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best approach plz

by RCV » Wed Oct 06, 2010 9:46 pm
On the number line shown, is zero halfway between a and b


<-----------------a---------b----c----------->

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.


it killed my 10 minutes and then I left it, what is the best approach?
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by shovan85 » Wed Oct 06, 2010 9:57 pm
IMO B

From 1 No way we can answer this so Not Sufficient.

From 2 |dist between a and c | = |distance between c and -b|

<-----------------a---------b----c----------->

How will it be possible. When a = -b this is possible.

Hence a = -b.

So zero must be the middle point of b and - b or we can say that of a and b. Sufficient.
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by sanju09 » Wed Oct 06, 2010 10:06 pm
shovan85 wrote:IMO B

From 1 No way we can answer this so Not Sufficient.

From 2 |dist between a and c | = |distance between c and -b|

<-----------------a---------b----c----------->

How will it be possible. When a = -b this is possible.

Hence a = -b.

So zero must be the middle point of b and - b or we can say that of a and b. Sufficient.
Your reasoning is not totally wrong that "for the distance between a and c to be equal to the distance between c and -b, a should be equal to -b." But you are missing one thing here that it will be true only if we know what type of number (+/-) b is, which we get to know only if we take statement 1 into account also. My answer is [spoiler]C[/spoiler]
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by shovan85 » Wed Oct 06, 2010 10:14 pm
sanju09 wrote:
shovan85 wrote:IMO B

From 1 No way we can answer this so Not Sufficient.

From 2 |dist between a and c | = |distance between c and -b|

<-----------------a---------b----c----------->

How will it be possible. When a = -b this is possible.

Hence a = -b.

So zero must be the middle point of b and - b or we can say that of a and b. Sufficient.
Your reasoning is not totally wrong that "for the distance between a and c to be equal to the distance between c and -b, a should be equal to -b." But you are missing one thing here that it will be true only if we know what type of number (+/-) b is, which we get to know only if we take statement 1 into account also. My answer is [spoiler]C[/spoiler]
Thanks but I thought of that scenario. Then again I saw the question which has a diagram. If we take the - sign then the order will be changed.
<--------------------------b----a-----c------>
So picked B. :)
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by shovan85 » Thu Oct 07, 2010 2:07 am
RCV wrote:On the number line shown, is zero halfway between a and b


<-----------------a---------b----c----------->

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.


it killed my 10 minutes and then I left it, what is the best approach?
Hi what is OA? If C then the image of the number line should not be there I guess as it shows the order a<b<c . Plz Reply.
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by sanju09 » Fri Oct 08, 2010 1:06 am
shovan85 wrote:
RCV wrote:On the number line shown, is zero halfway between a and b


<-----------------a---------b----c----------->

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.


it killed my 10 minutes and then I left it, what is the best approach?
Hi what is OA? If C then the image of the number line should not be there I guess as it shows the order a<b<c . Plz Reply.
a could be equal to -b, but it's not necessary, at least not if all a, b, and c are negative. We could have a = -6, b = -3 and c = -1, for example.
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by limestone » Fri Oct 08, 2010 3:27 am
@Sanju09 : shovan85 has just re-posted your topic in this link : https://www.beatthegmat.com/easy-but-tricky-t68043.html
And I have just confirmed your theory. It's correct.

Thanks for such an interesting and tricky question.

Ps: I'm not good at statistic and hope that sometimes you can post questions of this kind. ( I have seen a statistical DS sufficient question that asked me about standard deviation, sampling, ....). Picture enclosed for the problem I concern.
Image
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by sanju09 » Fri Oct 08, 2010 4:08 am
limestone wrote:@Sanju09 : shovan85 has just re-posted your topic in this link : https://www.beatthegmat.com/easy-but-tricky-t68043.html
And I have just confirmed your theory. It's correct.

Thanks for such an interesting and tricky question.

Ps: I'm not good at statistic and hope that sometimes you can post questions of this kind. ( I have seen a statistical DS sufficient question that asked me about standard deviation, sampling, ....). Picture enclosed for the problem I concern.
Image

The standard deviation is one of the statistical measures used to exemplify the distribution and central tendency of a set of data. The standard deviation is predominantly good for measuring the amount of disparity from the mean. On the GMAT, you probably will not necessitate calculating the standard deviation, but you are responsible for being perceptive what it means. The type of question that you might bump into is shown below:
If the average of 5 data points is 3.5, which new data point would result in the smallest standard deviation?
(A) 2
(B) 2.5
(C) 3
(D) 3.5
(E) 4

[spoiler]Source: https://www.800score.com[/spoiler]
The correct answer to this question is D. To play down the standard deviation, one should choose the value flanking to the present mean. Answer D consents us to choose a data point that equals the present mean, so it will add nothing to the sum of the squared deviations. Since the number of data points will be one more than before, the standard deviation will actually decrease slightly. There is no need to actually calculate the standard deviation on this problem.


For large samples and populations, the primary statistical measures used are the mean and standard deviation. The attached figure shows a typical normal distribution.

For a normal distribution (a sample or population which follows the typical bell-shaped curve as shown in the attachment):

"¢ 68% of the population lie within 1 standard deviation of the mean.

"¢ 95% of the population lie within 2 standard deviations of the mean, and

"¢ 99.7% lie within 3 standard deviations of the mean.

The other statistical measures (median, mode, and range) are considered by the mean and standard deviation.

For a large population, the 50th percentile (with a value equal to the mean) corresponds to the median for a small sample. Similarly, the mean and the 50th percentile characterize the mode of a large normal distribution; for a large population, the range is not employed for the reason that even at the "extremities" of the distribution, there is a finite likelihood of finding a data point. As an alternative, one describes the probability using the number of standard deviations away from the mean. The percentile scores on the GMAT are derived in this way.
Attachments
typicalND.gif
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by limestone » Fri Oct 08, 2010 5:24 am
Thanks again sanju09 for your explanation of what statistical concepts will be tested in GMAT.
And as I understand, one must score at least at 99.7 percentile in the GMAT to be considered as "extremity", huh? :D
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by sanju09 » Thu Mar 22, 2012 3:17 am
limestone wrote:Thanks again sanju09 for your explanation of what statistical concepts will be tested in GMAT.
And as I understand, one must score at least at 99.7 percentile in the GMAT to be considered as "extremity", huh? :D
Thank you limestone for the kind words, but a standard deviation question may appear at less percentile locations than 99.7, with lower level of difficulty, may be. We can't ignore standard deviation with this mere assumption. All the best
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by GMATGuruNY » Thu Mar 22, 2012 3:41 am
RCV wrote:On the number line shown, is zero halfway between a and b


<-----------------a---------b----c----------->

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.
HALFWAY BETWEEN two values is equal to the AVERAGE of the two values.
For the average of a and b to be 0, their SUM must be 0.

Question rephrased: Does a+b = 0?

Statement 1: b>0.
No information about a.
INSUFFICIENT.

Statement 2: The distance between c and a is the same as the distance between c and -b.
The DISTANCE between x and y = |x-y|.
Thus:
|c-a| = |c-(-b)|
|c-a| = |c+b|

Case 1:
c-a = c+b
0 = a+b.

Case 2:
a-c = c+b
a-b = 2c
c = (a-b)/2.
Since a<b on the number line, a-b<0.
Thus, c<0, implying that all 3 values in case 2 are negative:
a<b<c<0.

Since in the first case a+b=0, and in the second case a+b<0,
INSUFFICIENT.

Statements 1 and 2 combined:
Since b>0, case 2 is not possible.
Thus, only case 1 is possible, implying that a+b=0.
SUFFICIENT.

The correct answer is C.

A further exploration of Case 2:
Since c = (a-b)/2 and c>b, we get:
(a-b)/2 > b
a-b > 2b
a > 3b.
Since in case 2 a<b<c<0, the following combination of values would work:
b=-2, a=-4, c = (-4 - (-2))/2 = -1.
The distance between c=-1 and a=-4 is 3, and the distance between c=-1 and -b=2 is 3.
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by [email protected] » Sat May 10, 2014 9:36 pm
Statements 1 and 2 combined:
Since b>0, case 2 is not possible.
Thus, only case 1 is possible, implying that a+b=0.
SUFFICIENT.
How case 2 is not possible, if b>0?

Regards,
Mukherjee
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by GMATGuruNY » Sun May 11, 2014 2:16 am
[email protected] wrote:
Statements 1 and 2 combined:
Since b>0, case 2 is not possible.
Thus, only case 1 is possible, implying that a+b=0.
SUFFICIENT.
How case 2 is not possible, if b>0?

Regards,
Mukherjee
Case 2 requires that a<b<c<0.
In other words, all three values must be NEGATIVE.
Thus, Case 2 is not possible if b>0.
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by GMATGuruNY » Sun May 11, 2014 3:13 am
RCV wrote:On the number line shown, is zero halfway between a and b


<-----------------a---------b----c----------->

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.
Alternate approach:

The number line indicates that a<b<c.
Statement 1 is clearly INSUFFICIENT.
When evaluating statement 2, test one case that also satisfies both statements and one case that satisfies only statement 2.

Statement 2: The distance between c and a is the same as the distance between c and -b.

Case 1: b=1, implying that -b = -1
Since c must be the right of b, let c=2.
The following number line is yielded:
.....-b=-1.....0.....b=1.....c=2.....

Here, -b is 3 places from c.
Thus, a must also be 3 places from c.
Since a must be to the LEFT of c, a must be 3 PLACES TO THE LEFT OF C=2.
In other words, a=-1.
Thus, -b=a=-1, yielding the following number line:
.....-b=a=-1.....0.....b=1.....c=2.....
In this case, 0 is halfway between a and b.

Case 2: b=-1, implying that -b=1
Since a must be the left of b, let a=-2.
The following number line is yielded:
.....a=-2.....b=-1.....0.....-b=1.....

Here, for c to be equidistant from a and -b, c must be halfway between them.
Since there are 3 places between a and -b, c must be 1.5 places to the right of a, yielding the following number line:
.....a=-2.....b=-1.....c=-0.5.....0.....-b=1.....
In this case, 0 is not halfway between a and b.
INSUFFICIENT.

Statements combined:
Since statement 1 requires that b>0, Case 2 is not possible.
Case 1 implies that -- when both statements are satisfied -- 0 is halfway between a and b.
SUFFICIENT.

The correct answer is C.
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