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## Algebra: Linear Equation Test-4

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Question 1 |

1 yr ago, a mother was 4 times older to her son. After 6 yr. her age becomes more than double her son's age by 5 yr. The present ratio of their age will be

13 : 12 | |

3 : 1 | |

11 : 3 | |

25 : 7 |

Question 1 Explanation:

Let present age of the mother is m & age of her son is s year.

According to question:

(m -1)=4(s-1)

m = 4s-3â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦a

& (m+6)=2(s+6)+5

m = 2s + 11â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..b

From equation a and b

4s â€“ 3 = 2s +11

s = 7

and m = 25 years

According to question:

(m -1)=4(s-1)

m = 4s-3â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦a

& (m+6)=2(s+6)+5

m = 2s + 11â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..b

From equation a and b

4s â€“ 3 = 2s +11

s = 7

and m = 25 years

Question 2 |

Students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is

90 | |

94 | |

92 | |

96 |

Question 2 Explanation:

Suppose there are x rows & y students in each row

Then according to question

(x-2)(y+4)=xy &

(x+4)(y-4)=xy ,

Solving these equations

We get x=8 & y=12 then total students=96

Then according to question

(x-2)(y+4)=xy &

(x+4)(y-4)=xy ,

Solving these equations

We get x=8 & y=12 then total students=96

Question 3 |

$ \displaystyle If\,\,\frac{x}{\left( 2x+y+z \right)}=\frac{y}{\left( x+2y+z \right)}=\frac{z}{\left( x+y+2z \right)}=a,\,\,\,then\,\,\,find\,\,\,a,\,\,if\,\,\,x+y+z\ne 0$

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

$ \displaystyle \frac{1}{4}$ | |

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

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

Question 3 Explanation:

Put x=y=z=1 then a=1/4

Question 4 |

One-fourth of a herd of cows is in the forest. Twice the square root of the herd has gone to mountains and the remaining 15 are on the banks of a river. The total number of cows is

6 | |

100 | |

63 | |

36 |

Question 4 Explanation:

Let total no of cows is x,

then according to question

x-(x/4+2 âˆšx)=15

then 3x-8âˆšx+60=0

Now if we look at the options then we see that there are two answer

options which are perfect square and for the value of âˆšx we need

a perfect square so if the value of x is 100 then the equation will become

3 x 100 â€“ 8 x 10 + 60 = 300 â€“ 80 + 60

which is not equal to zero then the value of x is 36.

then according to question

x-(x/4+2 âˆšx)=15

then 3x-8âˆšx+60=0

Now if we look at the options then we see that there are two answer

options which are perfect square and for the value of âˆšx we need

a perfect square so if the value of x is 100 then the equation will become

3 x 100 â€“ 8 x 10 + 60 = 300 â€“ 80 + 60

which is not equal to zero then the value of x is 36.

Question 5 |

$ \displaystyle The\,\,\,\,value\,\,\,of\,\,\,\left( \frac{1}{{{x}^{2}}} \right)+\left( \frac{1}{{{y}^{2}}} \right),\,\,\,where\,\,x=2+\sqrt{3}\,\,\,\,and\,\,\,y=2-\sqrt{3},\,\,\,is$

14 | |

12 | |

10 | |

16 |

Question 5 Explanation:

$ \displaystyle \begin{array}{l}\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}=\frac{1}{{{\left( 2x+\sqrt{3} \right)}^{2}}}+\frac{1}{{{\left( 2-\sqrt{3} \right)}^{2}}}\\=\frac{1}{\left( 7+4\sqrt{3} \right)}+\frac{1}{\left( 7-4\sqrt{3} \right)}\\=\frac{14}{49-16\times 3}=14\end{array}$

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