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Before 3 years, dog A’s age was 4times of dog B’s age an

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Before 3 years, dog A’s age was 4times of dog B’s age an

by Max@Math Revolution » Wed Jun 22, 2016 7:50 pm
Before 3 years, dog A's age was 4times of dog B's age and after 4 years, dog A's age will be 3times of dog B's age. What is the difference of dog A's age and dog B's now?
A. 36
B. 42
C. 54
D. 60
E. 64

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by regor60 » Thu Jun 23, 2016 7:37 am
A is the Methuselah of dogs apparently A
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by DavidG@VeritasPrep » Thu Jun 23, 2016 8:00 am
regor60 wrote:A is the Methuselah of dogs apparently A
:) (Though the question never says they're still alive, raising all sorts of deep philosophical question regarding whether one can continue to age after death.)

Double-check your work on this one. (I'm assuming the question means "Three years ago, A's age was 4 times that of B, and four years from now, A's age will be 3 times that of B.)
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by Max@Math Revolution » Sun Jun 26, 2016 5:13 pm
A-3=4(B-3) and A+4=3(B+4), B=17 and A=59. Hence, the correct answer is B.
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by regor60 » Mon Jun 27, 2016 9:02 am
DavidG@VeritasPrep wrote:
regor60 wrote:A is the Methuselah of dogs apparently A
:) (Though the question never says they're still alive, raising all sorts of deep philosophical question regarding whether one can continue to age after death.)

Double-check your work on this one. (I'm assuming the question means "Three years ago, A's age was 4 times that of B, and four years from now, A's age will be 3 times that of B.)
yes read too fast
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by DavidG@VeritasPrep » Mon Jun 27, 2016 12:12 pm
regor60 wrote:
DavidG@VeritasPrep wrote:
regor60 wrote:A is the Methuselah of dogs apparently A
:) (Though the question never says they're still alive, raising all sorts of deep philosophical question regarding whether one can continue to age after death.)

Double-check your work on this one. (I'm assuming the question means "Three years ago, A's age was 4 times that of B, and four years from now, A's age will be 3 times that of B.)
yes read too fast
(You're in good company.)
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