SolveItNCERT · CBSE Boards

NCERT Exemplar · Class 9 Mathematics Number Systems

46 questions · 46 still being checked

EXERCISE 1.1 11–21 (part 2 of 6)

  1. Write the correct answer in each of the following:

    Exercise 11

    10×15\displaystyle \sqrt{10} \times \sqrt{15} is equal to (A) 65\displaystyle 6 \sqrt{5} (B) 56\displaystyle 5 \sqrt{6} (C) 25\displaystyle \sqrt{25} (D) 105\displaystyle 10 \sqrt{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
    (B)
    (B) \(\displaystyle 5\sqrt6\). \[\sqrt{10} \times \sqrt{15} = \sqrt{150} = \sqrt{25 \times 6} = 5\sqrt6 \]

  2. Exercise 12

    The number obtained on rationalising the denominator of 17−2\displaystyle \frac{1}{\sqrt{7}-2} is (A) 7+23\displaystyle \frac{\sqrt{7}+2}{3} (B) 7−23\displaystyle \frac{\sqrt{7}-2}{3} (C) 7+25\displaystyle \frac{\sqrt{7}+2}{5} (D) 7+245\displaystyle \frac{\sqrt{7}+2}{45}

    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
    (A)
    (A) \(\displaystyle \dfrac{\sqrt7+2}{3}\). \[\frac{1}{\sqrt7-2} \times \frac{\sqrt7+2}{\sqrt7+2} = \frac{\sqrt7+2}{(\sqrt7)^2-2^2} = \frac{\sqrt7+2}{7-4} = \frac{\sqrt7+2}{3} \]

  3. Exercise 13

    19−8\displaystyle \frac{1}{\sqrt{9}-\sqrt{8}} is equal to (A) 12(3−22)\displaystyle \frac{1}{2}(3-2 \sqrt{2}) (B) 13+22\displaystyle \frac{1}{3+2 \sqrt{2}} (C) 3−22\displaystyle 3-2 \sqrt{2} (D) 3+22\displaystyle 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
    (D)
    (D) \(\displaystyle 3+2\sqrt{2}\)Rationalise the denominator: \[\frac{1}{\sqrt9-\sqrt8}\cdot\frac{\sqrt9+\sqrt8}{\sqrt9+\sqrt8}=\frac{\sqrt9+\sqrt8}{9-8} \] \[=\sqrt9+\sqrt8=3+2\sqrt2 \]

  4. Exercise 14

    After rationalising the denominator of 733−22\displaystyle \frac{7}{3 \sqrt{3}-2 \sqrt{2}}, we get the denominator as (A) 13\displaystyle 13 (B) 19\displaystyle 19 (C) 5\displaystyle 5 (D) 35\displaystyle 35

    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
    (B)
    (B) \(\displaystyle 19\)Rationalise the denominator: \[\frac{7}{3\sqrt3-2\sqrt2}\cdot\frac{3\sqrt3+2\sqrt2}{3\sqrt3+2\sqrt2}=\frac{7(3\sqrt3+2\sqrt2)}{(3\sqrt3)^2-(2\sqrt2)^2} \] \[=\frac{7(3\sqrt3+2\sqrt2)}{27-8}=\frac{7(3\sqrt3+2\sqrt2)}{19} \]

  5. Exercise 15

    The value of 32+488+12\displaystyle \frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}} is equal to (A) 2\displaystyle \sqrt{2} (B) 2\displaystyle 2 (C) 4\displaystyle 4 (D) 8\displaystyle 8

    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
    (B)
    (B) \(\displaystyle 2\)\[\sqrt{32}=4\sqrt2,\quad \sqrt{48}=4\sqrt3,\quad \sqrt8=2\sqrt2,\quad \sqrt{12}=2\sqrt3 \] \[\frac{\sqrt{32}+\sqrt{48}}{\sqrt8+\sqrt{12}}=\frac{4(\sqrt2+\sqrt3)}{2(\sqrt2+\sqrt3)}=2 \]

  6. Exercise 16

    If 2=1.4142\displaystyle \sqrt{2}=1.4142, then 2−12+1\displaystyle \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} is equal to (A) 2.4142\displaystyle 2.4142 (B) 5.8282\displaystyle 5.8282 (C) 0.4142\displaystyle 0.4142 (D) 0.1718\displaystyle 0.1718

    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
    (C)
    (C) \(\displaystyle 0.4142\)\[\frac{\sqrt2-1}{\sqrt2+1}\cdot\frac{\sqrt2-1}{\sqrt2-1}=\frac{(\sqrt2-1)^2}{2-1}=(\sqrt2-1)^2 \] \[\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}=\sqrt2-1=1.4142-1=0.4142 \]

  7. Exercise 17

    2234\displaystyle \sqrt[4]{\sqrt[3]{2^{2}}} equals (A) 2−16\displaystyle 2^{-\frac{1}{6}} (B) 2−6\displaystyle 2^{-6} (C) 216\displaystyle 2^{\frac{1}{6}} (D) 26\displaystyle 2^{6}

    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
    (C)
    (C) \(\displaystyle 2^{\frac16}\)\[\sqrt[4]{\sqrt[3]{2^2}}=\left(2^{\frac23}\right)^{\frac14}=2^{\frac23\cdot\frac14}=2^{\frac16} \]

  8. Exercise 18

    The product 23⋅24⋅3212\displaystyle \sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} equals (A) 2\displaystyle \sqrt{2} (B) 2\displaystyle 2 (C) 212\displaystyle \sqrt[12]{2} (D) 3212\displaystyle \sqrt[12]{32}

    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
    (B)
    (B) \(\displaystyle 2\)\[\sqrt[3]{2}\cdot\sqrt[4]{2}\cdot\sqrt[12]{32}=2^{\frac13}\cdot2^{\frac14}\cdot2^{\frac{5}{12}} \] \[=2^{\frac13+\frac14+\frac{5}{12}}=2^{\frac{4+3+5}{12}}=2^1=2 \]

  9. Exercise 19

    Value of (81)−24\displaystyle \sqrt[4]{(81)^{-2}} is (A) 19\displaystyle \frac{1}{9} (B) 13\displaystyle \frac{1}{3} (C) 9\displaystyle 9 (D) 181\displaystyle \frac{1}{81}

    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
    (A)
    (A) \(\displaystyle \frac19\)\[\sqrt[4]{(81)^{-2}}=(81)^{-\frac24}=(81)^{-\frac12}=\frac{1}{\sqrt{81}} \] \[=\frac19 \]

  10. Exercise 20

    Value of (256)0.16×(256)0.09\displaystyle (256)^{0.16} \times(256)^{0.09} is (A) 4\displaystyle 4 (B) 16\displaystyle 16 (C) 64\displaystyle 64 (D) 256.25\displaystyle 256.25

    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
    (A)
    (A) \(\displaystyle 4\)\[(256)^{0.16}\times(256)^{0.09}=(256)^{0.25}=(256)^{\frac14} \] \[256=4^4 \implies (256)^{\frac14}=4 \]

  11. Exercise 21

    Which of the following is equal to x\displaystyle x ? (A) x127−x57\displaystyle x^{\frac{12}{7}}-x^{\frac{5}{7}} (B) (x4)1312\displaystyle \sqrt[12]{\left(x^{4}\right)^{\frac{1}{3}}} (C) (x3)23\displaystyle \left(\sqrt{x^{3}}\right)^{\frac{2}{3}} (D) x127×x712\displaystyle x^{\frac{12}{7}} \times x^{\frac{7}{12}}

    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
    (C)
    (C) \(\displaystyle \left(\sqrt{x^3}\right)^{\frac23}\)\[\left(\sqrt{x^3}\right)^{\frac23}=\left(x^{\frac32}\right)^{\frac23}=x^{\frac32\cdot\frac23}=x^1=x \] The other three options' exponents do not simplify to \(\displaystyle 1\).