- 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 36

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Question 1 |

$\begin{align}
& ifI=\frac{2}{3}\div \frac{7}{6}, \\
& II=2\div \left[ \left( 7\div 8 \right)\div (1\div 4) \right], \\
& III=\left[ 6\div \left( 7\div 5 \right) \right]\div 6, \\
& IV=3\div 2\left( 5\div 4 \right) \\
& then \\
\end{align}$

I and II are equal | |

I and IV are equal | |

I and III are equal | |

All are equal |

Question 1 Explanation:

$\begin{align}
& I.\,=\frac{2}{3}\times \frac{6}{7}=\frac{4}{7} \\
& II.\,\,=2\div \left[ \frac{7}{8}\times 4 \right] \\
& =2\div \frac{7}{2}=\frac{4}{7} \\
& III.\,\,\,=\left[ 6\div \frac{7}{5} \right]\div 6=\frac{30}{7}\div 6=\frac{5}{7} \\
& IV.\,\,=3\div 2\times \frac{5}{4}=3\div \frac{5}{2}=\frac{6}{5} \\
\end{align}$

Obviously, (I) and (II) are equal

Obviously, (I) and (II) are equal

Question 2 |

The value of $\frac{0.062\times 0.062\times 0.062+0.033\times 0.033\times 0.033}{0.062\times 0.062-0.062\times 0.033+0.033\times 0.033}$ is:

0.95 | |

0.095 | |

0.0095 | |

0.00095 |

Question 2 Explanation:

Let 0.062=x and 0.033= y

Therefore given expression

$\begin{align} & =\frac{{{x}^{3}}+{{y}^{3}}}{{{x}^{2}}-xy+{{y}^{2}}} \\ & =\frac{\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)}{{{x}^{2}}-xy+{{y}^{2}}} \\ & =x+y=0.062+0.033 \\ & =0.095 \\ \end{align}$

Therefore given expression

$\begin{align} & =\frac{{{x}^{3}}+{{y}^{3}}}{{{x}^{2}}-xy+{{y}^{2}}} \\ & =\frac{\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)}{{{x}^{2}}-xy+{{y}^{2}}} \\ & =x+y=0.062+0.033 \\ & =0.095 \\ \end{align}$

Question 3 |

The simplification of
$1.\overline{38}-3.\overline{15}+2.\overline{53}$
equals:

0.76 | |

$0.\overline{76}$ | |

2.64 | |

$2.\overline{64}$ |

Question 3 Explanation:

$\begin{align}
& 1.\overline{38}-3.\overline{15}+2.\overline{53} \\
& =1\frac{38}{99}-3\frac{15}{99}+2\frac{53}{99} \\
& =1+\frac{38}{99}-3-\frac{15}{99}+2+\frac{53}{99} \\
& =\left( 1-3+2 \right)+\left( \frac{38}{99}-\frac{15}{99}+\frac{53}{99} \right) \\
& =0+\left( \frac{38-15+53}{99} \right) \\
& =\frac{76}{99} \\
& =0.\overline{76} \\
\end{align}$

Question 4 |

The value of
\[\frac{0.8\times 0.8\times 0.8+0.1\times 0.1\times 0.1+0.7\times 0.7\times 0.7-3\times 0.8\times 0.1\times 0.7}{0.8\times 0.8+0.1\times 0.1+0.7\times 0.7-0.8\times 0.1-0.1\times 0.7-0.7\times 0.8}\] is

1.6 | |

0.6 | |

0.16 | |

1.0 |

Question 4 Explanation:

Let 0.8= x, 0.1= y and 0.7= z

Then, the given expression

$\begin{align} & \frac{x\times x\times x+y\times y\times y+z\times z\times z-3\times x\times y\times z}{x\times x+y\times y+z\times z-x\times y-y\times z-z\times x} \\ & =\frac{{{x}^{3}}+{{y}^{3}}+{{x}^{3}}-3xyz}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx} \\ & =\frac{\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx} \\ & =x+y+z \\ & =0.8+0.1+0.7 \\ & =1.6 \\ \end{align}$

Then, the given expression

$\begin{align} & \frac{x\times x\times x+y\times y\times y+z\times z\times z-3\times x\times y\times z}{x\times x+y\times y+z\times z-x\times y-y\times z-z\times x} \\ & =\frac{{{x}^{3}}+{{y}^{3}}+{{x}^{3}}-3xyz}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx} \\ & =\frac{\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx} \\ & =x+y+z \\ & =0.8+0.1+0.7 \\ & =1.6 \\ \end{align}$

Question 5 |

The simplified value of
$\left( 1+\frac{1}{3} \right)\,\left( 1+\frac{1}{4} \right)\,\left( 1+\frac{1}{5} \right)\,.....\left( 1+\frac{1}{99} \right)\,\left( 1+\frac{1}{100} \right)$
is

$\frac{3}{99}$ | |

$\frac{1}{101}$ | |

$\frac{101}{3}$ | |

$\frac{3}{100}$ |

Question 5 Explanation:

$\begin{align}
& \frac{4}{3}\times \frac{5}{4}\times \frac{6}{5}\times .......\times \frac{100}{99}\times \frac{101}{100} \\
& =\frac{101}{3} \\
\end{align}$

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