Question

Fill in the blanks:
(i) $9\frac{2}{3}+......=19$
(ii) $6\frac{1}{6}-?=\frac{29}{30}$
(iii) $7-5\frac{2}{3}=......$
(iv) $\frac{72}{90}$ reduced to simples form is ......
(v) $\frac{42}{54}=\frac{7}{\boxed{}}$

Answer 1

$$\begin{gathered} \left(i\right)9\frac{2}{3}+......=19 \\ ......=19-9\frac{2}{3} \\ ......=\frac{19}{1}-\frac{29}{3} \\ ......=\frac{57-29}{3} \\ ......=\frac{28}{3} \\ .....=9\frac{1}{3} \end{gathered}$$


(ii)
$$\begin{gathered} 6\frac{1}{6}-?=\frac{29}{30} \\ ?=6\frac{1}{6}-\frac{29}{30} \\ ?=\frac{37}{6}-\frac{29}{30} \\ ?=\frac{185-29}{30} \\ ?=\frac{{\cancel{156}}^{26}}{{\cancel{30}}_5} \\ ?=5\frac{1}{5} \end{gathered}$$


(iii)
$$\begin{gathered} 7-5\frac{2}{3} \\ =\frac{7}{1}-\frac{17}{3} \\ =\frac{21-17}{3} \\ =\frac{4}{3} \\ =1\frac{1}{3} \end{gathered}$$

(iv)

$$\begin{gathered} \mathrm{The}\mathrm{factors}\mathrm{of}72\mathrm{are}1,2,3,4,6,8,9,12,18,24,36,72. \\ \mathrm{The}\mathrm{factors}\mathrm{of}90\mathrm{are}1,2,3,5,6,9,10,15,18,30,45,90. \\ \mathrm{The}\mathrm{common}\mathrm{factors}\mathrm{of}72\mathrm{and}90\mathrm{are}1,2,3,6,9,18. \\ \mathrm{H}.\mathrm{C}.\mathrm{F}.\mathrm{of}72\mathrm{and}90\mathrm{is}18. \\ \frac{72÷18}{90÷18}=\frac{4}{5} \end{gathered}$$

(v)

$$\begin{gathered} \frac{42}{54}=\frac{7}{9} \\ \Rightarrow \frac{42÷6}{54÷6}=\frac{7}{9} \end{gathered}$$