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22 November, 06:26

Father is four times older than his son. In 14 years the father will be twice as old. What are their current ages?

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Answers (2)
  1. 22 November, 06:30
    0
    The son is 7 years old

    The father is 28 years old

    Step-by-step explanation:

    f = fathers age

    s = sons age

    f = 4s

    In 14 years (so we add 14 to the age)

    f+14 = 2 (s+14)

    We have 2 equations and 2 unknowns

    Replace every f with 4s (from the first equation)

    4s+14 = 2 (s+14)

    Distribute the 2

    4s+14 = 2s+28

    Subtract 2s from each side

    4s-2s+14 = 2s-2s+28

    2s+14 = 28

    Subtract 14 from each side

    2s-14-14 = 28-14

    2s = 14

    Divide by 2

    2s/2 = 14/2

    s = 7

    The son is 7 years old

    Now lets calculate the fathers age

    f = 4s

    f = 4*7

    f = 28

    The father is 28 years old
  2. 22 November, 06:52
    0
    ---

    Step-by-step explanation:

    The son's age is unknown, so call it: X

    The father's age, four times older, so that means: 4*X = 4X (I use * for multiply)

    After 16 years, we add 16 to both ages:

    Son: X+16

    Father: 4X+16, now, after 16yrs father's age 2x older than son:

    father→ 4X+16 = 2 (X+16) = 2X+32 (father twice older than the son age)

    4X + 16 = 2X+32 (after opening the brackets)

    4X-2X = 32 - 16 - --> we moved 2X to left, turned minus -, we moved 16 to the right turned minus -

    2X = 16

    X = 16/2 = 8

    Son's age is 8 yrs old

    Father (Sunil) age is 4x8 = 32

    After 16 yrs, Father's age twice older:

    8+16=24 (son)

    32+16 = 48 (Father, Sunil, twice older than son)
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