Maths ·Class 6 Maths

Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. Class 6

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· Jun 29, 2026 · Reviewed & updated Oct 02, 2026 · 1 min read

Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. Class 6

Question

Solution:

(a) $\frac{7}{2}=\frac{14}{4}=\frac{21}{6}=\frac{28}{8}=\frac{35}{10}$ and $\frac{3}{5}=\frac{6}{10}$

So, we have the required pair of fractions with same fractional units as [latex]\frac{35}{10}[/latex] and [latex]\frac{6}{10}[/latex].

(b) $\frac{8}{3}=\frac{16}{6}$ and $\frac{5}{6}=\frac{5}{6}$

So, required pair of fractions are [latex]\frac{16}{6}[/latex] and [latex]\frac{5}{6}[/latex].

(c) $\begin{aligned} & \frac{3}{4}=\frac{6}{8}=\frac{9}{12}=\frac{12}{16}=\frac{15}{20} \text { and } \\ & \frac{3}{5}=\frac{6}{10}=\frac{9}{15}=\frac{12}{20}\end{aligned}$

So, required pair of fractions are [latex]\frac{15}{20}[/latex] and [latex]\frac{12}{20}[/latex].

(d) $\frac{6}{7}=\frac{12}{14}=\frac{18}{21}=\frac{24}{28}=\frac{30}{35}$

and $\frac{8}{5}=\frac{16}{10}=\frac{24}{15}=\frac{32}{20}=\frac{40}{25}=\frac{48}{30}=\frac{56}{35}$

So, required pair of fractions are [latex]\frac{30}{35}[/latex] and [latex]\frac{56}{35}[/latex].

(e) [latex]\frac{9}{4}[/latex] = [latex]\frac{9}{4}[/latex] and [latex]\frac{5}{2}[/latex] = [latex]\frac{10}{4}[/latex]

So, required pair of fractions are [latex]\frac{9}{4}[/latex] and [latex]\frac{10}{4}[/latex].

(f) $\begin{aligned} \frac{1}{10} & =\frac{2}{20}=\frac{3}{30}=\frac{4}{40}=\frac{5}{50}=\frac{6}{60}=\frac{7}{70} \\ & =\frac{8}{80}=\frac{9}{90} \text { and } \frac{2}{9}=\frac{4}{18}=\frac{6}{27}=\frac{8}{36}=\frac{10}{45} \\ & =\frac{12}{54}=\frac{14}{63}=\frac{16}{72}=\frac{18}{81}=\frac{20}{90} .\end{aligned}$

So, required pair of fractions are [latex]\frac{9}{90}[/latex] and [latex]\frac{20}{90}[/latex].

(g) $\frac{8}{3}=\frac{16}{6}=\frac{24}{9}=\frac{32}{12}$ and $\frac{11}{4}=\frac{22}{8}=\frac{33}{12}$

So, required pair of fractions are [latex]\frac{32}{12}[/latex] and [latex]\frac{33}{12}[/latex].

(h) $\frac{13}{6}=\frac{26}{12}=\frac{39}{18}$ and $\frac{1}{9}=\frac{2}{18}$

So, required pair of fractions are [latex]\frac{39}{18}[/latex] and [latex]\frac{2}{18}[/latex].

Question 2.

Express the following fractions in lowest terms : Class 6

(a) [latex]\frac{17}{51}[/latex]

(b) [latex]\frac{64}{144}[/latex]

(d) [latex]\frac{525}{112}[/latex]

Solution:

(a) $\frac{17}{51}=\frac{17 \times 1}{17 \times 3}=\frac{1}{3}$.

(b) $\frac{64}{144}=\frac{16 \times 4}{16 \times 9}=\frac{4}{9}$.

(d) $\frac{525}{112}=\frac{7 \times 75}{7 \times 16}=\frac{75}{16}$.