SOLUTION: Four members of one family have a combined age of 135 years.
Jeff is twice as old as his daughter Ann, while Trudy is 5 years younger than her husband, Jeff.
Harry is a third
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-> SOLUTION: Four members of one family have a combined age of 135 years.
Jeff is twice as old as his daughter Ann, while Trudy is 5 years younger than her husband, Jeff.
Harry is a third
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Question 1132202: Four members of one family have a combined age of 135 years.
Jeff is twice as old as his daughter Ann, while Trudy is 5 years younger than her husband, Jeff.
Harry is a third of his mother Trudy's age.
How old are each of the four members of this family? Found 3 solutions by addingup, ankor@dixie-net.com, MathTherapy:Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! J + A + T + H = 135
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J = 2A
T = J - 5 = 2A - 5
H = 1/3T = 1/3(2A - 5)
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A + 2A + 2A - 5 + 1/3(2A - 5) = 135
5A - 5 + 2/3A - 1 2/3 = 135
5 2/3A - 6 2/3 = 135
5 2/3A = 141 2/3
A = 141 2/3/ 5 2/3 = 25
Ann is 25 years old.
Jeff is 25 x 2 = 50
Trudy is 2(25) - 5 = 45
Harry is 1/3(45) = 15
.
The sum of their ages:
25 + 50 + 45 + 15 = 135 Correct
You can put this solution on YOUR website! Four members of one family have a combined age of 135 years.
j + a + t + h = 135
Jeff is twice as old as his daughter Ann,
j = 2a
while Trudy is 5 years younger than her husband, Jeff.
j = t + 5
Harry is a third of his mother Trudy's age.
h = t
:
In the 2nd equation replace j with (t+5)
t + 5 = 2a
a =
:
In the first equation replace in j + a + t + h = 135
j = (t+5)
a =
h = t
(t+5) + + t + t = 135
Multiply by 6 cancel the denominators
6(t+5) + 3(t+5) + 6t + 2t = 6(135)
6t + 30 + 3t + 15 + 6t + 2t = 810
17t + 45 = 810
17t = 810 - 45
17t = 765
t = 45 yrs is Trudy's age
:
How old are each of the four members of this family?
j = 45 + 5
j = 50 yrs is Jeff's age
:
a = 50/2
a = 25 is Ann's age
and
h = (45)
h = 15 yrs is Harry's age
:
:
:
check 50 + 45 + 25 + 15 = 135
You can put this solution on YOUR website! Four members of one family have a combined age of 135 years.
Jeff is twice as old as his daughter Ann, while Trudy is 5 years younger than her husband, Jeff.
Harry is a third of his mother Trudy's age.
How old are each of the four members of this family?
Let Ann's age be A
Then Jeff's is 2A, Trudy's is 2A - 5, and Harry's is
We then get:
15A + 2A - 5 = 420 -------- Multiplying by LCD, 3
17A = 425
We then get:
