If the angles on the straight line above are multiples of 10

Once again, 0 is a multiple of 10.
Please explain how x cannot be 120 without arrogantly restating the stem.

You think that I am referring to the ANSWER of 0 being a multiple of 10...I am not. If that were true, wouldn't I be questioning the validity of 5 being an answer too?

Let me break it down according to the givens in the stem:
"the ANGLES on the straight line above are multiples of 10
1) 60 is a multiple of 10 (TRUE)
2) x=120 is a multiple of 10 (TRUE)
3) 2(0)=0 is a multiple of 10 (TRUE)

So I am asking what constraints in the stem am I failing to see to make A not a correct answer. I would say Jamboree needs to rephrase this question or give them kudos for disguising some constraint I am not aware of

I ? Carcass? arrogant ???

oh boy.........

Even though zero is a multiple of any number in this case it cannot be zero

1) If we have x+2y+60=180

x+2y=120

x +0=120

x=120

Then the question would not have any sense in asking you for the value of Y

2) Even though there is no information about the diagram and we do not know if it draws the scale or not haing the 2y angle to 0 is also impossible because IF we do have an angle even if it is close or approximate to zero CANNOT be zero because from the figure we DO have an angle.

I hope this helps

1) The GRE doesn't have to justify if what it's asking makes sense, just if an answer is logically sound.
1.1) Does it make sense that the GRE is asking us this question if they already know the answer?

2) Are you saying that an angle cannot be zero?

I hope this helps

Sure it can. Why not

However from the equation that we could easily set up we can infer that y is not zero

\(y=\frac{120-x}{2}\)

\(\frac{120-x}{2}\) is a number divisible by two

x \(\neq\) zero

so y \(\neq\) zero

Therefore we can conclude that zero is NOT among our answer choices for sure

Is zero(0) considered as a multiple of 10 or any number?? That information will be necessary to determine whether A will be an answer or not! Carcass

Ok. I ll try to clarify. Considering what I know and not being arrogant as someone said before. Ten years I help students here, on this amazing website/forum for free, and people label me arrogant. Mah

Is 0 a Multiple of $10 ?$
- Yes, 0 is a multiple of every integer, including 10.
- Mathematically, a number $k$ is a multiple of $n$ if $\(k=n \times m\)$ for some integer $m$.
- Since $\(0=10 \times 0,0\)$ is a multiple of 10 .

Re-evaluating the Problem with This in Mind
Given:
- Angles on a straight line: $\(60^{\circ}, x, 2 y\)$ (sum $\(=180^{\circ}\)$ ).
- All angles are multiples of 10.
- $\(x>90^{\circ}\)$.

From earlier:

$$
\(x=120-2 y\)
$$


Constraints:
1. $\(x>90^{\circ} \rightarrow y<15^{\circ}\)$.
2. $x$ and $2 y$ must be multiples of $\(10 \rightarrow y\)$ must be a multiple of 5 (since $\(2 y\)$ must be divisible by 10 ).

Testing Option A: $y=0$
- $\(2 y=0\)$ (valid multiple of 10 ).
- $\(x=120-0=120^{\circ}\)$ (valid, since $\(120>90^{\circ}\)$ and is a multiple of 10 ).
- Angles: $\(60^{\circ}, 120^{\circ}, 0^{\circ} \rightarrow\)$ Sum $\(=180^{\circ}\)$.

Conclusion:
- $\(y=0\)$ is mathematically valid because:
- It satisfies $\(y<15^{\circ}\)$.
- $\(2 y=0\)$ is a multiple of 10 .

$\(\circ x=120^{\circ}\)$ is a multiple of 10 and $\(>90^{\circ}\)$.

Should We Exclude $y=0$ ?
- Geometrically, an angle of $\(0^{\circ}\)$ means the two rays coincide (no opening).
- Some problems may implicitly assume angles are positive, but the problem statement does not explicitly exclude $\(0^{\circ}\)$.
- Since $\(\mathbf{0}\)$ is a multiple of $\(\mathbf{1 0}\)$, and no restrictions are given, Option $\(\mathbf{A}(\mathbf{0})\)$ must be considered valid.

Final Valid Options
From the given choices:
- A (0) - Valid (mathematically and logically, unless excluded).
- C (5) - Valid ( $\(y=5, x=110^{\circ}, 2 y=10^{\circ}\)$ ).
- D (10) - Valid ( $\(y=10, x=100^{\circ}, 2 y=20^{\circ}\)$ ).

B (2) and E (20) are invalid (as previously determined).
Final Answer
If $\(0^{\circ}\)$ is allowed, the correct options are:
\(A, C, D\)
However, if the problem assumes angles must be positive, then only:
\(C, D\)
Since the problem does not explicitly exclude $0^{\circ}$, the most complete answer is:
$\(A, C, D\)$
But if the context implies only positive angles, then:
$\(C, D\)$

(Most standardized tests would likely expect $\(\boldsymbol{C}\)$ and $\(\boldsymbol{D}\)$ only, but strictly speaking, $\(\boldsymbol{A}\)$ is also correct unless specified otherwise.)