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I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

U r right here, the sign of m depends on sign of Z. let say Z is +1 , then m> 3z and m< 4z, which simply means that m>3 and m<4, and m is positive.

Second case let say Z is -1 , then m> 3(-1) and m< 4(-1), which translates to m> -3 and m< -4, and m is negative. But if we draw these regions on a number line, there is no overlap, these regions do not give any solution.

m> -3 : m can be -2,1-,0,1.... m< -4 : m can be -5,-6,-7.....

SO effectively only first case z= +1 is valid case and so both z and m are positive..!!

I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

(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 same sign --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too --> \(positive+positive>0\). Sufficient.

1) m-3z>0 If z=1 m=4 m+z>0 whereas if z=-4 m=1 m+z<0. Hence Statement 1alone is not sufficient 2) 4z-m>0 If m=4 z=2 m+z>0 whereas if m=-4 and z=2 m+z<0. Hence statement 2 alone is not sufficient

Adding both 1 and 2 m-3z >0 4z-m>0 -------- z>0 Since m>3z m is also greater than 0 and Hence m+z>0 . So C
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(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.

I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

(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 same sign --> \(m-3z+4z-m>0\) --> \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too --> \(positive+positive>0\). Sufficient.

Answer: C.

Please help with last part (statement 1+ statement 2). I understand that if z>0 then 3z<m means that m>0 (since posiive number can't less than negative. But this is not true for condition 4z>m where m can be + or -. What i am missing?
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I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

Adding 1 and 2 we get => z>0 and m>3z=> m>0 too . hence C
_________________

Rule used here => WE CAN ADD TWO OR MORE INEQUALITIES if and only if They are in the same direction is Same < or > symbol . Never Subtract ,multiply or divide them as it will get messed up..
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