B is the answer..
1, insufficient. give no information abt the probability of getting the channel he likes
2,

ratio = 4: 1..

probability if getting teh channel he likes is 4/5
probability of getting the channel he dones like 1/5

the solution can be calculated like this = 4/5 + 1/5 *4/5 + 1/5 * 1/5 *4/5

Let's order the channels from A -> J, with A being Channel 1 and J being Channel 10.

Let's assign values to teach channel: 0 if Vitality doesn't like the channel, and 1 if he does.

So we're here so far:

\frac{}{A} \leq \frac{}{B}\leq \frac{}{C}\leq \frac{}{D}\leq \frac{}{E}\leq \frac{}{F}\leq \frac{}{G}\leq \frac{}{H}\leq \frac{}{I}\leq \frac{}{J}

Question:(Vitality stops before Channel 4?)
Question:(Vitality finds a channel he likes before 4?)
Question:(At least A, B or C is value 1?)
Question:(A + B + C > 0)?

(1) The multiple of the channels that he likes to the channels that he does not like is 16

X - channels he likes
Y - channels he does not like

X + Y = 10

XY = 16
X(10-X) = 16
10X - X^{2} = 16
X^{2} - 10X + 16 = 0
X^{2} - 2X - 8X + 16 = 0
X(X-2) - 8(X - 2) = 0
(X-2)(X-8) = 0
\longrightarrow X = 2 OR X = 8

So he likes either 2 channels or 8 channels.

Which would give us these possible arrangements:

Say he only likes 2 channels:
\frac{0}{A} \leq \frac{0}{B}\leq \frac{0}{C}\leq \frac{0}{D}\leq \frac{0}{E}\leq \frac{0}{F}\leq \frac{0}{G}\leq \frac{0}{H}\leq \frac{1}{I}\leq \frac{1}{J}

So in this case, he wouldn't stop before channel 4, so we have a NO answer.

But if X=8

Let's take 7 channels and put them from D onwards to J.
\frac{}{A} \leq \frac{}{B}\leq \frac{}{C}\leq \frac{1}{D}\leq \frac{1}{E}\leq \frac{1}{F}\leq \frac{1}{G}\leq \frac{1}{H}\leq \frac{1}{I}\leq \frac{1}{J}

Now we have 1 more channel he likes, and there is only 3 places we can put it in-- in A, B or C. So we have the 3 following arrangements.

A=1
\frac{1}{A} \leq \frac{0}{B}\leq \frac{0}{C}\leq \frac{1}{D}\leq \frac{1}{E}\leq \frac{1}{F}\leq \frac{1}{G}\leq \frac{1}{H}\leq \frac{1}{I}\leq \frac{1}{J}

We have a YES answer.

B=1
\frac{0}{A} \leq \frac{1}{B}\leq \frac{0}{C}\leq \frac{1}{D}\leq \frac{1}{E}\leq \frac{1}{F}\leq \frac{1}{G}\leq \frac{1}{H}\leq \frac{1}{I}\leq \frac{1}{J}

We have a YES answer.

C=1
\frac{0}{A} \leq \frac{0}{B}\leq \frac{1}{C}\leq \frac{1}{D}\leq \frac{1}{E}\leq \frac{1}{F}\leq \frac{1}{G}\leq \frac{1}{H}\leq \frac{1}{I}\leq \frac{1}{J}

We have a YES answer.


So YES & NO \longrightarrow INSUFFICIENT.

(2) The ratio of the channels that he likes to the channels that he does not like is 4:1[/quote]

X - channels he likes
Y - channels he does not like

X+Y=10
X/Y = 4/1

Substituting in:
Y = X/4
X+X/4=10
4X+X=40
5X=40
X=8.

This leaves us with the case that he likes 8 channels, and the answer will always be YES in this case (as the case is described in (1)).

\longrightarrow SUFFICIENT.


Final Answer, B.
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