Vitality is watching television and is trying to find a

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, \(B\).

hades,

do u use any software for typing these mathematical equations.. What is it ?

Google \(\Latex\)