Question 1059463: The sum of the ages of Mark's two sons is equal to his own age of 51.
21 years ago, Mark's older son was twice as old as the younger son.
How old was Mark when the sum of his sons' ages was half his age?
Found 3 solutions by jorel555, addingup, MathTherapy:
Answer by jorel555(1290)   (Show Source):
You can put this solution on YOUR website! Let Mark's older son be m, and his younger son be n. Then
m+n=51
m=51-n
and
m-21=2(n-21), So:
51-n-21=2n-42
3n=72
n=24
Let p be the number of years ago that m and n together made half his age. Then
51-p=2(24-p+27-p)
51-p=48+54-4p
3p=51
p=17
17 years ago, Mark was twice as old as his sons combined. ☺☺☺☺
Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! a+b = 51 so a = 51-b
a-21 = 2(b-21)
a-21 = 2b-21 substitute for a:
51-b-21 = 2b-21
72 = 3b flip it
3b = 72
b = 24 and a = 51-b = 51-24 = 27
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check: 21 years ago:
a = 27-21 = 6
b = 24-21 = 3 and 3 is 1/2 of 6 Correct answer.
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Mark is 51 and 51/2= 25.5. The boys are 3 years apart, as I just showed. Let's call the age of the youngest x, and the oldest will be x+3. So:
x+x+3 = 25.5
2x = 22.5
x = 11.25 the youngest was 11 1/4 years old.
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The youngest is now 24 and 24-11 1/4 = 12 3/4
Mark is now 51 and 51-12 3/4 = 38 1/4 This is how old Mark was, 38.25 or 38 1/4
:
John
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! The sum of the ages of Mark's two sons is equal to his own age of 51.
21 years ago, Mark's older son was twice as old as the younger son.
How old was Mark when the sum of his sons' ages was half his age?
Mark was:  when he was twice the sum of his sons' ages.
DON'T let anyone tell you otherwise!
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