqrt5+\sqrt6-\sqrt6+\sqrt7-\sqrt7+\sqrt8-\sqrt8+\sqrt9\\ \\ =-1+\sqrt9=-1+3=2$
Question 35 of 37

35. Question
1 point(s)
Find the values of $a $ and $b $ if : $\dfrac{7+3\sqrt5}{3+\sqrt5}-\dfrac{7-3\sqrt5}{3-\sqrt5}=a+\sqrt{5b} $
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Hint

$\dfrac{7+3\sqrt5}{3+\sqrt5}-\dfrac{7-3\sqrt5}{3-\sqrt5}\\ \\ =\dfrac{(7+3\sqrt5)(3-\sqrt5)}{9-5}-\dfrac{(7-3\sqrt5)(3+\sqrt5)}{9-5}\\ \\ \dfrac{=21-7\sqrt5+9\sqrt5-15-21-7\sqrt5+9\sqrt5+15}{4}\\ \\ =\dfrac{4\sqrt5}{4}=a+\sqrt5b\\ \\ \Rightarrow a=0,b=1$
Question 36 of 37

36. Question
1 point(s)
If $a=7-4\sqrt3 $, find the value of $\left(\sqrt a+\dfrac{1}{\sqrt a}\right) $
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Hint

$a=7-4\sqrt3\\ \\ \sqrt a+\dfrac{1}{\sqrt a}\\ \\ \dfrac{1}{a}=7+4\sqrt3\\ \\ a+\dfrac{1}{a}=7-4\sqrt3+7+4\sqrt3=14\\ \\ \left(\sqrt a+\dfrac{1}{\sqrt a}\right)^2=\sqrt a^2+\left(\dfrac{1}{\sqrt a}\right)^2+2\\ \\ =a+\dfrac{1}{a}+2=16\\ \\ \sqrt a+\dfrac{1}{\sqrt a}=4$
Question 37 of 37

37. Question
1 point(s)
Represent $\sqrt{9.3} $ on the number line.
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Hint

Steps:
1) Draw a line segment AB of length 9.3 units.
2) Extend the line by 1 unit more such that BC=1 unit .
3) Find the midpoint of AC.
4) Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.
5) Draw an arc DE such that BE = BD.
Therefore, BE = $\sqrt{9.3}$ units