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DS - Inequalities - Strategy help please

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DS - Inequalities - Strategy help please

by zueswoods » Wed Jan 18, 2012 8:21 am
Hi,

With the difficult Inequality DS questions where by you need to test different values ... does anyone have a rock hard strategy for testing the correct values.

I know you test fractions, zero, negs, positives, HOWEVER with a question like this (from MGMAT CAT):

Is X > Y ?

(1) root(x) > y

(2) X^3 > y


I would try 2 values for each statement (fractions and negs) and realize on their own BOTH are insufficient.

However when testing to see if they are sufficient together, i find it difficult to figure out (quickly) what values to test that will suit both conditions.

For example statement 1
i will try

y= 1/2, y^2= 1/4, x = 1/2 ... (x>y? no)

y = -2, Y^2= 4, x = 5 ... (x>y? yes)

NOT SUFF
statement 2

x = 2, x^3 = 8, y = 7 (x>y? no)

x= 1/2. x^3 = 1/8, y = 1/9 (x>y? yes)

NOT SUFF

(Now I know i need to test for both, however, I find it difficult (within 2 mins) to figure what values to use in order to suit both conditions, and try and figure out if there is only one possible answer)

The answer is C (both)

I will appreciate any help.

Thanks!




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by GMATGuruNY » Wed Jan 18, 2012 12:26 pm
zueswoods wrote:Hi,

With the difficult Inequality DS questions where by you need to test different values ... does anyone have a rock hard strategy for testing the correct values.

I know you test fractions, zero, negs, positives, HOWEVER with a question like this (from MGMAT CAT):

Is X > Y ?

(1) root(x) > y

(2) X^3 > y


I would try 2 values for each statement (fractions and negs) and realize on their own BOTH are insufficient.

However when testing to see if they are sufficient together, i find it difficult to figure out (quickly) what values to test that will suit both conditions.

For example statement 1
i will try

y= 1/2, y^2= 1/4, x = 1/2 ... (x>y? no)

y = -2, Y^2= 4, x = 5 ... (x>y? yes)

NOT SUFF
statement 2

x = 2, x^3 = 8, y = 7 (x>y? no)

x= 1/2. x^3 = 1/8, y = 1/9 (x>y? yes)

NOT SUFF

(Now I know i need to test for both, however, I find it difficult (within 2 mins) to figure what values to use in order to suit both conditions, and try and figure out if there is only one possible answer)

The answer is C (both)

I will appreciate any help.

Thanks!
Since we can't take the square root of a negative, statement 1 implies that x≥0.
Thus, when we combine the two statements, if y<0, we know that y<x.
Our concern is what happens when y≥0.
One approach is to memorize the shapes of some basic graphs:

Image

Only in the yellow region is y<√x and y<x³.
The entire yellow region is below the graph of y=x, implying that y<x throughout the entire region.
Thus, combining the two statements, we know that y<x.
SUFFICIENT.

The correct answer is C.

An alternate approach would be to use algebra to test the 3 cases: y=x, y>x, and y<x.

Case 1: y=x.
Statement 1: If y=x and y<√x, then x < √x.
Statement 2: If y=x and y<x³, then x < x³.
No value for x will work here: a number cannot be less than both its root and its cube.
Thus, y≠x.

Case 2: y>x.
Statement 1: If x<y and y<√x, then x < √x.
Statement 2: If x<y and y<x³, then x < x³.
No value for x will work here: a number cannot be less than both its root and its cube.
Thus, it is not possible that y>x.

Since it is not possible that y=x or that y>x, we know that y<x.
SUFFICIENT.
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by Ian Stewart » Wed Jan 18, 2012 3:37 pm
zueswoods wrote:Hi,

With the difficult Inequality DS questions where by you need to test different values ... does anyone have a rock hard strategy for testing the correct values.

I know you test fractions, zero, negs, positives, HOWEVER with a question like this (from MGMAT CAT):

Is X > Y ?

(1) root(x) > y

(2) X^3 > y
If the question mentions √x, then x clearly cannot be negative. When we combine the two statements, then if x is 0 or 1, clearly x > y from either statement. If x is between 0 and 1, then x^3 is smaller than x, and √x is greater than x. So in this case, combining the two statements, we know that √x > x > x^3 > y. If x > 1, then x^3 > x and √x < x, so in this case, combining the two statements, we know that x^3 > x > √x > y. So in every case we find x > y, and the two statements together are sufficient.
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by [email protected] » Sat Jan 21, 2012 5:21 am
honestly this sum at first instance is impossible to solve in 2 mins as you will not be able to come up with the values only...
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by golu23 » Fri Jan 27, 2012 1:15 am
hi zeuswood could u please eloborate on ur explanation i understand it a bit but not fully
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by zueswoods » Tue Feb 14, 2012 12:43 pm
Hi GmatguryNY

Hi again,

I wanted to make sure that I have got this technique down.

The question is:

Is x>y?

(1) x^2 > y
(2) root(x) < y

So I drew the graphs as follows

(1) y < x^2

Image




and we see that the yellow region could be above y=x or below so its insufficient

(2) y > root(x)

Image




still insuf.

Now when taking them both together

Image



We still see that the yellow region is above y=x and below y=x, therefore the answer is E.

Did i work through this correctly?

Thanks so much for your time, i was really strugling on these questions

Also, when will this technique not work?

I know with absolute values I will have to work through it and solve, but other than that?

zueswoods
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by [email protected] » Wed Feb 15, 2012 4:52 am
Is X > Y ?

(1) root(x) > y

(2) X^3 > y



Explanation:

Statement 1 Alone:

If you take statement 1, you find that the inequality does not stand when 0 < x < 1. i.e for fractions between 0 and 1


Statement 2 Alone:

If you take statement 2, you find that the inequality does not stand when x > 1;


Both 1 and 2:


Both the statements meet the required conditions...
Yes in order to understand properly, you can refer to the graphical inequalities above...



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by GMATGuruNY » Wed Feb 15, 2012 12:39 pm
zueswoods wrote:Hi GmatguryNY

Hi again,

I wanted to make sure that I have got this technique down.

The question is:

Is x>y?

(1) x^2 > y
(2) root(x) < y

So I drew the graphs as follows

(1) y < x^2

Image




and we see that the yellow region could be above y=x or below so its insufficient

(2) y > root(x)

Image




still insuf.

Now when taking them both together

Image



We still see that the yellow region is above y=x and below y=x, therefore the answer is E.

Did i work through this correctly?

Thanks so much for your time, i was really strugling on these questions

Also, when will this technique not work?

I know with absolute values I will have to work through it and solve, but other than that?

zueswoods
Nice work.

In my post above, I offered a graphical solution because y=x, y=x³ and y=√x are common graphs that can be drawn quickly.

The problem you cited also is about common graphs, so a graphical approach works well.
It might be quicker, however, simply to combine the inequalities and plug in:
Combining √x<y and y<x², we get √x < y < x².
If x=100, then 10<y<10,000.
Thus, it's possible that y<x, y=x, or y>x, indicating that the two statements combined are INSUFFICIENT.

Ideally, you'll be comfortable with a variety of approaches.
Some problems are best solved algebraically, others by plugging in values.
Graphing can be helpful, as can drawing a number line.
Many problems are best solved with a combination of techniques.

Here's another problem that I solved graphically:

https://www.beatthegmat.com/xy-plane-and ... 97228.html
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