- 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 47

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Question 1 |

$3.5\overline{625}$ is equal to

$3\frac{4394}{995}$ | |

$3\frac{292}{333}$ | |

$3\frac{562}{999}$ | |

$3\frac{878}{999}$ |

Question 1 Explanation:

\[\begin{align}
& 3.5\overline{625}=\frac{35625-35}{9990} \\
& =\frac{35590}{9990}=3\frac{562}{999} \\
\end{align}\]

Question 2 |

$0.65\times 6.5-2\times 6.5\times 0.35+0.35\times 3.5$ is equal to

9 | |

90 | |

9.9 | |

0.9 |

Question 2 Explanation:

Expression

$\begin{align} & 0.65\times 6.5-2\times 6.5\times 0.35+0.35\times 3.5 \\ & =0.65\times 0.65\times 10-2\times 0.65\times 0.35\times 10+0.35\times 0.35\times 10 \\ & =10\left( {{\left( 0.65 \right)}^{2}}-2\times 0.65\times 0.35+{{\left( 0.35 \right)}^{2}} \right) \\ & =10{{\left( 0.65-0.35 \right)}^{2}} \\ & =10\times 0.09 \\ & =0.9 \\ \end{align}$

$\begin{align} & 0.65\times 6.5-2\times 6.5\times 0.35+0.35\times 3.5 \\ & =0.65\times 0.65\times 10-2\times 0.65\times 0.35\times 10+0.35\times 0.35\times 10 \\ & =10\left( {{\left( 0.65 \right)}^{2}}-2\times 0.65\times 0.35+{{\left( 0.35 \right)}^{2}} \right) \\ & =10{{\left( 0.65-0.35 \right)}^{2}} \\ & =10\times 0.09 \\ & =0.9 \\ \end{align}$

Question 3 |

$\frac{{{\left( 7.89 \right)}^{2}}-{{\left( 4.56 \right)}^{2}}}{7.89+4.56}$
is simplified to

3.3 | |

3.33 | |

3.23 | |

3.26 |

Question 3 Explanation:

Let 7.89= a and 4.56 = b

Therefore expression

$\begin{align} & =\frac{{{a}^{2}}-{{b}^{2}}}{a+b} \\ & =\frac{\left( a-b \right)\left( a+b \right)}{a+b} \\ & =a-b=7.89-4.56=3.33 \\ \end{align}$

Therefore expression

$\begin{align} & =\frac{{{a}^{2}}-{{b}^{2}}}{a+b} \\ & =\frac{\left( a-b \right)\left( a+b \right)}{a+b} \\ & =a-b=7.89-4.56=3.33 \\ \end{align}$

Question 4 |

A sum of Rs.27.15 is made of 170 coins which are either 10 paise coins of 25 paise coins. The number of 10 paise coins is:

119 | |

125 | |

109 | |

96 |

Question 4 Explanation:

Let the number of 10 paise coins be x

Number of 25 paise coins = 180 - x

According to the question,

$\begin{align} & x\times 10+25\,\,\left( 170-x \right) \\ & =27.15\times 100 \\ & \Rightarrow 10x+4250-25x=2715 \\ & \Rightarrow 15x=4500-2715=1785 \\ & \Rightarrow x=\frac{1785}{15}=119 \\ & \therefore \,\,Number\,\,\,of\,\,10\,\,paise\,\,coins\,\,\,=119 \\ \end{align}$

Number of 25 paise coins = 180 - x

According to the question,

$\begin{align} & x\times 10+25\,\,\left( 170-x \right) \\ & =27.15\times 100 \\ & \Rightarrow 10x+4250-25x=2715 \\ & \Rightarrow 15x=4500-2715=1785 \\ & \Rightarrow x=\frac{1785}{15}=119 \\ & \therefore \,\,Number\,\,\,of\,\,10\,\,paise\,\,coins\,\,\,=119 \\ \end{align}$

Question 5 |

(25)

^{3}÷ (5)^{2 }=?5 | |

625 | |

125 | |

25 |

Question 5 Explanation:

$?=\frac{25\times 25\times 25}{5\times 5}=625$

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