- 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 59

Congratulations - you have completed *Basic Maths: Test 59*.

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Your answers are highlighted below.

Question 1 |

30 | |

140 | |

225 | |

9 |

Question 1 Explanation:

$\begin{align}
& \,\,\sqrt{1225}\times \frac{35}{100}\div 0.01={{\left( ? \right)}^{2}}\div 16 \\
& \Rightarrow 35\times \frac{35}{100}\times 100=\frac{{{?}^{2}}}{16} \\
& \Rightarrow {{?}^{2}}=35\times 35\times 16 \\
& ={{35}^{2}}\times {{4}^{2}} \\
& =?=35\times 4=140 \\
\end{align}$

Question 2 |

$\frac{225\times 225-121\times 121}{104}$ is equal to

(a) 420 | |

450 | |

350 | |

320 |

Question 2 Explanation:

$\begin{align}
& If\,\,225=a\,\,and\,\,\,121=b,\,\,then \\
& Expression \\
& =\frac{{{a}^{2}}-{{b}^{2}}}{a-b} \\
& \left[ a-b=225-121=104 \right] \\
& =\frac{\left( a+b \right)\left( a-b \right)}{\left( a-b \right)}=a+b \\
& =225+121 \\
& =350 \\
\end{align}$

Question 3 |

$\left[ 3\sqrt{63}-6\sqrt{\frac{7}{36}}-\sqrt{112} \right]$ is equal to

0 | |

4 | |

$4\sqrt{7}$ | |

$\sqrt{7}$ |

Question 3 Explanation:

$\begin{align}
& =3\sqrt{63}-6\sqrt{\frac{7}{36}}-\sqrt{112} \\
& =3\sqrt{9\times 7}-\sqrt{\frac{7\times 6\times 6}{6\times 6}}-\sqrt{16\times 7} \\
& =3\times 3\sqrt{7}-\sqrt{7}-4\sqrt{7}=4\sqrt{7} \\
\end{align}$

Question 4 |

$\left[ 7.3\times 7.3+2\times 7.3\times 2.7+2.7\times 2.7 \right]$ is equal to

1.69 | |

10 | |

75.69 | |

100 |

Question 4 Explanation:

$\begin{align}
& ={{\left( 7.3+2.7 \right)}^{2}} \\
& \left[ {{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}} \right] \\
& ={{10}^{2}} \\
& =100 \\
\end{align}$

Question 5 |

$\frac{{{\left( 2.56 \right)}^{3}}-{{\left( 1.28 \right)}^{3}}}{{{\left( 2.56 \right)}^{2}}+2.56\times 1.28+{{\left( 1.28 \right)}^{2}}}$ is equal to

1.28 | |

5.24 | |

1.26 | |

2.92 |

Question 5 Explanation:

$\begin{align}
& Let\,\,2.56=a\,\,and\,\,1.28=b \\
& Therefore\,\,\exp ression \\
& =\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}} \\
& =\frac{\left( a-b \right)\left( {{a}^{2}}+ab+b \right)}{{{a}^{2}}+ab+{{b}^{2}}} \\
& =a-b=2.56-1.28=1.28 \\
\end{align}$

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