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NCERT Exemplar · Class 10 Mathematics Real Numbers

39 questions · 39 still being checked

EXERCISE 1.3 11–14 (part 4 of 5)

  1. Exercise 11

    Show that 12n\displaystyle 12^n cannot end with the digit 0\displaystyle 0 or 5\displaystyle 5 for any natural number n\displaystyle n.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[12^n = (2^2\times3)^n = 2^{2n}\times3^n \] A number ending in $\displaystyle 0$ or $\displaystyle 5$ is divisible by 5. By the Fundamental Theorem of Arithmetic the prime factorisation of \(\displaystyle 12^n\) is unique and contains only the primes $\displaystyle 2$ and 3. \[5 \nmid 2^{2n}\times3^n \] So \(\displaystyle 5\) never divides \(\displaystyle 12^n\).Answer: \(\displaystyle 12^n\) cannot end in $\displaystyle 0$ or $\displaystyle 5$, for any natural number \(\displaystyle n\).

  2. Exercise 12

    On a morning walk, three persons step off together and their steps measure 40\displaystyle 40 cm, 42\displaystyle 42 cm and 45\displaystyle 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

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    NCERT’s answer
    $\displaystyle 2520$ cm
    The minimum common distance is \(\displaystyle \text{LCM}(40,42,45)\). \[40 = 2^3\times5, \quad 42 = 2\times3\times7, \quad 45 = 3^2\times5 \] \[\text{LCM} = 2^3\times3^2\times5\times7 = 2520 \]Answer: \(\displaystyle 2520\) cm, i.e. \(\displaystyle 25\) m \(\displaystyle 20\) cm.

  3. Exercise 13

    Write the denominator of the rational number 2575000\displaystyle \frac{257}{5000} in the form 2m×5n\displaystyle 2^m \times 5^n, where m,n\displaystyle m, n are non-negative integers. Hence, write its decimal expansion, without actual division.

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

    NCERT’s answer
    \(\displaystyle 2^3 .5^4, 0.0514\)
    \[5000 = 2^3\times5^4 \] So \(\displaystyle m=3,\ n=4\). \[\frac{257}{5000} = \frac{257}{2^3\times5^4} = \frac{257\times2}{2^4\times5^4} = \frac{514}{10^4} \]Answer: \(\displaystyle 5000=2^3\times5^4\); \(\displaystyle \dfrac{257}{5000}=0.0514\).

  4. Exercise 14

    Prove that p+q\displaystyle \sqrt{p}+\sqrt{q} is irrational, where p,q\displaystyle p, q are primes.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Suppose, for contradiction, \(\displaystyle \sqrt p+\sqrt q=r\) with \(\displaystyle r\) rational; \(\displaystyle r\neq0\) since \(\displaystyle p,q>0\). \[\sqrt p = r-\sqrt q \] \[p = r^2 - 2r\sqrt q + q \] \[\sqrt q = \frac{r^2+q-p}{2r} \] The right side is rational, so \(\displaystyle \sqrt q\) would be rational, contradicting that the square root of a prime is irrational. So the assumption is false.Answer: \(\displaystyle \sqrt p+\sqrt q\) is irrational.