Quick way to do these type of problems

Let David's present age = x
After 10years, x+10=4(Aaron+10)....(1)
20 years back, x-20=2(Ellen-20).....(2)
also, Ellen=x-7......(3)
Substituting (3) in (2) we get x=34
from (1) & x=34 we get Aaron=1year

so IMO A, cnt think of any other shorter method

giatch wrote:What is the quickest way to structure and solve these type of problems?

In ten years, David will be four times as
old as Aaron. Twenty years ago, David
was twice as old as Ellen. If David is
seven years older than Ellen, how old is
Aaron?
A. 1–5
B. 6–10
C. 11–15
D. 16–20
E. 21–25

you can always solve it with a time diagram but the concept is there are three time zones in this question and three persons (3x3) matrix

-20 years
Now
+10 years

+ 10 years:

D = 4x
A = X

- 20 years

D = 4x - 30
E=D/2 = 2x - 15

NOW

D = 4x-10
E = 2x+5
A = X-10

If David is seven years older than Ellen:

4x-10 = 7 + 2x+5

X = 11

and A = X-10 = 1

Writing this problem takes time but if you do all of these in a 3x3 matrix, you will solve it around 1 minute.


Hope it helps!

giatch wrote:What is the quickest way to structure and solve these type of problems?

In ten years, David will be four times as
old as Aaron. Twenty years ago, David
was twice as old as Ellen. If David is
seven years older than Ellen, how old is
Aaron?
A. 1–5
B. 6–10
C. 11–15
D. 16–20
E. 21–25

Two equations just have D and E in them, so let's start with those.

d = e + 7
d - 20 = 2(e - 20)

Since "d" is in the "a" equation, let's solve for d (i.e. sub in for e):

e = d - 7
d - 20 = 2(d - 7 - 20)
d - 20 = 2d - 54
54 - 20 = 2d - d
34 = d

d + 10 = 4(a + 10)
34 + 10 = 4a + 40
44 - 40 = 4a
4 = 4a
1 = a

If the answer choices had been exact numbers instead of ranges, we could have backsolved.

thanks guys!!