Basic Maths: Test 40

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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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