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

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Intern
Joined: 24 Jun 2017
Posts: 37
Location: Singapore
GMAT 1: 660 Q46 V34
GPA: 3.83
Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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17 Jul 2017, 17:06
1
5
00:00

Difficulty:

45% (medium)

Question Stats:

70% (02:06) correct 30% (02:03) wrong based on 126 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
Senior Manager
Joined: 28 Jun 2015
Posts: 297
Concentration: Finance
GPA: 3.5
Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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17 Jul 2017, 19:38
2
1
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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20 Jul 2017, 16:52
1
2
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

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Intern
Joined: 14 Oct 2016
Posts: 32
Location: India
WE: Sales (Energy and Utilities)
Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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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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Abhimanyu

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Posts: 162
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: Every time Cole tries to teach his old Collie dog, Cachaca, a new [#permalink]

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