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Real Numbers Test - 26

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Real Numbers Test - 26
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  • Question 1
    1 / -0
    If $$d$$ is the $$HCF$$ of 45 and 27, then $$x, y$$ satisfying $$d=27x+45y$$ are :
    Solution
    Applying Euclid's division lemma to $$27$$ and $$45$$, we get
    $$45 = 27 \times 1 + 18$$                        ...(1)
    $$27 = 18 \times 1 + 9$$                          ...(2)
    $$18 = 9 \times 2 + 0$$                            ...(3)
    Since the remainder is zero, therefore, last divisor $$9$$ is the $$HCF$$ of $$27$$ and $$45$$.

    From (2), we get
          $$9 = 27 - 18 \times 1$$
             $$ = 27 - \left(45 - 27 \times 1\right) \times 1 \ \ \ \ \ \ \left[using   (1)\right]$$
             $$ = 27 - 45 \times 1 + 27 \times 1 \times 1$$
             $$ = 27 - 45 + 27$$
             $$ = 54 - 45$$
    $$\Rightarrow 9 = 27 \times 2 - 45 \times 1$$                     ...(4)

    Comparing (4) with $$d = 27x + 45y$$, we get
    $$d = 9,  x = 2 \ and\  y = -1$$
  • Question 2
    1 / -0
    Use Euclid's division lemma to find the HCF : 
    65 and 495.
    Solution
    According to Euclid’s Division Lemma if we have two positive integers $$a$$ and $$b$$, then there exists unique integers $$q$$ and $$r$$ which satisfies the condition $$a = bq + r$$ where $$0 ≤ r < b$$.

    $$HCF$$ is the largest number which exactly divides two or more positive integers. 
    By Euclid's division lemma, we mean that on dividing both the integers $$a$$ and $$b$$ , the remainder is zero.

    The given integers are $$a=65$$ and $$b=495$$.

    Clearly $$495 > 65$$.

    So, we will apply Euclid’s division lemma to $$65$$ and $$495$$, we get,

    $$495 = (65\times 7) + 40$$

    Since the remainder $$40≠ 0$$. So we again apply the division lemma to the divisor $$65$$ and remainder $$40$$. We get,

    $$65 = (40\times 1) + 25$$

    Again the remainder $$25≠ 0$$, so applying the division lemma to the new divisor $$40$$ and remainder $$25$$. We get,

    $$40 = (25\times 1) + 15$$

    Now, again the remainder $$15≠ 0$$, so applying the division lemma to the new divisor $$25$$ and remainder $$15$$. We get,

    $$25 = (15\times 1) + 10$$

    Again the remainder $$10≠ 0$$, so applying the division lemma to the new divisor $$15$$ and remainder $$10$$. We get,

    $$15 = (10\times 1) + 5$$

    Again the remainder $$5≠ 0$$, so applying the division lemma to the new divisor $$10$$ and remainder $$5$$. We get,

    $$10 = (5\times 1) + 0$$

    Finally we get the remainder $$0$$ and the divisor is $$5$$.

    Hence, the $$HCF$$ of $$65$$ and $$495$$ is $$5$$.
  • Question 3
    1 / -0
    Use Euclid's division lemma to find the $$HCF$$ of the following numbers:
    $$8068$$ and $$12464$$
    Solution
    Here, $$a=12464$$ and $$b=8068$$

    So, by using Euclid's division lemma,
    $$a=bq+r$$ ; $$0<r<b$$

    $$\begin{aligned}{}12464 &= (8068)(1) + 4396\\8068 &= (4396)(1) + 3672\\4396& = (3672)(1) + 724\\3672 &= (724)(5) + 52\\724& = (52)(13) + 48\\52 &= (48)(1) + 4\\48 &= (4)(12) + 0\end{aligned}$$

    Hence, $$HCF$$ of $$8068$$ and $$12464$$ is equal to $$4.$$
  • Question 4
    1 / -0
    The decimal expansion of the rational number $$\dfrac {33}{2^2\cdot 5}$$ will terminate after:
    Solution
    $$\dfrac { 33 }{ 2^{ 2 }\cdot 5 } =\dfrac { 33\times5 }{ 4\times 5\times 5 } $$
            $$ =\dfrac { 165 }{ 100 } $$
            $$ =1.65$$.
    Thus the decimal expansion will terminate after two decimal places.
    Therefore, option $$B$$ is correct.
  • Question 5
    1 / -0
    Without doing any actual division, find which of the following rational numbers have terminating decimal representation :
    (i) $$\displaystyle \dfrac{7}{16}$$ (ii) $$\displaystyle \dfrac{23}{125}$$
    (iii) $$\displaystyle \dfrac{9}{14}$$ (iv) $$\displaystyle \dfrac{32}{45}$$
    (v) $$\displaystyle \dfrac{43}{50}$$ (vi) $$\displaystyle \dfrac{17}{40}$$
    (vii) $$\displaystyle \dfrac{61}{75}$$ (viii) $$\displaystyle \dfrac{123}{250}$$
    Solution
     The rational not having denominator as multiple of $$2^{m}\times 5^{n}$$ will be non terminating.
    (1) $$\dfrac{7}{16}=\dfrac{7}{2\times2\times2\times2}-$$ denominator is multiple of $$2^{m}\times 5^{n}$$ hence terminating.
    (2) $$\dfrac{23}{125}=\dfrac{23}{5\times5\times5}$$ --  denominator is multiple of $$2^{m}\times 5^{n}$$ hence terminating.
    (3) $$\dfrac{9}{14}=\dfrac{9}{7\times2}$$ -- denominator is not multiple of $$2^{m}\times 5^{n}$$ hence non-terminating
    (4)$$\dfrac{32}{45}=\dfrac{32}{3\times3\times5}$$-- denominator is not multiple of $$2^{m}\times 5^{n}$$ hence non-terminating.
    (5) $$\dfrac{43}{50}=\dfrac{43}{5\times5\times2}$$--  denominator is multiple of $$2^{m}\times 5^{n}$$ hence terminating.
    (6)$$\dfrac{17}{40}=\dfrac{17}{2\times2\times2\times5}$$ -- denominator is multiple of $$2^{m}\times 5^{n}$$ hence terminating.
    (7)$$\dfrac{61}{75}$$--  denominator is not multiple of $$2^{m}\times 5^{n}$$ hence non-terminating.
    (8)$$\dfrac{123}{250}=\dfrac{123}{5\times5\times5\times2}$$-- denominator is multiple of $$2^{m}\times 5^{n}$$ hence terminating.
    (i), (ii), (v), (vi) and (viii) will have  terminating decimal.
  • Question 6
    1 / -0
    The given pair of numbers $$ 231, 396$$ are __________ .
    Solution
     Euclid's division lemma,
    $$a=bq+r$$ ; $$0\le r<b$$
    Applying Euclid's Algorithim to $$231 , 395$$
    $$395 = 231\times 1+164$$
    $$231 = 164 \times 1+67$$
    $$164 = 67 \times 2+30$$
    $$67 = 30 \times 2 +7 $$
    $$30 = 7 \times 4 +2$$
    $$7 = 2 \times 3+1$$
    $$2 = 1 \times 2+0$$

    The given numbers have only $$1$$ as common factor.
    Hence, the given numbers are co-prime.
  • Question 7
    1 / -0
    Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
    $$\dfrac {29}{343}$$
    Solution
    Given, $$\displaystyle \frac {29}{343}= \frac {29}{7^{3}}$$
    As it is not in the form of $${ 2 }^{ m }\times { 5 }^{ n }$$.
    So, the rational number $$\displaystyle \frac {29}{343}$$ has a non terminating decimal expansion
  • Question 8
    1 / -0

    Directions For Questions

    With out actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

    ...view full instructions

    $$\frac {23}{2^35^2}$$
    Solution
    The given fraction, $$\displaystyle \frac {23}{2^{3}5^{2}}$$ has its denominator composed of only $$2$$'s and $$5$$'s only.
    Since the fraction is in the form of $${ 2 }^{ m }\times{ 5 }^{ n }$$ where ($$n=3,m=2$$),
    the rational number $$\dfrac {23}{2^{3}5^{2}}$$ will have a terminating decimal expansion.
  • Question 9
    1 / -0
    If these numbers form positive odd integer 6q+1, or 6q+3 or 6q+5 for some q then q belongs to:
    Solution
    We know that product  and sum of integers will give integer.
    Given  $$6q+1,   6q+3 , 6q+5$$  are integers so $$6q$$ will be integer and then q will be integer. 
    So correct answer will be option A.
  • Question 10
    1 / -0
     The square of any positive odd integer for some integer $$ m$$ is of the form 
    Solution
    We know that any positive odd integer $$n$$ is of the form $$4q + 1 $$ or $$4q + 3$$ where $$q$$ is some integer.

    When, $$n=4q+1$$
    $$n^2=(4q+1)^2$$     
    $$=16q^2+8q+1$$        
    $$=8q(2q+1)+1$$     
    $$=8m+1$$Where $$m=q(2q+1)$$
    $$=>n^2$$ is in the form $$8m+1$$

    When$$n=4q+3$$
    $$n^2=(4q+3)^2$$     
    $$=16q^2+24q+9$$     
    $$=16q^2+24q+8+1$$     
    $$=8(2q^2+3q+1)+1$$     
    $$=8m+1$$ Where $$m=(2q^2+3q+1)$$
    $$=>n^2$$ is in the form $$8m+1$$

    Hence,$$n^2$$ is in the form $$8m+1$$ if $$n$$ is an odd positive integer.
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