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Chapter 4

Fractions — Exercise 4(B)

Class - 6 RS Aggarwal Mathematics Solutions

Exercise 4(B)

Question 1

Which of the following fractions are in simplest form?

(i) 2140\dfrac{21}{40}

(ii) 3549\dfrac{35}{49}

(iii) 4254\dfrac{42}{54}

(iv) 6481\dfrac{64}{81}

(v) 5665\dfrac{56}{65}

(vi) 2392\dfrac{23}{92}

(vii) 102119\dfrac{102}{119}

(viii) 91114\dfrac{91}{114}

Answer

A fraction is in its simplest form when the highest common factor (HCF) of its numerator and denominator is 1.

(i) 2140\dfrac{21}{40}

First, we find the H.C.F. of 21 and 40.

21)40(‾1ac))−21‾x2=)19)21(‾1ac=sc))−19‾x2)+2x=2)19(‾9ac=sc+sa))−18‾x2)+2x=+dca1)2(‾2ac=sc+sa+sa))−2‾x2+3x−54)+2x)×\begin{array}{l} 21\overline{\smash{\big)}\quad 40\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-21} \\ \phantom{{x^2 =)}} 19 \overline{\smash{\big)}\quad 21\smash{\big(}} 1 \\ \phantom{ac = sc)}\phantom{)}\underline{-19} \\ \phantom{{x^2 )} + 2x =}2\overline{\smash{\big)}\quad 19 \smash{\big(}} 9 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-18} \\ \phantom{{x^2 )} + 2x = + dca}1\overline{\smash{\big)}\quad 2 \smash{\big(}} 2 \\ \phantom{ac = sc + sa + sa)}\phantom{)}\underline{-2} \\ \phantom{{x^2 + 3x - 54)} + 2x)}\times \\ \end{array}

So, HCF of 21 and 40 = 1.

Hence, 2140\dfrac{21}{40} is in simplest form.

(ii) 3549\dfrac{35}{49}

35)49(‾1ac))−35‾x2=)14)35(‾2ac=sc))−28‾x2)+2x=7)14(‾2ac=sc+sa))−14‾x2+3x−54)+×\begin{array}{l} 35\overline{\smash{\big)}\quad 49\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-35} \\ \phantom{{x^2 =)}} 14 \overline{\smash{\big)}\quad 35\smash{\big(}} 2 \\ \phantom{ac = sc)}\phantom{)}\underline{-28} \\ \phantom{{x^2 )} + 2x =}7\overline{\smash{\big)}\quad 14 \smash{\big(}} 2 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-14} \\ \phantom{{x^2 + 3x - 54)} + }\times \\ \end{array}

So, HCF of 35 and 49 = 7.

Hence, 3549\dfrac{35}{49} is not in simplest form.

(iii) 4254\dfrac{42}{54}

42)54(‾1ac))−42‾x2=)12)42(‾3ac=sc)))−36‾x2)+2x=6)12(‾2ac=sc+sa))−12‾x2+3x−54)+×\begin{array}{l} 42\overline{\smash{\big)}\quad 54\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-42} \\ \phantom{{x^2 =)}} 12 \overline{\smash{\big)}\quad 42\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-36} \\ \phantom{{x^2 )} + 2x =}6\overline{\smash{\big)}\quad 12 \smash{\big(}} 2 \\ \phantom{ac = sc + sa)}\phantom{)}\underline{-12} \\ \phantom{{x^2 + 3x - 54)} + }\times \\ \end{array}

So, HCF of 42 and 54 = 6.

Hence, 4254\dfrac{42}{54} is not in simplest form.

(iv) 6481\dfrac{64}{81}

64)81(‾1ac))−64‾x2=)17)64(‾3ac=sc)))−51‾x2)+2x=13)17(‾1ac=sc+sa))))−13‾x2)+2x=acdvu4)13(‾3ac=sc+sa+df))−12‾x2)+2x=acdv+df1)4(‾4ac=sc+sa+df+dc))−4‾x2+3x−54)+advnscj×\begin{array}{l} 64\overline{\smash{\big)}\quad 81\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-64} \\ \phantom{{x^2 =)}} 17 \overline{\smash{\big)}\quad 64\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-51} \\ \phantom{{x^2 )} + 2x =}13\overline{\smash{\big)}\quad 17 \smash{\big(}} 1 \\ \phantom{ac = sc + sa)))}\phantom{)}\underline{-13} \\ \phantom{{x^2 )} + 2x = acdvu}4\overline{\smash{\big)}\quad 13 \smash{\big(}} 3 \\ \phantom{ac = sc + sa+ df)}\phantom{)}\underline{-12} \\ \phantom{{x^2 )} + 2x = acdv + df}1\overline{\smash{\big)}\quad 4 \smash{\big(}} 4 \\ \phantom{ac = sc + sa+ df + dc)}\phantom{)}\underline{-4} \\ \phantom{{x^2 + 3x - 54)} + advnscj }\times \\ \end{array}

So, HCF of 64 and 81 = 1.

Hence, 6481\dfrac{64}{81} is in simplest form.

(v) 5665\dfrac{56}{65}

56)65(‾1ac))−56‾x2=)9)56(‾6ac=sc))−54‾x2)+2x=2)9(‾4ac=sc+sa)))−8‾x2)+2x=acdv1)2(‾2ac=sc+sa+df))−2‾x2+3x−54)+aax×\begin{array}{l} 56\overline{\smash{\big)}\quad 65\smash{\big(}} 1 \\ \phantom{ac)}\phantom{)}\underline{-56} \\ \phantom{{x^2 =)}} 9 \overline{\smash{\big)}\quad 56\smash{\big(}} 6 \\ \phantom{ac = sc)}\phantom{)}\underline{-54} \\ \phantom{{x^2 )} + 2x =}2\overline{\smash{\big)}\quad 9 \smash{\big(}} 4 \\ \phantom{ac = sc + sa))}\phantom{)}\underline{-8} \\ \phantom{{x^2 )} + 2x = acdv}1\overline{\smash{\big)}\quad 2 \smash{\big(}} 2 \\ \phantom{ac = sc + sa+ df)}\phantom{)}\underline{-2} \\ \phantom{{x^2 + 3x - 54)} + aax}\times \\ \end{array}

So, HCF of 56 and 65 = 1.

Hence, 5665\dfrac{56}{65} is in simplest form.

(vi) 2392\dfrac{23}{92}

23)92(‾4ac))−92‾x2+3×\begin{array}{l} 23\overline{\smash{\big)}\quad 92\smash{\big(}} 4 \\ \phantom{ac)}\phantom{)}\underline{-92} \\ \phantom{{x^2 + 3}}\times \\ \end{array}

So, HCF of 23 and 92 = 23.

Hence, 2392\dfrac{23}{92} is not in simplest form.

(vii) 102119\dfrac{102}{119}

102)119(‾1ac=)−102‾x2=aa)17)102(‾6ac=sc=s))−102‾x2+3x−54)x×\begin{array}{l} 102\overline{\smash{\big)}\quad 119\smash{\big(}} 1 \\ \phantom{ac =)}\phantom{}\underline{-102} \\ \phantom{{x^2 = aa)}} 17 \overline{\smash{\big)}\quad 102\smash{\big(}} 6 \\ \phantom{ac = sc =s)}\phantom{)}\underline{-102} \\ \phantom{{x^2 + 3x - 54)} x}\times \\ \end{array}

So, HCF of 102 and 119 = 17.

Hence, 102119\dfrac{102}{119} is not in simplest form.

(viii) 91114\dfrac{91}{114}

91)114(‾1ac+))−91‾x2=)))23)91(‾3ac=sc)))−69‾x2)+2x=22)23(‾1ac=sc+sa=))−22‾x2)+2x=ac+ad1)22(‾22ac=sc+sa+df−))−22‾x2+3x−54)+aax+s×\begin{array}{l} 91\overline{\smash{\big)}\quad 114\smash{\big(}} 1 \\ \phantom{ac +)}\phantom{)}\underline{-91} \\ \phantom{{x^2 =)))}} 23 \overline{\smash{\big)}\quad 91\smash{\big(}} 3 \\ \phantom{ac = sc))}\phantom{)}\underline{-69} \\ \phantom{{x^2 )} + 2x =}22\overline{\smash{\big)}\quad 23 \smash{\big(}} 1 \\ \phantom{ac = sc + sa =)}\phantom{)}\underline{-22} \\ \phantom{{x^2 )} + 2x = ac + ad}1\overline{\smash{\big)}\quad 22 \smash{\big(}} 22 \\ \phantom{ac = sc + sa+ df -)}\phantom{)}\underline{-22} \\ \phantom{{x^2 + 3x - 54)} + aax+ s} \times \\ \end{array}

So, HCF of 91 and 114 = 1.

Hence, 91114\dfrac{91}{114} is in simplest form.

Question 2

Reduce each of the following fraction to its lowest terms :

(i) 2736\dfrac{27}{36}

(ii) 4554\dfrac{45}{54}

(iii) 3895\dfrac{38}{95}

(iv) 5887\dfrac{58}{87}

(v) 85153\dfrac{85}{153}

(vi) 105168\dfrac{105}{168}

(vii) 117143\dfrac{117}{143}

(viii) 135150\dfrac{135}{150}

Answer

(i) Prime factorizing,

2736=3×3×32×2×3×3=34\dfrac{27}{36} = \dfrac{3 \times 3 \times 3}{2 \times 2 \times 3 \times 3} = \dfrac{3}{4}.

Hence, 2736\dfrac{27}{36} in lowest terms is 34\dfrac{3}{4}.

(ii) Prime factorizing,

4554=3×3×53×3×3×2=56\dfrac{45}{54} = \dfrac{3 \times 3 \times 5}{3 \times 3 \times 3 \times 2} = \dfrac{5}{6}.

Hence, 4554\dfrac{45}{54} in lowest terms is 56\dfrac{5}{6}.

(iii) Prime factorizing,

3895=2×195×19=25\dfrac{38}{95} = \dfrac{2 \times 19}{5 \times 19} = \dfrac{2}{5}.

Hence, 3895\dfrac{38}{95} in lowest terms is 25\dfrac{2}{5}.

(iv) Prime factorizing,

5887=2×293×29=23\dfrac{58}{87} = \dfrac{2 \times 29}{3 \times 29} = \dfrac{2}{3}.

Hence, 5887\dfrac{58}{87} in lowest terms is 23\dfrac{2}{3}.

(v) Prime factorizing,

85153=5×173×3×17=59\dfrac{85}{153} = \dfrac{5 \times 17}{3 \times 3 \times 17} = \dfrac{5}{9}.

Hence, 85153\dfrac{85}{153} in lowest terms is 59\dfrac{5}{9}.

(vi) Prime factorizing,

105168=3×5×72×2×2×3×7=58\dfrac{105}{168} = \dfrac{3 \times 5 \times 7}{2 \times 2 \times 2\times 3 \times 7} = \dfrac{5}{8}.

Hence, 105168\dfrac{105}{168} in lowest terms is 58\dfrac{5}{8}.

(vii) Prime factorizing,

117143=3×3×1311×13=911\dfrac{117}{143} = \dfrac{3 \times 3 \times 13}{11 \times 13} = \dfrac{9}{11}.

Hence, 117143\dfrac{117}{143} in lowest terms is 911\dfrac{9}{11}.

(viii) Prime factorizing,

135150=3×3×3×52×3×5×5=910\dfrac{135}{150} = \dfrac{3 \times 3 \times 3 \times 5}{2 \times 3 \times 5 \times 5} = \dfrac{9}{10}.

Hence, 135150\dfrac{135}{150} in lowest terms is 910\dfrac{9}{10}.

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