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Basic Maths: Test 40

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Question 1
The expression ${{729}^{0.9}}\times {{243}^{-0.08}}$ is equal to
A
81
B
243
C
323
D
729
Question 1 Explanation:
$\begin{align} & {{\left( 729 \right)}^{0.9}}\times {{\left( 243 \right)}^{-0.08}} \\ & ={{\left( {{3}^{6}} \right)}^{0.9}}\times {{\left( {{3}^{5}} \right)}^{-0.08}} \\ & ={{3}^{5.4}}\times {{3}^{-0.4}}={{3}^{5.4-0.4}} \\ & ={{3}^{5}}=243 \\ \end{align}$
Question 2
The value of $\sqrt[4]{63-36\sqrt{3}}\times \sqrt{6+3\sqrt{3}}$ is
A
$\sqrt[3]{3}$
B
$\sqrt[6]{3}$
C
$\sqrt{3}$
D
3
Question 2 Explanation:
$\begin{align} & Expressoin \\ & =\sqrt[4]{63-36\sqrt{3}}\times \sqrt{6+3\sqrt{3}} \\ & =\sqrt[4]{9\left( 7-4\sqrt{3} \right)}\times \sqrt{6+3\sqrt{3}} \\ & =\sqrt[4]{9\left( 4+3-2\times 2\times \sqrt{3} \right)}\times \sqrt{6+3\sqrt{3}} \\ & =\sqrt[4]{9{{\left( 2-\sqrt{3} \right)}^{2}}}\times \sqrt{6+3\sqrt{3}} \\ & =\sqrt[{}]{3\left( 2-\sqrt{3} \right)}\times \sqrt{6+3\sqrt{3}} \\ & =\sqrt[{}]{\left( 6-3\sqrt{3} \right)\left( 6+3\sqrt{3} \right)}=\sqrt[{}]{36-27} \\ & =\sqrt[{}]{9}=3 \\ \end{align}$
Question 3
The simplification of $\frac{1}{4}+\frac{1}{{{4}^{2}}}+\frac{1}{{{4}^{3}}}+\frac{1}{{{4}^{4}}}+\frac{1}{{{4}^{5}}}$up to three-places of decimals yields
A
0.133
B
0.163
C
0.333
D
0.713
Question 3 Explanation:
Expression $0.25+0.0625+0.015625+0.00390625+0.0009765625=0.333$
Question 4
$\frac{13.3\times 13.3\times 13.3+1}{13.3\times 13.3-13.3+1}$ is equal to:
A
15.3
B
14.3
C
13.3
D
12.3
Question 4 Explanation:
$\begin{align} & \frac{13.3\times 13.3\times 13.3+1}{13.3\times 13.3-13.3+1} \\ & Let\,\,13.3=a\,and\,\,1=b,\, \\ & Then, \\ & Expression\,\,=\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{2}}-ab+{{b}^{2}}} \\ & =\frac{\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)}{{{a}^{2}}-ab+{{b}^{2}}} \\ & =a+b=13.3+1=14.3 \\ \end{align}$
Question 5
$\frac{1.51\times 1.51-0.49\times 0.49}{1.51-0.49}$ is equal to:
A
0.20
B
20.00
C
2.00
D
22.00
Question 5 Explanation:
$\begin{align} & Let,\text{ }1.51=\text{ }a\text{ }and\text{ }0.49=\text{ }b \\ & Therefore\,\,\frac{{{a}^{2}}-{{b}^{2}}}{a-b} \\ & =\frac{\left( a+b \right)\left( a-b \right)}{\left( a-b \right)}=a+b \\ & Therefore\,\,1.51+0.49=2 \\ \end{align}$
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