Add the following fractions using Brahmagupta’s method: Class 6

Add the following fractions using Brahmagupta’s method: Class 6

Question 1.

Add the following fractions using Brahmagupta’s method: Class 6

Solution:

(a) $\frac{2}{7}+\frac{5}{7}+\frac{6}{7}=\frac{13}{7}=1 \frac{6}{7}$.

(b) $\frac{3}{4}=\frac{3 \times 3}{4 \times 3}=\frac{9}{12}$ and $\frac{1}{3}=\frac{1 \times 4}{3 \times 4}=\frac{4}{12}$.

So, $\frac{3}{4}+\frac{1}{3}=\frac{9}{12}+\frac{4}{12}=\frac{13}{12}=1 \frac{1}{12}$.

(c) $\frac{2}{3}=\frac{2 \times 2}{3 \times 2}=\frac{4}{6}$ and $\frac{5}{6}=\frac{5}{6}$.

So, $\frac{2}{3}+\frac{5}{6}=\frac{4}{6}+\frac{5}{6}=\frac{9}{6}=\frac{3}{2}=1 \frac{1}{2}$.

(d) $\frac{2}{3}=\frac{2 \times 7}{3 \times 7} \doteq \frac{14}{21}$ and $\frac{2}{7}=\frac{2 \times 3}{7 \times 3}=\frac{6}{21}$.

So, $\frac{2}{3}+\frac{2}{7}=\frac{14}{21}+\frac{6}{21}=\frac{20}{21}$.

(e) $\frac{3}{4}=\frac{3 \times 15}{4 \times 15}=\frac{45}{60}, \frac{1}{3}=\frac{1 \times 20}{3 \times 20}=\frac{20}{60}$ and $\frac{1}{5}=\frac{1 \times 12}{5 \times 12}=\frac{12}{60}$.

(60 is the smallest common multiple of 4, 3 and 5)

So, $\frac{3}{4}+\frac{1}{3}+\frac{1}{5}=\frac{45}{60}+\frac{20}{60}+\frac{12}{60}=\frac{77}{60}=1 \frac{17}{60}$.

(f) $\frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15}$ and $\frac{4}{5}=\frac{4 \times 3}{5 \times 3}=\frac{12}{15}$.

So, $\frac{2}{3}+\frac{4}{5}=\frac{10}{15}+\frac{12}{15}=\frac{22}{15}=1 \frac{7}{15}$.

$\begin{aligned} \text { (g) } \frac{4}{5}+\frac{2}{3} & =\frac{4 \times 3}{5 \times 3}+\frac{2 \times 5}{3 \times 5}=\frac{12}{15}+\frac{10}{15} \\ & =\frac{22}{15}=1 \frac{7}{15} . \\ \text { (h) } \frac{3}{5}+\frac{5}{8} & =\frac{3 \times 8}{5 \times 8}+\frac{5 \times 5}{8 \times 5}=\frac{24}{40}+\frac{25}{40} \\ & =\frac{49}{40}=1 \frac{9}{40} .\end{aligned}$

(i)

$\frac{9}{2}+\frac{5}{4}=\frac{9 \times 2}{2 \times 2}+\frac{5 \times 1}{4 \times 1}=\frac{18}{4}+\frac{5}{4}=\frac{23}{4}=5 \frac{3}{4}$.

$\begin{aligned}(j) \frac{8}{3}+\frac{2}{7} & =\frac{8 \times 7}{3 \times 7}+\frac{2 \times 3}{7 \times 3}=\frac{56}{21}+\frac{6}{21} \\ & =\frac{62}{21}=2 \frac{20}{21}\end{aligned}$

(k) $\frac{3}{4}+\frac{1}{3}+\frac{1}{5}=1 \frac{17}{60}$.

[Same as Part (e) above.]

$\begin{aligned}(l) \frac{2}{3}+\frac{4}{5}+\frac{3}{7} & =\frac{2 \times 35}{3 \times 35}+\frac{4 \times 21}{5 \times 21}+\frac{3 \times 15}{7 \times 15} \\ & =\frac{70}{105}+\frac{84}{105}+\frac{45}{105} \\ & =\frac{199}{105}=1 \frac{94}{105} . \\ (m) \frac{9}{2}+\frac{5}{4}+\frac{7}{6} & =\frac{9 \times 6}{2 \times 6}+\frac{5 \times 3}{4 \times 3}+\frac{7 \times 2}{6 \times 2} \\ & =\frac{54}{12}+\frac{15}{12}+\frac{14}{12}=\frac{83}{12}=6 \frac{11}{12} .\end{aligned}$