Question 1084615: FOUR YEARS AGO THE SUM OF THE AGESOF THE FOUR FAMILY MEMBERS WAS 84.TODAY THE DAUGHTER IS MARRIED OFF AND REPLACED BY DAUGHTER IN LAW.WITH THE RESULT SUM FO AGES IS 95.ASSUMING THERE IS NO DIFFERENCE FOR ALL OTHER WHAT IS THE DIFFERNCE IN AGES BETWEEN DAUGHTER AND DAUGHTER IN LAW
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the sum of their ages 4 years ago was 84.
today, the sum of their ages would have been 100.
after the loss of one daughter and the gain of one daughter in law, the sum of their ages is now 95.
this means that the daughter in law had to be 5 years younger than the daughter that was lost.
logically this makes sense.
solving it algebraically can be tricky.
the problem comes in if you let a new variable equal their ages today while the old variable equaled their ages 4 years ago.
you get weird answers that don't make sense.
i solved it this way.
4 years ago, the sum of their ages was 84.
today, the sum of their ages has to be 4 * 4 = 16 more which makes the sum of their ages today = 100 if none of the daughters was married off.
let x = the sum of the ages of the 3 daughters who weren't married off and let y = the age of the daughter who was married off.
you get x + y = 100
if you lose the daughter who was married off, but gain a daughter in law, and the sum of their ages is now 95, and you let z = the age of the daughter in law that was added, then you get:
x + z = 95
you have 2 equations that need to be solved simultaneously.
they are:
x + y = 100
x + z = 95
subtract the second equation from the first and you get y - z = 5
solve for y and you get y = z + 5
this means the daughter who was married off was 5 years older than the daughter in law.
the difference in their ages is 5 years.
the daughter in law is 5 years younger than the daughter who was married off.
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