Question

Solve:

(i) $\frac{5}{12}+\frac{1}{12}$
(ii) $\frac{3}{15}+\frac{7}{15}$
(iii) $\frac{3}{22}+\frac{7}{22}$
(iv) $\frac{1}{4}+\frac{0}{4}$
(v) $\frac{4}{13}+\frac{2}{13}+\frac{1}{13}$
(vi) $\frac{0}{15}+\frac{2}{15}+\frac{1}{15}$
(vii) $\frac{7}{31}-\frac{4}{31}+\frac{9}{31}$
(viii) $3\frac{2}{7}+\frac{1}{7}-2\frac{3}{7}$
(ix) $2\frac{1}{3}-1\frac{2}{3}+4\frac{1}{3}$
(x) $1-\frac{2}{3}+\frac{7}{3}$
(xi) $\frac{16}{7}-\frac{5}{7}+\frac{9}{7}$

Answer 1

(i)$$\begin{gathered} \frac{5}{12}+\frac{1}{12}=\frac{5+1}{12} \\ =\frac{6÷6}{12÷6}=\frac{1}{2}(\mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}\mathrm{their}\mathrm{HCF}) \end{gathered}$$

(ii)$$\begin{gathered} \frac{3}{15}+\frac{7}{15}=\frac{3+7}{15} \\ =\frac{10÷5}{15÷5}=\frac{2}{3}(\mathrm{Dividing}\mathrm{numerator}\hspace{0.17em}\&\mathrm{denominator}\mathrm{by}\mathrm{their}\mathrm{HCF}) \end{gathered}$$

(iii)$$\begin{gathered} \frac{3}{22}+\frac{7}{22}=\frac{3+7}{22} \\ =\frac{10÷2}{22÷2}=\frac{5}{11}(\mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}\mathrm{their}\mathrm{HCF}) \end{gathered}$$

(iv)$\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}$

(v)$\frac{4}{13}+\frac{2}{13}+\frac{1}{13}=\frac{4+2+1}{13}=\frac{7}{13}$

(vi)$$\begin{gathered} \frac{0}{15}+\frac{2}{15}+\frac{1}{15}=\frac{0+2+1}{15} \\ =\frac{3÷3}{15÷3}=\frac{1}{5}(\mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}\mathrm{their}\mathrm{HCF}) \end{gathered}$$

(vii)$\frac{7}{31}-\frac{4}{31}+\frac{9}{31}=\frac{7-4+9}{31}=\frac{12}{31}$

(viii)$3\frac{2}{7}+\frac{1}{7}-2\frac{3}{7}=\frac{23}{7}+\frac{1}{7}-\frac{17}{7}=\frac{23+1-17}{7}=\frac{7}{7}=1$
(ix)$$\begin{gathered} 2\frac{1}{3}-1\frac{2}{3}+4\frac{1}{3}=\frac{7}{3}-\frac{5}{3}+\frac{13}{3} \\ =\frac{7-5+13}{3}=\frac{15÷3}{3÷3}=5(\mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}\mathrm{their}\mathrm{HCF}) \end{gathered}$$

(x)$1-\frac{2}{3}+\frac{7}{3}=\frac{3-2+7}{3}=\frac{8}{3}$

(xi)$\frac{16}{7}-\frac{5}{7}+\frac{9}{7}=\frac{16-5+9}{7}=\frac{20}{7}$