DS Quant Review #89 Inequalities

[tonebeeze]

What is the best approach regarding solving multivariable inequalities? Thanks


If x + y + z > -, is z >1?

1. z> x + y + 1

2. x + y + 1 < 0

[Night reader]

What is the best approach regarding solving multivariable inequalities? Thanks


If x + y + z > - ..., is z >1?

1. z> x + y + 1

2. x + y + 1 < 0

one difference between inequalities and equations is the sign => instead of = we may have <, >, =<, >= so four alternatives
x+y+z>- ... can be rewritten as x+y+z+... >0
then (1) z>x+y+1 OR z-x-y-1>0 [+] x+y+z+...>0 is equivalent to 2z-1+...>0 and z>1/2-...
so depending on (...) z can be less or greater than 1 or even equal to 1 Not Sufficient
(2) x+y+1<0 OR -x-y-1>0 [+] x+y+z+...>0 is equivalent to z-1+...>0 and z>1-...
depending on (...) z can be less or greater than 1 or even equal to 1 Not Sufficient
Combining st (1&2) we get z-x-y-1>0 [-] -x-y-1>0 and z>0, so z can be again less, greater or equal to 1 Not Sufficient
Our answer is defined with the value and sign of (...)
If (...)<-1/2 then pick D
If (...)=<0 then pick B
If (...)>0 then pick E

so what's (...)?

[MAAJ]

Remember that you can combine inequalities (You can add them, and also you can multiply them but only when both sides are positive. You CANT subtract them neither divide'm)

So here we go... :

If x + y + z > 0 is z > 1?

1) z > x + y + 1

Rephrase question stem -> z > -x - y
ADD this inequality to the statement 1:

z > - x - y
z > x + y +1
2z > 1 SO z > 1/2
This is not sufficient

2) x + y + 1 < 0

Rephrase question stem -> x + y > -z (then multiply by -1) -x - y < z
Rephrase the stament 2 -> x + y < -1
ADD this two inequalities:

- x - y < z
x + y < -1
0 < z - 1 SO z > 1
This is sufficient

Correct Answer: [spoiler](B)[/spoiler]

[tonebeeze]

What is the best approach regarding solving multivariable inequalities? Thanks


If x + y + z > - ..., is z >1?

1. z> x + y + 1

2. x + y + 1 < 0

one difference between inequalities and equations is the sign => instead of = we may have <, >, =<, >= so four alternatives
x+y+z>- ... can be rewritten as x+y+z+... >0
then (1) z>x+y+1 OR z-x-y-1>0 [+] x+y+z+...>0 is equivalent to 2z-1+...>0 and z>1/2-...
so depending on (...) z can be less or greater than 1 or even equal to 1 Not Sufficient
(2) x+y+1<0 OR -x-y-1>0 [+] x+y+z+...>0 is equivalent to z-1+...>0 and z>1-...
depending on (...) z can be less or greater than 1 or even equal to 1 Not Sufficient
Combining st (1&2) we get z-x-y-1>0 [-] -x-y-1>0 and z>0, so z can be again less, greater or equal to 1 Not Sufficient
Our answer is defined with the value and sign of (...)
If (...)<-1/2 then pick D
If (...)=<0 then pick B
If (...)>0 then pick E

so what's (...)?

OA: B

[Night reader]

Hi Tonebeeze, if you substitute the ... blank space with 0 or any other number you can get the solution tips from the previous post.