Problem Solving

I have to disagree with the post above; I find it very risky to try to apply some distorted version of the 'n linear equations/n unknowns' rule to similar questions at the higher level of the GMAT. Of course sometimes this will give the right answer, but as often as not it will land a test taker in a trap. In the question above, it does not ask for the values of our unknowns, but rather for their sum; we may need less information to find some combination of x and y than to find either x or y alone. Second, neither of our equations are linear, so the rule doesn't even apply. Third, while we cannot have negative values in geometry, that does not mean we can ignore the fact that our equations are non-linear, as the first question below illustrates. Were you to apply such a 'rule' to either of the following questions, for example, you would not get the right answer:

If the lengths of three sides of a right triangle are x, y and 10, where 10 is the length of the hypotenuse, what is the value of x?

1) x + y = 14
2) xy = 48

Despite the fact that we have two equations, two unknowns when we consider statement 1 alone, we still can't determine whether x is equal to 6 or to 8. The same is true when we use Statement 2 alone. Even combining the statements, we have three distinct equations, two unknowns, and still cannot find x. The answer is E.

If the lengths of the three sides of a triangle are x, y and 7, what is the perimeter of the triangle?

1) x = 10 - y
2) x - y = 2



This is a rather simple example, but illustrates that we don't always need to know the values of x or y alone to find the value of x+y; that is, we don't always need two distinct equations in our two unknowns if we are asked to find some combination of those unknowns. Using Statement 1, we have only one equation, two unknowns, but we can find x+y; the answer is A.

Questions 123, 150 and 168 in the DS section of OG12, to give just three official examples among many that I have seen, seem specifically designed to trap test takers who simply count equations and unknowns. Questions 150 and 168 in particular ask about combinations of unknowns; we don't need to find the value of each unknown to answer these questions.

Back to the original question:

Chick wrote:The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimetres.

(2) The 2 legs of the triangle are of equal length.

I do think the algebraic solution in the OG is worth looking at here. The (x+y)^2 = x^2 + 2xy + y^2 relationship appears very frequently on the GMAT, and if you ever see any two of the expressions x+y, xy and x^2 + y^2 in a question, you'll be able to use this relationship (and similarly, if you see two of x-y, xy and x^2 + y^2 you'll be able to use the fact that (x-y)^2 = x^2 - 2xy + y^2).

There are certainly other ways to look at the problem. As an alternative to the algebraic solution in the OG, you might notice that, if x and y are the lengths of the legs of the triangle, then x^2 + y^2 = 100 and xy = 50, so x^2 + y^2 = 2xy, and x^2 - 2xy + y^2 = 0. Factoring, (x-y)^2 = 0, so x = y, and we have the information given in Statement 2.

I can see a couple of purely geometric (non-algebraic) ways to solve the problem, but I think an algebraic approach may be simpler here. Anyway, one geometric solution: draw a circle around our right triangle. Since we have a right triangle, the hypotenuse must be a diameter. So the radius of our circle is 5. Now draw a height from the hypotenuse to the opposite vertex. Since the area of the triangle is 25, the height is of length 5. So this height must also be a radius of the circle, and therefore emanates from the midpoint of the hypotenuse. From here, we must have an isosceles triangle, and the result follows. Much easier to explain with a diagram!