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## Algebra: Basics Test-3

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Question 1 |

$ \displaystyle if\,\,\,\frac{a}{3}=\frac{b}{4}=\frac{c}{7}\,\,then\,\,\frac{a+b+c}{c}\,\,is\,\,equal\,\,\,to:$

2 | |

4 | |

6 | |

7 |

Question 1 Explanation:

$ \displaystyle \begin{array}{l}\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=1\,(let)\\a=3k,\,=4k\,,c=7k\\\therefore \frac{a+b+c}{c}=\frac{3k+4k+7k}{7k}\\=\frac{14k}{7k}=2\end{array}$

Question 2 |

$ \displaystyle f\,\,\frac{144}{0.144}=\frac{14.4}{p}$,then the value of p is

144 | |

0.0144 | |

1.44 | |

14.4 |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}\frac{144}{0.144}=\frac{14.4}{p}\\\Rightarrow 144\times p=14.4\times 0.144\\\Rightarrow p=\frac{14.4\times 0.144}{144}\\=\frac{144\times 144}{144\times 10000}=0.0144\end{array}$

Question 3 |

If 1 < c < 2, then the value of

$ \displaystyle \sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( c-3 \right)}^{2}}}\,$ is

$ \displaystyle \sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( c-3 \right)}^{2}}}\,$ is

3 | |

2x−4 | |

1 | |

2 |

Question 3 Explanation:

Since 1< c < 2, we have

c−1 > 0 and

c−3 < 0

or, 3 –c > 0

$ \displaystyle \therefore \sqrt{{{\left( c-1 \right)}^{2}}}+\sqrt{{{\left( 3-c \right)}^{2}}}\,$

$ \displaystyle \begin{array}{l}=\sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( 3-c \right)}^{2}}}\\\left[ \because \,{{\left( c-3 \right)}^{2}}={{\left( 3-c \right)}^{2}}\, \right]\\=c-1+3-c=2\end{array}$

c−1 > 0 and

c−3 < 0

or, 3 –c > 0

$ \displaystyle \therefore \sqrt{{{\left( c-1 \right)}^{2}}}+\sqrt{{{\left( 3-c \right)}^{2}}}\,$

$ \displaystyle \begin{array}{l}=\sqrt{{{\left( c-1 \right)}^{2}}}\,+\sqrt{{{\left( 3-c \right)}^{2}}}\\\left[ \because \,{{\left( c-3 \right)}^{2}}={{\left( 3-c \right)}^{2}}\, \right]\\=c-1+3-c=2\end{array}$

Question 4 |

p x q = (p × q) +q, then 5 x 7 equals to

40 | |

42 | |

30 | |

32 |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}p\times r=\left( p\times q \right)+q\\\therefore 5\times 7=\left( 5\times 7 \right)+7=35+7=42\end{array}$

Question 5 |

If f = 0.1039, then the value of $ \displaystyle \sqrt{4{{f}^{2}}-4f+1\,\,}\,$ +3f is

1.1039 | |

2.1039 | |

0.1039 | |

0.2078 |

Question 5 Explanation:

f= 0.1039 (given)

$ \displaystyle \begin{array}{l}Now,\,\sqrt{4{{f}^{2}}-4f+1+3f}\,\\=\sqrt{{{\left( 1-2f \right)}^{2}}}\,+3f\\=1-2f+3f\\=1+f=1+0.1039\\=1.1039\end{array}$

$ \displaystyle \begin{array}{l}Now,\,\sqrt{4{{f}^{2}}-4f+1+3f}\,\\=\sqrt{{{\left( 1-2f \right)}^{2}}}\,+3f\\=1-2f+3f\\=1+f=1+0.1039\\=1.1039\end{array}$

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