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This is the second of our many quizzes on rational numbers. Click on this to access more such competitive quizzes on rational and irrational numbers.

Rational Numbers Quiz 2

mathematics

 If $p = \sqrt 7 $,$q = \sqrt {343} $and $r = \sqrt {2401} $, then which of the following is rational number

 Evaluate and find which number can be simplified in the form $\dfrac{p}{q}$.

  $\text{As, }pq = \sqrt {7 \times 343}  = \sqrt {2401}  = {({7^4})^{\frac{1}{2}}} = {7^2}\text{ which this rational }\\pr = \sqrt {7 \times 2401}  = \sqrt {16807}  = {({7^5})^{\frac{1}{2}}} = {7^{\frac{5}{2}}}\text{ which is not rational }\\ qr = \sqrt {343 \times 2401}  = \sqrt {{7^7}}  = {({7^7})^{\frac{1}{2}}} = {7^{\frac{7}{2}}}\text{ which is not rational. }\\\text{Hence, “pq” will be a rational number.}$

 Evaluate $\dfrac{1}{{1 + \dfrac{{\dfrac{3}{4}}}{{1 + \dfrac{3}{4} + \dfrac{{\dfrac{3}{4}}}{{1 - \dfrac{3}{4}}}}}}}$

 Take the LCM of both the terms that you are adding.

  $\dfrac{1}{{1 + \dfrac{{\dfrac{3}{4}}}{{1 + \dfrac{3}{4} + \dfrac{{\dfrac{3}{4}}}{{1 - \dfrac{3}{4}}}}}}}\\  \Rightarrow \dfrac{1}{{1 + \dfrac{{\dfrac{3}{4}}}{{1 + \dfrac{{19}}{4}}}}} \\  \Rightarrow \dfrac{1}{{1 + \dfrac{{\dfrac{3}{4}}}{{\dfrac{{19}}{4}}}}} \\  \Rightarrow \dfrac{1}{{1 + \dfrac{3}{{19}}}} \\  \Rightarrow \dfrac{{19}}{{22}}$ 

 Arrange A, B, C, D in increasing order, where A, B, C, D are $\dfrac{{ - 5}}{6},0,\dfrac{3}{8},\dfrac{6}{4}$ respectively.

 Evaluate all the options and arrange them in increasing order of magnitude.

 $\text{Decimal value of numbers is: }\\ A =  - \dfrac{5}{6} =  - 0.83 \\  B = 0 \\  C = \dfrac{3}{8} = 0.37 \\  D = \dfrac{6}{4} = 1.5\\\text{Hence, correct order of arrangement will be }D > C > B > A$

 If $a = {( - 1)^{305}}$ and $b = {( - 1)^{306}}$, then $a - b = ?$ 

 As we know that $( - a)^{\text{odd number}} =  - a$ and $( - a)^{\text{even number}} = a$, Use this and evaluate the expression. 

 $a = {( - 1)^{305}} =  - 1\text{ because }{( - a)^{\text{odd number}}} =  - a\\ \text{ and }b = {( - 1)^{306}} = 1,\text{ because }{( - a)^{\text{even number}}} = a\\ \text{ So, } (a - b) = \left[ { - 1 - \left( 1 \right)} \right] \\  \Rightarrow \left( { - 1 - 1} \right) = ( - 2)$ 

 If $\dfrac{{9 + 6\sqrt 2 }}{{9 - 6\sqrt 2 }} = a + b\sqrt 2 $, find the value of $a + b$

 Rationalize the denominator and use the identities $  \\ (a + b)(a + b) = {(a + b)^2} \\ (a + b)(a - b) = {a^2} - {b^2}$ 

 $\dfrac{{9 + 6\sqrt 2 }}{{9 - 6\sqrt 2 }} = a + b\sqrt 2 \\ \text{ Taking LHS and rationalize the denominator}\\ \dfrac{{(9 + 6\sqrt 2 ) \times (9 + 6\sqrt 2 )}}{{(9 - 6\sqrt 2 ) \times (9 + 6\sqrt 2 )}}  \\ \Rightarrow \dfrac{{\left[ {{{(9 + 6\sqrt 2 )}^2}} \right]}}{{\left[ {{{(9)}^2} - {{(6\sqrt 2 )}^2}} \right]}} \\ \Rightarrow \dfrac{{(81 + 72 + 108\sqrt 2 )}}{{81 - 72}} \\ \Rightarrow \dfrac{{153}}{9} + \dfrac{{108}}{9}\sqrt 2  \\ \therefore a = \dfrac{{153}}{9} = 17,b = \dfrac{{108}}{9} = 12 \\\text{ Hence, }a + b = 17 + 12 = 29$