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Easy but Tricky

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by shovan85 » Fri Oct 08, 2010 12:32 am
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.

OA: C
Last edited by shovan85 on Sun Oct 17, 2010 11:12 am, edited 1 time in total.
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Source: — Data Sufficiency |
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by limestone » Fri Oct 08, 2010 1:16 am
1. O can either be:

0<a<b
a<0<b

Then insuff.

2.
The distance between c and a is the same as the distance between c and -b.
<--------------a------b----------c--------------(-b)------------>

The distance between c and b is smaller than the distance between c and a ( as given information)
Then the distance between c and b is smaller than the distance between c and -b
However, the distance between 0 and b is equal to the distance between 0 and -b.

Then b<c<0<-b as shown below:

<-------------a---b------------c---0-------------(-b)------->

To make it clearer,
let's call length of "bc" : x
that of "-bc": y
Ob =O-b = z
As length of O-b=length of ca and length of ca > than length of cb,
then z>x, or O is to the right of c.

If 0>c then obviously:
0>c>b>a

Then Zero is not halfway between a and b.

Suff.

Pick B
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by shovan85 » Fri Oct 08, 2010 2:23 am
limestone wrote:1. O can either be:

0<a<b
a<0<b

Then insuff.

2.
The distance between c and a is the same as the distance between c and -b.
<--------------a------b----------c--------------(-b)------------>

The distance between c and b is smaller than the distance between c and a ( as given information)
Then the distance between c and b is smaller than the distance between c and -b
However, the distance between 0 and b is equal to the distance between 0 and -b.

Then b<c<0<-b as shown below:

<-------------a---b------------c---0-------------(-b)------->

To make it clearer,
let's call length of "bc" : x
that of "-bc": y
Ob =O-b = z
As length of O-b=length of ca and length of ca > than length of cb,
then z>x, or O is to the right of c.

If 0>c then obviously:
0>c>b>a

Then Zero is not halfway between a and b.

Suff.

Pick B
I agree. But Sanju09 has different theory here https://www.beatthegmat.com/best-approac ... 67985.html.
Can you please have a look there?
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by limestone » Fri Oct 08, 2010 3:15 am
I have just looked at Sanju09's theory. And it's correct.

I'll give out 2 example that make (2) insuf :

1st one:

a = -4; b =-2, c=-1, -b = 2

The distance between a and c is 3 units of length.
The distance between -b and c is 3 units of length too.

In this case a<b<0

2nd one:

a = -4, b = 4, c = 1, -b = -4

The distance between a and c is 5 units of length.
The distance between -b and c is 5 units of length too.

In this case a<0<b

From the above examples, 2 alone is not suf.

1 alone is obviously not suf.

However, when combining with 2, (1) will eliminate the case that a<b<0 ( where b is to the left of zero). Thus the combined condition forces a and b to such an order of: a<0<b. Then 1 & 2 together is suff.

C should be the answer.

Thanks Sanju09 for an interesting and tricky question and shovan85 for reposting this question.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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