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OG Quant Review #55 & #80

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OG Quant Review #55 & #80

by 480ocean » Wed Feb 04, 2009 8:08 am
#55

If (t-8) is a factor of t^2-kt-48, then k = ?

a. 16
b. 12
c. 2
d. 6
e. 14

#80

If x is to be chosen at random from the set (1,2,3,4) and y is to be chosen at random from the set (5,6,7), what is the probability that xy will be even?

a. 1/6
b. 1/3
c. 1/2
d. 2/3
e. 5/6

Thank you in advance for your help with these questions.
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by DeepakR » Wed Feb 04, 2009 8:38 am
P1#
Since t-8 is a factor put t=8 in the given equation,
t^2-kt-48= 0 hence 64-k8-48=0. So k=2. Ans.) C

P2#
If x is to be chosen at random from the set (1,2,3,4) and y is to be chosen at random from the set (5,6,7), what is the probability that xy will be even?

Since xy = even hence x=even and y=even.

Probablility = P(choosing even number from x) AND Prob(choosing even number from y)= (2/4)*(1/3)=1/6=Ans.) A

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by mdavis » Wed Feb 04, 2009 9:28 am
An even multiplied by an odd will also equal an even.

I started by finding the probability of choosing x from the set and choosing y from the set. So choosing x is 1/4 and y is 1/3. That equals 1/12 when you multiply them.

I then counted the combinations that would then equal xy being even.

1*6
2*5
2*6
2*7
3*6
4*5
4*6
4*7

So 8 possibilities. Add 1/12 8 times and you get 8/12 or when reduced, 2/3. Answer is D.

Does anyone have a faster way to do this?
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by DeepakR » Wed Feb 04, 2009 9:40 am
Yes you are correct. I missed that one!!. It would be something like below,

P= (Both x and y even) or (X even and Y odd) or (X odd and Y even)
P=(2/4*1/3) + (2/4*2/3) + (2/4*1/3) = 20/24 = 2/3 D.)

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by gaggleofgirls » Wed Feb 04, 2009 4:08 pm
Another way to do the first problem (#55) is

If you know (t-8) is a factor, then (t-8)(t+x) = t^2-kt-48
48/8 = 6 so (t-8)(t+6) = t^2-kt-48
t^2-2t-48 = t^2-kt-48
k=2

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by 480ocean » Wed Feb 04, 2009 6:50 pm
DeepakR, Thank you! Your explanation for #55 was a million times easier than the explanation in the book.

Thanks for all the help on #80 too!
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