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Answer :
i) Clearly, \(\frac{36}{100} \) can be written as 0.36.
Therefore, it is a terminating decimal.
ii) Dividing 1 by 11, we get$$ \require{enclose}
\begin{array}{rll}
0.0909 && \hbox{(Explanations)} \\[-3pt]
11 \enclose{longdiv}{100}\kern-.2ex \\[-3pt]
\underline{99\phantom{00}} && \hbox{($9 \times 11 = 99$)} \\[-3pt]
\phantom{0}100 && \hbox{($100 - 99 = 1$)} \\[-3pt]
\underline{\phantom{0}99} && \hbox{($9 \times 11 = 99$)} \\[-3pt]
\phantom{00}1
\end{array}$$Hence, \(\frac{1}{11} \) can be written as 0.090909...
Therefore, it is a non-terminating decimal.
iii) We have, \(4\frac{1}{8}\) = \(\frac{4×8+1}{8}\) = \(\frac{33}{8}\)
Dividing 33 by 8, we get,
$$ \require{enclose}
\begin{array}{rll}
4.125 && \hbox{(Explanations)} \\[-3pt]
8 \enclose{longdiv}{33}\kern-.2ex \\[-3pt]
\underline{32\phantom{00}} && \hbox{($8 \times 4 = 32$)} \\[-3pt]
\phantom{0}10 && \hbox{($33 - 32 = 1$)} \\[-3pt]
\underline{8\phantom{00}} && \hbox{($8 \times 1 = 8$)} \\[-3pt]
\phantom{0}20 && \hbox{($10 - 8 = 2$)} \\[-3pt]
\underline{16\phantom{00}} && \hbox{($8 \times 2 = 16$)} \\[-3pt]
\phantom{0}40 && \hbox{($20 - 16 = 4$)} \\[-3pt]
\underline{\phantom{0}40} && \hbox{($8 \times 5 = 40$)} \\[-3pt]
\phantom{00}0
\end{array}$$Hence, \(4\frac{1}{8}\) can be written as 4.125
Therefore, it is a terminating decimal.
iv) We have, \(\frac{3}{13} \)
Dividing 3 by 13, we get,
$$ \require{enclose}
\begin{array}{rll}
0.230769 && \hbox{(Explanations)} \\[-3pt]
13 \enclose{longdiv}{30}\kern-.2ex \\[-3pt]
\underline{26\phantom{00}} && \hbox{($2 \times 13 = 26$)} \\[-3pt]
\phantom{0}40 && \hbox{($30 - 26 = 4$)} \\[-3pt]
\underline{39\phantom{00}} && \hbox{($3 \times 13 = 39$)} \\[-3pt]
\phantom{0}100 && \hbox{($40 - 39 = 1$)} \\[-3pt]
\underline{91\phantom{00}} && \hbox{($7 \times 13 = 91$)} \\[-3pt]
\phantom{0}90 && \hbox{($100 - 91 = 4$)} \\[-3pt]
\underline{78\phantom{00}} && \hbox{($6 \times 13 = 78$)} \\[-3pt]
\phantom{0}120 && \hbox{($90 - 78 = 12$)} \\[-3pt]
\underline{\phantom{0}117} && \hbox{($9 \times 13 = 117$)} \\[-3pt]
\phantom{00}3
\end{array}$$Hence, \(\frac{3}{13} \) can be written as 0.230769
Therefore, it is a non-terminating decimal.
v) We have, \(\frac{2}{11} \)
Dividing 2 by 11, we get,
$$ \require{enclose}
\begin{array}{rll}
0.1818 && \hbox{(Explanations)} \\[-3pt]
11 \enclose{longdiv}{20}\kern-.2ex \\[-3pt]
\underline{11\phantom{00}} && \hbox{($1 \times 11 = 11$)} \\[-3pt]
\phantom{0}90 && \hbox{($20 - 11 = 9$)} \\[-3pt]
\underline{88\phantom{00}} && \hbox{($8 \times 11 = 11$)} \\[-3pt]
\phantom{0}20 && \hbox{($90 - 88 = 2$)} \\[-3pt]
\underline{11\phantom{00}} && \hbox{($1 \times 11 = 11$)} \\[-3pt]
\phantom{0}90 && \hbox{($20 - 11 = 9$)} \\[-3pt]
\underline{\phantom{0}88} && \hbox{($8 \times 11 = 88$)} \\[-3pt]
\phantom{00}2
\end{array}$$Hence, \(\frac{2}{11} \) can be written as 0.181818...
Therefore, it is a non-terminating decimal.
vi) We have, \(\frac{329}{400} \)
Dividing 329 by 400, we get,
$$ \require{enclose}
\begin{array}{rll}
0.8225 && \hbox{(Explanations)} \\[-3pt]
400 \enclose{longdiv}{3290}\kern-.2ex \\[-3pt]
\underline{3200\phantom{00}} && \hbox{($8 \times 400 = 3200$)} \\[-3pt]
\phantom{0}900 && \hbox{($3290 - 3200 = 9$)} \\[-3pt]
\underline{800\phantom{00}} && \hbox{($2 \times 400 = 800$)} \\[-3pt]
\phantom{0}1000 && \hbox{($900 - 800 = 100$)} \\[-3pt]
\underline{800\phantom{00}} && \hbox{($2 \times 400 = 800$)} \\[-3pt]
\phantom{0}2000 && \hbox{($1000 - 800 = 200$)} \\[-3pt]
\underline{\phantom{0}2000} && \hbox{($5 \times 400 = 400$)} \\[-3pt]
\phantom{00}0
\end{array}$$Hence, \(\frac{329}{400} \) can be written as 0.8225
Therefore, it is a terminating decimal.