GRE Prep 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If in the triangle shown above all three interior angles are multiples
[#permalink]
26 Apr 2023, 00:52
26 Apr 2023, 00:52
1
Expert Reply
4
Bookmarks
00:00
Question Stats:
38% (02:52) correct
61% (03:04) wrong based on 18 sessions
Hide Show timer Statistics
Attachment:
GRE If in the triangle shown above all three interior angles.jpg [ 12.08 KiB | Viewed 2012 times ]
If in the triangle shown above all three interior angles are multiples of 10 and angle ACB is neither the largest nor the smallest angle, how many distinct values of x are possible?
0
1
2
3
More than 3
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE GEOMETRY
Re: If in the triangle shown above all three interior angles are multiples
[#permalink]
28 May 2023, 01:24
28 May 2023, 01:24
1
Expert Reply
OE
Given that,
∠CAB = 2x
∠ACD = 3x
∠ACB = 180 – 3x ………………. [∠ACD + ∠ACB = 180]
Let ∠ABC = y
In ⊿ ACB,
∠ACB + ∠BAC + ∠ ABC = 1800
180 – 3x + 2x + y = 180
-x + y = 0
x = y
Therefore, ∠ ABC = x0
As ∠ACB is neither the largest nor the smallest angle,
So, ∠ABC < ∠ACB < ∠CAB
x < 180 – 3x < 2x
Simplifying this inequality
x < 180 – 3x
4x < 180
x < 45°
180 – 3x < 2x
5x > 180
x > 36°
36 < x < 45
We are given that all the interior angles are multiples of 10, so x being a multiple of 10
can only be 40.
So, x = 40°
Ans. (B)
Given that,
∠CAB = 2x
∠ACD = 3x
∠ACB = 180 – 3x ………………. [∠ACD + ∠ACB = 180]
Let ∠ABC = y
In ⊿ ACB,
∠ACB + ∠BAC + ∠ ABC = 1800
180 – 3x + 2x + y = 180
-x + y = 0
x = y
Therefore, ∠ ABC = x0
As ∠ACB is neither the largest nor the smallest angle,
So, ∠ABC < ∠ACB < ∠CAB
x < 180 – 3x < 2x
Simplifying this inequality
x < 180 – 3x
4x < 180
x < 45°
180 – 3x < 2x
5x > 180
x > 36°
36 < x < 45
We are given that all the interior angles are multiples of 10, so x being a multiple of 10
can only be 40.
So, x = 40°
Ans. (B)
Re: If in the triangle shown above all three interior angles are multiples
[#permalink]
08 Sep 2023, 08:42
08 Sep 2023, 08:42
1
Alternative to the algebra, you can plug in numbers. You know that all interior angles are multiples of 10, so x needs to be a multiple of 10. There's an upper limit of 3x to be less than 180, so the max that x can equal is 60 (180/3=60).
This leaves only a couple options for x: x = 10, 20, 30, 40 or 50.
If x = 10:
angle ACD = 30, ACB = 150 (not good, because this would make it the largest angle)
If x = 20:
angle ACD = 60, ACB = 120, CAB = 40, CBA = 30 (not good, ACB is the largest angle)
If x = 30:
ACD = 90, ACB = 90, CAB = 60, CBA = 30 (not good)
x = 40:
ACD = 120, ACB = 60, CAB = 80, CBA = 40 (good)
x = 50:
ACD = 150, ACB = 30, CAB = 100, CBA = 50 (not good, ACB is the smallest)
This means ACB is only the middle angle as a multiple of 10 one time. Choice B is correct.
This leaves only a couple options for x: x = 10, 20, 30, 40 or 50.
If x = 10:
angle ACD = 30, ACB = 150 (not good, because this would make it the largest angle)
If x = 20:
angle ACD = 60, ACB = 120, CAB = 40, CBA = 30 (not good, ACB is the largest angle)
If x = 30:
ACD = 90, ACB = 90, CAB = 60, CBA = 30 (not good)
x = 40:
ACD = 120, ACB = 60, CAB = 80, CBA = 40 (good)
x = 50:
ACD = 150, ACB = 30, CAB = 100, CBA = 50 (not good, ACB is the smallest)
This means ACB is only the middle angle as a multiple of 10 one time. Choice B is correct.
