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## Algebra Level 2 Test 5

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Question 1 |

If log

_{7}log_{5}Â {âˆš(x + 5) + âˆšx}= 01 | |

0 | |

2 | |

None of these |

Question 1 Explanation:

log

log âˆš(x + 5)+ âˆšxÂ = 7

or

âˆš(x + 5)+ âˆšxÂ = 5

Therefore 2âˆšxÂ = 0. or

x = 0.

_{7}log_{5}{ âˆš(x + 5) + âˆšxÂ }= 0log âˆš(x + 5)+ âˆšxÂ = 7

^{0}= 1,or

âˆš(x + 5)+ âˆšxÂ = 5

^{1 }= 5.Therefore 2âˆšxÂ = 0. or

x = 0.

Question 2 |

If a + b + c = 0, where a â‰ b â‰ c, then {a

^{2}/(2a^{2}+bc)} + {b^{2}/(2b^{2}+ac)} +{c^{2}/(2c^{2}+ab)}zero | |

1 | |

â€“1 | |

abc |

Question 2 Explanation:

To solve this question, we use the help of values.

Assuming such values of a, b & c so that a + b + c =0

and a â‰ b â‰ c, we can find the value of the expression.

Assuming a = 1, b = -1 and c = 0 , we find that Â {a

Â½ + Â½ + 0 = 1.

Hence we get our answer without any sweat.

Assuming such values of a, b & c so that a + b + c =0

and a â‰ b â‰ c, we can find the value of the expression.

Assuming a = 1, b = -1 and c = 0 , we find that Â {a

^{2}/(2a^{2}+bc)} + {b^{2}/(2b^{2}+ac)} +{c^{2}/(2c^{2}+ab)}Â½ + Â½ + 0 = 1.

Hence we get our answer without any sweat.

Question 3 |

If the harmonic mean between two positive numbers is to their geometric mean as 12 : 13; then the numbers could be in the ratio

12 : 13 | |

1/12 : 1/13 | |

4 : 9 | |

2 : 3 |

Question 3 Explanation:

Since the harmonic mean of two numbers x and y is 2xy/(x + y) and the geometric mean is âˆšxy.

So from the question 2xy /(x + y)/âˆšxy = 12/13,

Further solving this

Squaring both sides we get

Therefore 2 x

But we want the values for x : y so

= 2xy/(x + y) = 12/13

= 2âˆšxy / (x + y ) âˆšxy

= {(x + y ) âˆšxy /Â 2âˆšxy}

For further solving we have to get back in our school days for the concept of component and dividendo

2âˆšx/2âˆšy = 6/4 = x / y = 9/4 or y / x = 4/9

So from the question 2xy /(x + y)/âˆšxy = 12/13,

Further solving this

Squaring both sides we get

Therefore 2 x

^{3}y^{3}/ (x + y)^{2}= 144/169.But we want the values for x : y so

= 2xy/(x + y) = 12/13

= 2âˆšxy / (x + y ) âˆšxy

= {(x + y ) âˆšxy /Â 2âˆšxy}

For further solving we have to get back in our school days for the concept of component and dividendo

2âˆšx/2âˆšy = 6/4 = x / y = 9/4 or y / x = 4/9

Question 4 |

If one root of x

^{2}+ px + 12= is Â 4, while the equation x^{2}+ px + q = has equal roots, then the value of q is49/4 | |

4/49 | |

4 | |

Â¼ |

Question 4 Explanation:

If one root of x

4

From thr above expression

p = -7.

Put this value in the second given equation x

Therefore If the roots are equal then b

4q = 49

q = 49/4

^{2}+ px + 12 = 0 is 4, then4

^{2 }+ 4p + 12 = 0,From thr above expression

p = -7.

Put this value in the second given equation x

^{2}â€“ 7x + q = 0 has equal roots.Therefore If the roots are equal then b

^{2 }Â = 4ac4q = 49

q = 49/4

Question 5 |

Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?

7 | |

64 | |

56 | |

Cannot be determined |

Question 5 Explanation:

Letâ€™s take an example to understand a trick that can help us solve this question.

Let us suppose we have a series 1 + 2 + 3 and the middle term is 2.

We are asked for the sum of this series so a simple shortcut that can be used is that:

sum of series = middle term x number of terms

i.e the sum of the supposed series is 2 x 3 = 6

Using the above, we are given by the fourth term which is the middle term of the series. So the sum of the series is 8 x 7 = 56.

Let us suppose we have a series 1 + 2 + 3 and the middle term is 2.

We are asked for the sum of this series so a simple shortcut that can be used is that:

sum of series = middle term x number of terms

i.e the sum of the supposed series is 2 x 3 = 6

Using the above, we are given by the fourth term which is the middle term of the series. So the sum of the series is 8 x 7 = 56.

Question 6 |

What is the value of m which satisfies 3m

^{2}â€“ 21m + 30 < 0?m < 2 or m > 5 | |

m > 2 c. | |

2 < m < 5; | |

Both a and c |

Question 6 Explanation:

3m

Dividing by 3

Therefore m

NowÂ on solving this quadratic equation

(m â€“ 5)(m â€“ 2) < 0.

Now there are two cases

Either (m â€“ 5) < 0 and (m â€“ 2) > 0 or (m â€“ 2) < 0 and (m â€“5) > 0.

Hence, either m < 5 and m > 2, i.e. 2< m < 5 or m < 2 and m > 5.

^{2}â€“ 21m + 30 < 0Dividing by 3

Therefore m

^{2}â€“ 7m + 10 < 0,NowÂ on solving this quadratic equation

(m â€“ 5)(m â€“ 2) < 0.

Now there are two cases

Either (m â€“ 5) < 0 and (m â€“ 2) > 0 or (m â€“ 2) < 0 and (m â€“5) > 0.

Hence, either m < 5 and m > 2, i.e. 2< m < 5 or m < 2 and m > 5.

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