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Quick way to do these type of problems

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by giatch » Tue Oct 28, 2008 1:37 pm
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
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Source: — Problem Solving |

by earth@work » Tue Oct 28, 2008 1:48 pm
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
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Re: Quick way to do these type of problems

by logitech » Tue Oct 28, 2008 2:26 pm
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!
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
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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.
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by giatch » Tue Oct 28, 2008 7:25 pm
thanks guys!!
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