SOLUTION: The oldest child in a family is 5 years older than her brother. The mother is 1 year younger than twice the sum of her kid's ages. The father's age equals the mother's age plus the
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Question 802717: The oldest child in a family is 5 years older than her brother. The mother is 1 year younger than twice the sum of her kid's ages. The father's age equals the mother's age plus the son's age. If the sum of all 4 ages is three times a prime number and the mom was younger than 50 when she had her 2nd child how old is the daughter? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! let d = age of the daughter
let b = age of the brother
let m = age of the mother
let f = age of the father
let p = prime number referred to in the problem
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Write an equation for each statement
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The oldest child in a family is 5 years older than her brother.
d = b + 5
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The mother is 1 year younger than twice the sum of her kid's ages.
m = 2(d+b) - 1
m = 2d + 2b - 1
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The father's age equals the mother's age plus the son's age.
f = m + b
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If the sum of all 4 ages is three times a prime number
d + b + m + f = 3p
replace m with (2d+2b-1)
d + b + (2d+2b-1) + f = 3p
3d + 3b + f = 3p+1
Replace f with (m+b)
3d + 3b + (m+b) = 3p+1
3d + 4b + m = 3p+1
replace m with (2d+2b-1) again
3d + 4b + (2d+2b-1) = 3p+1
5d + 6b = 3p + 1 + 1
5d + 6b = 3p + 2
Replace d with (b+5)
5(b+5) + 6b = 3p + 2
5b + 25 + 6b = 3p + 2
11b = 3p + 2 - 25
11b = 3p - 23
rearrange to
3p = 11b + 23
divide by 3
p = b +
Find a value for b that makes p a prime number
when b = 8, (brother's age) and p = 37, sum of all 4 ages: 3*37 = 111
then
d = 8+5
d = 13 yrs, daughter's age
m = 2(8+13)) - 1
m = 41 yrs is Mom's age
f = 41 + 8
f = 49 yrs is Dad's age
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See if that adds up 8 + 13 + 41 + 49 = 111
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Sorry about the first attempt, got hung up on 5th grade math!
