- 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 31

Congratulations - you have completed *Basic Maths: Test 31*.

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Question 1 |

Simplify: $\frac{{{\left( 4.5 \right)}^{3}}+{{\left( 1.7 \right)}^{3}}+{{\left( 2.8 \right)}^{3}}-3\times 4.5\times 1.7\times 2.8}{{{\left( 4.5 \right)}^{2}}+{{\left( 1.7 \right)}^{2}}+{{\left( 2.8 \right)}^{2}}-4.5\times 1.7-1.7\times 2.8-2.8\times 4.5}$

9 | |

11 | |

12 | |

3 |

Question 1 Explanation:

$\frac{{{\left( 4.5 \right)}^{3}}+{{\left( 1.7 \right)}^{3}}+{{\left( 2.8 \right)}^{3}}-3\times 4.5\times 1.7\times 2.8}{{{\left( 4.5 \right)}^{2}}+{{\left( 1.7 \right)}^{2}}+{{\left( 2.8 \right)}^{2}}-4.5\times 1.7-1.7\times 2.8-2.8\times 4.5}$
$\begin{align}
& =\frac{\left( 4.5+1.7+2.8 \right)\,\left( {{4.5}^{2}}+{{1.7}^{2}}+{{2.8}^{2}}-4.5\times 1.7-1.7\times 2.8-2.8\times 4.5 \right)}{{{\left( 4.5 \right)}^{2}}+{{\left( 1.7 \right)}^{2}}+{{\left( 2.8 \right)}^{2}}-4.5\times 1.7-1.7\times 2.8-2.8\times 4.5} \\
& =4.5+1.7+2.8=9 \\
\end{align}$

Question 2 |

The value of $0.006\times 0.07\times 0.026\div \left( 0.13\times 0.0003 \right)$is:

2.8 | |

0.28 | |

0.028 | |

2.08 |

Question 2 Explanation:

$\begin{align}
& 0.006\times 0.07\times 0.026\div \left( 0.13\times 0.0003 \right) \\
& =0.006\times 0.07\times 0.026\div \left( 0.000039 \right) \\
& =0.006\times 0.07\times \frac{0.026}{0.000039} \\
& =\frac{0.00001092}{0.000039}=0.28 \\
\end{align}$.

Question 3 |

Find the sum of the following:
\[\left[ \frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208} \right]\div {{\left( \frac{3}{4} \right)}^{2}}\]

$\frac{1}{2}$ | |

1 | |

$\frac{1}{9}$ | |

$\frac{1}{2520}$ |

Question 3 Explanation:

$\begin{align}
& \left[ \frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208} \right]\div {{\left( \frac{3}{4} \right)}^{2}} \\
& =\left[ \frac{1}{4\times 7}+\frac{1}{7\times 10}+\frac{1}{10\times 13}+\frac{1}{13\times 16} \right]\times {{\left( \frac{4}{3} \right)}^{2}} \\
& =\frac{1}{3}\left[ \frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16} \right]\times \frac{16}{9} \\
& =\frac{1}{3}\left[ \frac{1}{4}-\frac{1}{16} \right]\times \frac{16}{9} \\
& =\frac{1}{3}\times \frac{12}{64}\times \frac{16}{9}=\frac{1}{9} \\
\end{align}$

Question 4 |

Simplify:
$\frac{2.68\times 36+2.68\times 24}{{{\left( 6.34 \right)}^{2}}-{{\left( 3.66 \right)}^{2}}}$

7.2 | |

8 | |

6 | |

10 |

Question 4 Explanation:

$\begin{align}
& =\frac{2.68\left( 60 \right)}{\left( 6.34+3.66 \right)\,\left( 6.34-3.66 \right)} \\
& =\frac{2.68(60)}{10\times 2.68}=6 \\
\end{align}$

Question 5 |

The value of $\frac{{{\left( 0.05 \right)}^{2}}-{{\left( 0.02 \right)}^{2}}}{0.05-0.02}$is :

0.4 | |

0.007 | |

0.07 | |

0.04 |

Question 5 Explanation:

$\begin{align}
& \frac{{{\left( 0.05 \right)}^{2}}-{{\left( 0.02 \right)}^{2}}}{0.05-0.02} \\
& \left[ U\sin g\,\,\,{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \right] \\
& =\frac{\left( 0.05+0.02 \right)\,\left( 0.05-0.02 \right)}{0.05-0.02} \\
& =0.05+0.02=0.07 \\
\end{align}$

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