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## Algebra Level 2 Test 3

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Question 1 |

If the roots of the equation (c

^{2}- ab) x^{2}- 2 (a^{2}- bc) x + (b^{2}- ac) = 0 for a is not equal to 0 are real and equal, then the value of a^{3}+ b^{3 }+ c^{3}isabc | |

3abc | |

zero | |

None of these |

Question 1 Explanation:

We are given by that the roots are real and equal, that means we have

= b

Therefore the equation becomes

[-2(a

On solving we have

4a (a

Therefore the value of a

= b

^{2}- 4ac = 0Therefore the equation becomes

[-2(a

^{2}- cb) ]^{2}- 4 (c^{2}- ba) (b^{2}- ac) = 0On solving we have

4a (a

^{3}+ b^{3 }+ c^{3}â€“ 3abc) = 0Therefore the value of a

^{3}+ b^{3 }+ c^{3}= 3abcQuestion 2 |

If 2x

^{2 }-7xy + 3y^{2}= 0, then the value of x: y is3 : 2 | |

2 : 3 | |

3 : 1 and 1 : 2 | |

5 : 6 |

Question 2 Explanation:

2x

2 (x/y)

x/y = {-b + ^(b

= {7 + ^(49 â€“ 24)} / 2 x 2 = (7+ 5) / 4 = 3, Â½

Therefore the value of x/y = 3/1 and x/y = Â½

^{2 }-7xy + 3y^{2}= 02 (x/y)

^{2}â€“ 7(x/y) +3 = 0x/y = {-b + ^(b

^{2}â€“ 4ac) } / 2a= {7 + ^(49 â€“ 24)} / 2 x 2 = (7+ 5) / 4 = 3, Â½

Therefore the value of x/y = 3/1 and x/y = Â½

Question 3 |

If (a + b + 2c + 3d) (a - b â€“ 2c + 3d) = (a - b + 2c - 3d) x (a + b â€“ 2c - 3d), then 2bc is equal to

3ad | |

3/2 | |

a ^{2}d^{2} | |

3a/2d |

Question 3 Explanation:

(a + b + 2c + 3d) (a - b â€“ 2c + 3d) = (a - b + 2c - 3d) x (a + b â€“ 2c - 3d)

Therefore [(a+b) + (2c -3d) x (a + b â€“ 2c - 3d)

= Â Â [(a-b) + (2c -3d)] x [(a + b â€“ (2c - 3d)]

=(a+b)(a-b) â€“ (a+b)(2c â€“ 3d) + (a â€“ b)(2c â€“ 3d) â€“ (2c +3d)(2c -3d)

OnÂ further solving

= 2ac +3ad -2bc -3bd = 2ac - 3ad + 2bc - 3bd

= 2bc = 3ad

Therefore [(a+b) + (2c -3d) x (a + b â€“ 2c - 3d)

= Â Â [(a-b) + (2c -3d)] x [(a + b â€“ (2c - 3d)]

=(a+b)(a-b) â€“ (a+b)(2c â€“ 3d) + (a â€“ b)(2c â€“ 3d) â€“ (2c +3d)(2c -3d)

OnÂ further solving

= 2ac +3ad -2bc -3bd = 2ac - 3ad + 2bc - 3bd

= 2bc = 3ad

Question 4 |

If X=4ab/(a+b), a+-b, then the value of (x +2a)/(x â€“ 2a) + (x +2b)/(x â€“ 2b) is

0 | |

1 | |

2 | |

None of these |

Question 4 Explanation:

(x +2a)/(x â€“ 2a) + (x +2b)/(x â€“ 2b

Put values of x in the expression we will get

[{(6ab + 2a

= {2a(3b + a)/2a(b â€“ a)} + {2b (3b+a) / 2b (a â€“ b)}

(3b + a)/ (b â€“ a) â€“ (3b + a)/ (b â€“ a) = 0

Put values of x in the expression we will get

[{(6ab + 2a

^{2})/(a+b)} / {(2ab â€“ 2a^{2})/(a+b)}] + [{(6b^{2}+ 2ab)/(a+b)} / {(2ab â€“ 2b^{2})/(a+b)}]= {2a(3b + a)/2a(b â€“ a)} + {2b (3b+a) / 2b (a â€“ b)}

(3b + a)/ (b â€“ a) â€“ (3b + a)/ (b â€“ a) = 0

Question 5 |

Richa's age is1/6

^{th}of Natasha 'sage. Natasha 's age will be twice of Sona 's age after 10 yr. If Sona' s eight birthday was celebrated two yr ago, then the present age of Richa Â must be5 yr | |

10 yr | |

15 yr | |

20 yr |

Question 5 Explanation:

Given that Age of Richa = 1/6 age of Natasha

Present age of Sona = 10 years

That means

Natasha + 10 = 2( Sona + 10)

Therefore Natasha + 10 = 40

Natasha = 30

Therefore the age of Richa = 5 years

Present age of Sona = 10 years

That means

Natasha + 10 = 2( Sona + 10)

Therefore Natasha + 10 = 40

Natasha = 30

Therefore the age of Richa = 5 years

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