- 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 32

Congratulations - you have completed *Basic Maths: Test 32*.

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Your answers are highlighted below.

Question 1 |

$\left( 7.5\times 7.5-37.5+2.5\times 2.5 \right)$ is equal to:

15 | |

5 | |

25 | |

35 |

Question 1 Explanation:

$\begin{align}
& \left( 7.5\times 7.5-37.5+2.5\times 2.5 \right) \\
& =\left[ {{\left( 7.5 \right)}^{2}}-2\times 7.5\times 2.5+{{\left( 2.5 \right)}^{2}} \right] \\
& ={{\left( 7.5-2.5 \right)}^{2}}={{\left( 5 \right)}^{2}}=25 \\
\end{align}$

Question 2 |

$\left( \sqrt{192}-\sqrt{147} \right)\div \sqrt{24}$ is equal to:

$\sqrt{6}$ | |

$\sqrt{3}/2$ | |

$\sqrt{2}/4$ | |

$\sqrt{6}/2$ |

Question 2 Explanation:

$\begin{align}
& \left( \sqrt{192}-\sqrt{147} \right)\div \sqrt{24} \\
& =\frac{\sqrt{192}-\sqrt{147}}{\sqrt{24}} \\
& =\frac{8\sqrt{3}-7\sqrt{3}}{2\sqrt{6}}=\frac{\sqrt{3}}{2\sqrt{2}\sqrt{3}}=\frac{1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\
& =\frac{\sqrt{2}}{4} \\
\end{align}$

Question 3 |

The value of

\[\frac{\sqrt{63}-\sqrt{99}}{\sqrt{28}-\sqrt{44}}\]is:

\[\frac{\sqrt{63}-\sqrt{99}}{\sqrt{28}-\sqrt{44}}\]is:

$\frac{3}{4}$ | |

$\frac{2}{3}$ | |

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

$1\frac{2}{3}$ |

Question 3 Explanation:

$\begin{align}
& \frac{\sqrt{63}-\sqrt{99}}{\sqrt{28}-\sqrt{44}} \\
& =\frac{\sqrt{9\times 7}-\sqrt{9\times 11}}{\sqrt{4\times 7}-\sqrt{4\times 11}} \\
& =\frac{3\sqrt{7}-3\sqrt{11}}{2\sqrt{7}-2\sqrt{11}}=\frac{3\left( \sqrt{7}-\sqrt{11} \right)}{2\left( \sqrt{7}-\sqrt{11} \right)} \\
& =\frac{3}{2}=1\frac{1}{2} \\
\end{align}$

Question 4 |

$\sqrt{12+\sqrt{12+\sqrt{12+.......}}}$is equal to:

6⅔ | |

3
| |

$3\frac{1}{2}$ | |

4 |

Question 4 Explanation:

$\begin{align}
& Let\,\,x\,=\,\sqrt{12+\sqrt{12+\sqrt{12+.......}}} \\
& or,\,x=\sqrt{12+x} \\
& or,\,{{x}^{2}}=12+x \\
& or,\,{{x}^{2}}-x-12=0 \\
& or,\,{{x}^{2}}-4x+3x-12=0 \\
& or\,x\left( x-4 \right)+3\left( x-4 \right)=0 \\
& or,\,\left( x+3 \right)\,\left( x-4 \right)=0 \\
& Therefore\,\,x=-3\,\,or\,x=4 \\
\end{align}$

But the given expression is positive Hence, $x\ne -3$

But the given expression is positive Hence, $x\ne -3$

Question 5 |

$\left( 1.5\times 1.5+16.5+5.5\times 5.5 \right)$ is equal to:

99 | |

49
| |

100 | |

36 |

Question 5 Explanation:

$\begin{align}
& ?=\left( 1.5\times 1.5+16.5+5.5\times 5.5 \right) \\
& =\left[ {{\left( 1.5 \right)}^{2}}+2\times 1.5\times 5.5+{{\left( 5.5 \right)}^{2}} \right] \\
& ={{\left[ 1.5+5.5 \right]}^{2}}={{\left( 7 \right)}^{2}}=49 \\
\end{align}$

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