Really hard math riddle?

c – present age of car

r – present age of radio

For better understanding I replace the sentence:

“A car is twice as old as its radio was when the car was as old as the radio is.”

by two equivalent sentences:

“A car is twice as old as its radio was x years ago.

X years ago the car was as old as the radio is.”

Now we have system of three linear equations to be solved.

c = 2 (r - x)

c - x = r

c + r = 49

Solution is

x = 7

c = 28

r = 21

The oldtimer car is 28 years old and the radio is 21 years old.

(The car is probably my Buick LeSabre which a have sold 20 years ago :-)

-

Let x be the car

Let y be the radio

Let a be the gap of time

A car is twice as old as its radio was :

x=2(y-a)

The car was as old as the radio is:

x-a=y

The sum of their ages is 49 years:

x+y=49

So:

x=2(y-a)

x-a=y=>2(y-a)-a=y => 2y-3a=y =>y=3a

x=2(y-a) =>x=4a

x+y=49 =>3a+4a=49 => 7a=49 =>a=7

Now:

x=2(y-7)

x-7=y =>2y-14-7=y =>y=21

x=2(y-7) =>x=2*(21-7)=28

so the car is 28 years old, and the radio is 21

24.5?

isnt it 75% off of original age of 49?

So like 36.75 - 49 = 12.25 x 2 = 24.5

so maybe the raido is 12.25 and the car is 24.5

man IDK ,lol