Math, help...?

Question

when jess was 40 yrs old, her son was twice her daughter's age. jess will be twice her son's age when her daughter is 28 yrs old. how old will jess be when her daughter is 20 yrs old?

the answer is 56yrs, dun gimme the wrong one okay! oh and explain why,

and no algebra pls.

I SAID NO ALGEBRA,

Answer 1

let x be years from first to second

when jess is 40 s = 2d

40 + x = 2(s + x) = 2s + 2x = 4d + 2x
so 4d + x = 40
d + x = 28
3d = 12
d = 4, s =8

so when the daughter is 20 = 4 + 16
Jess is 40 + 16 = 56

Answer 2

If the answer is 56 then that means there is a 36 year difference. That means when Jess is 40, her daughter would be 4. Which would make her son 8. (If j=40 then s=2d or if j=40 then 8=2(4)) When her daughter is 28 her son will be 32 and she will be 64. (If d=28 then j=2s or if d=28 then 64=2(32))

Answer 3

let's have a table for values...

Jess 40 2y
son 2x y
daughter x 28

the difference of the second situation and the first should be the same because every year, their ages will be all increased by one..then...

28 - x = y - 2x = 2y - 40

separate the equations...

28 - x = y - 2x and y - 2x = 2y - 40
28 + x = y 40 - 2x = y

y = y

28 + x = 40 - 2x
3x = 12
x = 4

then substituting the value of 4 to x in the above table...

Jess 40 2y
son 8 y
daughter 4 28

then if the daughter is 20, there will be a difference of 16, therefore we should add 16 to jess' age to find her age.
so,
40 + 16 = 56 years old

Answer 4

Let s be son's age, and d be daughters age when jess is 40.
And x be number of years that pass when jess is twice her son's age .
SO, 40+x=2(s+x)
And x+d=28.
And s=2d.
SO, substitute,
40+(28-d)=2(2d+(28-d))
68-d=4d+56-2d
12=3d
d=4 .
Now, x=28-4=24.
So, jess is 40+24=64 when she's twice her son's age and daughter is 28.
Now she will be 64-8=56 when daughter is 20.