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Jim is twice as old as Stephanie, who, four years ago, was three times

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Carcass
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Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   13 Apr 2021, 05:51
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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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B
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   13 Apr 2021, 05:58
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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

Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book


Let's solve this question by testing the answer choices, so . . .

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


Test answer choice B
If Stephanie is PRESENTLY 10 years old, then JIM is PRESENTLY 20 years old
FOUR YEARS AGO, Stephanie as 6 years old, which means Kate was 2 years old FOUR YEARS AGO
So, FIVE YEARS FROM NOW, Stephanie will be 15, Jim will be 25, and Kate will be 11
Sum of the ages = 15 + 25 + 11 = 51
It works!!

Answer: B

Cheers,
Brent
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   01 Aug 2021, 07:35
GreenlightTestPrep wrote:
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


Let's solve this question by testing the answer choices, so . . .

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


Test answer choice B
If Stephanie is PRESENTLY 10 years old, then JIM is PRESENTLY 20 years old
FOUR YEARS AGO, Stephanie as 6 years old, which means Kate was 2 years old FOUR YEARS AGO
So, FIVE YEARS FROM NOW, Stephanie will be 15, Jim will be 25, and Kate will be 11
Sum of the ages = 15 + 25 + 11 = 51
It works!!

Answer: B

Cheers,
Brent


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] New post   01 Aug 2021, 07:52
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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] New post   01 Aug 2021, 07:56
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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] New post   01 Aug 2021, 07:57
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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] New post   01 Aug 2021, 08:09
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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] New post   01 Aug 2021, 09:52
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] New post   16 Jul 2022, 07:37
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] New post   17 Jul 2022, 04:59
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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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   13 Nov 2024, 01:06
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