Gujarat BoardEnglish MediumSTD 7MATHSData Handling2 Marks
Question
Find atleast 5 numbers between $\frac{1}{2}$ and $\frac{1}{3}$.
✓
Answer
Given numbers are $\frac{1}{2}$ and $\frac{1}{3}$. 1 st number $=\frac{\text { Sum of given numbers }}{\text { Number of digits }}=\frac{\frac{1}{2}+\frac{1}{3}}{2}=\frac{\frac{3+2}{6}}{2}$ $=\frac{5}{6 \times 2}=\frac{5}{12} \quad[\because$ LCM of 2 and $3=6]$ $\therefore \quad \frac{1}{2}>\frac{5}{12}>\frac{1}{3}$ 2 nd number $=\frac{\text { Given number }+1 \text { st number }}{2}$ $ =\frac{\frac{1}{2}+\frac{5}{12}}{2}=\frac{\frac{6+5}{12}}{2}=\frac{11}{12 \times 2}=\frac{11}{24} $ $ \therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{5}{12}>\frac{1}{3} $ Similarly, 3rd number $=\frac{\frac{5}{12}+\frac{1}{3}}{2}=\frac{\frac{5+4}{12}}{2}=\frac{9}{12 \times 2}=\frac{9}{24}$ $ \begin{array}{l} \therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{5}{12}>\frac{9}{24}>\frac{1}{3} \\ \text { 4th number }=\frac{\frac{5}{12}+\frac{11}{24}}{2}=\frac{\frac{10+11}{24}}{2}=\frac{21}{24 \times 2}=\frac{21}{48} \\ \therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{21}{48}>\frac{5}{12}>\frac{9}{24}>\frac{1}{3} \\ \text { 5th number }=\frac{\frac{5}{12}+\frac{9}{24}}{2}=\frac{\frac{10+9}{24}}{2}=\frac{10+9}{24 \times 2}=\frac{19}{48} \\ \therefore \quad \frac{1}{2}>\frac{11}{24}>\frac{21}{48}>\frac{5}{12}>\frac{19}{48}>\frac{9}{24}>\frac{1}{3} \end{array} $ Hence, 5 numbers between $\frac{1}{2}$ and $\frac{1}{3}$ are $ \frac{11}{24}, \frac{21}{48}, \frac{5}{12}, \frac{19}{48} \text { and } \frac{9}{24} . $
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