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## Algebra: Basics Test-2

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Question 1 |

If $ \displaystyle \frac{2x+y}{x+4y}=3$,then find the value of $ \displaystyle \frac{p+q}{p+2q}$

$ \displaystyle \frac{15}{9}$ | |

$ \displaystyle \frac{12}{9}$ | |

$ \displaystyle \frac{10}{9}$ | |

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

Question 1 Explanation:

$ \displaystyle \begin{array}{l}\frac{2p+q}{p+4q}=3\,\,\\\Rightarrow 2p+q=3p+12q\\\Rightarrow 3p-2p=q-12q\\\Rightarrow p=-11q\\Then,\,\,\frac{p+q}{p+2q}=\frac{-11q+q}{-11q+2q}\\=\frac{-10q}{-9q}=\frac{10}{9}\end{array}$

Question 2 |

If a =4, 965 b =2.343 and c = 2.622, then the value of a

^{3}âˆ’b^{3}âˆ’c^{3}-3xyz is equal to0 | |

âˆ’3 | |

1 | |

1.9 ^{3} |

Question 2 Explanation:

We are given by the values of x ,y and z i.e.

a =4, 965 b=2.343 and c = 2.622

If a + b + c = 0

Then a

$ \displaystyle \begin{array}{l}\therefore \,When\,\,a-b-c=0,\\{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=3abc\\i.e.,\,{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=0\end{array}$

$ \displaystyle \begin{array}{l}Here,\\a=4.965,b=2.343,c=2.6222\\\therefore a-b-c=4.965-2.343-2.622=0\\Hence,\,{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=0\end{array}$

a =4, 965 b=2.343 and c = 2.622

If a + b + c = 0

Then a

^{3}+ b^{3}+ c = 3xyz$ \displaystyle \begin{array}{l}\therefore \,When\,\,a-b-c=0,\\{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=3abc\\i.e.,\,{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=0\end{array}$

$ \displaystyle \begin{array}{l}Here,\\a=4.965,b=2.343,c=2.6222\\\therefore a-b-c=4.965-2.343-2.622=0\\Hence,\,{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=0\end{array}$

Question 3 |

p x q= p+ q+ pq, then 3Ã—4 âˆ’2Ã—3 is equal to:

6 | |

4 | |

8 | |

9 |

Question 3 Explanation:

We have

$ \displaystyle \begin{array}{l}p\times q=p+q+pq\\\therefore 3\times 4-2\times 3\\=\left( 3+4+3\times 4 \right)-\left( 2+3+2\times 3 \right)\\=\left( 7+12 \right)-\left( 5+6 \right)=19-11=8\end{array}$

$ \displaystyle \begin{array}{l}p\times q=p+q+pq\\\therefore 3\times 4-2\times 3\\=\left( 3+4+3\times 4 \right)-\left( 2+3+2\times 3 \right)\\=\left( 7+12 \right)-\left( 5+6 \right)=19-11=8\end{array}$

Question 4 |

If p= 1.21, q= 2.12 and r = âˆ’3.33 then, the value of p

^{3}+ q^{3}+r^{3}-3pqr is1 | |

2 | |

3 | |

0 |

Question 4 Explanation:

We have p= 1.21, q= 2.12 and r = âˆ’3.33

$ \displaystyle \begin{array}{l}p+q+r\,\,=1.21+2.12-3.33=0\\{{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr=0\end{array}$

$ \displaystyle \begin{array}{l}p+q+r\,\,=1.21+2.12-3.33=0\\{{p}^{3}}+{{q}^{3}}+{{r}^{3}}-3pqr=0\end{array}$

Question 5 |

If l Ã— m= 3l +2m, then 2 Ã—3 + 3Ã—4 is equal to:

29 | |

27 | |

39 | |

37 |

Question 5 Explanation:

We are given by the information

l Ã— m = 3l +2m

$ \displaystyle \begin{array}{l}\because l\times m=3l+2m\\2\times 3+3\times 4\\=3\times 2+2\times 3+3\times 3+2\times 4\\=6+6+9+8=29\end{array}$

l Ã— m = 3l +2m

$ \displaystyle \begin{array}{l}\because l\times m=3l+2m\\2\times 3+3\times 4\\=3\times 2+2\times 3+3\times 3+2\times 4\\=6+6+9+8=29\end{array}$

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