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# In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago

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In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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26 Dec 2014, 07:52
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Tough and Tricky questions: Word Problems.

In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago Carolyn was twice as old as Sergio was. How old is Sergio now, in years?

A. 20
B. 24
C. 32
D. 35
E. 40

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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26 Dec 2014, 08:12
Quote:
In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago Carolyn was twice as old as Sergio was. How old is Sergio now, in years?

A. 20
B. 24
C. 32
D. 35
E. 40

C = S + 10
C - 30 = 2 * (S - 30)

We can substitute C in the second equation with the first:

S + 10 - 30 = 2 * (S - 30)
=> S = 40

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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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26 Dec 2014, 17:08
Bunuel wrote:

Tough and Tricky questions: Word Problems.

In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago Carolyn was twice as old as Sergio was. How old is Sergio now, in years?

A. 20
B. 24
C. 32
D. 35
E. 40

Kudos for a correct solution.

Sergio Age now = S
Carolin S Age now = C

10 years from now C = S+10 --- (1)

30 Years before (C-30) = 2 (S-30)
C = 2S-30 --(2)

(2) - (1) 0 = S - 40
S = 40

Age of Sergio is 40

Ans E
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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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26 Dec 2014, 22:40
1
KUDOS
Eliminate A and B because age cannot be negative (after subtracting 30)

start from C

Ser: now - 32, 30 years ago - 2, in 10 years - 42

Car: now-34, 30 years ago - 4, in 10 years - 44 (eliminate C)

go D

Ser: now - 35, 30 years ago - 5, in 10 years - 45

Car: now - 40, 30 years ago - 10, in 10 years - 50 (eliminate D)

it is E

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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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26 Dec 2014, 23:30
let sergio's age be S and Carolyn age be C.

Now 10 years from now S will be equal to Carolyn's present age. Hence S+10 =C

30 years earlier Carolyn was twice as old as sergio

hence C-30=2 (s-30) ==> Substituting C=S+10 in this we get

S+10-30 = 2s-60 =>s=40.. hence answer is E.

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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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31 Dec 2014, 20:45
Answer = E. 40

.......................... Sergio ................... Carolyn

Current age ........... x ........................... x+10

30 yrs before ......... x-30 ........................ x-20

Given that 2(x-30) = x-20

x = 40
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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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08 Jan 2015, 08:11
Bunuel wrote:

Tough and Tricky questions: Word Problems.

In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago Carolyn was twice as old as Sergio was. How old is Sergio now, in years?

A. 20
B. 24
C. 32
D. 35
E. 40

Kudos for a correct solution.

OFFICIAL SOLUTION:

(E) Let S = Sergio's current age and C = Carolyn's current age.

First, rewrite the two statements in algebraic form.

The first statement translates into:
S + 10 = C.

The second statement translates into:
C – 30 = 2(S – 30).

Because we now have two equations, we can solve for either of the variables. Substitute S + 10 in the first equation for C in the second, and solve:
(S + 10) – 30 = 2(S – 30)
S – 20 = 2S – 60,
S + 40 = 2S
S = 40.

The correct answer is choice (E).
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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago [#permalink]

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13 Dec 2017, 06:09
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Re: In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago   [#permalink] 13 Dec 2017, 06:09
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# In 10 years, Sergio will be as old as Carolyn is now. Thirty years ago

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