- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Algebra: Basics Test-4

Congratulations - you have completed

*Algebra: Basics Test-4*.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 |

If p = 0.25, q = âˆ’0.05, r = 0.5, then the value of

$ \displaystyle \frac{{{p}^{2}}-{{q}^{2}}-{{r}^{2}}-2pq}{{{p}^{2}}+{{q}^{2}}-2qr-{{r}^{2}}}$

$ \displaystyle \frac{{{p}^{2}}-{{q}^{2}}-{{r}^{2}}-2pq}{{{p}^{2}}+{{q}^{2}}-2qr-{{r}^{2}}}$

$ \displaystyle \frac{7}{8}$ | |

$ \displaystyle \frac{8}{9}$ | |

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

$ \displaystyle \frac{5}{8}$ |

Question 1 Explanation:

$ \displaystyle \begin{array}{l}\frac{{{p}^{2}}-{{q}^{2}}-{{r}^{2}}-2qr}{{{p}^{2}}+{{q}^{2}}-2pq-{{r}^{2}}}\\=\frac{{{p}^{2}}-\left( {{q}^{2}}+{{r}^{2}}+2qr \right)}{\left( {{p}^{2}}+{{q}^{2}}-2pq \right)-{{r}^{2}}}\\=\frac{{{p}^{2}}-{{\left( q+r \right)}^{2}}}{{{\left( p-q \right)}^{2}}-{{r}^{2}}}\\=\frac{\left( p+q+r \right)\left( p-q-r \right)}{\left( p-q+r \right)\left( p-q-r \right)}\\=\frac{p+q+r}{p-q+r}=\frac{0.25-0.05+0.5}{0.25+0.05+0.5}\\=\frac{0.7}{0.8}=\frac{7}{8}\end{array}$

Question 2 |

If a x b = a

^{2}+ b^{2}â€“ab, then the value of 9 Ã— 11 is103 | |

113 | |

119 | |

129 |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}a\times b={{a}^{2}}+{{b}^{2}}-ab\,\left( Given \right)\\\Rightarrow 9\times 11={{9}^{2}}+{{11}^{2}}-3\times 11\\=81+121-99\\=202-99=103\end{array}$

Question 3 |

If p = 999, then the value of

$ \displaystyle 3\sqrt{p\left( {{p}^{2}}+3p+3 \right)+1}$

$ \displaystyle 3\sqrt{p\left( {{p}^{2}}+3p+3 \right)+1}$

1000 | |

999 | |

998 | |

1002 |

Question 3 Explanation:

p= 999 (Given)

$ \displaystyle \begin{array}{l}Now,\,3\sqrt{p\left( {{p}^{2}}+3p+3 \right)+1}\,\\3\sqrt{{{p}^{3}}+3{{p}^{2}}+3p+1}\,\\=3\sqrt{{{\left( p+1 \right)}^{3}}}\,=p+1\\=333+1=1000\end{array}$

$ \displaystyle \begin{array}{l}Now,\,3\sqrt{p\left( {{p}^{2}}+3p+3 \right)+1}\,\\3\sqrt{{{p}^{3}}+3{{p}^{2}}+3p+1}\,\\=3\sqrt{{{\left( p+1 \right)}^{3}}}\,=p+1\\=333+1=1000\end{array}$

Question 4 |

If 5

^{5a + 5}=1, then a equals -1 | |

-2 | |

3/5 | |

-4/5 |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}{{5}^{5a+5}}=1\\\Rightarrow {{5}^{5a}}\times {{5}^{5}}\Rightarrow {{5}^{5a}}=\frac{1}{{{5}^{5}}}\\\Rightarrow {{5}^{5a}}={{5}^{-5}}\Rightarrow 5a-5\\\Rightarrow a=-1\end{array}$

Question 5 |

If 3

^{r+3}+7=250, then r is equal to 4 | |

6 | |

2 | |

8 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}{{3}^{r+3}}+7=250\\\Rightarrow {{3}^{r+3}}=243\,\,\,\,\,\,\,\,\Rightarrow {{3}^{r+3}}={{3}^{5}}\\\Rightarrow r+3=5\Rightarrow r=2\end{array}$

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |