GPREP DS

[abhasjha]

[[email protected]]

Hi abhasjha,

This is actually a great "concept" question; you don't have to do much "math" to solve it, if you recognize the geometry formulas involved.

From the picture and the information in the prompt, we know that we're dealing with a square (with side length = 4) and that the two shaded triangles are IDENTICAL (each is a right triangle, one side is 4 and we're told that LM = PQ).

We're asked for the length of PQ.

Fact 1: We're given the length of PR.

This means that we have 2 sides of a right triangle, so we CAN figure out the third side (using the Pythagorean Theorem). We CAN find the value of PQ, but I'm not going to waste time doing that math.
Fact 1 is SUFFICIENT.

Fact 2: The ratio of the unshaded region to the shaded region is 2:1

Since we have a square, we know the total area is 4x4 = 16, so we COULD figure out the actual area of both the unshaded and shaded regions. From there, we take the area of the shaded region and cut it IN HALF (since that region consists of 2 identical triangles). Then we'll have the AREA of 1 of the triangles; we CAN use the triangle area formula A = (B)(4)/2 to find the missing side PQ. While we could do lots of math here, none of that math is required.
Fact 2 is SUFFICIENT.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

Points M and P lie on square LNQR, and LM = PQ. What is the length of the line segment PQ?

(1) PR = (4/3)√10
(2) The ratio of the area of the unshaded region to the total area of the shaded region is 2 to 1

Statement 1: PR = (4/3)√10
Since QR=4, PR = (4/3)√10, and PQ² + QR² = PR², we can solve for PQ.
SUFFICIENT.

Statement 2: The ratio of the area of the unshaded region to the total area of the shaded region is 2 to 1.
Since the two statements cannot contradict each other, the case given in statement 1 -- QR=4 and PR = (4/3)√10 -- must also satisfy statement 2.
Check whether this is the ONLY case that will satisfy statement 2.

Plugging QR=4 and PR = (4/3)√10 into PQ² + QR² = PR², we get:
PQ² + 4² = [(4/3)√10]²
PQ² + 16 = 160/9
PQ² = 160/9 - 144/9
PQ = 4/3.

In this case:
Area of ∆PQR = (1/2)(PQ)(QR) = (1/2)(4/3)(4) = 8/3.
Since LM = PQ, ∆LMN = ∆PQR = 8/3.
Thus:
The total area of the shaded region = ∆LMN + ∆PQR = 8/3 + 8/3 = 16/3.
The total area of the unshaded region = LNQR - shaded region = 16 - 16/3 = 32/3.
Resulting ratio:
unshaded : shaded = (32/3) : (16/3) = 2:1.

The value given in statement 1 -- PR = (4/3)√10 -- yields the 2:1 ratio required by statement 2.
If the length of PR increases or decreases, the ratio of the two areas will NOT be 2:1.
Thus, to satisfy statement 2, it must be true that PR = (4/3)√10, implying that PQ = 4/3.
SUFFICIENT.

The correct answer is D.