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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}$

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

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

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

D | $ \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}−b^{3}−c^{3}-3xyz is equal toA | 0 |

B | −3 |

C | 1 |

D | 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 3×4 −2×3 is equal to:

A | 6 |

B | 4 |

C | 8 |

D | 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 = −3.33 then, the value of p

^{3}+ q^{3}+r^{3}-3pqr isA | 1 |

B | 2 |

C | 3 |

D | 0 |

Question 4 Explanation:

We have p= 1.21, q= 2.12 and r = −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 × m= 3l +2m, then 2 ×3 + 3×4 is equal to:

A | 29 |

B | 27 |

C | 39 |

D | 37 |

Question 5 Explanation:

We are given by the information

l × 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 × 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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