SolveItNCERT · CBSE Boards

NCERT Exemplar · Class 9 Mathematics Number Systems

46 questions · 46 still being checked

EXERCISE 1.4 1–7 (part 6 of 6)

  1. Exercise 1

    Express 0.6+0.7‾+0.47‾\displaystyle 0.6+0 . \overline{7}+0.4 \overline{7} in the form pq\displaystyle \frac{p}{q}, where p\displaystyle p and q\displaystyle q are integers and q≠0\displaystyle q \neq 0.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle \frac{167}{90}$
    Let \(\displaystyle x = 0.\overline{7} \) and \(\displaystyle y = 0.4\overline{7} \).\[10x - x = 7.\overline{7} - 0.\overline{7} = 7 \implies x = \frac{7}{9} \]\[100y - 10y = 47.\overline{7} - 4.\overline{7} = 43 \implies y = \frac{43}{90} \]\[0.6 + x + y = \frac{3}{5} + \frac{7}{9} + \frac{43}{90} = \frac{54+70+43}{90} \]Answer: \(\displaystyle \dfrac{167}{90} \)

  2. Exercise 2

    Simplify: 7310+3−256+5−3215+32\displaystyle \frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 1$
    Rationalise each term separately.\[\frac{7\sqrt3}{\sqrt{10}+\sqrt3}\cdot\frac{\sqrt{10}-\sqrt3}{\sqrt{10}-\sqrt3} = \frac{7\sqrt3(\sqrt{10}-\sqrt3)}{10-3} = \sqrt{30}-3 \]\[\frac{2\sqrt5}{\sqrt6+\sqrt5}\cdot\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5} = \frac{2\sqrt5(\sqrt6-\sqrt5)}{6-5} = 2\sqrt{30}-10 \]\[\frac{3\sqrt2}{\sqrt{15}+3\sqrt2}\cdot\frac{\sqrt{15}-3\sqrt2}{\sqrt{15}-3\sqrt2} = \frac{3\sqrt2(\sqrt{15}-3\sqrt2)}{15-18} = 6-\sqrt{30} \]\[(\sqrt{30}-3)-(2\sqrt{30}-10)-(6-\sqrt{30}) = 1 \]Answer: \(\displaystyle 1\)

  3. Exercise 3

    If 2=1.414,3=1.732\displaystyle \sqrt{2}=1.414, \sqrt{3}=1.732, then find the value of 433−22+333+22\displaystyle \frac{4}{3 \sqrt{3}-2 \sqrt{2}}+\frac{3}{3 \sqrt{3}+2 \sqrt{2}}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    2.$\displaystyle 063$
    \[(3\sqrt3)^2-(2\sqrt2)^2 = 27-8=19 \]\[\frac{4}{3\sqrt3-2\sqrt2}+\frac{3}{3\sqrt3+2\sqrt2} = \frac{4(3\sqrt3+2\sqrt2)+3(3\sqrt3-2\sqrt2)}{19} = \frac{21\sqrt3+2\sqrt2}{19} \]Substitute \(\displaystyle \sqrt3=1.732,\ \sqrt2=1.414 \):\[\frac{21(1.732)+2(1.414)}{19} = \frac{36.372+2.828}{19} = \frac{39.200}{19} \]Answer: \(\displaystyle \approx 2.063\)

  4. Exercise 4

    If a=3+52\displaystyle a=\frac{3+\sqrt{5}}{2}, then find the value of a2+1a2\displaystyle a^{2}+\frac{1}{a^{2}}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 7$
    \[a=\frac{3+\sqrt5}{2} \implies \frac1a=\frac{2}{3+\sqrt5}\cdot\frac{3-\sqrt5}{3-\sqrt5}=\frac{3-\sqrt5}{2} \]\[a+\frac1a = \frac{3+\sqrt5}{2}+\frac{3-\sqrt5}{2}=3 \]\[a^2+\frac{1}{a^2}=\left(a+\frac1a\right)^2-2 = 3^2-2 \]Answer: \(\displaystyle 7\)

  5. Exercise 5

    If x=3+23−2\displaystyle x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} and y=3−23+2\displaystyle y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}, then find the value of x2+y2\displaystyle x^{2}+y^{2}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 98$
    Note \(\displaystyle y=\dfrac1x \), since \(\displaystyle xy=1 \).\[x=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\frac{(\sqrt3+\sqrt2)^2}{3-2}=5+2\sqrt6 \]\[y=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\cdot\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}=\frac{(\sqrt3-\sqrt2)^2}{3-2}=5-2\sqrt6 \]\[x+y=10, \qquad xy=(5+2\sqrt6)(5-2\sqrt6)=25-24=1 \]\[x^2+y^2=(x+y)^2-2xy=10^2-2(1) \]Answer: \(\displaystyle 98\)

  6. Exercise 6

    Simplify : (256)−(4−32)\displaystyle (256)^{-\left(4^{\frac{-3}{2}}\right)}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle \frac{1}{2}$
    \[4^{-3/2} = \left(2^2\right)^{-3/2} = 2^{-3} = \frac18 \]\[(256)^{-\left(4^{-3/2}\right)} = (2^8)^{-1/8} = 2^{8\cdot(-1/8)} = 2^{-1} \]Answer: \(\displaystyle \dfrac12 \)

  7. Exercise 7

    Find the value of 4(216)−23+1(256)−34+2(243)−15\displaystyle \frac{4}{(216)^{-\frac{2}{3}}}+\frac{1}{(256)^{-\frac{3}{4}}}+\frac{2}{(243)^{-\frac{1}{5}}}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 214$
    Using \(\displaystyle a^{-n}=\dfrac{1}{a^{n}}\), each term becomes a positive power of a perfect base. \[\frac{4}{(216)^{-2/3}}=4\cdot(216)^{2/3}=4\cdot(6^{3})^{2/3}=4\cdot 6^{2}=144 \] \[\frac{1}{(256)^{-3/4}}=(256)^{3/4}=(2^{8})^{3/4}=2^{6}=64 \] \[\frac{2}{(243)^{-1/5}}=2\cdot(243)^{1/5}=2\cdot(3^{5})^{1/5}=2\cdot 3=6 \] Adding the three terms, \[144+64+6=214 \] Answer: \(\displaystyle 214\)