SOLUTION: When John is as old as his father is now, his sister will be twice as old as she is now, and the age of his father will be twice the age john will be when his sister is as old as h
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Question 159094: When John is as old as his father is now, his sister will be twice as old as she is now, and the age of his father will be twice the age john will be when his sister is as old as his father is now.
The total age of the three is a century.
How old is everyone now?
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
took a while but i think i have the answer.
let S = sister's age now.
let J = john's age now.
let F = father's age now.
total age of the three is 100 so S + J + F = 100 (equation 1)*****
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first part of the problem states that when john is as old as his father is now, his sister will be twice as old as she is now.
let when that happens be in A years.
J + A = F (equation 2)*****
S + A = 2*S (equation 3)*****
subtracting equation 3 from equation 2 to remove the A, we get
J-S = F-2*S
adding 2*S to both sides of the equation, we get
J+S = F
subtracting J from both sides of the equation, we get
S = F-J
substituting in equation 1 above, we get
S + J + F = F - J + J + F = 100
the J cancels out and we are left with 2*F = 100
dividing both sides of the equation by 100, we get
F = 50 (equation 4)*****
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second part of the problem states that the age of his father will be twice the age john will be when his sister is as old as his father is now.
let B = the number of years in which that will happen.
then,
F + B = 2*(J+B) = 2*J + 2*B (equation 5)*****
S + B = F (equation 6)*****
solving for F in equation 5, we get
F + B = 2*J + 2*B
F = 2*J + 2*B - B
F = 2*J + B
since equation 6 also equals F we can make 2*J + B = S + B
subtracting B from both sides of this equation, we get
2*J = S
which is the same as
S = 2*J (equation 7)*****
since equation 1 states that F + S + J = 100 and equation 4 states that F = 50 and equation 7 states that S = 2*J, equation 1 becomes
50 + 2*J + J = 100
subtracting 50 from both sides and adding the J's together, we get
3*J = 50
J = 50/3 (equation 8)*****
equation 7 states that S = 2*J, so S must be 100/3.
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we now have 3 ages and need to test to see if they're accurate.
we have
F = 50
J = 50/3
S = 100/3
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first part of the problem states that when john is as old as his father is now, his sister will be twice as old as she is now.
that led to equations 2 and 3.
since john's age is 50/3 and his father's age is 50, the first part of the first part of the problem can be solved for A.
50/3 + A = 50
multiplying both sides of the equation by 3, we get
50 + 3*A = 150
subtracting 50 from both sides of the equation, we get
3*A = 100
A = 100/3
so john will be as old as his father is now in 100/3 years.
substituting in the original equation, we get
50/3 + 100/3 = 50
150/3 = 50
50 = 50 so first part of the first part of the problem checks out.
second part of the first part of the problem states that in A years his sister will be twice as old as she is now.
if A = 100/3 and his sister is 100/3 years old now, then in 100/3 years his sister will be 200/3 years old which is twice as old as she is now.
second part of the first part of the problem checks out.
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second part of the problem states that the age of his father will be twice the age john will be when his sister is as old as his father is now.
let B equal the number of years when that happens.
first part of the second part of the problem states that in B years, father will be twice as old as john will be then.
equation for this was F+B = 2*J+2*B (equation 5)
since we know that F = 50 and J = 50/3, this equation becomes
50 + B = 2*(50/3) + 2*B
subtracting B from both sides of the equation, we get
50 = 2*(50/3) + B
subtracting 2*(50/3) from both sides of the equation, we get
50 - 2*(50/3) = B
multiplying both sides of the equation by 3, we get
150 - 2*50 = 3*B
50 = 3*B
B = 50/3
so in 50/3 years, john's father will be twice as old as john is then.
substituting for B in equation 5, we get
50 + 50/3 = 2*(50/3) + 2*(50/3)
multiplying both sides of the equation by 3, we get
150 + 50 = 2*50 + 2*50
200 = 100 + 100
200 = 200
so first part of the second part of the problem checks out.
second part of the second part of the problem states that in B years, sister will be as old as her father is now.
equation for this is S + B = F
substituting for known values of S and B and F, we get
100/3 + 50/3 = 50
150/3 = 50
50 = 50
second part of the second part of the problem checks out.
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answer to the problem is
Father = 50 years old now.
Sister = 100/3 = 33 and 1/3 years old now.
John = 50/3 = 16 and 2/3 years old now.
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