Question

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

Answer 1

Answer:

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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)

Answer 2

Answer:

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