three years ago a father was four times as old as his daughter is now.

Three years ago a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the present ages of the daughter and the father.
in Algebra 2 Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Three years ago a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the present ages of the daughter and the father.

Let F be the current age of the father.

Let D be the current age of the daughter.

Then,

F * D = 430

Also,

(F - 3) = D * 4      (The father's age, 3 yrs ago, was 4 times the daughter's current age)

F = 4D + 3

Substituting for F = 4D + 3 into FD = 430,

(4D + 3)D = 430

4D^2 + 3D - 430 = 0

(4D + 43)(D - 10) = 0

D = -43/4 (ignore this solution for age since it is negative)

D = 10

So, daughter's age is 10, Father's age is 43 ​

by Level 11 User (81.5k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!

Most popular questions within the last 3000 days

  1. 50+50-25X0+2+2= (19)
  2. it takes 10 road workers 5 days to complete a road repair when working 2 hours a day.Working at the same pace,how many days will it take 2 people working 5 hours a day to finish it? (3)
  3. Solve 15 -1(12÷4+1)? (4)
  4. 8:4(5+2) (3)
  5. How can i represent 1/3; 3/3; 7/3; 10/3 and 15/3 on a number line? (1)

Most popular tags

87,516 questions
100,279 answers
2,420 comments
732,173 users