- This is an assessment test.
- These tests focus on the basics of Maths and are meant to indicate your preparation level for the subject.
- Kindly take the tests in this series with a pre-defined schedule.

## Basic Maths: Test 40

Congratulations - you have completed *Basic Maths: Test 40*.

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Question 1 |

The expression ${{729}^{0.9}}\times {{243}^{-0.08}}$ is equal to

81 | |

243 | |

323 | |

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

$\sqrt[3]{3}$ | |

$\sqrt[6]{3}$ | |

$\sqrt{3}$ | |

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

0.133 | |

0.163 | |

0.333 | |

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:

15.3 | |

14.3 | |

13.3 | |

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:

0.20 | |

20.00 | |

2.00 | |

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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