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# Vitality is watching television and is trying to find a

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Manager
Joined: 13 May 2009
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Vitality is watching television and is trying to find a [#permalink]

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02 Jun 2009, 23:57
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Vitality is watching television and is trying to find a channel that he likes. He starts at channel 1, and if he likes it, he stops. Otherwise, he continues to channel 2, and repeats this process up until channel 10. Does Vitality stop before channel 4?

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

(2) The ratio of the channels that he likes to the channels that he does not like is $$4:1$$
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Last edited by Hades on 03 Jun 2009, 15:41, edited 1 time in total.
Senior Manager
Joined: 15 Jan 2008
Posts: 275

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03 Jun 2009, 00:06
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
Manager
Joined: 13 May 2009
Posts: 191

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03 Jun 2009, 14:15
Correct but incorrect logic
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Manager
Joined: 13 May 2009
Posts: 191

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03 Jun 2009, 15:35
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
$$\frac{1}{A} \leq \frac{0}{B}\leq \frac{0}{C}\leq \frac{1}{D}\leq \frac{1}{E}\leq \frac{1}{F}\leq \frac{1}{G}\leq \frac{1}{H}\leq \frac{1}{I}\leq \frac{1}{J}$$

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

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

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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Senior Manager
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03 Jun 2009, 21:53

do u use any software for typing these mathematical equations.. What is it ?
Manager
Joined: 13 May 2009
Posts: 191

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03 Jun 2009, 22:59
Google $$\Latex$$
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Re: Tough DS 8   [#permalink] 03 Jun 2009, 22:59
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