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To me, I prefer to draw an XY plan in order to conclude faster .... Let understand m as x and z as y.

m+z > 0 ?
<=> z > -m ? >>> This question ask us if the points (m,z) are above the line z = -m. In the Fig 1, I draw in green the area we are looking for.

From 1 m-3z > 0
<=> z < m/3 >>> It's all points (m,z) below the line z=m/3

Obviously, by looking at the Fig 2, we can conclude that we have points in the green area where z > -m and points in the red one where z < - m.

INSUFF.

From 2 4z-m > 0
<=> z > m/4 >>> It's all points (m,z) above the line z=m/4

Obviously, by looking at the Fig 3, we can conclude that we have points in the green area where z > -m and points in the red one where z < - m.

INSUFF.

Both (1) and (2) We can conlude with the graph as well, but I prefer here to use some alegbra

We have the system:
o m-3z > 0 (A)
o 4z-m > 0 (B)

(A) + (B) <=> (m-3z) + (4z-m) > 0
<=> z > 0

As well,
(A) <=> m > 3*z
=> m > 3*z > z > 0 as z > 0 then, 3z > z.

Thus,
m>0 and z>0
=> m + z > 0.

SUFF.

Attachments

Fig1_z sup to -m.gif [ 3.99 KiB | Viewed 6460 times ]

Fig2_z inf to m div 3.gif [ 4.32 KiB | Viewed 6455 times ]

Fig3_z sup to m div 4.gif [ 4.08 KiB | Viewed 6457 times ]

why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0 if z=1, m=1,then 4z-m = 3 > 0 if z=1, m= -10 so 4z-m = 14 > 0......am I missing something? _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

Is m+z > 0? (1) m-3z > 0 (2) 4z-m > 0 [#permalink]
02 Feb 2011, 01:54

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puneetj wrote:

why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0 if z=1, m=1,then 4z-m = 3 > 0 if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?

terp26's reasoning for (1) and (2) is not correct: you cannot write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by \(z\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 0?

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

Z is positive when combining both the statements - understood. Using statement (1)m-3z > 0 you proved that m>3z = positive - understood

How do you prove statement (2) the way you proved statement 1? or should we bother proving it at all? 4z>m which means 4 * some positive number > m. Does that prove anything about m?

Please confirm. _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

Z is positive when combining both the statements - understood. Using statement (1)m-3z > 0 you proved that m>3z = positive - understood

How do you prove statement (2) the way you proved statement 1? or should we bother proving it at all? 4z>m which means 4 * some positive number > m. Does that prove anything about m?

Please confirm.

On the GMAT, two data sufficiency statements always provide TRUE information. So: (1) m-3z>0 and (2) 4z-m>0 are given to be true, you shouldn't prove them. You should check whether m+z>0 is true, which is done when we take these two (true) statements together: by adding them we get that z>0 and then looking on (1) with this info we get that m>0 too.

why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0 if z=1, m=1,then 4z-m = 3 > 0 if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?

terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by \(z\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 0?

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

Hi BUnuel I agree to your solution and even i approached the problem in the same fashion. But i have got a doubt ...I understand that whn we combine both the statements we get....Z is positive and when we substitute in the equation m>3z...we end up saying that m must also be greater than 0. On the contrary when we substitute any positive value for z in statement 2 ; we end up getting both negative and positive value for m.... Since we have 4z-m>0 ie m<4z Now if we take say z=1....m<4 that means we can have negative values too for z.....

why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0 if z=1, m=1,then 4z-m = 3 > 0 if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?

terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by \(z\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 0?

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

Hi BUnuel I agree to your solution and even i approached the problem in the same fashion. But i have got a doubt ...I understand that whn we combine both the statements we get....Z is positive and when we substitute in the equation m>3z...we end up saying that m must also be greater than 0. On the contrary when we substitute any positive value for z in statement 2 ; we end up getting both negative and positive value for m.... Since we have 4z-m>0 ie m<4z Now if we take say z=1....m<4 that means we can have negative values too for z.....

Need your help...

Archit

Yes, from z>0 and m<4z we cannot get that m must be positive and that's exactly why we use m>3z to get that.

why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0 if z=1, m=1,then 4z-m = 3 > 0 if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?

terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by \(z\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 0?

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0 if z=1, m=1,then 4z-m = 3 > 0 if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?

terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by \(z\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 0?

(1) m - 3z > 0. Insufficient on its own. (2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) --> \(m+z=positive+positive>0\). Sufficient.

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