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Age Prob

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by Sreema » Tue Nov 25, 2008 6:53 pm
in three years, janice will be three times as old as her daughter. Six years ago her age was her daughter's age squared. How old is Janice.

a.18
b.36
c.40
d.42
e.45


I understand this prob with substitution. I would like to know if it can be solved with equations.

Ans. D
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by dmateer25 » Tue Nov 25, 2008 7:34 pm
j + 3 = 3(y+3)
j + 3 = 3y + 9..................(i)

j - 6 = (y - 6)^2
j - 6 = y^2 -12y + 36 .......(ii)

j - 6 = y^2 -12y + 36
-j - 3 = -3y - 9
_______________________
-9 = y^2 - 15y +27

0= y^2 - 15y + 36
0 = (y-12)(y-3)
y=12 | y=3

So the daughters age is either 12 or 3.

You can plug in to find the correct one. Or you can just realize the daughter cannot be 3 since one of the equations involves "6 years ago."

So plugging 12 in

j + 3 = 3(12) + 9
j = 42
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by deepamohn » Tue Nov 25, 2008 8:22 pm
---------------------- Janice . Daughter-----------------
Age of Janice before 6 yrs = J
Age of Daughter before 6 yrs = x

Janice Current age = J+6
Daughter Current age= x+6

Janice after 3 yrs = J+9
Daughtr after 3 yrs= x+9


given J+9 = 3(x+9) ----> 1 and J= x^2 ---->2

x^2-3x-18=0
x= 6 0r -3 (age cannot be -ve) Therefore x= 6

Janice age before 6 yrs = 6^2 = 36
Current age = 36+6= 42

hope this helps.
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by Sreema » Wed Nov 26, 2008 7:35 am
That was a much easier approach thanks deepamohn
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