Vitality is watching television and is trying to find a

Question

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

Neochronic · Answer

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

Hades · Answer

Correct but incorrect logic

Hades · Answer

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

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 .

Neochronic · Answer

hades,

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

Hades · Answer

Google  \Latex

Vitality is watching television and is trying to find a

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