Wow, thank you for that very fast and detailed response! Did you have to create that, or was this asked before?

Thanks again!

acon wrote:Can anyone explain this? I don't understand why they say the answer is 1.

We actually have an isosceles right triangle OPQ with O containing the right-angled vertex of it. Things are given on a real rectangular coordinate plane in such a way that we can easily understand that O is (0, 0) and s is positive. One can apply basics of coordinate geometry to come down to the following equations:

s^2 + t^2 = 4 and

(s + √3)^2 + (t - 1)^2 = 8, which reduces to t = s √3

And therefore s^2 = 1 or s = ±1, NO...

Just 1

Well cheers to a fellow British Colombian. Sunshine coast is a beautiful spot! I'm down in Victoria, but grew up in Vancouver also. I checked out your site, it seems you offer a great service!

I am writing the GMAT tomorrow, and today is my first and last real day of preparations. Would have liked to have been more prepared, but time constraints don't permit much.

The following are the only other two questions that I haven't been able to figure out on my own from the GMAT Prep.

Their answer in this first one is 16, which I would have thought to be correct had the question been 32 + ...,



This one I am entirely lost in..



If anyone is able to help it would be greatly appreciated!

Hi acon,

All the best for tomorrow!!!

1. let equal sides of iso. tri. be x. Then perimeter = 2x + √2x

So 2x + √2x = 16 + 16√2. Now the equation does not hold when √2x = 16√2. So √2x = 16.
Now multiplying both sides by √2 => 2x = 16√2 => x = 8√2
Hypo = √2x = 8√2 * 2 = 16.

2. for iso. tri. QRS, let equal angles be a. for iso. tri. QRS, let equal angles be b.

a+b+x = 180 & x + P = 180. but P = 90 => a+b=90

Hence x = 90

acon wrote:Well cheers to a fellow British Colombian. Sunshine coast is a beautiful spot! I'm down in Victoria, but grew up in Vancouver also. I checked out your site, it seems you offer a great service!

I am writing the GMAT tomorrow, and today is my first and last real day of preparations. Would have liked to have been more prepared, but time constraints don't permit much.

The following are the only other two questions that I haven't been able to figure out on my own from the GMAT Prep.

Their answer in this first one is 16, which I would have thought to be correct had the question been 32 + ...,



This one I am entirely lost in..



If anyone is able to help it would be greatly appreciated!

We must keep in wits some special triangles that the GMAT tests very recurrently, more than ever 45-45-90 and 30-60-90 ones. Your first problem is the case of a 45-45-90 triangle whose sides opposite those angles in order bear the ratio 1:1:√2. Hence, if a be each leg of the isosceles right triangle, then a √2 will be its hypotenuse and 2 a + a √2 its perimeter, which is given as 16 + 16 √2 in your problem.

Take 2 a + a √2 = 16 + 16 √2

Or a (2 + √2) = 16 + 16 √2

Or a = (16 + 16 √2)/( 2 + √2)

But we want its hypotenuse, which is a √2

Hence a √2 = (√2) (16 + 16 √2)/ (2 + √2) = (16 √2 + 32)/ (2 + √2) = 16 (2 + √2)/ (2 + √2) = 16.

Take B.


For your second problem, let's make few assumptions first, like as if ∠PRT = α so that ∠PTR = 90 - α .

Now (1) allows us to let ∠RQS = ∠RSQ = 1/2 (180 - α) = 90 - (α/2).

Whereas ∠TSU = 180 - {x+90 - (α/2)} = 90 - x + (α/2)

And of course ∠TUS = 180 - {90 - x + (α/2) + 90 - α} = 2 x - (α/2)

We can now come inside the quadrilateral PQSU that will contain the four interior angles as 90, 90 + (α/2), x, and 180 - 2 x + (α/2) such that

90 + 90 + (α/2) + x + 180 - 2 x + (α/2) = 360

Or simply, x = α. So x can be known only if α is known. Insufficient.

Now switch to statement (2). Even this information alone plus our valid assumptions take us to the same situation as it was with statement (1) alone. Hence it is too insufficient.

Now take the two statements together, I mean take ∠TSU = ∠TUS or 90 - x + (α/2) = 2 x - (α/2) with x = α, or we can now have 90 - x + (x/2) = 2 x - (x/2) or x = 45. SUFFICIENT

C.

All the best acon!