# Manhattan GMAT Challenge Problem of the Week – 1 October 2012:

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## Question

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

## Answer

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*) = 4*t* + 8 (since the Akkadian is now 4 × 2 = 8 dog-years old on January 1, 2012)

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

Livonian’s age *L*(*t*) = 7*t*

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

14*t* > 9t + 13

5*t* > 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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## 2 comments

Jesse on October 2nd, 2012 at 7:18 am

Is there an easier way to explain this problem? I do not understand the first paragraph.

Abhinav on October 4th, 2012 at 1:39 am

Just take human reference as 1 year, so basically

A is increasing in progression of 4

K in 5

and

L in 7

in 2015,

A = 20

K=20

L=21

hence A+k=2*L.

Hope its clear.