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## Algebra Level 3 Test 4

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*Algebra Level 3 Test 4*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%% Your answers are highlighted below.

Question 1 |

The total number of integers pairs (x, y) satisfying the equation x + y = xy is

A | 0 |

B | 1 |

C | 2 |

D | None of the above. |

Question 1 Explanation:

Given equation is x + y = xy

â‡’ xy â€“ x â€“ y + 1 = 1

â‡’ (x â€“ 1)(y â€“ 1) = 1

x â€“ 1= 1

andÂ y âˆ’1= 1or

x âˆ’1= â€“1

y â€“ 1= â€“1

Clearly (0, 0) and (2, 2) are the only pairs that will satisfy the equation

â‡’ xy â€“ x â€“ y + 1 = 1

â‡’ (x â€“ 1)(y â€“ 1) = 1

x â€“ 1= 1

andÂ y âˆ’1= 1or

x âˆ’1= â€“1

y â€“ 1= â€“1

Clearly (0, 0) and (2, 2) are the only pairs that will satisfy the equation

Question 2 |

If | b |â‰¥ 1 and x = âˆ’ | a | b, then which one of the following is necessarily true?

A | a â€“ xb < 0 |

B | a â€“ xb â‰¥ 0 |

C | a â€“ xb > 0 |

D | a â€“ xb â‰¤ 0 |

Question 2 Explanation:

x = â€“|a| b

Now a â€“ xb = a â€“ (â€“ |a| b) b

= a + |a|b

Therefore Â a â€“ xb = a + ab

OR

a â€“ xb

= a â€“ ab

= a(1 + b

Consider first case:

As a â‰¥ 0 and |b| â‰¥ 1, therefore (1 + b

Therefore a (1 + b

Therefore a â€“ xb â‰¥ 0

Consider second case.

As a < 0 and |b| â‰¥ 1, therefore (1 â€“ b

Therefore a (1 â€“ b

Therefore, in both cases a â€“ xb â‰¥ 0.

Now a â€“ xb = a â€“ (â€“ |a| b) b

= a + |a|b

^{2}Therefore Â a â€“ xb = a + ab

^{2}â€¦a â‰¥ 0OR

a â€“ xb

= a â€“ ab

^{2}â€¦a < 0= a(1 + b

^{2}) = a(1 â€“ b^{2})Consider first case:

As a â‰¥ 0 and |b| â‰¥ 1, therefore (1 + b

^{2}) is positive.Therefore a (1 + b

^{2}) â‰¥ 0Therefore a â€“ xb â‰¥ 0

Consider second case.

As a < 0 and |b| â‰¥ 1, therefore (1 â€“ b

^{2}) â‰¤ 0Therefore a (1 â€“ b

^{2}) â‰¥ 0 (Since â€“ve Ã— -ve = +ve and 1 â€“ b^{2}can be zero also), i.e. a â€“ xb â‰¥ 0Therefore, in both cases a â€“ xb â‰¥ 0.

Question 3 |

If 13x + 1 < 2z and z + 3 = 5y

^{2}, thenA | x is necessarily less than y |

B | x is necessarily greater than y |

C | x is necessarily equal to y |

D | None of the above is necessarily true |

Question 3 Explanation:

13x + 1 < 2z and z + 3 = 5y

13x + 1 < 2 (5y

13x + 1< 10y

13x + 7 < 10y

20 < 10y

Therefore y

so option d is the right answer

^{2}13x + 1 < 2 (5y

^{2}âˆ’ 3)13x + 1< 10y

^{2}âˆ’ 613x + 7 < 10y

^{2}put x = 120 < 10y

^{2}and y^{2}> 2Therefore y

^{2}> 2 = Â (y^{2}âˆ’ 2) > 0so option d is the right answer

Question 4 |

If n is such that 36 â‰¤ n â‰¤ 72, Then a = {(n

^{2}+ 2âˆšn(n+ 4) +16)} / (n+ 4âˆšn +4) satisfiesA | 20 < x < 54 |

B | 23 < x < 58 |

C | 25 < x < 64 |

D | 28 < x < 60 |

Question 4 Explanation:

36 â‰¤ n â‰¤ 72

We are given by

a = {(n

Put a = 36.

And get the value 28 which is the least value

We are given by

a = {(n

^{2}+ 2âˆšn(n+ 4) +16)} / (n+ 4âˆšn +4)Put a = 36.

And get the value 28 which is the least value

Question 5 |

Consider the sets Tn = {n, n +1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,â€¦, 96. How many of these sets contain 6 or any integral multiple thereof (i.e. any one of the numbers 6, 12, 18, â€¦)?

A | 80 |

B | 81 |

C | 82 |

D | 83 |

Question 5 Explanation:

From the question we can observe that 6 will appear in 5

sets T2, T3, T4, T5 and T6.

Similarly, 12 will also appear in 5 sets

But the multiple of 6 appears in the next 5

that is in T12 Thus, each multiple of 6 will appear in 5 distinct sets.

Therefore we can say that there will be 16 multiplies of 6 till 96 .

Therefore 16 multiples of 6 will appear in 16 Ã— 5 = 80 sets.

sets T2, T3, T4, T5 and T6.

Similarly, 12 will also appear in 5 sets

But the multiple of 6 appears in the next 5

^{th}setthat is in T12 Thus, each multiple of 6 will appear in 5 distinct sets.

Therefore we can say that there will be 16 multiplies of 6 till 96 .

Therefore 16 multiples of 6 will appear in 16 Ã— 5 = 80 sets.

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