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NCERT Exemplar · Class 9 Mathematics Number Systems

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EXERCISE 1.1 1–10 (part 1 of 6)

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

    Exercise 1

    Every rational number is (A) a natural number (B) an integer (C) a real number (D) a whole number

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    NCERT’s answer
    (C)
    (C) a real number. \[\mathbb{Q} \subset \mathbb{R} \] A rational number such as \(\displaystyle \tfrac12\) is not natural, not an integer, and not a whole number, so (A), (B), (D) fail.

  2. Exercise 2

    Between two rational numbers (A) there is no rational number (B) there is exactly one rational number (C) there are infinitely many rational numbers (D) there are only rational numbers and no irrational numbers

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    NCERT’s answer
    (C)
    (C) there are infinitely many rational numbers. \[a < \frac{a+b}{2} < b \quad \text{for rationals } a<b \] Repeating this mean construction between \(\displaystyle a\) and each new midpoint never stops, so no two distinct rationals are ever adjacent.

  3. Exercise 3

    Decimal representation of a rational number cannot be (A) terminating (B) non-terminating (C) non-terminating repeating (D) non-terminating non-repeating

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    NCERT’s answer
    (D)
    (D) non-terminating non-repeating. Every rational number's decimal expansion either terminates or repeats; a non-terminating non-repeating decimal is exactly the signature of an irrational number, so a rational can never have one.

  4. Exercise 4

    The product of any two irrational numbers is (A) always an irrational number (B) always a rational number (C) always an integer (D) sometimes rational, sometimes irrational

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    NCERT’s answer
    (D)
    (D) sometimes rational, sometimes irrational. \[\sqrt2 \times \sqrt2 = 2 \quad \text{(rational)} \] \[\sqrt2 \times \sqrt3 = \sqrt6 \quad \text{(irrational)} \] Both outcomes occur, so no single case covers every pair.

  5. Exercise 5

    The decimal expansion of the number 2\displaystyle \sqrt{2} is (A) a finite decimal (B) 1.41421\displaystyle 1.41421 (C) non-terminating recurring (D) non-terminating non-recurring

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    NCERT’s answer
    (D)
    (D) non-terminating non-recurring. \(\displaystyle \sqrt2\) is irrational, and every irrational number's decimal expansion is non-terminating and non-recurring; \(\displaystyle 1.41421\ldots\) never settles into a repeating block.

  6. Exercise 6

    Which of the following is irrational? (A) 49\displaystyle \sqrt{\frac{4}{9}} (B) 123\displaystyle \frac{\sqrt{12}}{\sqrt{3}} (C) 7\displaystyle \sqrt{7} (D) 81\displaystyle \sqrt{81}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle \sqrt7\). \[\sqrt{\tfrac49} = \tfrac23, \qquad \frac{\sqrt{12}}{\sqrt3} = \sqrt4 = 2, \qquad \sqrt{81} = 9 \] all rational, while \(\displaystyle 7\) is not a perfect square, so \(\displaystyle \sqrt7\) is irrational.

  7. Exercise 7

    Which of the following is irrational? (A) 0.14\displaystyle 0.14 (B) 0.1416‾\displaystyle 0.14 \overline{16} (C) 0.1416‾\displaystyle 0 . \overline{1416} (D) 0.4014001400014\displaystyle 0.4014001400014…

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 0.4014001400014\ldots\). \[0.14, \quad 0.14\overline{16}, \quad 0.\overline{1416} \] all terminate or repeat, hence are rational; the fourth decimal never repeats a fixed block, so it is irrational.

  8. Exercise 8

    A rational number between 2\displaystyle \sqrt{2} and 3\displaystyle \sqrt{3} is (A) 2+32\displaystyle \frac{\sqrt{2}+\sqrt{3}}{2} (B) 2⋅32\displaystyle \frac{\sqrt{2} \cdot \sqrt{3}}{2} (C) 1.5\displaystyle 1.5 (D) 1.8\displaystyle 1.8

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 1.5\). \[\sqrt2 \approx 1.414, \qquad \sqrt3 \approx 1.732 \] \[1.414 < 1.5 < 1.732 \] and \(\displaystyle 1.5\) is rational; \(\displaystyle 1.8\) exceeds \(\displaystyle \sqrt3\), while (A) and (B) are irrational.

  9. Exercise 9

    The value of 1.999…\displaystyle 1.999 \ldots 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, is (A) 1910\displaystyle \frac{19}{10} (B) 19991000\displaystyle \frac{1999}{1000} (C) 2\displaystyle 2 (D) 19\displaystyle \frac{1}{9}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 2\). \[x = 1.999\ldots \] \[10x = 19.999\ldots \] \[10x - x = 19.999\ldots - 1.999\ldots \] \[9x = 18 \implies x = 2 \]

  10. Exercise 10

    23+3\displaystyle 2 \sqrt{3}+\sqrt{3} is equal to (A) 26\displaystyle 2 \sqrt{6} (B) 6\displaystyle 6 (C) 33\displaystyle 3 \sqrt{3} (D) 46\displaystyle 4 \sqrt{6}

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 3\sqrt3\). \[2\sqrt3 + \sqrt3 = (2+1)\sqrt3 = 3\sqrt3 \]