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How do you solve this problem? In 2005, a man and his son were looking at a collection of paintings. The young boy looked at the first dates on the two paintings and noticed that the first year was exactly twice the second. His father noticed that the exactly twice the second. His father noticed that the second painting was exactly twice the second. His father noticed that the second painting was exactly three times the age of the first. How old was the first painting in 2005? ⚑ Flag

by Emily.Hi. at September 06, 2011

PLEASE, PLEASE, PLEASE ANSWER THIS QUESTION AS SOON AS POSSIBLE!!! It's a hard question but I know there are a lot of smart people out there in Brightstorm so please answer this question... Thanks!!!

i'm sorry but your problem doesn't make sense at all :(

juliao. September 06, 2011

:'( my brain died

nnn133 September 07, 2011

Okay, I'll give it a shot. Painting A was dated 1604 and Painting B was dated 802. If (in 2005) the first painting is three times older than the second, then: 3*(2005-A)=(2005-B) And if the numerical value of the date on A is twice that on B, then: A=2B So we have an equation for A in terms of B, let’s substitute it into the first equation and solve for B 3*(2005-2*B)=2005-B =>6015-6B=2005-B =>4010=5B =>4010/5=B =>802=B And two times 802 is 1604. And the age of B is 2005-802=1203years which is three times the age of A: 2005-1604=401years (in 2005) Good times

SerTorm September 08, 2011
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