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## Algebra: Basics Test-6

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*Algebra: Basics Test-6*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

$ \displaystyle if\,\,\,{{7}^{g}}\,\,\,=\frac{1}{343},$<br
then the value of g is

1/3 | |

-1/2 | |

-3 | |

1/7 |

Question 1 Explanation:

7

7

7

g = -3

^{g}= 1/3437

^{g}= 1/7^{3}7

^{g}= 7^{-3}g = -3

Question 2 |

If * is an operation such that d*f= 3d +2f, then 2*3+3*4 is equal to:

39 | |

25 | |

27 | |

29 |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}Here,\,\,\,d*f=3d+2f\\Therefore\,\,\,\,\,2*3+3*4\\=\left( 3\times 2+2\times 3 \right)+\left( 3\times 3+2\times 4 \right)\\=12+17=29\end{array}$

Question 3 |

If 0.13 Ã·d

^{2}=13, then d is equal to10.001 | |

0.101 | |

0.1000 | |

10.011 |

Question 3 Explanation:

$ \displaystyle \begin{array}{l}0.13\div {{d}^{2}}=13\\\Rightarrow \frac{0.13}{{{d}^{2}}}=13\\\Rightarrow {{d}^{2}}=\frac{0.13}{13}=\frac{1}{100}\\\Rightarrow d=\frac{1}{10}=0.1\end{array}$

Question 4 |

If P=4.36, Q=2.39 and R=1.97, then the value of P

^{3}-Q^{3}-R^{3}-3PQR-1 | |

-1.3 | |

2 | |

0 |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}Here,\,\,\,P-Q-R\\=4.36-2.39-1.97=0\\Therefore\,\,\,\,\,{{P}^{3}}-{{Q}^{3}}-{{R}^{3}}=3PQR\\\Rightarrow {{P}^{3}}-{{Q}^{3}}-{{R}^{3}}-3PQR=0\end{array}$

Question 5 |

For what value (s) of p is

$ \displaystyle q+\frac{1}{4}\sqrt{q}+{{p}^{2}}$ a perfect square?

$ \displaystyle q+\frac{1}{4}\sqrt{q}+{{p}^{2}}$ a perfect square?

$ \displaystyle \pm \frac{1}{18}$ | |

$ \displaystyle \pm \frac{1}{8}$ | |

$ \displaystyle -\frac{1}{5}$ | |

$ \displaystyle \frac{1}{4}$ |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}q+\frac{1}{4}\sqrt{q}+{{p}^{2}}\\={{\left( \sqrt{q} \right)}^{2}}+2.\sqrt{q.}\frac{1}{8}+{{\left( p \right)}^{2}}\\clearly\,\,\,q=\frac{1}{8}\\Then,\,\,\,\,\exp ression={{\left( \sqrt{q}+\frac{1}{8} \right)}^{2}}\end{array}$

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