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I'm thinking this is one of the tough GMAT questions.

Assuming IaI & IbI are 3 digit numbers with the digit in the units and hundreds place being the same, aren't they actually asking: If a < b internally ? example 121 and 131 should be the same as 2 < 3? also, 151 and 141 should be same as 5 < 4 ? 1) b < -a => a + b < 0 (INSUFFIC...

Relative speeds are always tricky. Unless specified or explicitly asked for you measure speeds with respect to something that is still. Even though the belt is moving at 3ft/s, its irrelevant. What is Bill's rate of movement? (his speed + the speed of the belt) with respect to something that is stat...

Oops, I think I mis-read the question. I will try to solve it again assuming IaI means absolute value of a and not Integer A Integer. Sorry about the mis-lead.

I had an elaborate solution but I learned this from another friend yesterday. Really Cool.

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Thanks

I think, there is a small flaw in your approach for the 2nd argument. A quadrilateral whose diagonals are perpendicular bisectors doesn't have to be a rhombus. It is a parallelogram for sure but not a rhombus. Remember, a rhombus is a quadrilateral with all sides of the same length. It doesn't talk ...

Sorry, I completely forgot about this... This is what I think... 1. b < -a implies that a + b < 0 (you get this when you add a to either sides of the inequality). This implies either, i) both a and b are -ve (cannot say if |a| > |b| here because we are not sure if b is more -ve (far down on...

Sorry again! I meant E is the answer choice: Statements 1 and 2 TOGETHER are NOT sufficient

Assuming that ? is not an arithmetic operator (+, -, * / etc)

let us say ? is p, what is asked is:
xy*p = x * (x -y)
p = x * (x - y) / xy

=> p = x/y -1, in other words ? = x/y -1

putting this value back; x( x/y-1)y == x( x-y) ;)

Seems like a funny one to solve...

The answer is 16. What the problem meant is that the triangle is a right isosceles triangle. So, if the 2 sides are 'x' and 'x' the hypotenuse would be x * sqrt(2) [use the pythagoras theorem to get that] perimeter of the right isosceles triangle would be 2x + sqrt(2) * x = 16 + 16 * sqrt(2) [given]...

I think Ron's solution could be simplified even more. Ron's solution considers a rectangle but it would be simpler to consider a trapezium. Let's say we draw a perpendicular line to x -axis from R and that intersects x-axis at M. Then all we need to calculate is: A(Trapezium OQRM) - A(Right Triangle...

This seems like a real tough one. Hmmm...I could try an explanation but I'm not very sure at this point. h(100) + 1 = (2 * 4 * 6 * ... * 98 * 100) + 1 Taking 2 as a common factor: = 2 ^ 50 * (1 * 2 * 3 * ... * 47 * 48 * 49 * 50) = 2 ^ 50 * 50! We could observe that the above value contains all the p...

a) m + n < 0 => m < -n (INSUFFICIENT) which does not reveal much because if m and n are both equal to -1 then m + n < 0 and m == n but if m and n are -2 and -1 then m + n < 0 and m != n (not equal to) So, this is a MAYBE case. b) mn < 0 (SUFFICIENT) => (m != 0 and n !=0 and eith...