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On the number line shown above, the tick marks are equally spaced. Which of the following statements about the numbers x, y, and z must be true?
I. xyz < 0
II. x + z = y
III. z(y - x) > 0
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
Attachment:
2015-05-12_1519.png
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Bunuel
Based on the number line shown, we can determine that \(x\) is negative, while \(y\) and \(z\) are positive. Additionally, since the tick marks are equally spaced, the distance between \(x\) and 0 is the same as the distance between \(y\) and 0, and the distance between \(z\) and 0 is twice that distance. For instance, we can assume \(x = -1\), \(y = 1\), and \(z = 2\).
I. \(xyz < 0\)
Given that \(x\) is negative and both \(y\) and \(z\) are positive, this statement must be true.
II. \(x + z = y\)
Using the example values of \(x = -1\), \(y = 1\), and \(z = 2\), we can confirm that this statement must also be true.
III. \(z(y - x) > 0\)
Similarly, with the example values of \(x = -1\), \(y = 1\), and \(z = 2\), we can confirm that this statement must be true as well.
Answer: E
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General Discussion
As the numbers are equally spaced one can see that
x = -y ; y = y and z = 2y
Now statement I is always true as xyz is nothing but -(y * y *2y) which is always negative. So I must be true.
Statement II x + z = -y + 2y = y ; So x+z =y So Statement II must be true.
Statement III : z(y-x) = 2y(y - -y) = 2(2y) which is always positive. Hence III must be true.
So I, II & III must be true.
Answer E
Ambarish
x = -y ; y = y and z = 2y
Now statement I is always true as xyz is nothing but -(y * y *2y) which is always negative. So I must be true.
Statement II x + z = -y + 2y = y ; So x+z =y So Statement II must be true.
Statement III : z(y-x) = 2y(y - -y) = 2(2y) which is always positive. Hence III must be true.
So I, II & III must be true.
Answer E
Ambarish
askhere
Sorry, there was a typo in II. It reads: x + z = y, not II. x + z = z.
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Bunuel
Ans: E
We solve this quickly, as soon as we put the values for x,y and z. as it is given they are equally spaced.
we can put x=-1, y=1 and z=2
we have our answer I, II and III all are correct[ANS: E]
EMPOWERgmatRichC
Expert
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,775
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert reply
Hi All,
Roman Numeral questions are relatively rare on Test Day (you'll probably see just one in the Quant section), but they almost always come with some type of built-in 'shortcut' so that you can avoid some of the implied work.
Here, we're given a number line and told that the "ticks" for X, Y and Z are EVENLY SPACED. This means that the distance between any two ticks is the SAME. We're asked which of the three statements MUST be true, which really means "which of these is ALWAYS TRUE no matter how many different examples we can come up with?"
From the picture, there is an immediate set of deductions that can be made about the 3 variables:
X is NEGATIVE (it's to the "left" of 0)
Y and Z are POSITIVE (they're to the "right" of 0)
RN 1: XYZ < 0
With the information we noted above....
(X)(Y)(Z) = (-)(+)(+) = ALWAYS Negative
Roman Numeral 1 is ALWAYS TRUE
Eliminate Answers B and C.
RN 2: X+Z = Y
Noting the picture, we know that....
X = "-1 tick"
Y = "+1 tick"
Z = "+2 ticks"
X+Z = "-1 tick" + "2 ticks" = "+1 tick"
This equals the value of Y (which is "+1 tick")
Roman Numeral 2 is ALWAYS TRUE.
Eliminate Answers A and D.
We don't even need to check Roman Numeral 3.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
Roman Numeral questions are relatively rare on Test Day (you'll probably see just one in the Quant section), but they almost always come with some type of built-in 'shortcut' so that you can avoid some of the implied work.
Here, we're given a number line and told that the "ticks" for X, Y and Z are EVENLY SPACED. This means that the distance between any two ticks is the SAME. We're asked which of the three statements MUST be true, which really means "which of these is ALWAYS TRUE no matter how many different examples we can come up with?"
From the picture, there is an immediate set of deductions that can be made about the 3 variables:
X is NEGATIVE (it's to the "left" of 0)
Y and Z are POSITIVE (they're to the "right" of 0)
RN 1: XYZ < 0
With the information we noted above....
(X)(Y)(Z) = (-)(+)(+) = ALWAYS Negative
Roman Numeral 1 is ALWAYS TRUE
Eliminate Answers B and C.
RN 2: X+Z = Y
Noting the picture, we know that....
X = "-1 tick"
Y = "+1 tick"
Z = "+2 ticks"
X+Z = "-1 tick" + "2 ticks" = "+1 tick"
This equals the value of Y (which is "+1 tick")
Roman Numeral 2 is ALWAYS TRUE.
Eliminate Answers A and D.
We don't even need to check Roman Numeral 3.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
Posts: 386
assign any 3 values that satisfy the equally spaces condition such as x = -1, y= 1, z = 2 and test for all 3 options we can see that all e options are for given figure
correct answer - E
correct answer - E
I guessed E but then I started thinking what if this is a trick question and x is +ve and y and z are -ve.
Is it safe to assume that that won't be the case? Bunuel
Is it safe to assume that that won't be the case? Bunuel
EMPOWERgmatRichC
Expert
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,775
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert reply
TarPhi
Hi TarPhi,
On a Number Line, values always INCREASE as you move to the RIGHT and DECREASE as you move to the LEFT.
Thus, any values to the RIGHT of the '0' on a Number line are POSITIVE and any values to the LEFT of the '0' are NEGATIVE.
GMAT assassins aren't born, they're made,
Rich
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Hope it helps.
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