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Basic Maths: Test 19

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Question 1
$ \displaystyle \left( 0.\overline{11}+0.\overline{22} \right)\times 2$
A
$ \displaystyle 1.\overline{9}$
B
3
C
2/3
D
$ \displaystyle 0.\overline{3}$
Question 1 Explanation:
Expression $ \displaystyle \begin{array}{l}=\left( 0.\overline{11}+0.\overline{22} \right)\times 2\\=\left( \frac{11}{99}+\frac{22}{99} \right)\times 2\\=\frac{33}{99}\times 2\\=2/3\end{array}$
Question 2
$ \displaystyle \sqrt{0.014\times 0.14a}=0.014\times 0.14\sqrt{b},\,\,find\,\,\,the\,\,\,value\,\,\,of\,\,\,\frac{a}{b}.$
A
0.000196
B
0.00196
C
0.0196
D
0.196
Question 2 Explanation:
$\displaystyle \begin{array}{l}=\sqrt{0.014\times 0.14a}=0.014\times 0.14\sqrt{b}\\On\,\,squaring\,\,\,both\,\,sides,\\0.014\times 0.14a={{\left( 0.014 \right)}^{2}}\times {{\left( 0.14 \right)}^{2}}b\\therefore\,\,\,\frac{a}{b}=0.014\times 0.14=0.00196\end{array}$
Question 3
$ \displaystyle \left\{ \frac{{{\left( 0.1 \right)}^{2}}-{{\left( 0.01 \right)}^{2}}}{0.0001}+2 \right\}$
is equal to
A
1010
B
110
C
101
D
100
Question 3 Explanation:
$ \displaystyle \begin{array}{l}=\frac{0.01-0.0001}{0.0001}+2=\frac{0.0099}{0.0001}+2\\=99+2=101\end{array}$
Question 4
$ \displaystyle \sqrt{1\frac{1}{4}\times \frac{64}{125}\times 3.24}$
is equal to
A
$ \displaystyle 1\frac{1}{25}$
B
$ \displaystyle \frac{36}{25}$
C
$ \displaystyle \frac{23}{25}$
D
$ \displaystyle \frac{21}{25}$
Question 4 Explanation:
$ \displaystyle \begin{array}{l}=\sqrt{\frac{5}{4}\times \frac{64}{125}\times 3.24}\\=\sqrt{\frac{16}{25}\times \frac{324}{100}}\\=\frac{4}{5}\times \frac{18}{10}\\=\frac{36}{25}\end{array}$
Question 5
$ \displaystyle \frac{\left( 5+5+5+5 \right)\div 4}{3+3+3+3\div 2}$
A
1
B
$ \displaystyle \frac{3}{10}$
C
$ \displaystyle \frac{4}{9}$
D
$ \displaystyle \frac{10}{21}$
Question 5 Explanation:
$ \displaystyle =\frac{20\div 4}{9+3\div 2}=\frac{10}{21}$
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