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If the successive tick marks shown on the number line above

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Manager
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Re: If the successive tick marks shown on the number line above [#permalink]
I have a question for this type of DS problem. If one is absolutely sure that statement 1 is sufficient, during a exam under time pressure, could we plug statement one's given variable into statement 2 to test out?

For example:

Statement 1) Given x=1/2
"I'm sure that it is possible to get y if we have 0 and X=1/2 on the number line. So sufficient."

Statement 2) Given y-x=2/3
"Since I'm confident that statement one is sufficient, let's try plugging in 1/2 for x."

so:
y-x=2/3

y-(1/2)=2/3 "From here I know that we can get y" Suf.

y=7/6

Is this reasoning okay?

Can we do this
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Re: If the successive tick marks shown on the number line above [#permalink]
1
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DelSingh wrote:
I have a question for this type of DS problem. If one is absolutely sure that statement 1 is sufficient, during a exam under time pressure, could we plug statement one's given variable into statement 2 to test out?

For example:

Statement 1) Given x=1/2
"I'm sure that it is possible to get y if we have 0 and X=1/2 on the number line. So sufficient."

Statement 2) Given y-x=2/3
"Since I'm confident that statement one is sufficient, let's try plugging in 1/2 for x."

so:
y-x=2/3

y-(1/2)=2/3 "From here I know that we can get y" Suf.

y=7/6

Is this reasoning okay?

Can we do this

For me this must be a bad idea during the exam, because in stress, minor mistakes are very common, so, for me, Such practices must be avoided.
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Re: If the successive tick marks shown on the number line above [#permalink]
1
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If the successive tick marks shown on the number line above are equally spaced and if x and y are the numbers designating the end points of intervals as shown, what is the value of y?

(1) x = 1/2
(2) y - x = 2/3

Attachment:
Number line.png

This is how I solved

1) X= 1/2 means X = 0.5 on the no line
Notice there are 3 intervals between 0 and 0.5
ie 0.5/3 = 0.16 per interval as the spacing between each interval is constant

Now to find the value of y we know that the space between X and Y is 4 more intervals after X
ie 0.5 + 4*0.16 to arrive at value of Y
Therefore its sufficient

Y-X = 2/3
which means Y-X = 0.66 and the interval between X and Y has 4 intervals so each interval = 0.16 .
With this info we can calculate the intervals between 0 and Y = 7 intervals
Value of Y = 0.165 * 7 = sufficient

ANs = D
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Re: If the successive tick marks shown on the number line above [#permalink]