- 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 43

Congratulations - you have completed *Basic Maths: Test 43*.

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Question 1 |

The simplified value of
\[{{\left[ {{\left( 0.444 \right)}^{3}}+{{\left( 0.555 \right)}^{3}}-{{\left( 0.999 \right)}^{3}}+3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right) \right]}^{3}}\]
is:

0.999 | |

0 | |

0.888 | |

0.111 |

Question 1 Explanation:

If a +b +c= 0,

$\begin{align} & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\ & Here,\,0.444+0.555+\left( -0.999 \right)=0 \\ & Therefore\,\,{{\left( 0.444 \right)}^{3}}+{{\left( 0.555 \right)}^{3}}+{{\left( -0.999 \right)}^{3}} \\ & =-3\times 0.444\times 0.555\times 0.999 \\ & Therefore\,\,\,\exp ression \\ & =\left[ -3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right)+3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right) \right]=0 \\ \end{align}$

$\begin{align} & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\ & Here,\,0.444+0.555+\left( -0.999 \right)=0 \\ & Therefore\,\,{{\left( 0.444 \right)}^{3}}+{{\left( 0.555 \right)}^{3}}+{{\left( -0.999 \right)}^{3}} \\ & =-3\times 0.444\times 0.555\times 0.999 \\ & Therefore\,\,\,\exp ression \\ & =\left[ -3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right)+3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right) \right]=0 \\ \end{align}$

Question 2 |

$\begin{align}
& \frac{{{\left( 2.53-4.07 \right)}^{2}}}{\left( 4.07-3.15 \right)\left( 3.15-2.53 \right)} \\
& +\frac{{{\left( 4.07-3.15 \right)}^{2}}}{\left( 3.15-2.53 \right)\left( 2.53-4.07 \right)} \\
& +\frac{{{\left( 3.15-2.53 \right)}^{2}}}{\left( 2.53-4.07 \right)\left( 4.07-3.15 \right)} \\
\end{align}$
is simplified to

0 | |

1 | |

2 | |

3 |

Question 2 Explanation:

$\begin{align}
& Let\,\,\left( 2.53-4.07 \right)\,=a\,\,\left( 4.07-3.15 \right)=b\,\, \\
& and\,\,\left( 3.15-2.53 \right)=c\,\,Therefore\,\,a+b+c=0 \\
& Therefore\,\,\,\exp ression \\
& =\frac{{{a}^{2}}}{bc}+\frac{{{b}^{2}}}{ac}+\frac{{{c}^{2}}}{ab} \\
& =\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}=\frac{3abc}{abc}=3 \\
& \left[ if\,\,a+b+c=0,\,{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \right] \\
\end{align}$

Question 3 |

$\frac{13}{15}\times \frac{13}{15}+\frac{3}{15}\times \frac{3}{15}-\frac{13}{15}\times \frac{6}{15}$
is equal to

1 | |

2/3 | |

4/9 | |

4/3 |

Question 3 Explanation:

$\begin{align}
& =\frac{13}{15}\times \frac{13}{15}+\frac{3}{15}\times \frac{3}{15}-2\times \frac{13}{15}\times \frac{3}{15} \\
& ={{\left( \frac{13}{15}-\frac{3}{15} \right)}^{2}} \\
& ={{\left( \frac{13-3}{15} \right)}^{2}} \\
& ={{\left( \frac{10}{15} \right)}^{2}} \\
& =\frac{4}{9} \\
\end{align}$

Question 4 |

The value of
$\frac{0.893\times 0.893-0.107\times 0.107}{0.893-0.107}$
is:

0.408 | |

0.59 | |

0.592 | |

1 |

Question 4 Explanation:

Let 0.893= a

And 0.107= b

Therefore expression

$\begin{align} & =\frac{a\times a-b\times b}{a-b} \\ & =\frac{{{a}^{2}}-{{b}^{2}}}{a-b} \\ & =\frac{\left( a+b \right)\,\left( a-b \right)}{a-b} \\ & =a+b=0.893+0.107 \\ & =1 \\ \end{align}$

And 0.107= b

Therefore expression

$\begin{align} & =\frac{a\times a-b\times b}{a-b} \\ & =\frac{{{a}^{2}}-{{b}^{2}}}{a-b} \\ & =\frac{\left( a+b \right)\,\left( a-b \right)}{a-b} \\ & =a+b=0.893+0.107 \\ & =1 \\ \end{align}$

Question 5 |

${{\left( 4\frac{5}{17}+\frac{17}{73} \right)}^{2}}-{{\left( 4\frac{5}{17}-\frac{17}{73} \right)}^{2}}$
is equal to:

1 | |

2 | |

3 | |

4 |

Question 5 Explanation:

$\begin{align}
& Let\,4\frac{5}{17}\,\,=a\,\,\,and\,\,\frac{17}{73}=b. \\
& Therefore\,\,\,\exp ression \\
& ={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}} \\
& =\left( {{a}^{2}}+{{b}^{2}}+2ab \right)-\left( {{a}^{2}}+{{b}^{2}}-2ab \right) \\
& =4ab \\
& =4\times 4\frac{5}{17}\times \frac{17}{73} \\
& =4\times \frac{73}{17}\times \frac{17}{73} \\
& =4 \\
\end{align}$

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