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NCERT Exemplar · Class 10 Mathematics Real Numbers

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EXERCISE 1.2 1–10 (part 2 of 5)

  1. Exercise 1

    Write whether every positive integer can be of the form 4q+2\displaystyle 4 q+2, where q\displaystyle q is an integer. Justify your answer.

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    NCERT’s answer
    No, because an integer can be written in the form \(\displaystyle 4 q, 4 q+1,4 q+2,4 q+3\).
    No. \[n = 4q + r, \quad r \in \{0,1,2,3\} \quad \text{(Euclid's division lemma, divisor 4)} \] Only \(\displaystyle r=2\) gives the form \(\displaystyle 4q+2\); the other three residues are excluded. \[1 = 4(0)+1 \] \(\displaystyle 1\) is a positive integer not of that form.

  2. Exercise 2

    "The product of two consecutive positive integers is divisible by 2\displaystyle 2". Is this statement true or false? Give reasons.

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    NCERT’s answer
    True, because \(\displaystyle n(n+1)\) will always be even, as one out of \(\displaystyle n\) or \(\displaystyle (n+1)\) must be even.
    True. \[n(n+1), \quad n = 2q \ \text{or} \ 2q+1 \quad \text{(Euclid's division lemma, divisor 2)} \] \[n=2q: \quad n(n+1) = 2q(2q+1) \] \[n=2q+1: \quad n(n+1) = (2q+1)\cdot 2(q+1) \] Both cases carry a factor of \(\displaystyle 2\).

  3. Exercise 3

    "The product of three consecutive positive integers is divisible by 6\displaystyle 6". Is this statement true or false"? Justify your answer.

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    NCERT’s answer
    True, because \(\displaystyle n(n+1)(n+2)\) will always be divisible by $\displaystyle 6$, as atleast one of the factors will be divisible by $\displaystyle 2$ and atleast one of the factors will be divisible by 3.
    True. \[n(n+1)(n+2), \quad n = 3q,\ 3q+1,\ 3q+2 \quad \text{(Euclid's division lemma, divisor 3)} \] \[n=3q: \quad 3 \mid n \] \[n=3q+1: \quad n+2 = 3(q+1) \] \[n=3q+2: \quad n+1 = 3(q+1) \] One of \(\displaystyle n, n+1\) is even, so \(\displaystyle 2 \mid n(n+1)(n+2)\) in every case. \[\gcd(2,3) = 1 \ \Rightarrow\ 6 \mid n(n+1)(n+2) \]

  4. Exercise 4

    Write whether the square of any positive integer can be of the form 3m+2\displaystyle 3 m+2, where m\displaystyle m is a natural number. Justify your answer.

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    NCERT’s answer
    No. Since any positive integer can be written as \(\displaystyle 3 q, 3 q+1,3 q+2\), therefore, square will be \(\displaystyle 9 q^2=3 m, 9 q^2+6 q+1=3\left(3 q^2+2 q\right)+1=3 m+1\), \(\displaystyle 9 q^2+12 q+3+1=3 m+1\).
    No. \[a = 3q,\ 3q+1,\ 3q+2 \quad \text{(Euclid's division lemma, divisor 3)} \] \[(3q)^2 = 3(3q^2) \] \[(3q+1)^2 = 3(3q^2+2q)+1 \] \[(3q+2)^2 = 3(3q^2+4q+1)+1 \] Every case leaves remainder \(\displaystyle 0\) or \(\displaystyle 1\), never \(\displaystyle 2\).

  5. Exercise 5

    A positive integer is of the form 3q+1,q\displaystyle 3 q+1, q being a natural number. Can you write its square in any form other than 3m+1\displaystyle 3 m+1, i.e., 3m\displaystyle 3 m or 3m+2\displaystyle 3 m+2 for some integer m\displaystyle m? Justify your answer.

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    NCERT’s answer
    No. \(\displaystyle (3 q+1)^2=9 q^2+6 q+1=3\left(3 q^2+2 q\right)=3 m+1\).
    No. \[a = 3q+1 \] \[a^2 = 9q^2+6q+1 = 3(3q^2+2q)+1 \] \[m = 3q^2+2q \quad \Rightarrow \quad a^2 = 3m+1 \] The square is always of the form \(\displaystyle 3m+1\), never \(\displaystyle 3m\) or \(\displaystyle 3m+2\).

  6. Exercise 6

    The numbers 525\displaystyle 525 and 3000\displaystyle 3000 are both divisible only by 3,5,15,25\displaystyle 3,5,15,25 and 75. What is HCF (525,3000)\displaystyle (525, 3000)? Justify your answer.

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    NCERT’s answer
    \(\displaystyle \mathrm{HCF}=75\), as HCF is the highest common factor.
    The common divisors of $\displaystyle 525$ and $\displaystyle 3000$ are exactly \(\displaystyle 3, 5, 15, 25, 75\); the HCF is the greatest of these. \[\text{HCF}(525, 3000) = 75 \] Answer: \(\displaystyle 75\)

  7. Exercise 7

    Explain why 3×5×7+7\displaystyle 3 \times 5 \times 7+7 is a composite number.

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    NCERT’s answer
    \(\displaystyle 3 \times 5 \times 7+7=7(3 \times 5+1)=7(16)\), which has more than two factors.
    \[3 \times 5 \times 7 + 7 = 7(3 \times 5 + 1) = 7 \times 16 \] Both factors exceed \(\displaystyle 1\), so the number has divisors besides \(\displaystyle 1\) and itself. Answer: composite, equal to \(\displaystyle 7 \times 16 = 112\)

  8. Exercise 8

    Can two numbers have 18\displaystyle 18 as their HCF and 380\displaystyle 380 as their LCM? Give reasons.

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    NCERT’s answer
    No, because HCF ($\displaystyle 18$) does not divide LCM ($\displaystyle 380$).
    \[\text{HCF} \mid \text{LCM} \quad \text{(property of HCF, LCM)} \] \[380 = 18 \times 21 + 2 \] $\displaystyle 18$ does not divide $\displaystyle 380$ exactly, so no two numbers can have HCF \(\displaystyle 18\) and LCM \(\displaystyle 380\). Answer: No such pair exists.

  9. Exercise 9

    Without actually performing the long division, find if 98710500\displaystyle \frac{987}{10500} will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

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    NCERT’s answer
    Terminating decimal expansion, because \(\displaystyle \frac{987}{10500}=\frac{47}{500}\) and \(\displaystyle 500=5^3 \times 2^2\) \[\left[\frac{987}{10500}=\frac{329}{3500}=\frac{329}{2^2 .5^3 .7}=\frac{47}{2^2 5^3}=.094 .\right] \]
    \[987 = 3 \times 7 \times 47, \quad 10500 = 2^2 \times 3 \times 5^3 \times 7 \] \[\gcd(987,10500) = 3 \times 7 = 21 \] \[\frac{987}{10500} = \frac{47}{500}, \quad 500 = 2^2 \times 5^3 \] In lowest terms the denominator has only \(\displaystyle 2\) and \(\displaystyle 5\) as prime factors. Answer: terminating decimal expansion.

  10. Exercise 10

    A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q\displaystyle q, when this number is expressed in the form pq\displaystyle \frac{p}{q} ? Give reasons.

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    NCERT’s answer
    Since $\displaystyle 327.7081$ is a terminating decimal number, so \(\displaystyle q\) must be of the form \(\displaystyle 2^m .5^n\); \(\displaystyle m, n\) are natural numbers.
    \[327.7081 = \frac{3277081}{10000}, \quad 10000 = 2^4 \times 5^4 \] \(\displaystyle 3277081\) is odd and does not end in \(\displaystyle 0\) or \(\displaystyle 5\), so it shares no factor with \(\displaystyle 10000\); the fraction is already in lowest terms. Answer: \(\displaystyle q\)'s only prime factors are \(\displaystyle 2\) and \(\displaystyle 5\).