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## Algebra Level 3 Test 3

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Question 1 |

There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is

34 | |

38 | |

36 | |

32 |

Question 1 Explanation:

Let the number of horizontal layers in the pile is = p

The number of balls in each layer in such a way that the

first layer contains one ball, second layer contains 3 balls,

whereas the third ball contains 6 balls and so on ∑p(p+1)/2 = 8436

½ [∑p

= {p(p+1)(2p+1)}/12 + { p(p+1)}/4 = 8436

= n(n+1)[(2p+1)/12] = 8436

= {p(p+1)(2p+1)}/6 = 8436

On solving this we have p = 36

The number of balls in each layer in such a way that the

first layer contains one ball, second layer contains 3 balls,

whereas the third ball contains 6 balls and so on ∑p(p+1)/2 = 8436

½ [∑p

^{2}+ ∑p] = 8436= {p(p+1)(2p+1)}/12 + { p(p+1)}/4 = 8436

= n(n+1)[(2p+1)/12] = 8436

= {p(p+1)(2p+1)}/6 = 8436

On solving this we have p = 36

Question 2 |

If the product of n positive real numbers is unity, then their sum is necessarily

a multiple of n | |

equal to (n +1/n) | |

never less than n | |

a positive integer |

Question 2 Explanation:

The best way to do this is to take some value and verify.

Let’s take the values as 2,1/2,1 so the sum of three of them is 7/2 with n = 3

Let’s take the values as 2,1/2,1/4 so the sum of three of them is 11/4 with n = 3

Therefore options A ,B and C are ruled out and the correct answer for the question is option number (d)

Let’s take the values as 2,1/2,1 so the sum of three of them is 7/2 with n = 3

Let’s take the values as 2,1/2,1/4 so the sum of three of them is 11/4 with n = 3

Therefore options A ,B and C are ruled out and the correct answer for the question is option number (d)

Question 3 |

Given that −1≤ v ≤ 1, −2 ≤ u ≤ −0.5 and −2 ≤ z ≤ −0.5 and w = vz /u, then which of the following is necessarily true?

−0.5 ≤ w ≤1 | |

−4 ≤ w ≤ 4 | |

−4 ≤ w ≤ 2 | |

−2 ≤ w ≤ −0.5 |

Question 3 Explanation:

U is always negative. Hence, for us to have a minimum value of vz/u , vz should be positive.

Also for the least value, the numerator has to be the maximum positive value and the denominator has to be the smallest negative value.

In other words, vz has to be 2 and u has to be –0.5.

Hence the minimum value of vz/u = -2/0.5 = -4.

For us to get the maximum value, vz has to be the smallest negative value and u has to be the highest negative value.

Thus, vz has to be –2 and u has to be –0.5. Hence the maximum value of vz/u = -2/-0.5 = 4

Also for the least value, the numerator has to be the maximum positive value and the denominator has to be the smallest negative value.

In other words, vz has to be 2 and u has to be –0.5.

Hence the minimum value of vz/u = -2/0.5 = -4.

For us to get the maximum value, vz has to be the smallest negative value and u has to be the highest negative value.

Thus, vz has to be –2 and u has to be –0.5. Hence the maximum value of vz/u = -2/-0.5 = 4

Question 4 |

If x, y, z are distinct positive real numbers the [{x

^{2}(y + z) y^{2}(x+ z) z^{2}(x+ y)}/xyz] would begreater than 4. | |

greater than 5. | |

greater than 6 | |

None of the above. |

Question 4 Explanation:

Here x, y, z are distinct positive real number

[{x

z

This can be written as x/y + x/z + y/x + y/z + z/x + z/y

This is also can be modified as

(x/y + y/z) + (y/z + z/y ) + (z/x + x/z)

Here in school days we did a property of numbers that if the numbers are distinct then the value of a/b + b/a > 2

So here the value of the expression must be greater than 2+2+2 > 6

[{x

^{2}(y + z) y^{2}(x+ z)z

^{2}(x+ y)}/xyz]This can be written as x/y + x/z + y/x + y/z + z/x + z/y

This is also can be modified as

(x/y + y/z) + (y/z + z/y ) + (z/x + x/z)

Here in school days we did a property of numbers that if the numbers are distinct then the value of a/b + b/a > 2

So here the value of the expression must be greater than 2+2+2 > 6

Question 5 |

In a certain examination paper, there are n questions. For j = 1,2 …n, there are 2

^{n–j}students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is12 | |

11 | |

10 | |

9 |

Question 5 Explanation:

Let us consider that there are only 3 questions.

Therefore the number of students that have done 1 or more questions wrongly are 2

The number of students that have done 2 or more

questions wrongly are 2

Thus total number of answers which are wrong = 4 + 2 + 1 = 7 = 2

But we are given by the wrong answers in our question,

the total number of wrong answers = 4095 = 2

Therefore the number of students that have done 1 or more questions wrongly are 2

^{3–1}= 4,The number of students that have done 2 or more

questions wrongly are 2

^{3–2}= 2 And the number of students that must have done all 3 wrongly are 2^{3–3}= 1.Thus total number of answers which are wrong = 4 + 2 + 1 = 7 = 2

^{3 }– 1 = 2^{n}– 1.But we are given by the wrong answers in our question,

the total number of wrong answers = 4095 = 2

^{12 }– 1. Thus n = 12.
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