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Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
Expert Reply
Translate

\(J=2S\)

\(S-4=3K\)

\(J+5+S+5+K+5=51\)

\(J+S+K+15=51\)

\(J+S+K=36\)

Substitute

\(2S+S+K=36\)

\(3S+\frac{S-4}{3}=36\)

\(9S+S-4=108\)

\(10S=112\)

\(S=11.3\)

So the closest option must be B which is equal to 10
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
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Carcass wrote:
Jim is twice as old as Stephanie, who, four years ago, was three times as old as Kate. If, five years from now, the sum of their ages will be 51, how old is Stephanie ?

A. 6
B. 10
C. 14
D. 20
E. 24

Here's an algebraic approach.

Let x = Stephanie's present age.

James is twice as old as Stephanie
So 2x = James' present age.

4 years ago, Stephanie's was 3 times as old as Kate
In other words, 4 years ago, Kate's age was 1/3 of Stephanie's age.
4 years ago, Stephanie'sage was x-4, so Kate's age 4 years ago, was (x-4)/3
So, Kate's present age = (x-4)/3 + 4

In 5 years . . .
Stephanie's age = x + 5
James' age = 2x + 5
Kate's age = (x-4)/3 + 4 + 5

5 years from now, the sum of their ages will be 51
So (x + 5) + (2x + 5) + (x-4)/3 + 4 + 5 = 51
Simplify: 3x + (x-4)/3 + 19 = 51
Subtract 19 from both sides: 3x + (x-4)/3 = 32
Multiply both sides by 3: 9x + (x-4) = 96
Solve . . . x = 10

Answer: B
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
2
Jim is twice as old as Stephanie

\(J = 2S\)

Four years ago, Stephanie was three times as old as Kate

\(S-4 = 3(K-4)\)

\(S = 3K - 8\)

five years from now, the sum of their ages will be \(51\)

\(J + 5 + S + 5 + K + 5 = 51\)

\(2S + S + K = 36\)

\(3(3K-8) + K = 36\)

\(10K = 60\)

\(K = 6\)

\(S = 10\)

mrk9414 wrote:
can you please explain to me through solving method so that I can understand behind logic
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Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
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Expert Reply
OMG

One time I attempt a question via algebra......another 3 top-notch explanations.......

Like honey bees :P

I made a mistake balanced the equation \(S−4=3(K−4)\)

:roll:
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Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
1
J=Jim, S=Stephanie and K=Kate.
Now ----> J=2S
4 years back ----> S-4=3(K-4)
5 years forward ----> J+S+K+3*5=51

Solve for S by substitution/combination of three equations

J+S+K=36 AND 2S+S+(S+8)/3=36 ----> 3S*3+S+8=36*3, 10S=108-8, S=10, Answer is B
S-4=3K-12, K=(S+8)/3

Carcass wrote:
Jim is twice as old as Stephanie, who, four years ago, was three times as old as Kate. If, five years from now, the sum of their ages will be 51, how old is Stephanie ?

A. 6
B. 10
C. 14
D. 20
E. 24

Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
Carcass wrote:
Translate

\(J=2S\)

\(S-4=3K\)

\(J+5+S+5+K+5=51\)

\(J+S+K+15=51\)

\(J+S+K=36\)

Substitute

\(2S+S+K=36\)

\(3S+\frac{S-4}{3}=36\)

\(9S+S-4=108\)

\(10S=112\)

\(S=11.3\)

So the closest option must be B which is equal to 10


In the S-4=3K
we will need to subtract 4 from K as well,
so the modified eq will be
S-4= 3K-4
which will yield exact 10 for S
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink]
Quote:
Test answer choice A
If Stephanie is PRESENTLY 6 years old, then JIM is PRESENTLY 20 years old
FOUR YEARS AGO, Stephanie as 2 years old, which means Kate was 2/3 years old FOUR YEARS AGO
Stop!!
Since it's highly unlikely that Kate's age is a fraction, we can move on


Sir, I think Jim's present age is an error here. Please check it out!
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