Math problem

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Math problem

by priyasaibaba » Thu Apr 22, 2010 12:13 pm
Can anyone help to solve this problem?

Is X close to 10^(-2) or 10^(-3)?
(1) x is closer to 10^-4 than to 10^-1.
(2) x is closer to 10^-3 than to 10^-1.

A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


Thanks,
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by thephoenix » Thu Apr 22, 2010 7:52 pm
IMO A

s1) on number line x will have a value more than the avg of 10^-4 and 10^-1 which is actually closer to 10^-3
suff
s2) insuff

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by Rahul@gurome » Thu Apr 22, 2010 9:36 pm
Let X = 10^y.
So the question becomes Is y close to -2 or to -3?
(1) y is closer to -4 than to -1.
(2) Y is closer to -3 than to -1.

Consider statement (1) alone.
If y is closer to - 4 than to -1, then there can be two cases.
Either y < -4 or - 4<y<-1.
If y < -4 obviously it will be closer to -3 than to -2.
If -4<y<-1 and y is closer to - 4, it means distance between y and -4 on the number line is less than the distance between y and -1.
Now distance between -4 and -1 is -1-(-4) = 3.
Half of this is 1.5.
So y-(-4) < 1.5
Or y+4<1.5
Or y < -2.5.
Take an example. Let y = -2.6 which is less than -2.5.
Distance between -2.6 and -2 is -2- (-2.6) = 0.6
Distance between -2.6 and -3 is -2.6 - (-3) = 0.4
So we see that -2.6 is closer to -3 than to -2.
So we can conclude that any number closer to -4 than to -1 is also closer to -3 than to -2.
So (1) is sufficient.
Consider (2)
Again there can be two cases.
First that y < -3 and second -3<y<-1.
In the first case it will be closer to -3 than to -2.
In the second case we will have that distance between y and -3 is less than the distance between y and -1.
Distance between -3 and -1 = 2
Half of this is 1.
So y-(-3) < 1
Or y<-2
Take y = -2.2 for example.
Dist between -2.2 and -2 is 0.2. and between -2.2 and -3 is 0.8.
Here y is closer to -2 than to -3.
Since nothing definite can be said we say that (2) is not sufficient.

The correct answer is A.
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by neodemion » Fri Apr 23, 2010 1:05 am
I might be totally mistaken, but I believe that both of the previous responses are wrong.

if you draw the situation:

--|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|
0.1 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0

having that:
10^-1 = 0.1
10^-2 = 0.01
10^-3 = 0.001
10^-4 = 0.0001

You can see that the last three points are located within the 0.01 - 0 range.. therefore even 0.03 is closer to any of them than to 10^-1... I believe that we have not enought data to solve the problem. Both of them are not sufficient.

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by neodemion » Fri Apr 23, 2010 1:15 am
Probably you can also see it better with integers..

the problem is the same as:

is X closer to 100 or to 10?

(a) x is closer to 1 than to 1000

(b) x is closer to 10 than to 1000

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by Nilakshi1001 » Fri Apr 23, 2010 2:02 am
My answer:

To find x is closer to which value, let's take the average of 0.01 and 0.001 which is 0.0055, meaning 5.5X10^-3.

Taking first condition, x is closer to 10^-4 than to 10^-1. The average of 0.0001 and 0.1 is 0.1001/2, i.e. 0.05005 which is 5.005X10^-2. Condition 1 gives us x<5.005X10^-2. Thus, the condition is insufficient.

Second condition gives us x is closer to 10^-3 than to 10^-1. Using this we get x < 0.101/2 = 0.0505 = 5.05X10^-2. the condition is insufficient.

.......... 10^-4....................10^-3..........................10^-2.......................10^-1.......
---------------------------------5.005X10^-2-------------------------------
============5.05X10^-2============

----------5.5X10^-3---------

using both the statement also, we cannot find out the answer.

Thus, my answer is E.

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by ajith » Fri Apr 23, 2010 5:26 am
Moved to DS forum!
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by Fiver » Fri Apr 23, 2010 6:13 am
Agree with A.

Is x close to 1/100 or to 1/1000?

St1] x is closer to 10^-4 than to 10^-1.
The value midway of 10^-1 and 10^-4 is 10^-2.5 and to comply to st1] x has to be > 10^-2.5
This makes x closer to 10^-3 than to 10^-2.
Suff.

St2] x is closer to 10^-3 than to 10^-1.
Applying the same logic as above x>10^-2.
Hence x could be for e.g. 10^-2.1, which makes it closer to 10^-2 or any value greater than 10^-2.5, which makes it closer to 10^-3.
Insuff.

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by neodemion » Fri Apr 23, 2010 6:45 am
Fiver, the midway of 10^-1, which is equal to 0.1, and 10^-4, which is equal to 0.0001 is not 10^-2.5, which is equal to 0.0032. The midway is (0.1 + 0.0001) / 2 = 0.05005

Besides, the simplification proposed by Rahul is wrong.

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by Fiver » Fri Apr 23, 2010 7:31 am
neodemion wrote:Fiver, the midway of 10^-1, which is equal to 0.1, and 10^-4, which is equal to 0.0001 is not 10^-2.5, which is equal to 0.0032. The midway is (0.1 + 0.0001) / 2 = 0.05005

Besides, the simplification proposed by Rahul is wrong.
You are right. My approach is fundamentally flawed.

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by tohellandback » Fri Apr 23, 2010 11:42 am
IMO E

1) INSUFF: the point x can lie in region m,9m,90m and also 900m upto <500m

2) INSUFF:same reasoning..the point x can lie in region m,9m,90m and also 900m upto <495m

combined:same reasoning
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by ansumania » Sun Apr 25, 2010 7:14 pm
hi,

is there a convenient way of analyzing the data?

pl. provide...

regards,

Ansuman