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A certain family has 3 sons: Richard is 6 years older than David, and

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A certain family has 3 sons: Richard is 6 years older than David, and  [#permalink]

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New post 17 Feb 2015, 03:42
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Question Stats:

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A certain family has 3 sons: Richard is 6 years older than David, and David is 8 years older than Scott. If in 8 years, Richard will be twice as old as Scott, then how old was David 4 years ago?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
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Re: A certain family has 3 sons: Richard is 6 years older than David, and  [#permalink]

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New post 17 Feb 2015, 04:46
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Stardust Chris wrote:
A certain family has 3 sons: Richard is 6 years older than David, and David is 8 years older than Scott. If in 8 years, Richard will be twice as old as Scott, then how old was David 4 years ago?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16


Let's say
Age of Richard is "R"
Age of David is "D"
Age of Scott is "S"

Now
Richard is 6 years older than David,
i.e. R = D +6

David is 8 years older than Scott
i.e. D = S +8

If in 8 years, Richard will be twice as old as Scott
i.e. R+8 = 2x(S+8)
i.e. R+8 = 2S + 16
i.e. R = 2S+8

But R = D+6 = (S+8)+6 = S+14

therefore, 2S + 8 = S +14
i.e. S = 6
i.e. R = 20
i.e. D = 14

Now,
how old was David 4 years ago?

i.e. D-4 = 14-4 = 10 years

Answer: Option
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Re: A certain family has 3 sons: Richard is 6 years older than David, and  [#permalink]

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New post 20 May 2017, 02:33
R is 6 years older than D, D is 8 years older than S
So R is (6+8)=14 years older than S, so we can say that R = S + 14

After 8 years, age of Scott = S+8, age of Richard = S+8+14 = S+22 (difference will always be 14 years as long as both are alive)
Given: S+22 = 2*(S+8)
Solve to get S=6

Thus currently S is 6 years, 4 years ago S would have been 2 years old, and David must have been 2+8 = 10 years old (difference between their ages would have been same then as now)

Hence answer is B
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Re: A certain family has 3 sons: Richard is 6 years older than David, and  [#permalink]

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New post 30 Jun 2018, 08:17
You shd draw a model to visualize their ages.

Scott + 14 + 8 = 2 (Scott + 8)
Scott = 6 years
David = 14 years.
4 years ago, David was 10 years old.

Answer : B
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Re: A certain family has 3 sons: Richard is 6 years older than David, and  [#permalink]

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New post 13 Oct 2018, 18:55
Stardust Chris wrote:
A certain family has 3 sons: Richard is 6 years older than David, and David is 8 years older than Scott. If in 8 years, Richard will be twice as old as Scott, then how old was David 4 years ago?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16


We can create the equations:

R = D + 6

and

D = S + 8

So we have R = (S + 8) + 6 or R = S + 14. We are also given that in 8 years, Richard will be twice as old as Scott. We express Richard’s age in 8 years as (R + 8) and Scott’s age in 8 years as (S + 8), so we have:

R + 8 = 2(S + 8)

S + 14 + 8 = 2S + 16

6 = S

Therefore, Scott is currently 6 years old and David is 6 + 8 = 14 years old. So 4 years ago, David was 10 years old.

Answer: B
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A certain family has 3 sons: Richard is 6 years older than David, and  [#permalink]

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New post 03 Feb 2019, 16:55
Drawing a table for this helps a lot. Solving with 1 variable:

"Richard is 6 years older than David, and David is 8 years older than Scott."

NOW
R= x+6
D= x
S= x-8

+8 years
R= x+14
D= x+8
S= x

"If in 8 years, Richard will be twice as old as Scott"
x+14 = 2x
x = 14

"how old was David 4 years ago?"
David is NOW x, or 14 years old so 14-4 = 10, B
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A certain family has 3 sons: Richard is 6 years older than David, and   [#permalink] 03 Feb 2019, 16:55
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