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NCERT Exemplar · Class 10 Mathematics Real Numbers

39 questions · 39 still being checked

EXERCISE 1.1 1–10 (part 1 of 5)

  1. Choose the correct answer from the given four options in the following questions:

    Exercise 1

    For some integer m\displaystyle m, every even integer is of the form
    (A)
    m\displaystyle m (B) m+1\displaystyle m+1
    (C)
    2m\displaystyle 2 m
    (D)
    2m+1\displaystyle 2 m+1

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 2m\)\[\text{even integer} = 2m, \quad m \in \mathbb{Z} \]

  2. Exercise 2

    For some integer q\displaystyle q, every odd integer is of the form
    (A)
    q\displaystyle q (B) q+1\displaystyle q+1
    (C)
    2q\displaystyle 2 q
    (D)
    2q+1\displaystyle 2 q+1

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 2q+1\)\[\text{odd integer} = 2q+1, \quad q \in \mathbb{Z} \]

  3. Exercise 3

    n2−1\displaystyle n^2-1 is divisible by 8\displaystyle 8, if n\displaystyle n is
    (A)
    an integer
    (B)
    a natural number
    (C)
    an odd integer
    (D)
    an even integer

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    NCERT’s answer
    (C)
    (C) an odd integer\[n = 2k+1 \] \[n^2 - 1 = 4k(k+1) \] \[k(k+1) \equiv 0 \pmod{2} \] \[n^2 - 1 \equiv 0 \pmod{8} \]

  4. Exercise 4

    If the HCF of 65\displaystyle 65 and 117\displaystyle 117 is expressible in the form 65m−117\displaystyle 65 m-117, then the value of m\displaystyle m is
    (A)
    4\displaystyle 4 (B) 2\displaystyle 2 (C) 1\displaystyle 1 (D) 3\displaystyle 3

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 2\)\[117 = 65(1) + 52 \] \[65 = 52(1) + 13 \] \[52 = 13(4) + 0 \] \[\text{HCF}(65,117) = 13 \] \[65m - 117 = 13 \implies m = 2 \]

  5. Exercise 5

    The largest number which divides 70\displaystyle 70 and 125\displaystyle 125, leaving remainders 5\displaystyle 5 and 8\displaystyle 8, respectively, is
    (A)
    13\displaystyle 13
    (B)
    65\displaystyle 65
    (C)
    875\displaystyle 875
    (D)
    1750\displaystyle 1750

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 13\)\[70 - 5 = 65, \quad 125 - 8 = 117 \] \[117 = 65(1) + 52 \] \[65 = 52(1) + 13 \] \[52 = 13(4) + 0 \] \[\text{HCF}(65,117) = 13 \]

  6. Exercise 6

    If two positive integers a\displaystyle a and b\displaystyle b are written as a=x3y2\displaystyle a=x^3 y^2 and b=xy3;x,y\displaystyle b=x y^3 ; x, y are prime numbers, then HCF⁡(a,b)\displaystyle \operatorname{HCF}(a, b) is
    (A)
    xy\displaystyle x y
    (B)
    xy2\displaystyle x y^2
    (C)
    x3y3\displaystyle x^3 y^3
    (D)
    x2y2\displaystyle x^2 y^2

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    NCERT’s answer
    (B)
    (B) \(\displaystyle xy^2\) \[a = x^3y^2, \qquad b = xy^3 \] \[\operatorname{HCF}(a,b) = x^{\min(3,1)}\,y^{\min(2,3)} = xy^2 \]

  7. Exercise 7

    If two positive integers p\displaystyle p and q\displaystyle q can be expressed as p=ab2\displaystyle p=a b^2 and q=a3b;a,b\displaystyle q=a^3 b ; a, b being prime numbers, then LCM⁡(p,q)\displaystyle \operatorname{LCM}(p, q) is
    (A)
    ab\displaystyle a b
    (B)
    a2b2\displaystyle a^2 b^2
    (C)
    a3b2\displaystyle a^3 b^2
    (D)
    a3b3\displaystyle a^3 b^3

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    NCERT’s answer
    (C)
    (C) \(\displaystyle a^3b^2\) \[p = ab^2, \qquad q = a^3b \] \[\operatorname{LCM}(p,q) = a^{\max(1,3)}\,b^{\max(2,1)} = a^3b^2 \]

  8. Exercise 8

    The product of a non-zero rational and an irrational number is
    (A)
    always irrational
    (B)
    always rational
    (C)
    rational or irrational
    (D)
    one

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    NCERT’s answer
    (A)
    (A) always irrational Assume \(\displaystyle rs\) is rational for nonzero rational \(\displaystyle r\) and irrational \(\displaystyle s\). \[s = \frac{rs}{r} \] The right side is a ratio of rationals, so rational — contradicting \(\displaystyle s\) irrational.

  9. Exercise 9

    The least number that is divisible by all the numbers from 1\displaystyle 1 to 10\displaystyle 10 (both inclusive) is
    (A)
    10\displaystyle 10
    (B)
    100\displaystyle 100
    (C)
    504\displaystyle 504
    (D)
    2520\displaystyle 2520

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 2520\) \[1,2,3,4{=}2^2,5,6{=}2\cdot3,7,8{=}2^3,9{=}3^2,10{=}2\cdot5 \] \[\operatorname{LCM}(1,\dots,10) = 2^3\cdot3^2\cdot5\cdot7 = 2520 \]

  10. Exercise 10

    The decimal expansion of the rational number 145871250\displaystyle \frac{14587}{1250} will terminate after:
    (A)
    one decimal place
    (B)
    two decimal places
    (C)
    three decimal places
    (D)
    four decimal places

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    NCERT’s answer
    (D)
    (D) four decimal places \[\frac{14587}{1250} = \frac{14587}{2\cdot5^4} = \frac{14587\times2^3}{2^4\cdot5^4} = \frac{116696}{10^4} = 11.6696 \]