Sabrina took a short cut

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Sabrina took a short cut

by sanju09 » Wed Dec 23, 2009 11:15 pm
Instead of walking along two adjacent sides of a rectangular field, Sabrina took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is:

(A) 1:4
(B) 3:8
(C) 1:2
(D) 2:3
(E) 3:4
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by Stuart@KaplanGMAT » Wed Dec 23, 2009 11:46 pm
sanju09 wrote:Instead of walking along two adjacent sides of a rectangular field, Sabrina took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is:

(A) 1:4
(B) 3:8
(C) 1:2
(D) 2:3
(E) 3:4
Let's call our 3 sides S, L and H (short, long, hypotenuse).

We know that H = S + L - (1/2)L,

since the shortcut (hypotenuse) is 1/2 a long side shorter than the trip around the edges.

So,

H = S + (1/2)L

and

H^2 = S^2 + L^2

subbing in for H:

(S + (1/2)L)^2 = S^2 + L^2

S^2 + SL + (1/4)L^2 = S^2 + L^2

SL = (3/4)L^2

S = (3/4)L^2/L

S = (3/4)L

S/L = 3/4.... choose (E).

Of course, we also could have done this by backsolving. For example, if we started with C (just because that's simplest):

S = 1; L = 2

So H^2 = 1 + 4
H = root5

clearly this can't be the right answer, since we need all of our sides to be integers. Once we realize that, we think "hey, what's the simplest integer side triangle?" and we answer ourselves "3/4/5!"

Backsolving for E:

S=3, L = 4, H = 5
5 = 3 + (1/2)4
5 = 5... bingo!
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by papgust » Thu Dec 24, 2009 12:44 am
Stuart, awesome strategy! Thanks

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by sanju09 » Thu Dec 24, 2009 3:48 am
clearly this can't be the right answer, since we need all of our sides to be integers.
Why do we need all our sides to be integer here, Stuart?
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by Stuart@KaplanGMAT » Thu Dec 24, 2009 4:00 am
sanju09 wrote:
clearly this can't be the right answer, since we need all of our sides to be integers.
Why do we need all our sides to be integer here, Stuart?
Since H = S + (1/2)L, we're almost certain that H, S and L will all be the same type of number, for two reasons: first, because this is the GMAT and we expect the numbers to work out nicely; and second, because the ratios in the answers are all integers.

So, if S and L both look like integers (based on the choices), H is almost certainly going to be an integer as well.
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by sanju09 » Thu Dec 24, 2009 4:49 am
Stuart Kovinsky wrote:
sanju09 wrote:
clearly this can't be the right answer, since we need all of our sides to be integers.
Why do we need all our sides to be integer here, Stuart?
Since H = S + (1/2)L, we're almost certain that H, S and L will all be the same type of number, for two reasons: first, because this is the GMAT and we expect the numbers to work out nicely; and second, because the ratios in the answers are all integers.

So, if S and L both look like integers (based on the choices), H is almost certainly going to be an integer as well.
Won't you consider calling it "rationals" rather than "integers", Stuart? :roll:
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by maihuna » Thu Dec 24, 2009 10:03 am
sanju09 wrote:Instead of walking along two adjacent sides of a rectangular field, Sabrina took a short cut along the diagonal and saved a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is:

(A) 1:4 (B) 3:8 (C) 1:2 (D) 2:3 (E) 3:4
With sides L, S: diagonal = sq root(L^2+S^2) longer path = (L)

Given : sq root(L^2+S^2) = L+S-L/2 = L/2 + S
or L^2 + S^2 = L^2/4 + LS + S^2
or L^2 = L^2/4 + LS
or 3/4L = S
or S/L = 3/4

or E
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by Stuart@KaplanGMAT » Thu Dec 24, 2009 6:10 pm
sanju09 wrote:
Stuart Kovinsky wrote:
sanju09 wrote:
clearly this can't be the right answer, since we need all of our sides to be integers.
Why do we need all our sides to be integer here, Stuart?
Since H = S + (1/2)L, we're almost certain that H, S and L will all be the same type of number, for two reasons: first, because this is the GMAT and we expect the numbers to work out nicely; and second, because the ratios in the answers are all integers.

So, if S and L both look like integers (based on the choices), H is almost certainly going to be an integer as well.
Won't you consider calling it "rationals" rather than "integers", Stuart? :roll:
Because on the GMAT it's very likely that they're going to work out to be integers, not just rationals.

We could use irrational numbers as well (for example 3root2, 4root2 and 5root2 fit all the info, since they're still in a 3:4:5 ratio), but why make life more complicated than it needs to be?

They key is to recognize that all the numbers will be the same type, not to focus on which type that will be.
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