GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Dec 2019, 09:09 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59730
In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

1
4 00:00

Difficulty:   75% (hard)

Question Stats: 68% (02:52) correct 32% (03:10) wrong based on 142 sessions

### HideShow timer Statistics In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

(A) 40
(B) 60
(C) 64
(D) 72
(E) 80

Attachment: 2017-08-18_1027_001.png [ 14.35 KiB | Viewed 5548 times ]

_________________
Current Student P
Joined: 18 Aug 2016
Posts: 593
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

Bunuel wrote: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

(A) 40
(B) 60
(C) 64
(D) 72
(E) 80

Attachment:
2017-08-18_1027_001.png

angle BAC = angle BCA = z
angle BCD = angle DCA = z/2
180 = x+z+z/2 = x+3z/2.....(i)
in triangle BDC
180 = 180-x+y+z/2
x-y-z/2=0...........................(ii)
in triangle ABC
180=z+z+y
180=2z+y..........................(iii)

we know y=x/3
x=3y
2y=z/2
4y=z

2z+y=180
2z+z/4=180
9z/4=180
z=80, y=20 and x=60
E
_________________
We must try to achieve the best within us

Thanks
Luckisnoexcuse
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3747
Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

1
1
Bunuel wrote: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

(A) 40
(B) 60
(C) 64
(D) 72
(E) 80

Attachment:
The attachment 2017-08-18_1027_001.png is no longer available

Attachment: ABCxyztriangle.png [ 24.49 KiB | Viewed 4532 times ]

Derivation of figure's annotations:

AB = BC: ∠ A = ∠ C
CD bisects ∠ C: two angles formed by the bisector are equal
∠ A = ∠ C, and ∠ A = $$z$$. So ∠ C = $$z$$
Angles at ∠ C formed by bisector = $$z/2$$

Isolate one variable:

1) Given: $$y = \frac{x}{3}$$. $$x = 3y$$

$$x$$ shows up in the figure only once; replace it with $$3y$$ in figure 2

∠ BDC and 3y are on a straight line. ∠ BDC = $$(180 - 3y)$$

2) All three triangles' angles measures now are expressed in terms of y and z. Find one variable in terms of the other.

$$∆ ABC: 2z + y = 180$$

$$∆ ADC: z + \frac{z}{2} + 3y = 180$$

$$∆ BDC: y + (180 - 3y) + \frac{z}{2} = 180$$

3) Use ∆ BDC first: with 180 on both sides of the equation, RHS will = 0. That's a good start.

$$y + (180 - 3y) + \frac{z}{2} = 180$$

$$y + 180 - 3y + \frac{z}{2} = 180$$

$$-2y + \frac{z}{2}= 0$$

$$\frac{z}{2} = 2y$$,

$$z = 4y$$

4) Find $$y$$, where $$z = 4y$$. ∆ ABC is simplest for substitution

$$2z + y = 180$$

$$2(4y) + y = 180$$

$$9y = 180$$

$$y = 20$$

5) If $$y = 20$$, from $$z = 4y$$, $$z = 80$$

_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.

Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has -- Margaret Mead
Intern  B
Joined: 17 Apr 2017
Posts: 10
Location: India
Schools: NTU '20 (S)
GMAT 1: 720 Q48 V40 WE: Analyst (Consulting)
In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

Given that AB=BC,
Therefore Angle C = Angle A = z
Also, CD is the angular bisector for angle C, so Angle BCD = Angle ACD = z/2
and y= x/3

As we know Angle ADC+ Angle BDC = 180 (Straight angle)
x+Angle BDC =180
Angle BDC = 180-x

Now that we have all angles
Solve for Z using sum of angles property in triangles ADC and BDC
In ADC, $$z+x+z/2 =180$$
=> $$x+3z/2 =180$$ --(1)

In BDC, $$x/3+180-x+z/2=180$$
=> $$z/2 =2x/3$$
=> $$x =3z/4$$ --(2)
use value of x from (2) in (1)
$$3z/4 + 3z/2 = 180$$
Therefore z=80

Ans: E
Intern  B
Joined: 08 Jun 2015
Posts: 19
Location: United States
Concentration: Finance, Economics
WE: Engineering (Other)
Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

In ∆ ABC z+y+c=180 or 2z+y=180 as AB=BC so z=c (Iso. triangle). So we can write it as 12z+2x=1080.....(1)

In ∆ ADC, z+x+c/2 = 180 or 3z+2x=360......(2)

Solving 1 & 2 we get z=80, So IMO "E" is correct answer.
Manager  B
Joined: 19 Jan 2018
Posts: 84
Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

Bunuel wrote: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

(A) 40
(B) 60
(C) 64
(D) 72
(E) 80

Attachment:
2017-08-18_1027_001.png

For these type of questions, I like to derive the equations so I can isolate the variable of choice, which is Z
From the Information above, we know that AB = BC, which means that ∠BAC = ∠BCA
∠BAC = Z, then ∠BCA = Z

∠ABC +∠BAC +∠BCA =180, or Y+Z+Z=180 Equation 1

∠ADC is equal to x, or 3y, and ∠DCA is equal to 0.5z because Line CD bisects the triangle, which means it divides ∠BCA by 2.

3Y+Z+0.5Z=180 Equation 2

Now that we have 2 unique equations, we can isolate Z and solve it.
Z = 80
Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8701
Location: United States (CA)
Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  [#permalink]

### Show Tags

Bunuel wrote: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

(A) 40
(B) 60
(C) 64
(D) 72
(E) 80

Attachment:
2017-08-18_1027_001.png

First, we see that angle CDB is (180 - x), so angle BCD is:

180 - y - (180 - x) = x - y

Since y = x/3, we can substitute this for y, obtaining x - x/3 = 2x/3 for angle CDB

Since angle C is bisected, by CD, we know that angle ACD is also 2x/3.

Thus, angle C is 4x/3, which is also equal to z (because triangle ABC is isosceles).

So, finally, we have:

4x/3 + 4x/3 + x/3 = 180

9x/3 = 180

3x = 180

x = 60

Recall that 4x/3 equals angle z; thus z = 4(60)/3 = 4*20 = 80.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =   [#permalink] 28 Nov 2019, 17:43
Display posts from previous: Sort by

# In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =  