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Inequality with absolute value

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by goyalsau » Fri Dec 17, 2010 4:33 am
minar wrote:What is the quickest approach to solve it? Please help...


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The OA is C

Question is | x - 1/5 | < 3/5

When there is mod we always take two cases

Ist Case : x - 1/5 < 3/5 ,,,,,,,,, x < 4/5

IInd Case : - x + 1/5 < 3/5 ,,,,,,,, - x < 2/5 , Multiplying by -ve , x > -2/5

Problem can be re written as is -2/5 < x < 4/5

Ist Statement ) -1/2 < x < 4/5

-1/2 is -.50 or in simple terms X is greater than -ive 50 or less than +ve 80 ( 1/2 is 50% & 4/5 is 80% ) It's easy to deal with numbers rather than fractions

Question ask is it less -ve 40 ,
Between -50 and -40 there are 9 values , x can be any value

So insufficient

IInd statement ) x > -1/7

Or we can say x is greater than -ve 14.28 but we don't know about +ve range

So insufficient

Combining statements together

From 1st x -1/2 < x < 4/5 , From IInd x > -1/7

We can say X is definitely in between -1/7 < x < 4/5

Hence Sufficient

Answer C
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by minar » Fri Dec 17, 2010 4:37 am
superb!
thanks a lot...
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by fskilnik@GMATH » Fri Dec 17, 2010 7:55 am
minar wrote:What is the quickest approach to solve it? Please help...
The question may be rephrased as: distance in the real number line between x and 1/5 is less than 3/5 or equivalently (adding and subtracting 3/5 of 1/5):

FOCUS: Is -2/5 < x < 4/5 ?


(1) BIFURCATES:

Take x=0 to get a "Yes"
Can we find x such that -0.5 = -1/2 < x <= -2/5 = -0.4 ? Sure: take x = -0.45 (or even -0.4 itself) to get a "No".

(2) BIFURCATES:

Take x = 0 to get a "Yes"
Take x = 1 to get a "No"

(1+2) We must have -1/7 < x < 4/5

From the fact that 2/5 > 1/7 then -2/5 < -1/7 and that means -2/5 < -1/7 < x < 4/5 therefore DECIDES.

The answer is C .

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by anshumishra » Fri Dec 17, 2010 9:15 am
Here is the pictorial representation of the scenario :

For a two dimensional scenario :
So whenever you see |z-a| = r (which means for all values of z(x,y) , the distance between z and a is constant...so what is this ? circle..right ? So all the points on the circumference of the circle satisfy this )
Think it as a circle with center (a,0) and radius r.

|z-a| < r => All the points within the circle
|z-a| > r => All the points outside of the circle

We have one dimension involved here :

we have equation only given in terms of x and a, so it reduces to a straight line (specifically x-axis). That is easier, forget about anything apart from x-axis (See the pic). We have to ensure that x lies between A and B.

Statement 1 : Implies the line segment : X(-1/2,0) to A(4/5,0) (INSUFF)

Statement 2 : Implies the line segment : W(-1/7,0) to infinity towards right (INSUFF)

Clearly statement 1+ statement 2 => W(-1/7,0) to A(4/5,0) which lies within the boundary i.e A and B (Hence sufficient).
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by fskilnik@GMATH » Fri Dec 17, 2010 9:21 am
Well done, anshumishra. Well done!

That´s exactly what is going on, but I guess people here would benefit even more if your figure were unidimensional only... thinking about open INTERVALS (not open BALLS) is more GMAT-oriented. But, anyway, congrats!

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by anshumishra » Fri Dec 17, 2010 9:29 am
fskilnik wrote:Well done, anshumishra. Well done!

That´s exactly what is going on, but I guess people here would benefit even more if your figure were unidimensional only... thinking about open INTERVALS (not open BALLS) is more GMAT-oriented. But, anyway, congrats!

Regards,
Fabio.
Thanks fskilnik !

Hopefully people get something out of this discussion. Since you had already discussed the uni-dimensional approach perfectly , I thought let me add a little bit more general approach (Also thought, it might help in some of the problems when co-ordinate geometry (circle) is applied with Mods --- Not sure whether GMAT does that -- but for me visualizing make things easier and faster).

I totally agree that thinking uni-dimensional here (or all mod x problems), is more than enough for these kind of problems.

Thanks
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by fskilnik@GMATH » Fri Dec 17, 2010 9:34 am
I am absolutely sure it will be a pleasure to find you in other BTG posts, anshumishra, therefore I hope to have this opportunity in the near future!

All the best,
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by tomada » Fri Dec 17, 2010 11:27 am
Nicely explained!

goyalsau wrote:
minar wrote:What is the quickest approach to solve it? Please help...


Image


The OA is C

Question is | x - 1/5 | < 3/5

When there is mod we always take two cases

Ist Case : x - 1/5 < 3/5 ,,,,,,,,, x < 4/5

IInd Case : - x + 1/5 < 3/5 ,,,,,,,, - x < 2/5 , Multiplying by -ve , x > -2/5

Problem can be re written as is -2/5 < x < 4/5

Ist Statement ) -1/2 < x < 4/5

-1/2 is -.50 or in simple terms X is greater than -ive 50 or less than +ve 80 ( 1/2 is 50% & 4/5 is 80% ) It's easy to deal with numbers rather than fractions

Question ask is it less -ve 40 ,
Between -50 and -40 there are 9 values , x can be any value

So insufficient

IInd statement ) x > -1/7

Or we can say x is greater than -ve 14.28 but we don't know about +ve range

So insufficient

Combining statements together

From 1st x -1/2 < x < 4/5 , From IInd x > -1/7

We can say X is definitely in between -1/7 < x < 4/5

Hence Sufficient

Answer C
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by tomada » Fri Dec 17, 2010 11:30 am
Nicely explained!

goyalsau wrote:
minar wrote:What is the quickest approach to solve it? Please help...


Image


The OA is C

Question is | x - 1/5 | < 3/5

When there is mod we always take two cases

Ist Case : x - 1/5 < 3/5 ,,,,,,,,, x < 4/5

IInd Case : - x + 1/5 < 3/5 ,,,,,,,, - x < 2/5 , Multiplying by -ve , x > -2/5

Problem can be re written as is -2/5 < x < 4/5

Ist Statement ) -1/2 < x < 4/5

-1/2 is -.50 or in simple terms X is greater than -ive 50 or less than +ve 80 ( 1/2 is 50% & 4/5 is 80% ) It's easy to deal with numbers rather than fractions

Question ask is it less -ve 40 ,
Between -50 and -40 there are 9 values , x can be any value

So insufficient

IInd statement ) x > -1/7

Or we can say x is greater than -ve 14.28 but we don't know about +ve range

So insufficient

Combining statements together

From 1st x -1/2 < x < 4/5 , From IInd x > -1/7

We can say X is definitely in between -1/7 < x < 4/5

Hence Sufficient

Answer C
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by Night reader » Fri Dec 17, 2010 1:24 pm
minar wrote:What is the quickest approach to solve it? Please help...


Image


The OA is C
simplify the problem |x-1/5| <3/5 => x-1/5 > -3/5 and x-1/5 <3/5 => -2/5<x<4/5

st(1) -1/2<x<4/5 leaving out interval on the xy-plane {-1/2; -2/5} Not sufficient
st(2) x>-1/7 suggests that x is greater than all numbers to the right from -1/7 Not sufficient
Combining st(1&2) => if x>-1/7 and x>-1/2 the interval for x falls within (-) infinity... -1/2 ...-2/5 ... -1/7 with this we clear the left side inequality values for -2/5<x; the right side is cleared with st(1)

Choice C
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by goyalsau » Fri Dec 17, 2010 8:01 pm
anshumishra wrote:Here is the pictorial representation of the scenario :

For a two dimensional scenario :
So whenever you see |z-a| = r (which means for all values of z(x,y) , the distance between z and a is constant...so what is this ? circle..right ? So all the points on the circumference of the circle satisfy this )
Think it as a circle with center (a,0) and radius r.

|z-a| < r => All the points within the circle
|z-a| > r => All the points outside of the circle
HI! anshu, This is a completely New approach for solve a problem, thanks for sharing But i have some doubts with your approach......... Please explain them..

Question was | z - a | < r { you changed it to |z-a| = r } Why is that so ????

When you says all values of z ( x, y ) and think the center of the circle as ( a,0) Its like calculating distance between two coordinates,

What is the reasoning behind this argument. ?????????
Saurabh Goyal
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by anshumishra » Fri Dec 17, 2010 8:22 pm
goyalsau wrote:
anshumishra wrote:Here is the pictorial representation of the scenario :

For a two dimensional scenario :
So whenever you see |z-a| = r (which means for all values of z(x,y) , the distance between z and a is constant...so what is this ? circle..right ? So all the points on the circumference of the circle satisfy this )
Think it as a circle with center (a,0) and radius r.

|z-a| < r => All the points within the circle
|z-a| > r => All the points outside of the circle
HI! anshu, This is a completely New approach for solve a problem, thanks for sharing But i have some doubts with your approach......... Please explain them..

Question was | z - a | < r { you changed it to |z-a| = r } Why is that so ????

When you says all values of z ( x, y ) and think the center of the circle as ( a,0) Its like calculating distance between two coordinates,

What is the reasoning behind this argument. ?????????
Hi goyalsau,

Glad that you asked!
Question was | z - a | < r { you changed it to |z-a| = r } Why is that so ????
I tried to explain that |z-a| = r , represents a circle with center at (a,0) and radius "r". So, all the points on the circumference of this circle satisfy the equation.

When the equation is : |z - a| < r , that means all the points within the circle (BUT NOT ON CIRCUMFERENCE) satisfies this equation. I started with |z -a| = r to just introduce with concept.

Similarly if the equation is |z - a| > r, all the values in the universe satisfy this equation except that are on the circumference or within the area of the circle.
When you says all values of z ( x, y ) and think the center of the circle as ( a,0) Its like calculating distance between two coordinates,

What is the reasoning behind this argument. ?????????
The reasoning behind the argument was that, I wanted the reader to himself deduce that this equation would lead to a circle. For e.g. think about it , from any static point (it may be [a,0] as here, or anything actually), if you have to find all the points in x-y ordinates, which are at equal distance , what would that curve be ? .......... Circle , right ?

Please note, the problem here was uni-dimesnsional (as only x co-ordinate was needed ),so we converged to the number line and only points of interest were those on X-coordinates (such as, O, A, B, etc..).

Let me know if you have any other query.
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