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## Algebra: Basics Test-1

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Question 1 |

If aÃ— b = 2a-3b+ab, then 3 Ã— 5 +5 x 3 is equal to:

A | 20 |

B | 23 |

C | 24 |

D | 22 |

Question 1 Explanation:

We are given by the statement that

$ \displaystyle \begin{array}{l}a\times b=2a-3b+ab\\\Rightarrow 3\times 5=2\times 3-3\times 5+3\times 5=6\\5\times 3=2\times 5-3\times 3+3\times 5\\=10-9+15=16\\\therefore \,\,3\times 5+5\times 3=6+16=22\end{array}$

$ \displaystyle \begin{array}{l}a\times b=2a-3b+ab\\\Rightarrow 3\times 5=2\times 3-3\times 5+3\times 5=6\\5\times 3=2\times 5-3\times 3+3\times 5\\=10-9+15=16\\\therefore \,\,3\times 5+5\times 3=6+16=22\end{array}$

Question 2 |

If p Ã—q = p Ã— q + p/q, the value of 8 Ã—2 are:

A | 13 |

B | 14 |

C | 15 |

D | 16 |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}p\,\,\times \,q=p+q+\frac{p}{q}\\\therefore 8\times 2=8+2+\frac{8}{2}\\=10+4=14\end{array}$

Question 3 |

Two numbers a and b ( a > b) are such that their sum is equal to three times their difference.Then value of $ \displaystyle \frac{3ab}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$Â will be:

A | $ \displaystyle 1\frac{1}{2}$ |

B | $ \displaystyle 1\frac{2}{3}$ |

C | $ \displaystyle 1\frac{1}{3}$ |

D | 1 |

Question 3 Explanation:

From the given statement we can say that

$ \displaystyle \begin{array}{l}\left( a+b \right)=3\left( a-b \right)=3a-3b\\\Rightarrow 3b+b=3a-ra\\\Rightarrow 2a=4b\\\Rightarrow a=2b\\\Rightarrow \frac{a}{b}=\frac{2}{1}\\\Rightarrow \frac{a}{b}=\frac{1}{2}\\\therefore \,a=2,\,b=1\\\frac{3ab}{2\left( {{a}^{2}}-{{b}^{2}} \right)}=\frac{3\times 2\times 1}{2\times \left( 4-1 \right)}=\frac{6}{6}=1\end{array}$

$ \displaystyle \begin{array}{l}\left( a+b \right)=3\left( a-b \right)=3a-3b\\\Rightarrow 3b+b=3a-ra\\\Rightarrow 2a=4b\\\Rightarrow a=2b\\\Rightarrow \frac{a}{b}=\frac{2}{1}\\\Rightarrow \frac{a}{b}=\frac{1}{2}\\\therefore \,a=2,\,b=1\\\frac{3ab}{2\left( {{a}^{2}}-{{b}^{2}} \right)}=\frac{3\times 2\times 1}{2\times \left( 4-1 \right)}=\frac{6}{6}=1\end{array}$

Question 4 |

The value of <br>
$ \displaystyle \left( 1+\frac{1}{p} \right)\left( 1\frac{1}{p+1} \right)\left( 1+\frac{1}{p+2} \right)\left( 1+\frac{1}{p+3} \right)$

is

is

A | $ \displaystyle 1+\frac{1}{p+4}$ |

B | $ \displaystyle p+4$ |

C | $ \displaystyle \frac{p+4}{p}$ |

D | $ \displaystyle \frac{1}{p}$ |

Question 4 Explanation:

We can solve the given fraction as

$ \displaystyle \begin{array}{l}=\left( 1+\frac{1}{p} \right)\left( 1+\frac{1}{p+1} \right)\left( 1+\frac{1}{p+2} \right)\left( 1+\frac{1}{p+3} \right)\\=\frac{p+1}{p}\times \frac{p+2}{p+1}\times \frac{p+3}{p+2}\times \frac{p+4}{p+3}\\=\frac{p+4}{p}\end{array}$

$ \displaystyle \begin{array}{l}=\left( 1+\frac{1}{p} \right)\left( 1+\frac{1}{p+1} \right)\left( 1+\frac{1}{p+2} \right)\left( 1+\frac{1}{p+3} \right)\\=\frac{p+1}{p}\times \frac{p+2}{p+1}\times \frac{p+3}{p+2}\times \frac{p+4}{p+3}\\=\frac{p+4}{p}\end{array}$

Question 5 |

If a Ã— b = 2 (a + b), then 5 Ã—2 is equal to:

A | 16 |

B | 12 |

C | 14 |

D | 18 |

Question 5 Explanation:

Given statement is

$ \displaystyle \begin{array}{l}a\times b=2\left( a+b \right)\\\therefore \,5\times 2=2\left( 5+2 \right)\,\,\\=2\times 7=14\end{array}$

$ \displaystyle \begin{array}{l}a\times b=2\left( a+b \right)\\\therefore \,5\times 2=2\left( 5+2 \right)\,\,\\=2\times 7=14\end{array}$

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