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NCERT Exemplar · Class 9 Mathematics Number Systems

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EXERCISE 1.3 1–10 (part 4 of 6)

  1. Exercise 1

    Find which of the variables x,y,z\displaystyle x, y, z and u\displaystyle u represent rational numbers and which irrational numbers : (i) x2=5\displaystyle x^{2}=5 (ii) y2=9\displaystyle y^{2}=9 (iii) z2=.04\displaystyle z^{2}=.04 (iv) u2=174\displaystyle u^{2}=\frac{17}{4}

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    NCERT’s answer
    Rational numbers: (ii), (iii) Irrational numbers: (i), (iv)
    \[x^{2}=5 \implies x=\pm\sqrt5 \quad\text{(5 not a perfect square)} \] \[y^{2}=9 \implies y=\pm3 \] \[z^{2}=0.04=\frac{1}{25} \implies z=\pm\frac15 \] \[u^{2}=\frac{17}{4} \implies u=\pm\frac{\sqrt{17}}{2} \quad\text{(17 not a perfect square)} \]\(\displaystyle x\) and \(\displaystyle u\) are irrational; \(\displaystyle y\) and \(\displaystyle z\) are rational.Answer: \(\displaystyle x,u\) irrational; \(\displaystyle y=\pm3,\ z=\pm\dfrac15\) rational.

  2. Exercise 2

    Find three rational numbers between (i) -1\displaystyle 1 and -2\displaystyle 2 (ii) 0.1\displaystyle 0.1 and 0.11\displaystyle 0.11 (iii) 57\displaystyle \frac{5}{7} and 67\displaystyle \frac{6}{7} (iv) 14\displaystyle \frac{1}{4} and 15\displaystyle \frac{1}{5}

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    \[-2<-\frac{19}{10}<-\frac{18}{10}<-\frac{17}{10}<-1 \] \[0.100<0.101<0.105<0.108<0.110 \] \[\frac57=\frac{20}{28}<\frac{21}{28}<\frac{22}{28}<\frac{23}{28}<\frac{24}{28}=\frac67 \] \[\frac15=\frac{20}{100}<\frac{21}{100}<\frac{22}{100}<\frac{23}{100}<\frac{25}{100}=\frac14 \]Answer: (i) \(\displaystyle -\frac{19}{10},-\frac{18}{10},-\frac{17}{10}\) (ii) \(\displaystyle 0.101,0.105,0.108\) (iii) \(\displaystyle \frac{21}{28},\frac{22}{28},\frac{23}{28}\) (iv) \(\displaystyle \frac{21}{100},\frac{22}{100},\frac{23}{100}\).NCERT prints: (i) \(\displaystyle -1.1,-1.2,-1.3\) (ii) \(\displaystyle 0.101,0.102,0.103\) (iii) \(\displaystyle \frac{51}{70},\frac{52}{70},\frac{53}{70}\) (iv) \(\displaystyle \frac{9}{40},\frac{17}{80},\frac{19}{80}\) — an equally valid different choice for this open-ended question; the chains above check just as well.

  3. Exercise 3

    Insert a rational number and an irrational number between the following : (i) 2\displaystyle 2 and 3\displaystyle 3 (ii) 0\displaystyle 0 and 0.1\displaystyle 0.1 (iii) 13\displaystyle \frac{1}{3} and 12\displaystyle \frac{1}{2} (iv) −25\displaystyle \frac{-2}{5} and 12\displaystyle \frac{1}{2} (v) 0.15\displaystyle 0.15 and 0.16\displaystyle 0.16 (vi) 2\displaystyle \sqrt{2} and 3\displaystyle \sqrt{3} (vii) 2.357\displaystyle 2.357 and 3.121\displaystyle 3.121 (viii) . 0001\displaystyle 0001 and . 001\displaystyle 001 (ix) 3.623623\displaystyle 3.623623 and 0.484848\displaystyle 0.484848 (x) 6.375289\displaystyle 6.375289 and 6.375738\displaystyle 6.375738

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    NCERT’s answer
    (i)
    2.$\displaystyle 1$, $\displaystyle 2.040040004$ ... (ii) $\displaystyle 0.03,0.007000700007, \ldots$ (iii) $\displaystyle \frac{5}{12}, 0.414114111 \ldots$ (iv) $\displaystyle 0,0.151151115 \ldots$ (v) $\displaystyle 0.151,0.151551555 \ldots$ (vi) $\displaystyle 1.5$, $\displaystyle 1.585585558$ ... (vii) $\displaystyle 3$, 3.101101110... (viii) $\displaystyle 0.00011$, . $\displaystyle 0001131331333$ ... (ix) $\displaystyle 1,1.909009000 \ldots$ (x) $\displaystyle 6.3753$, $\displaystyle 6.375414114111$ ...
    Rational: a number strictly between \(\displaystyle a\) and \(\displaystyle b\) -- usually the midpoint \(\displaystyle \frac{a+b}2\), or any convenient decimal/integer inside the interval. Irrational, where no simpler surd fits: \(\displaystyle a+(\sqrt2-1)(b-a)\), since \(\displaystyle 0<\sqrt2-1<1\).\[\text{(i)}\quad 2.5\in\mathbb Q, \qquad 2^2=4<5<9=3^2\implies 2<\sqrt5<3 \] \[\text{(ii)}\quad 0.05\in\mathbb Q, \qquad 0<0.1(\sqrt2-1)<0.1 \] \[\text{(iii)}\quad \frac{5}{12}\in\mathbb Q, \qquad \frac19<\frac15<\frac14\implies \frac13<\frac{1}{\sqrt5}<\frac12 \] \[\text{(iv)}\quad \frac{1}{20}\in\mathbb Q, \qquad -\frac25<\frac{\sqrt2}{10}<\frac12 \] \[\text{(v)}\quad 0.155\in\mathbb Q, \qquad 0.15<0.15+0.01(\sqrt2-1)<0.16 \] \[\text{(vi)}\quad \frac32\in\mathbb Q, \qquad 2<2.5<3\implies \sqrt2<\sqrt{2.5}<\sqrt3 \] \[\text{(vii)}\quad 2.739\in\mathbb Q, \qquad 2.357^2<7<3.121^2\implies 2.357<\sqrt7<3.121 \] \[\text{(viii)}\quad 0.00055\in\mathbb Q, \qquad 0.0001<0.0001+0.0009(\sqrt2-1)<0.001 \] \[\text{(ix)}\quad 2\in\mathbb Q, \qquad 0.484848<2<\sqrt6<3<3.623623\quad(2^2=4<6<9=3^2) \] \[\text{(x)}\quad 6.3755135\in\mathbb Q, \qquad 6.375289<6.375289+0.000449(\sqrt2-1)<6.375738 \]Answer: (i) \(\displaystyle 2.5,\sqrt5\) (ii) \(\displaystyle 0.05,\,0.1(\sqrt2-1)\) (iii) \(\displaystyle \frac5{12},\frac1{\sqrt5}\) (iv) \(\displaystyle \frac1{20},\frac{\sqrt2}{10}\) (v) \(\displaystyle 0.155,\,0.15+0.01(\sqrt2-1)\) (vi) \(\displaystyle \frac32,\sqrt{2.5}\) (vii) \(\displaystyle 2.739,\sqrt7\) (viii) \(\displaystyle 0.00055,\,0.0001+0.0009(\sqrt2-1)\) (ix) \(\displaystyle 2,\sqrt6\) (x) \(\displaystyle 6.3755135,\,6.375289+0.000449(\sqrt2-1)\).

  4. Exercise 4

    Represent the following numbers on the number line : 7,7.2,−32,−1257,7.2, \frac{-3}{2}, \frac{-12}{5}

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

    \[P=-\frac{12}{5}=-2.4,\qquad Q=-\frac32=-1.5,\qquad R=7,\qquad S=7.2 \] R and S are only \(\displaystyle 0.2\) apart, too close to mark at the scale that fits P, Q; the \(\displaystyle 6.8\)-to-\(\displaystyle 7.6\) stretch is redrawn at a finer scale.NCERT_Solution_Class9_Maths_Exemplar_Ch1_Ex1-3_Q4Answer: \(\displaystyle P(-2.4),\ Q(-1.5),\ R(7),\ S(7.2)\) marked in that order on the number line.

  5. Exercise 5

    Locate 5,10\displaystyle \sqrt{5}, \sqrt{10} and 17\displaystyle \sqrt{17} on the number line.

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

    1. Mark \(\displaystyle O\) as \(\displaystyle 0\); for \(\displaystyle m=2,3,4\) mark \(\displaystyle A_m\) with \(\displaystyle OA_m=m\). 2. At each \(\displaystyle A_m\) draw \(\displaystyle A_mB_m\perp\) the number line, \(\displaystyle A_mB_m=1\). \[OB_m=\sqrt{m^2+1}\quad\text{(Pythagoras)} \] 3. With centre \(\displaystyle O\), draw arcs of radius \(\displaystyle OB_2,OB_3,OB_4\) cutting the number line at \(\displaystyle P_5,P_{10},P_{17}\); then \(\displaystyle OP_5=OB_2,\ OP_{10}=OB_3,\ OP_{17}=OB_4\).NCERT_Solution_Class9_Maths_Exemplar_Ch1_Ex1-3_Q5\[m=2:\ OB_2=\sqrt5,\qquad m=3:\ OB_3=\sqrt{10},\qquad m=4:\ OB_4=\sqrt{17} \]Answer: \(\displaystyle P_5,P_{10},P_{17}\) are the points \(\displaystyle OP=\sqrt5,\sqrt{10},\sqrt{17}\), each swung from a right triangle of legs \(\displaystyle m\) and \(\displaystyle 1\).

  6. Exercise 6

    Represent geometrically the following numbers on the number line : (i) 4.5\displaystyle \sqrt{4.5} (ii) 5.6\displaystyle \sqrt{5.6} (iii) 8.1\displaystyle \sqrt{8.1} (iv) 2.3\displaystyle \sqrt{2.3}

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

    1. Take \(\displaystyle O\) at \(\displaystyle 0\) on the number line; mark \(\displaystyle A\) with \(\displaystyle OA=x\), then \(\displaystyle B\) with \(\displaystyle AB=1\). 2. Bisect \(\displaystyle OB\) at \(\displaystyle C\); draw a semicircle, centre \(\displaystyle C\), radius \(\displaystyle CO\,(=CB)\). 3. Draw \(\displaystyle AD\perp OB\) meeting the semicircle at \(\displaystyle D\). \[AD^{2}=OA\cdot AB=x\quad\text{(right angle in a semicircle)} \] 4. With centre \(\displaystyle O\), radius \(\displaystyle AD\), draw an arc cutting the line at \(\displaystyle E\); since \(\displaystyle O\) is \(\displaystyle 0\), \(\displaystyle OE=AD=\sqrt x\).NCERT_Solution_Class9_Maths_Exemplar_Ch1_Ex1-3_Q6\[x=4.5:\ \sqrt{4.5}\approx2.121 \qquad x=5.6:\ \sqrt{5.6}\approx2.366 \] \[x=8.1:\ \sqrt{8.1}\approx2.846 \qquad x=2.3:\ \sqrt{2.3}\approx1.517 \]Answer: the same construction, with \(\displaystyle O\) at \(\displaystyle 0\), \(\displaystyle OA=x,\ AB=1\), places \(\displaystyle \sqrt{4.5},\sqrt{5.6},\sqrt{8.1},\sqrt{2.3}\) on the line.

  7. Exercise 7

    Express the following 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 : (i) 0.2\displaystyle 0.2 (ii) 0.888…\displaystyle 0.888 \ldots (iii) 5.2‾\displaystyle 5 . \overline{2} (iv) 0.001‾\displaystyle 0 . \overline{001} (v) 0.2555…\displaystyle 0.2555 \ldots (vi) 0.134‾\displaystyle 0.1 \overline{34} (vii) .00323232... (viii) .404040...

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

    NCERT’s answer
    (i)
    $\displaystyle \frac{1}{5}$ (ii) $\displaystyle \frac{8}{9}$ (iii) $\displaystyle \frac{47}{9}$ (iv) $\displaystyle \frac{1}{999}$ (v) $\displaystyle \frac{23}{90}$ (vi) $\displaystyle \frac{133}{990}$ (vii) $\displaystyle \frac{8}{2475}$ (viii) $\displaystyle \frac{40}{99}$
    \[\text{(i)}\ 0.2=\frac{2}{10}=\frac15 \] \[\text{(ii)}\ x=0.\overline8\implies 10x-x=8\implies x=\frac89 \] \[\text{(iii)}\ x=5.\overline2\implies 10x-x=47\implies x=\frac{47}{9} \] \[\text{(iv)}\ x=0.\overline{001}\implies 1000x-x=1\implies x=\frac{1}{999} \] \[\text{(v)}\ x=0.2\overline5\implies 100x-10x=23\implies x=\frac{23}{90} \] \[\text{(vi)}\ x=0.1\overline{34}\implies 1000x-10x=133\implies x=\frac{133}{990} \] \[\text{(vii)}\ x=0.00\overline{32}\implies 9900x=32\implies x=\frac{32}{9900}=\frac{8}{2475} \] \[\text{(viii)}\ x=0.\overline{40}\implies 99x=40\implies x=\frac{40}{99} \]Answer: \(\displaystyle \frac15,\ \frac89,\ \frac{47}9,\ \frac1{999},\ \frac{23}{90},\ \frac{133}{990},\ \frac{8}{2475},\ \frac{40}{99}\).

  8. Exercise 8

    Show that 0.142857142857…=17\displaystyle 0.142857142857 \ldots=\frac{1}{7}

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

    \[x=0.\overline{142857}\implies 10^{6}x-x=142857 \] \[999999x=142857\implies x=\frac{142857}{999999} \] \[142857\times7=999999\implies x=\frac17 \]Answer: \(\displaystyle 0.\overline{142857}=\dfrac17\).

  9. Exercise 9

    Simplify the following: (i) 45−320+45\displaystyle \sqrt{45}-3 \sqrt{20}+4 \sqrt{5} (ii) 248+549\displaystyle \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9} (iii) 124×67\displaystyle \sqrt[4]{12} \times \sqrt[7]{6} (iv) 428÷37÷73\displaystyle 4 \sqrt{28} \div 3 \sqrt{7} \div \sqrt[3]{7} (v) 33+227+73\displaystyle 3 \sqrt{3}+2 \sqrt{27}+\frac{7}{\sqrt{3}} (vi) (3−2)2\displaystyle (\sqrt{3}-\sqrt{2})^{2} (vii) 814−82163+15325+225\displaystyle \sqrt[4]{81}-8 \sqrt[3]{216}+15 \sqrt[5]{32}+\sqrt{225} (viii) 38+12\displaystyle \frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}} (ix) 233−36\displaystyle \frac{2 \sqrt{3}}{3}-\frac{\sqrt{3}}{6}

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    (i)
    \[\sqrt{45}=3\sqrt5,\quad \sqrt{20}=2\sqrt5 \]
    \[\sqrt{45}-3\sqrt{20}+4\sqrt5 = 3\sqrt5-6\sqrt5+4\sqrt5=\sqrt5 \]
    (ii)
    \[\sqrt{24}=2\sqrt6,\quad \sqrt{54}=3\sqrt6 \]
    \[\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9} = \frac{\sqrt6}{4}+\frac{\sqrt6}{3}=\frac{7\sqrt6}{12} \]
    (iii)
    \[\sqrt[4]{12}=12^{7/28},\quad \sqrt[7]{6}=6^{4/28} \]
    \[\sqrt[4]{12}\times\sqrt[7]{6} = \sqrt[28]{12^{7}\cdot6^{4}} = \sqrt[28]{2^{18}\cdot3^{11}} \]
    NCERT's printed answer \(\displaystyle 168\sqrt2\) is the value of \(\displaystyle 4\sqrt{12}\times7\sqrt6\), not of the fourth-root/seventh-root expression printed here.
    (iv)
    \[4\sqrt{28}=8\sqrt7 \]
    \[8\sqrt7\div3\sqrt7 = \frac{8}{3} \]
    \[\frac{8}{3}\div\sqrt[3]{7} = \frac{8}{3\sqrt[3]{7}} = \frac{8\sqrt[3]{49}}{21} \]
    NCERT's printed answer \(\displaystyle \frac83\) is the value of \(\displaystyle 4\sqrt{28}\div3\sqrt7\) alone, without the \(\displaystyle \div\sqrt[3]7\) printed here.
    (v)
    \[2\sqrt{27}=6\sqrt3,\quad \frac{7}{\sqrt3}=\frac{7\sqrt3}{3} \]
    \[3\sqrt3+6\sqrt3+\frac{7\sqrt3}{3} = 9\sqrt3+\frac{7\sqrt3}{3}=\frac{34\sqrt3}{3} \]
    (vi)
    \[(\sqrt3-\sqrt2)^2 = 3-2\sqrt6+2 = 5-2\sqrt6 \]
    (vii)
    \[\sqrt[4]{81}=3,\ \sqrt[3]{216}=6,\ \sqrt[5]{32}=2,\ \sqrt{225}=15 \]
    \[3-8(6)+15(2)+15 = 3-48+30+15=0 \]
    (viii)
    \[\sqrt8=2\sqrt2 \]
    \[\frac{3}{2\sqrt2}+\frac{1}{\sqrt2} = \frac{3\sqrt2}{4}+\frac{2\sqrt2}{4}=\frac{5\sqrt2}{4} \]
    (ix)
    \[\frac{2\sqrt3}{3}-\frac{\sqrt3}{6} = \frac{4\sqrt3-\sqrt3}{6}=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2} \]
    Answer: (i) \(\displaystyle \sqrt5\) (ii) \(\displaystyle \tfrac{7\sqrt6}{12}\) (iii) \(\displaystyle \sqrt[28]{2^{18}3^{11}}\) (iv) \(\displaystyle \tfrac{8\sqrt[3]{49}}{21}\) (v) \(\displaystyle \tfrac{34\sqrt3}{3}\) (vi) \(\displaystyle 5-2\sqrt6\) (vii) \(\displaystyle 0\) (viii) \(\displaystyle \tfrac{5\sqrt2}{4}\) (ix) \(\displaystyle \tfrac{\sqrt3}{2}\)
    NCERT prints: (iii) \(\displaystyle 168\sqrt2\) (iv) \(\displaystyle \frac83\) — but the stem as printed asks for \(\displaystyle \sqrt[4]{12}\times\sqrt[7]{6}\) and \(\displaystyle 4\sqrt{28}\div3\sqrt7\div\sqrt[3]7\); those printed values belong instead to \(\displaystyle 4\sqrt{12}\times7\sqrt6\) and \(\displaystyle 4\sqrt{28}\div3\sqrt7\) alone.

  10. Exercise 10

    Rationalise the denominator of the following: (i) 233\displaystyle \frac{2}{3 \sqrt{3}} (ii) 403\displaystyle \frac{\sqrt{40}}{\sqrt{3}} (iii) 3+242\displaystyle \frac{3+\sqrt{2}}{4 \sqrt{2}} (iv) 1641−5\displaystyle \frac{16}{\sqrt{41}-5} (v) 2+32−3\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}} (vi) 62+3\displaystyle \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} (vii) 3+23−2\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} (viii) 35+35−3\displaystyle \frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} (ix) 43+5248+18\displaystyle \frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}

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

    NCERT’s answer
    (i)
    $\displaystyle \frac{2}{9} \sqrt{3}$ (ii) $\displaystyle \frac{2}{3} \sqrt{30}$ (iii) $\displaystyle \frac{2+3 \sqrt{2}}{8}$ (iv) $\displaystyle \sqrt{41}+5$ (v) $\displaystyle 7+4 \sqrt{3}$ (vi) $\displaystyle 3 \sqrt{2}-2 \sqrt{3}$ (vii) $\displaystyle 5+2 \sqrt{6}$ (viii) $\displaystyle 9+2 \sqrt{15}$ (ix) $\displaystyle \frac{9+4 \sqrt{6}}{15}$
    (i)
    \[\frac{2}{3\sqrt3}\times\frac{\sqrt3}{\sqrt3} = \frac{2\sqrt3}{9} \]
    (ii)
    \[\sqrt{40}=2\sqrt{10} \]
    \[\frac{2\sqrt{10}}{\sqrt3}\times\frac{\sqrt3}{\sqrt3} = \frac{2\sqrt{30}}{3} \]
    (iii)
    \[\frac{3+\sqrt2}{4\sqrt2}\times\frac{\sqrt2}{\sqrt2} = \frac{3\sqrt2+2}{8} \]
    (iv)
    \[\frac{16}{\sqrt{41}-5}\times\frac{\sqrt{41}+5}{\sqrt{41}+5} = \frac{16(\sqrt{41}+5)}{41-25}=\sqrt{41}+5 \]
    (v)
    \[\frac{2+\sqrt3}{2-\sqrt3}\times\frac{2+\sqrt3}{2+\sqrt3} = \frac{(2+\sqrt3)^2}{4-3}=7+4\sqrt3 \]
    (vi)
    \[\frac{\sqrt6}{\sqrt2+\sqrt3}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2} = \frac{\sqrt{18}-\sqrt{12}}{3-2}=3\sqrt2-2\sqrt3 \]
    (vii)
    \[\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2} = \frac{(\sqrt3+\sqrt2)^2}{3-2}=5+2\sqrt6 \]
    (viii)
    \[\frac{3\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3} = \frac{18+4\sqrt{15}}{5-3}=9+2\sqrt{15} \]
    (ix)
    \[\sqrt{48}=4\sqrt3,\ \sqrt{18}=3\sqrt2 \]
    \[\frac{4\sqrt3+5\sqrt2}{4\sqrt3+3\sqrt2}\times\frac{4\sqrt3-3\sqrt2}{4\sqrt3-3\sqrt2} = \frac{18+8\sqrt6}{30}=\frac{9+4\sqrt6}{15} \]
    Answer: (i) \(\displaystyle \tfrac{2\sqrt3}{9}\) (ii) \(\displaystyle \tfrac{2\sqrt{30}}{3}\) (iii) \(\displaystyle \tfrac{3\sqrt2+2}{8}\) (iv) \(\displaystyle \sqrt{41}+5\) (v) \(\displaystyle 7+4\sqrt3\) (vi) \(\displaystyle 3\sqrt2-2\sqrt3\) (vii) \(\displaystyle 5+2\sqrt6\) (viii) \(\displaystyle 9+2\sqrt{15}\) (ix) \(\displaystyle \tfrac{9+4\sqrt6}{15}\)