a man has 4 sons of different ages, each of the older 3 sons is twice the age of the oldest of his younger brothers. The fathers age is twice the age of the sum of his sons ages, the sum of the fathers and sons ages is 45. What is the oldest sons age?
2006-12-03 10:16:42 · 6 answers · asked by lynda c 1 in Science & Mathematics ➔ Mathematics
x - father's age
y - sum of the sons
=>
x = 2y
x+y = 45
=>
x-2y = 0
x+y = 45 ..-
---------------
3y = 45
y = 15
=>
x+y = 45
x+15 = 45
x = 30
The father is 30 y.o.
z - age of the youngest son
z+2z+4z+8z = 15
15z = 15
z = 1
Oldest son:
8z = 8
Answer: the oldest son is 8 years old.
₢
2006-12-03 10:19:02 · answer #1 · answered by Luiz S 7 · 0⤊ 0⤋
The oldest son is 8 yrs. old. To get this answer, use 15 (the sum of the sons ages) and 30 (how old the father is) to get 45. Then, use this equation: 15=x+1/2x+1/4x+1/8x. After you divide 15 by 15/8, you get 8. Plug 8 into the equation and you get 8+4+2+1 which is equal to 15. Thus, the oldest son is 8 yrs. old
2006-12-03 19:13:52 · answer #2 · answered by The Postulator 5 · 0⤊ 0⤋
oldest son: 8
the sons are 1, 2, 4, 8
and the dad is 30
2006-12-03 18:19:58 · answer #3 · answered by Roxanne 3 · 0⤊ 0⤋
Set up equations:
a = oldest son
b = 2nd
c = 3rd
d = 4th
e = father's
a = 3b
b = 3c
c = 3d
e = 2(a+b+c+d)
a + b + c + d +e = 45
Now all you have to do is substitute and solve for a.
2006-12-03 18:22:04 · answer #4 · answered by AibohphobiA 4 · 0⤊ 0⤋
8..
Youngest: 1
Second: 2
Third son: 4
Oldest: 8
Sum of ages=15
Father: 15x2= 30
Sum of father and sons ages=45
2006-12-03 18:20:38 · answer #5 · answered by Anonymous · 0⤊ 0⤋
a=2b=4c=8d
b=2c=4d
c=2d
d=?
f=2(a+b+c+d)=2(8d+4d+2d+d)=30d
45=a+b+c+d+f=8d+4d+2d+d+30d
d=1
a=8
The oldest son is eight years old.
2006-12-03 18:21:59 · answer #6 · answered by grigri9 2 · 0⤊ 0⤋