- 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 52

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Question 1 |

The simplified value of
$\left( 1-\frac{1}{5} \right)\,\left( 1-\frac{1}{6} \right)\,\left( 1-\frac{1}{7} \right)\,......\left( 1-\frac{1}{49} \right)\,\left( 1-\frac{1}{50} \right)$<br>
is

$\frac{2}{25}$ | |

$\frac{1}{25}$ | |

$\frac{1}{50}$ | |

$\frac{1}{100}$ |

Question 1 Explanation:

$\begin{align}
& =\left( 1-\frac{1}{5} \right)\,\left( 1-\frac{1}{6} \right)\,\left( 1-\frac{1}{7} \right)\,......\left( 1-\frac{1}{49} \right)\,\left( 1-\frac{1}{50} \right) \\
& =\left( \frac{5-1}{5} \right)\,\left( \frac{6-1}{6} \right)\,\left( \frac{7-1}{7} \right)\,......\left( \frac{49-1}{49} \right)\,\left( \frac{50-1}{50} \right) \\
& =\frac{4}{5}\times \frac{5}{6}\times \frac{6}{7}\times ......\times \frac{48}{49}\times \frac{49}{50} \\
& =\frac{4}{50} \\
& =\frac{2}{25} \\
\end{align}$

Question 2 |

$\frac{125{{\left( 1.3 \right)}^{3}}+1}{{{\left( 6.5 \right)}^{2}}-5.5}$ is equal to

2.75 | |

9/5 | |

4.75 | |

7.5 |

Question 2 Explanation:

$\begin{align}
& =\frac{125{{\left( 1.3 \right)}^{3}}+1}{{{\left( 6.5 \right)}^{2}}-5.5} \\
& =\frac{{{\left( 5\times 1.3 \right)}^{3}}+1}{{{\left( 6.5 \right)}^{2}}-6.5\times 1+{{1}^{2}}} \\
& =\frac{{{\left( 6.5 \right)}^{3}}+1}{{{\left( 6.5 \right)}^{2}}-6.5\times 1+{{1}^{2}}} \\
& \left[ {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \right] \\
& =6.5+1 \\
& =7.5 \\
\end{align}$

Question 3 |

The value of
$\frac{{{\left( 1.987-4.523 \right)}^{2}}+{{\left( 1.987+4.523 \right)}^{2}}}{1.987\times 1.987+4.523\times 4.523}$
is

4 | |

2 | |

2.199 | |

3.195 |

Question 3 Explanation:

Let 1.987=a and 4.523=b

Therefore expression

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

Therefore expression

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

Question 4 |

$\left( 5555\div 110 \right)+\left( 925\div 75 \right)+\left( 5168\div 51 \right)=?$

164.16 | |

174.26 | |

184.16 | |

194.26 |

Question 4 Explanation:

$\begin{align}
& ?=\frac{5555}{110}+\frac{925}{75}+\frac{5168}{51} \\
& =50.5+12.33+101.33 \\
& =164.16 \\
\end{align}$

Question 5 |

$\sqrt{32041}\times \sqrt{1681}-{{\left( 71 \right)}^{2}}={{\left( ? \right)}^{2}}+{{\left( 23 \right)}^{2}}$(? Is approximately equal to)

47 | |

42 | |

44 | |

48 |

Question 5 Explanation:

$\begin{align}
& \sqrt{32041}\times \sqrt{1681}-{{\left( 71 \right)}^{2}}={{\left( ? \right)}^{2}}+{{\left( 23 \right)}^{2}} \\
& \Rightarrow 179\times 41-5041={{?}^{2}}+529 \\
& \Rightarrow 2298={{?}^{2}}+529 \\
& \Rightarrow {{?}^{2}}=2298-529=1769 \\
& \Rightarrow ?=\sqrt{1769}=42.06 \\
\end{align}$

Approximately equals to 42

Approximately equals to 42

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