- 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 44

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Question 1 |

$\frac{{{\left( 3.2 \right)}^{3}}+0.008}{\left( {{3.2}^{2}}-0.0256+0.04 \right)}$
is equal to:

3.60 | |

3.00 | |

3.33 | |

3.40 |

Question 1 Explanation:

$\begin{align}
& \frac{{{\left( 3.2 \right)}^{3}}+{{\left( 0.2 \right)}^{3}}}{{{\left( 3.2 \right)}^{2}}-3.2\times 0.2+0.2\times 0.2} \\
& Let\,\,\,3.2=a\,\,\,\,and\,\,0.2=b \\
& Therefore\,\,\,\exp ression \\
& =\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{2}}-ab+{{b}^{2}}} \\
& =a+b \\
& =3.2+0.2 \\
& =3.4 \\
\end{align}$

Question 2 |

The value of \[\frac{0.4\times 0.4\times 0.4+0.06\times 0.06\times 0.06}{0.8\times 0.8\times 0.8+0.12\times 0.12\times 0.12}\] is:

0.0125 | |

0.125 | |

0.25 | |

0.5 |

Question 2 Explanation:

Let 0.4= a Therefore 0.8= 2a

And 0.06= b therefore 0.12= 2b

Therefore expression

$\begin{align} & =\frac{{{a}^{3}}+{{b}^{3}}}{8{{a}^{3}}+8{{b}^{3}}} \\ & =\frac{{{a}^{3}}+{{b}^{3}}}{8\left( {{a}^{3}}+{{b}^{3}} \right)} \\ & =\frac{1}{8}=0.125 \\ \end{align}$

And 0.06= b therefore 0.12= 2b

Therefore expression

$\begin{align} & =\frac{{{a}^{3}}+{{b}^{3}}}{8{{a}^{3}}+8{{b}^{3}}} \\ & =\frac{{{a}^{3}}+{{b}^{3}}}{8\left( {{a}^{3}}+{{b}^{3}} \right)} \\ & =\frac{1}{8}=0.125 \\ \end{align}$

Question 3 |

The value of
$\frac{7.15\times 7.15\times 7.15-3.97\times 3.97\times 3.97}{7.15\times 7.15+7.15\times 3.97+3.97\times 3.97}$
in simplified form is:

3.15 | |

3.18 | |

7.15 | |

2.18 |

Question 3 Explanation:

Let 7.15= a and 3.97= b

Therefore expression

$\begin{align} & =\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}} \\ & =\frac{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{{{a}^{2}}+ab+{{b}^{2}}} \\ & =a-b=7.15-3.97=3.18 \\ \end{align}$

Therefore expression

$\begin{align} & =\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}} \\ & =\frac{\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)}{{{a}^{2}}+ab+{{b}^{2}}} \\ & =a-b=7.15-3.97=3.18 \\ \end{align}$

Question 4 |

The value of
$\frac{{{\left( 4.5 \right)}^{3}}+{{\left( 3.7 \right)}^{3}}+{{\left( 1.8 \right)}^{3}}-3\times 4.5\times 3.7\times 1.8}{{{\left( 4.5 \right)}^{2}}+{{\left( 3.7 \right)}^{2}}+{{\left( 1.8 \right)}^{2}}-4.5\times 3.7-3.7\times 1.8-1.8\times 4.5}$
is:

0 | |

1 | |

10 | |

30 |

Question 4 Explanation:

Let 4.5= a, 3.7= b, 1.8= c

Therefore expression

$\begin{align} & =\frac{{{a}^{3}}+{{b}^{3}}{{+}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca} \\ & =a+b+c \\ & =4.5+3.7+1.8 \\ & =10 \\ \end{align}$

Therefore expression

$\begin{align} & =\frac{{{a}^{3}}+{{b}^{3}}{{+}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca} \\ & =a+b+c \\ & =4.5+3.7+1.8 \\ & =10 \\ \end{align}$

Question 5 |

The value of $\overline{0.5}+\overline{0.7}+0.\overline{57}$ is:

$1.\overline{87}$ | |

$1.\overline{90}$ | |

$1.\overline{82}$ | |

$0.\overline{98}$ |

Question 5 Explanation:

$\begin{align}
& \overline{0.5}+\overline{0.7}+0.\overline{57} \\
& =\frac{5}{9}+\frac{7}{9}+\frac{57}{99} \\
& =\frac{55+77+57}{99} \\
& =\frac{189}{99}=1\frac{90}{99} \\
& =1.\overline{90} \\
\end{align}$

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