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Different breeds of dogs get older at different rates in “dog years.”

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Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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Different breeds of dogs get older at different rates in “dog years.” Livonian wolfhounds age 7 times as fast as humans, whereas Khazarian terriers age 5 times as fast and Akkadian retrievers age 4 times as fast. If Dan bought a newborn Akkadian on January 1, 2010, a newborn Khazarian 1 year later, and a newborn Livonian 1 year after that, in what year will the sum of the dog-year ages of the Akkadian and the Khazarian first be exceeded by twice the age of the Livonian in dog years, rounding all ages down to the nearest integer?

A. 2013
B. 2014
C. 2015
D. 2016
E. 2017

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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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Bunuel wrote:
Different breeds of dogs get older at different rates in “dog years.” Livonian wolfhounds age 7 times as fast as humans, whereas Khazarian terriers age 5 times as fast and Akkadian retrievers age 4 times as fast. If Dan bought a newborn Akkadian on January 1, 2010, a newborn Khazarian 1 year later, and a newborn Livonian 1 year after that, in what year will the sum of the dog-year ages of the Akkadian and the Khazarian first be exceeded by twice the age of the Livonian in dog years, rounding all ages down to the nearest integer?

A. 2013
B. 2014
C. 2015
D. 2016
E. 2017

Kudos for a correct solution.


Let's \(x\) denotes number of years.
Then \(L = 7*(x-2)\); \(K = 5*(x-1)\) and \(A = 4x\)

And we need to find \(x\) when \(K+A<2L\)
\(5*(x-1) + 4x < 2*(7*(x-2))\)
\(5x - 5 + 4x < 14x-28\)
\(23 < 5x\)
Ans as we know \(x\) is integer so \(x\) should be \(5\)

\(2010+5 = 2015\)
Answer is C
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 29 Apr 2015, 06:02
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You can solve it by making a table using the values given in the question.

The year when 2L exceeds Sum of A & K is 2015

Hence - C
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I'm going with C:

I made a chart starting at 2012:
A+K=13
L=0.

Then I added 9 to a+k for each year and 14 to L for each year.

2013:AK=22 L=14
2014:AK=31 L=28
2015:AK=40 L=42

thus, 2015 is the correct answer. => C
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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Livonian wolfhounds age 7 times as fast as humans: 7h
Khazarian terriers age 5 times as fast: 5h
Akkadian retrievers age 4 times as fast: 4h

Only in year 2015 does
Akkadian + Khazarian be less than twice the age of a Livonian
20+20 < 2*21
40<42

Answer: C

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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 29 Apr 2015, 23:01
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Bunuel wrote:
Different breeds of dogs get older at different rates in “dog years.” Livonian wolfhounds age 7 times as fast as humans, whereas Khazarian terriers age 5 times as fast and Akkadian retrievers age 4 times as fast. If Dan bought a newborn Akkadian on January 1, 2010, a newborn Khazarian 1 year later, and a newborn Livonian 1 year after that, in what year will the sum of the dog-year ages of the Akkadian and the Khazarian first be exceeded by twice the age of the Livonian in dog years, rounding all ages down to the nearest integer?

A. 2013
B. 2014
C. 2015
D. 2016
E. 2017


Ans: C
in one year A will age 4 years, K will age 5 years, L will age 7 years

sol: A+k 2L
------------------------
2013 22 14
2014 31 28
2015 40 42

2015 is the answer
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 05:00
After reconsideration, wouldn't the actual event where 2L=K+A come during 2014? At the beginning of 2015 we know that 2L>K+A, so therefore the event should have happened at 2014.
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 06:07
MarkusKarl wrote:
After reconsideration, wouldn't the actual event where 2L=K+A come during 2014? At the beginning of 2015 we know that 2L>K+A, so therefore the event should have happened at 2014.


All events occur at 1 January. So technically it's 2015 year.
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 06:58
Harley1980 wrote:
MarkusKarl wrote:
After reconsideration, wouldn't the actual event where 2L=K+A come during 2014? At the beginning of 2015 we know that 2L>K+A, so therefore the event should have happened at 2014.


All events occur at 1 January. So technically it's 2015 year.


At January 1st 2015, 2L=42 and K+A=40, as such 2L has been more than K+A prior to January 1st 2015.

What am I missing?
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 07:08
MarkusKarl wrote:
Harley1980 wrote:
MarkusKarl wrote:
After reconsideration, wouldn't the actual event where 2L=K+A come during 2014? At the beginning of 2015 we know that 2L>K+A, so therefore the event should have happened at 2014.


All events occur at 1 January. So technically it's 2015 year.


At January 1st 2015, 2L=42 and K+A=40, as such 2L has been more than K+A prior to January 1st 2015.

What am I missing?


31 december 2014, 2L=28 and K+A=31

1st January day of birthday all dogs
At January 1st 2015, 2L=42 and K+A=40

So only at 1 January 2L became bigger than K+A
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 07:20
Harley1980 wrote:

31 december 2014, 2L=28 and K+A=31

1st January day of birthday all dogs
At January 1st 2015, 2L=42 and K+A=40

So only at 1 January 2L became bigger than K+A


This would mean that the dogs age only once a year and that time is discreet, not continuous. How am I to interpret the question to get this definition?
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New post 30 Apr 2015, 07:22
Wouldn't L be 364/365*7+14 at the next to last day of 2014? (just to clarify how I interpret the continuous flow of age)
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Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 07:49
MarkusKarl wrote:
Wouldn't L be 364/365*7+14 at the next to last day of 2014? (just to clarify how I interpret the continuous flow of age)



I'm not sure, but I think it's common rule. For example when some shop make age restrictions on alcohol sale. They do not use continuous flow of age. There is two possible variants: you are 21 and you can bye product or you fewer than 21 and you can't bye this product.
And it doesn't matter that today you are 20 and tomorrow will be 21.
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New post 30 Apr 2015, 08:07
Harley1980 wrote:
MarkusKarl wrote:
Wouldn't L be 364/365*7+14 at the next to last day of 2014? (just to clarify how I interpret the continuous flow of age)



I'm not sure, but I think it's common rule. For example when some shop make age restrictions on alcohol sale. They do not use continuous flow of age. There is two possible variants: you are 21 and you can bye product or you fewer than 21 and you can't bye this product.
And it doesn't matter that today you are 20 and tomorrow will be 21.


I get your point, but I don't think that this is really addressing my concern.

I will simply hope that the GMAT will not include such a question when I do my test. :)
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 16:08
MarkusKarl wrote:
Harley1980 wrote:
MarkusKarl wrote:
Wouldn't L be 364/365*7+14 at the next to last day of 2014? (just to clarify how I interpret the continuous flow of age)



I'm not sure, but I think it's common rule. For example when some shop make age restrictions on alcohol sale. They do not use continuous flow of age. There is two possible variants: you are 21 and you can bye product or you fewer than 21 and you can't bye this product.
And it doesn't matter that today you are 20 and tomorrow will be 21.


I get your point, but I don't think that this is really addressing my concern.

I will simply hope that the GMAT will not include such a question when I do my test. :)


My take on this is "question says rounding all age down to there nearest integer value, so I rounded the age of Humans also, because it is also included in the question and we are comparing all ages in terms of X*human age. 2014,2015,2016 are the rounded age calculation points.
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Re: Different breeds of dogs get older at different rates in “dog years.” [#permalink]

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New post 30 Apr 2015, 23:48
dkumar2012 wrote:

My take on this is "question says rounding all age down to there nearest integer value, so I rounded the age of Humans also, because it is also included in the question and we are comparing all ages in terms of X*human age. 2014,2015,2016 are the rounded age calculation points.



I interpreted that part differently than you as well. That's what actually made me want to pick 2014 after reconsidering.

I decided to round all ages at the end, and then I was at the end of the year 2014, rounding 2014 down instead of up it would yield 2014 as the answer.

But I'm dropping this now, I am afraid I am confusing people even more with this discussion.
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Bunuel wrote:
Different breeds of dogs get older at different rates in “dog years.” Livonian wolfhounds age 7 times as fast as humans, whereas Khazarian terriers age 5 times as fast and Akkadian retrievers age 4 times as fast. If Dan bought a newborn Akkadian on January 1, 2010, a newborn Khazarian 1 year later, and a newborn Livonian 1 year after that, in what year will the sum of the dog-year ages of the Akkadian and the Khazarian first be exceeded by twice the age of the Livonian in dog years, rounding all ages down to the nearest integer?

A. 2013
B. 2014
C. 2015
D. 2016
E. 2017

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Create a variable to represent calendar (human) years since January 1, 2012 (the date of the final purchase). If t = 0 in 2012 and 1 in 2013, then t = calendar year since 2012.

Now you can write functions to give you dog-ages for each breed as a function of t.

Akkadian’s age A(t) = 4t + 8 (since the Akkadian is now 4 × 2 = 8 dog-years old on January 1, 2012)

Khazarian’s age K(t) = 5t + 5 (the second 5 is for the one calendar year, or 5 dog-years for the Khazarian, since 2011)

Livonian’s age L(t) = 7t

The sum of A and K is 9t + 13. Twice L is 14t. We are looking for the t at which the sum (9t + 13) is first exceeded by 14t. So the inequality we’re looking for is this:

14t > 9t + 13
5t > 13
t > 2.6

The smallest integer t for which this is true is 3, so the calendar year = 2012 + 3 = 2015.

The correct answer is C.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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