- 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 34

Congratulations - you have completed *Basic Maths: Test 34*.

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Question 1 |

\[\frac{1}{60}+\frac{1}{90}+\frac{1}{126}+\frac{1}{168}+\frac{1}{216}+\frac{1}{270}+\frac{1}{330}+\frac{1}{396}\]

is equal to:

is equal to:

$\frac{1}{18}$ | |

$\frac{1}{12}$ | |

$\frac{1}{6}$ | |

$\frac{1}{10}$ |

Question 1 Explanation:

\[\begin{align}
& =\frac{1}{60}+\frac{1}{90}+\frac{1}{126}+\frac{1}{168}+\frac{1}{216}+\frac{1}{270}+\frac{1}{330}+\frac{1}{396} \\
& =\frac{1}{3}\left[ \frac{1}{4\times 5}+\frac{1}{5\times 6}+\frac{1}{6\times 7}+.....+\frac{1}{11\times 12} \right] \\
& =\frac{1}{3}\left[ \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{12} \right] \\
& =\frac{1}{12}-\frac{1}{36}=\frac{3-1}{36}=\frac{2}{36}=\frac{1}{18} \\
\end{align}\]

Question 2 |

Simplify

$\frac{10}{21}-\left[ \frac{3}{7}+\left\{ \frac{1}{7}+\left( \frac{1}{3}-\overline{\frac{10}{35}+\frac{1}{7}} \right) \right\} \right]$

$\frac{10}{21}-\left[ \frac{3}{7}+\left\{ \frac{1}{7}+\left( \frac{1}{3}-\overline{\frac{10}{35}+\frac{1}{7}} \right) \right\} \right]$

0 | |

1 | |

$\frac{9}{20}$ | |

$\frac{9}{10}$ |

Question 2 Explanation:

$\begin{align}
& \frac{10}{21}-\left[ \frac{3}{7}+\left\{ \frac{1}{7}+\left( \frac{1}{3}-\overline{\frac{10}{35}+\frac{1}{7}} \right) \right\} \right] \\
& =\frac{10}{21}-\left[ \frac{3}{7}+\left\{ \frac{1}{7}+\left( \frac{1}{3}-\frac{3}{7} \right) \right\} \right] \\
& =\frac{10}{21}-\left[ \frac{3}{7}+\left\{ \frac{1}{7}+\left( -\frac{2}{21} \right) \right\} \right] \\
& =\frac{10}{21}-\left[ \frac{3}{7}+\left\{ \frac{1}{21} \right\} \right] \\
& \frac{10}{21}-\left[ \frac{3}{7}+\frac{1}{21} \right]=\frac{10}{21}-\frac{10}{21}=0 \\
\end{align}$

Question 3 |

Simplify

\[\frac{0.6\overline{8}\div 6.2}{3.2\overline{13}-0.7\overline{68}}\,\]

is equal to:

\[\frac{0.6\overline{8}\div 6.2}{3.2\overline{13}-0.7\overline{68}}\,\]

is equal to:

0.454 | |

0.045 | |

0.063 | |

0.054 |

Question 3 Explanation:

$\begin{align}
& =\frac{0.6\overline{8}\div 6.2}{3.2\overline{13}-0.7\overline{68}}\,=\frac{\frac{68-6}{90}\div 6.2}{3\frac{211}{990}-\frac{768-7}{990}} \\
& =\frac{\frac{62}{90}\div 6.2}{3\frac{211}{990}-\frac{761}{990}}\,\,\,\,\,\,\,\,\,=\,\,\frac{\frac{62}{90}\div 6.2}{\frac{2420}{990}} \\
& \\
& =\frac{62}{90\times 6.2}\times \frac{990}{2420}\,\,\,\,=\frac{11}{242}=0.045 \\
\end{align}$

Question 4 |

The value of

$\sqrt{\frac{{{\left( 5.2 \right)}^{2}}+{{\left( 52.2 \right)}^{2}}+{{\left( 522.2 \right)}^{2}}}{{{\left( 0.52 \right)}^{2}}+{{\left( 5.22 \right)}^{2}}+{{\left( 52.22 \right)}^{2}}}}$is

$\sqrt{\frac{{{\left( 5.2 \right)}^{2}}+{{\left( 52.2 \right)}^{2}}+{{\left( 522.2 \right)}^{2}}}{{{\left( 0.52 \right)}^{2}}+{{\left( 5.22 \right)}^{2}}+{{\left( 52.22 \right)}^{2}}}}$is

0.1 | |

1.1 | |

10 | |

100 |

Question 4 Explanation:

$\begin{align}
& \sqrt{\frac{{{\left( 5.2 \right)}^{2}}+{{\left( 52.2 \right)}^{2}}+{{\left( 522.2 \right)}^{2}}}{{{\left( 0.52 \right)}^{2}}+{{\left( 5.22 \right)}^{2}}+{{\left( 52.22 \right)}^{2}}}} \\
& =\sqrt{\frac{{{\left( 10\times 0.52 \right)}^{2}}+{{\left( 10\times 5.22 \right)}^{2}}+{{\left( 10\times 52.22 \right)}^{2}}}{{{\left( 0.52 \right)}^{2}}+{{\left( 5.22 \right)}^{2}}+{{\left( 52.22 \right)}^{2}}}} \\
& =\sqrt{100}=10 \\
\end{align}$

Question 5 |

For what value of *, statement

$\left[ \frac{\left( * \right)}{29}\times \frac{\left( * \right)}{261} \right]=1$is correct?

$\left[ \frac{\left( * \right)}{29}\times \frac{\left( * \right)}{261} \right]=1$is correct?

261 | |

87 | |

63 | |

21 |

Question 5 Explanation:

$\begin{align}
& \left[ \frac{\left( H \right)}{29}\times \frac{\left( H \right)}{261} \right]=1 \\
& \Rightarrow {{\left( H \right)}^{2}}=29\times 261 \\
& \Rightarrow H=\sqrt{29\times 261}=87 \\
\end{align}$

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