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