x, y, z, inequalities ....

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x, y, z, inequalities ....

by yass20015 » Mon Aug 10, 2015 11:29 am
If x, y and z are positive numbers, is z between x and y ?
1) x< 2z < y
2) 2x< z < 2y

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by Brent@GMATPrepNow » Mon Aug 10, 2015 11:38 am
If x, y, and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y
Target question: Is z between x and y?

Statement 1: x < 2z < y
There are several set values of x, y and z that satisfy this condition. Here are two:
Case a: x = 3, y = 10, and z = 2, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 2x < z < 2y
There are several set values of x, y and z that satisfy this condition. Here are two:
Case a: x = 1, y = 2, and z = 3, in which case z is NOT between x and y
Case b: x = 1, y = 10, and z = 3, in which case z IS between x and y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x < 2z < y
Statement 2: 2x < z < 2y
Since the two inequalities are facing the same direction, we can ADD them to get:
3x < 3z < 3y
Divide all three parts by 3 to get: x < z < y
As we can see, z IS definitely between x and y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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Brent
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by Max@Math Revolution » Fri Aug 21, 2015 7:16 pm
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If x, y and z are positive numbers, is z between x and y ?
1) x< 2z < y
2) 2x< z < 2y


==> Since there are 3 variables (x,y,z) in the original condition, we need 3 equations to match the number of variables and since we have only 2 equations overall (1 each in (1), (2)), E is likely the answer. Using both (1) & (2)3x<3z<3y thus x<y<z, and the solution to the question is yes. Therefore the answer is C. Using both (1) & (2) together can save us time.


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