GMAT Prep 1

[Guest]

Any ideas on how to solve the attached problem?

Thanks in advance.

[Guest]

OP^2= 3 + 1 --> OP = 2

So the hypotenus PQ = 2*sqrt(2)

PQ should be also equal to the length or "projections of P an Q" on the X axis, which is = s - (- sqrt(3)) --> s + sqrt(3)

now you have

s + sqrt(3) = 2*sqrt(2)

s = 2*sqrt(2) - sqrt(3)

s = 2 * 1.42 - 1.73 (approximate)

s ~ 1 (closest answer)

===========================

long way

s^2 + t^2 = 4 ------------------- (1)

(s + (sqrt(3))^2 + (t+1)^2 = 8 ----- (2)

expand (2)

s^2 + 3 + 2.s. (sqrt(3)) + t^2 +1 - 2t = 8 ------- (3)

use (1) in (3) and re arrange

4 + 3 + 1 + 2.s. (sqrt(3)) - 2t = 8

2.s. (sqrt(3)) = 2t

s. (sqrt(3)) = t ------------ (4)

put (4) in (1)

s^2 . 3 + s^2 = 4

4. s^2 = 4

s = 1 or -1

(choose according to the quad + answer choices)

[RonPurewal]

Geez Louise, that's a deceptive picture...

If you know that root(3) is approximately 1.7, you could re-create the diagram to scale. Your new drawing would put point P closer to the x-axis than to the y-axis, and, if you drew the right angle faithfully, would put point Q closer to the y-axis than to the x-axis.
Since it appears that you used 'symmetry' (illusory symmetry about the y-axis, that is, inferred from the tricky diagram) to infer the solution you wrote, you could use a similar inference here (and it would be correct this time!). Point P is 1.7 units to the LEFT, so point Q must be 1.7 units UP; point P is 1 unit UP, so point Q must be 1 unit to the RIGHT. The latter gives you the value of 's'.

Notice that, if you ROTATE the figure 90 degrees anti-clockwise, point Q goes right on top of the original location of point P. (Even if your diagram is terrible, you still know this because of the right angle that's marked.) You can rotate the diagram, mark the root(3) and 1 on the diagram (they're the same as the coordinates of P after the rotation: root(3) or 1.7 units to the left, 1 unit up). After you rotate the diagram back, you'll have marked the x-coordinate = 1.