Consider the two triangles (not necessarily right) xyz and

Question

Question
Consider the two triangles (not necessarily right) xyz and srq.
If x â€“ q = s â€“ y, what is the value of z?

1) xq + sy + sx + yq = zr

2) zq â€“ ry = rx â€“ zs

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

thearch · Answer

E for me
in the stem: x+y=q+s

Statement (1)
zr=(q+s)(x+y)=> zr=(q+s)^2 or zr=(x+y)^2
insufficient

Statement (2)
r(x+y)=z(s+q)
r=z
nothing about actual z value

combining...mmmm nothing new for me
I am not sure if knowing that z=x+y is sufficient

Tyr · Answer

Each statement alone is insufficient, trivial.
Let's see what can be done with (1) + (2)

xq + sy + sx + yq = zr <=>

z = (x+y)(s+q)/r

zq â€“ ry = rx â€“ zs <=>

z = (x+y)r/(s+q)

now taking into account that x â€“ q = s â€“ y <=> x+y = s+q

z = r and z = (x+y)^2 /r

r^2 = (x+y)^2

z = r = (x+y) = (s+q)

we still have no idea what Z is.

E

greenandwise · Answer

Hey guys,
 All I'm getting from the first statement is:

(s+q)^2/r = z and not a value so A is not sufficient


From the second statement I'm getting:

r = z, but not a value

If I plug in the second into the first I get:

z^2 = (s+q)^2

And since I don't have a value for either my answer is E

krisrini · Answer

I think the answer is C.

From statement 1 we get 

(x+y) (q+s) = zr

From statement 2 we get z(q+s) = r(x+y)

Combing both of them we get 

z^2 = (x+y)^2

or z = x + y

in triangle xyz, z = 180 - (x+y)

Therefore, z = 180 - z

Hence 2z = 180 or z = 90

Please let me know if my answer and approach are correct.

Tyr · Answer

Are you sure we are talking about angles?
I though xyz and qrs are  sides 

GMATT73 you need to state clearly what part of a triangle do you mean by xyz

july05 · Answer

i also thought they were angles. why mention triangle then?

pb_india · Answer

1 => simplifying we get, z= x+y Insuff
2. we get Z=r, which we already know

1 + 2 => Insuff. Hence E

GMATT73 · Answer

OA is A. For the detailed OE go to: https://www.manhattangmat.com/ChallProbL ... cfm?ID=186

thearch · Answer

In this case it is sufficient since x y and z are angles and not sides (ETS do state this in standard questions). Knowing that z=x+y and that x+y+z=180 leads us to an actual value

Tyr · Answer

What's wrong with 3 sides?
That needs to be clearly stated.
Angles sure add two more equalities x+y+z=180=q+s+r
Take into an account x+y = s+q => z=r

Now 1 becomes z = (x+y)(s+q)/r, z^2 = (x+y)^2, z = x+y, z = 90

ian7777 · Answer

I think it's A:

We get all these equations just off the main one. Starting with x-q=s-y. Since that mixes up the angles, rearrange it: x+y=s+q. 

First of all, we've got all the triangles:

x+y+z=180
q+r+s=180

If x+y=s+q, then z must equal r, since both triangles must equal 180.

That's a lot of info.

The first statement says that xq+sy+sx+yq=zr.

That one is suspicious. It reads just like a polynomial. In fact, with a little manipulating, we get: (x+y)(s+q)=zr. That's great, because we know that x+y=s+q, and we know that z=r. So we can rewrite that like this:
(x+y)^2=z^2, which means that x+y=z.

Do the math and z will equal 90.

B has too many variables to do the same thing, so there's no way to break it down and see what's happening.

HongHu · Answer

x-q=s-y
(x-q)-(s-y)=0
x+y=q+s
=>
z=r

1) (x+y)(q+s)=zr
(x+y)^2=z^2
x+y=z
z=90

Sufficient


2) zq â€“ ry = rx â€“ zs
r(x+y)=z(q+s)
z(q+s)=z(q+s)
No additional information is added to the stem. 

A.

Consider the two triangles (not necessarily right) xyz and

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