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If x and z are positive numbers less than 16, is z greater than the

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If x and z are positive numbers less than 16, is z greater than the  [#permalink]

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New post 17 Dec 2017, 23:38
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If x and z are positive numbers less than 16, is z greater than the average (arithmetic mean) of x and 16 ?

(1) z > x, and both x and z are perfect squares.
(2) x ≠ 1

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Re: If x and z are positive numbers less than 16, is z greater than the  [#permalink]

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New post 17 Dec 2017, 23:41
Bunuel wrote:
If x and z are positive numbers less than 16, is z greater than the average (arithmetic mean) of x and 16 ?

(1) z > x, and both x and z are perfect squares.
(2) x ≠ 1


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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x and z are positive numbers less than 16, is z greater than the  [#permalink]

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New post 18 Dec 2017, 02:03
(1) x and z can take the following values respectively:
case 1:
z=4 x=1 average between x and 16=8,5
z<average
case 2:
z=9 x=1 average between x and 16=8,5

z>average
case 3:
z=9 x=4 average between x and 16=10
z<average
Insufficient
(2) no information about z
Insuffcient
(1) + (2)
case 3:
z=9 x=4 average between x and 16=10
z<average

Answer is NO
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Re: If x and z are positive numbers less than 16, is z greater than the  [#permalink]

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New post 18 Dec 2017, 09:01
Bunuel wrote:
If x and z are positive numbers less than 16, is z greater than the average (arithmetic mean) of x and 16 ?

(1) z > x, and both x and z are perfect squares.
(2) x ≠ 1


Given: \(0 < x; z < 16\)

Question: Is \(z > \frac{(x+16)}{2}\) ?

(1) z > x, and both x and z are perfect squares. Possible perfect squares {1,4,9}. If z > x then, x = 1 z = 4 or 9 and average = (1+16)/2 = 8.5. If z = 9 > average, and if z = 4 < average, insufficient.

(2) x ≠ 1. So, x can be any number other than 1 from 0 to 16. If x = 4 then z = 5 or 11 and average = (4+16)/2 = 10: if z = 5 < average, and if z = 11 > average, insufficient.

(1+2) x = 4, z = 9 and average = (4+16)/2 = 10. So z = 9 < average, sufficient.

(C) is the answer.
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Re: If x and z are positive numbers less than 16, is z greater than the  [#permalink]

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New post 19 Dec 2017, 02:23
Bunuel wrote:
If x and z are positive numbers less than 16, is z greater than the average (arithmetic mean) of x and 16 ?

(1) z > x, and both x and z are perfect squares.
(2) x ≠ 1


Average of x and 16 will lie exactly in the middle of them (eg, average of 2 and 16 is 9, and 9 is 7 steps away from both 2 and 16. Average of 3 and 16 is 9.5, and 9.5 is 6.5 steps away from both 3 and 16). So we have to determine whether z is greater than the number which lies in the middle of x and 16.

(1) Both x, z are perfect squares and z>x but both less than 16. So positive perfect squares less than 16 are: 1, 4, 9. If x=1, then average of x and 16 = 8.5. z could be 4 and thus less than 8.5 or z could be 9 and thus greater than 8.5. So Insufficient.

(2) x is not 1, but nothing given about z. Clearly Insufficient.

Combining the two statements, if x is not 1, then x and z can be perfect squares among 4 & 9 only. x < z so only possibility is x=4, z=9. Since we have unique values we can find mean of x and 16 and determine whether z is greater than that or not. Sufficient.

Hence C answer
Re: If x and z are positive numbers less than 16, is z greater than the &nbs [#permalink] 19 Dec 2017, 02:23
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