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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

0 | |

1 | |

2 | |

None of the above. |

Question 1 Explanation:

Given equation is x + y = xy

⇒ xy – x – y + 1 = 1

⇒ (x – 1)(y – 1) = 1

x – 1= 1

and y −1= 1or

x −1= –1

y – 1= –1

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

⇒ xy – x – y + 1 = 1

⇒ (x – 1)(y – 1) = 1

x – 1= 1

and y −1= 1or

x −1= –1

y – 1= –1

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

Question 2 |

If | b |≥ 1 and x = − | a | b , then which one of the following is necessarily true?

a – xb < 0 | |

a – xb ≥ 0 | |

a – xb > 0 | |

a – xb ≤ 0 |

Question 2 Explanation:

x = –|a| b

Now a – xb = a – (– |a| b) b

= a + |a|b

Therefore a – xb = a + ab

OR

a – xb

= a – ab

= a(1 + b

Consider first case:

As a ≥ 0 and |b| ≥ 1, therefore (1 + b

Therefore a (1 + b

Therefore a – xb ≥ 0

Consider second case.

As a < 0 and |b| ≥ 1, therefore (1 – b

Therefore a (1 – b

Therefore, in both cases a – xb ≥ 0.

Now a – xb = a – (– |a| b) b

= a + |a|b

^{2}Therefore a – xb = a + ab

^{2}…a ≥ 0OR

a – xb

= a – ab

^{2}…a < 0= a(1 + b

^{2}) = a(1 – b^{2})Consider first case:

As a ≥ 0 and |b| ≥ 1, therefore (1 + b

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

^{2}) ≥ 0Therefore a – xb ≥ 0

Consider second case.

As a < 0 and |b| ≥ 1, therefore (1 – b

^{2}) ≤ 0Therefore a (1 – b

^{2}) ≥ 0 (Since –ve × -ve = +ve and 1 – b^{2}can be zero also), i.e. a – xb ≥ 0Therefore, in both cases a – xb ≥ 0.

Question 3 |

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

^{2}, thenx is necessarily less than y | |

x is necessarily greater than y | |

x is necessarily equal to y | |

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}− 3)13x + 1< 10y

^{2}− 613x + 7 < 10y

^{2}put x = 120 < 10y

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

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

Question 4 |

If n is such that 36 ≤ n ≤ 72 , Then a = {(n

^{2}+ 2√n(n+ 4) +16)} / (n+ 4√n +4) satisfies20 < x < 54 | |

23 < x < 58 | |

25 < x < 64 | |

28 < x < 60 |

Question 4 Explanation:

36 ≤ n ≤ 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}+ 2√n(n+ 4) +16)} / (n+ 4√n +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,…, 96. How many of these sets contain 6 or any integral multiple thereof (i.e. any one of the numbers 6, 12, 18, …)?

80 | |

81 | |

82 | |

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 × 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 × 5 = 80 sets.

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