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Every time Cole tries to teach his old Collie dog, Cachaca, a new

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Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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New post 17 Jul 2017, 17:06
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Difficulty:

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Question Stats:

66% (01:42) correct 34% (02:03) wrong based on 101 sessions

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Every time Cole tries to teach his old Collie dog, Cachaca, a new trick, it takes 3 minutes more than half the time it took to teach it the previous trick. If it took Cole m minutes to teach Cachaca the latest trick, and m>5, how long did it take him to teach Cachaca the trick before the previous trick?

A 4m-18
B 4m+9
C 2m+9
D 2m +12
E m-6
[Reveal] Spoiler: OA

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Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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New post 17 Jul 2017, 19:38
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giuliab3 wrote:
Every time Cole tries to teach his old Collie dog, Cachaca, a new trick, it takes 3 minutes more than half the time it took to teach it the previous trick. If it took Cole m minutes to teach Cachaca the latest trick, and m>5, how long did it take him to teach Cachaca the trick before the previous trick?

A 4m-18
B 4m+9
C 2m+9
D 2m +12
E m-6


Let the latest trick, the trick before that, and the trick before that be denoted as \(m, n\), and \(p\), respectively.

Since \(m>5\), let us assume that \(m = 6\).

We have, \(m = 6 = \frac{n}{2} + 3.\)

\(n+6 = 2m\)

\(n = 2m-6.\)

Since \(n = \frac{p}{2} + 3\), we have

\(2m - 6 = \frac{p}{2} + 3\)

\(2m - 6 = \frac{p + 6}{2}\)

\(4m - 12 = p + 6\)

\(p = 4m - 18\). Ans - A.
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Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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New post 20 Jul 2017, 16:52
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giuliab3 wrote:
Every time Cole tries to teach his old Collie dog, Cachaca, a new trick, it takes 3 minutes more than half the time it took to teach it the previous trick. If it took Cole m minutes to teach Cachaca the latest trick, and m>5, how long did it take him to teach Cachaca the trick before the previous trick?

A 4m-18
B 4m+9
C 2m+9
D 2m +12
E m-6


We can let the time for the previous trick (i.e., the trick before the latest trick) = p; thus, we have:

½p + 3 = m

p + 6 = 2m

p = 2m - 6

Now we let the time for the trick before the previous trick = t; thus, we have:

½t + 3 = p

½t + 3 = 2m - 6

t + 6 = 4m - 12

t = 4m - 18

Answer: A
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Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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New post 09 Sep 2017, 21:08
If we can figure out how each term in seq is going to be formed then its just a 2 Step Solution .
X(n)= X(n-1)/2 +3
X(n-1)= 2X(n) -6
X(n-2)=2X(n-1) -6
X(n-2)=2{X(n) -6} -6
X(n-2)=4X(n) -18

Because X(n)=m
X(n-2)= 4m -18
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Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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New post 17 Oct 2017, 07:18
Use answer choice
It says m>5. Let m=6.
Plugin this in the answer choice.
4m=18=4*6-18=24-18=6.
Next time taken=half of 6+3=6
Next time taken=half of 6+3=6
Answer A.

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Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new   [#permalink] 17 Oct 2017, 07:18
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