In the figure shown, what is the value of x?

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Hi

for geom it is often more convenient to name angles your own way to avoid excessive abstraction: a, b, c, d (please the attached file) - because many of the angles get cancelled out during calculations and overall the less astractive the more comprehensible.

We are given that angle RPT=90. We derive that QSTU is a rectangle where one of the angles = x = 360 - 90 - c - d. We also understand that x = 180 - a - b.

ANSWER C:

ST 1: QR=RS hence a=a=angle RQS=angle RSQ - insufficient, because we do not know how other angles relate to each other and to x. At this point we understand that knowing b would suffice the question.

ST 2: ST=UT hence b=b - insufficient for the same reasons stated above. Clearly this statement gives just the same piece of information as the statement 1 does. We also understand it is a good thing to combine both statements.

ST 2 + ST 1: at this point we can stop and choose C because the stimulus gives you a bold hint that one of the angles of the triangle equals 90, x is within the rectangle where one of the angles is 90, 2 of the smaller triangles are isosceles, those base angles in the triangles are adjacent to x - having taken a short glimpse of all those factors you could figure out halfway that this is pretty much enough for a geom question like that to avoid answer E and choose C.

However, this is the solution for C:

1. x=180-(a+b) and x=360-(90+c+d)
2. a=180-c and b=180-d
3. combine (1) and (2): 360-90-c-d = 180 - (180-c) - (180-d)
4. c+d=225
5. x=360-(90+225)=45

Well this is how I would solve this.

Original statement:

What is x?

Without jumping into statements, we can clearly see that x = 180 - ARSQ - ATSU.

Statement 1:

Statement 1 says that triangle QRS is isosceles. with ARSQ = ARQS

therefore we know that ARSQ = (180 - ASRQ)/2.

However, without information about triangle TSU cannot solve the above equation about x.

Insufficient

Statement 2:

Statement 2 says that triangle TSU is isosceles with ATSU = (180 - ASTU)/2.
Similar to statement 1, not enough information about triangle RSQ to find x.

Insufficient.

Statement 1 + 2:

From statement 1: we know that ARSQ = (180-ASRQ)/2
From statement 2: we know that ATSU = (180-ASTU)/2

Plugging these information into the x = 180 - ARSQ - ATSU we see that

x = 180 - (180-ASRQ)/2 - (180-ASTU)/2 = 180 - (360/2) + (ASRQ + ASTU)/2 = (ASRQ + ASTU)/2

Since we know points RTP makes a right triangle, we know that ASRQ + ASTU = 90.

x = 45.

Sufficient.

My answer would be C.

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Statement 1: Gives us nothing , since we don't know any other angles or property to compute other angles

Statement 2: Gives us nothing , since we don't know any other angles or property to compute other angles

Combined: We know that \(\angle\)RQS and \(\angle\)RSQ are same, since have also equal sides RS=RQ. Also angles \(\angle\)TSU and \(\angle\)TUS are same since ST=TU

We will name angles as below for simplicity;

\(\angle\) RQS=Z, \(\angle\)RSQ=Z, \(\angle\)QRS=B

\(\angle\)TSU=W, \(\angle\)TUS=W, \(\angle\)STU=C

So, we know that: \(90^{\circ}\)+B+C=180 meaning, B+C=\(90^{\circ}\)

I we add up angles of both triangles of QRS and STU: \(\angle\)B+2 *\(\angle\)Z+\(\angle\)C+2 *\(\angle\)W=\(360^{\circ}\)

Since we know \(\angle\)B+\(\angle\)C=\(90^{\circ}\), we subtract it: 2 *\(\angle\)Z+2 *\(\angle\)W=\(270^{\circ}\)

We can divide both sides by 2 and get : \(\angle\)Z+\(\angle\)W=\(135^{\circ}\)

To find \(\angle\)SQP and \(\angle\)SUP
360-\(\angle\)(Z+W)=\(225^{\circ}\)

\(\angle\)SQPU=\(360^{\circ}\)
90+225+x=\(360^{\circ}\)

X=\(45^{\circ}\)

Both statements together are enough. Answer C

Two notes on this:
1. An isosceles triangle has at least two sides of the same length, which means that if a triangle is equilateral it could still be considered as isosceles.

2. Triangle QSR is not necessarily 45-45-90 or 60-60-60. Why cannot it be say, 15-15-150?

How do I know it is not actually equilateral?

The biggest difficulty that one may face on this question is to stop oneself from thinking about complex ideas to solve the question. Once we get to the answer, you will realize that even 700 level questions on the GMAT test basic concepts very often.

From the diagram, we recognize that triangle PRT is a right-angled triangle, right-angled at P. Therefore, angles R and T should add up to 90 degrees.

From statement I alone, length of QR = length of RS. Therefore, angle RQS = angle RSQ = y (say)
There is no information about the other angles or the other line segments. Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, length of ST = length of TU. Therefore, angle TSU = angle TUS = z (say).
There is no information about the other angles or the other line segments. Statement II alone is insufficient. Answer option B can be eliminated. Possible answer options are C or E.

Combining statements I and II, we have the following:

From statement I alone, angle RQS = angle RSQ = y. Therefore, angle QRS = 180 – 2y (from the interior angle property of a triangle).
From statement II alone, angle TSU = angle TUS = z. Therefore, angle STU = 180 – 2z (from the interior angle property of a triangle).

IN triangle PRT, angle PRT + angle PTR = 90 degrees (angle TPR = 90 degrees).

Since angle PRT = angle QRS and angle PTR = angle STU, we can substitute the magnitudes of these angles in the equation above.
Therefore, 180 – 2y + 180 – 2z = 90. Simplifying, we obtain y+z = 135.

In the given diagram, angle RSQ + angle QSU + angle TSU = 180 (linearly adjacent angles add up to 180 degrees)
Substituting the values, we have y + x + z = 180. Since y + z = 135, x = 45 degrees.

The combination of statements is sufficient to answer the question. Answer option E can be eliminated.

The correct answer option is C.

If the diagram appears complex, it is not necessary that the solution should consist of complex concepts. That's the learning from this 700 level Geometry question.

Hope that helps!
Aravind B T

All angles in equilateral triangle are 60 degrees, thus a right triangle (a triangle with one 90-degree angle) cannot be equilateral.

Does this make sense?

Ok. Here is what I don't get.
To solve this question it is assumed that the two triangles RQS and SUT are isosceles triangles. This assumption is based on the condition QS=QR. My problem however is that an equilateral triangle QS=QR=SR would also satisfy the condition QS=QR. In that way the angles could be either 45-45-90 or 60-60-60. So what am I missing that gives proof that these triangles are Isosceles? As far as I know one should never make assumptions based on the graphs in the GMAT.

thanks for quick response

Hi Bunuel, I'm trying to understand the working after \(x+\angle{QSR}+\angle{UST}=180\) but I can't see how you reach the answer. Can you help clarify for me?

For (1)+(2) we have \(x+\angle{QSR}+\angle{UST}=180\).

From (1): \(\angle{QSR}=\frac{180-\angle{R}}{2}\).

From (2): \(\angle{UST}=\frac{180-\angle{T}}{2}\).

Substitute these into \(x+\angle{QSR}+\angle{UST}=180\) to get \(x+\frac{180-\angle{R}}{2}+\frac{180-\angle{T}}{2}=180\). This leads to \(x+\frac{360-(\angle{R}+\angle{T})}{2}=180\). Since \(\angle{R}+\angle{T}=90\), then \(x+\frac{360-90}{2}=180\) --> \(x=45\).

Hope it helps.

Thanks Bunuel, perfect solution. For more practice, can you please provide links for similar problems, thanks.

Bunuel VeritasKarishma chetan2u IanStewart

Experts..

Looking at statements 1 and 2 , I could immediately figure out there's an incentre with s,q and u as its tangents.

But i could not get into calculations and deduce Angle (Qst)

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