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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   27 Apr 2012, 11:26
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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

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I need an help to translate words into math.

So. J= 2 S

S-4=3 (k-4)

S+5........here is the part that I do not catch. Some one can suggest me something in the right direction ??

thanks
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Bunuel
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   27 Apr 2012, 11:37
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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

I need an help to translate words into math.

So. J= 2 S

S-4=3 (k-4)

S+5........here is the part that I do not catch. Some one can suggest me something in the right direction ??

thanks


Jim is twice as old as Stephanie --> J=2S;

Stephanie four years ago, was three times as old as Kate --> S-4=3(K-4) --> K=(S+8)/3 (it would be better if it were "Stephanie four years ago, was three times as old as Kate was four years ago");

Five years from now, the sum of their ages will be 51 --> (J+5)+(S+5)+(K+5)=51 --> (2S+5)+(S+5)+((S+8)/3+5)=51 --> S=10.

Answer: B.
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BrentGMATPrepNow
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   20 Apr 2019, 06:14
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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


A student asked me to 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   28 Nov 2015, 10:51
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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 ?




From the above table we have -

(6k+13) + (3k+9) + (k+9) = 51

10k + 31 = 51

10k = 20

So, k = 2

We know , age of Stephanie now is 3k + 4 =>3*2 + 4 =10

Hence answer is (B)


PS : For such age related problems ( including age x yrs from now) the best method is coming to present age from x years back.
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   29 Mar 2018, 16:53
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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



The fastest approach here might be to plug in the answer choices.

Having said that, 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   27 Apr 2012, 17:14
Was quite evident (not for me at the moment) that we solved for S and search for J and K (the other two variables).....and the rest is clear :)

Thanks Mod. You are a landmark
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   27 Apr 2012, 18:39
carcass wrote:
Was quite evident (not for me at the moment) that we solved for S and search for J and K (the other two variables).....and the rest is clear :)

Thanks Mod. You are a landmark

Are you serious :roll:
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   28 Apr 2012, 08:43
what do you mean ??'
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   02 Jul 2013, 01:16
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

DS questions on Arithmetic: search.php?search_id=tag&tag_id=30
PS questions on Arithmetic: search.php?search_id=tag&tag_id=51
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   Updated on: 24 Nov 2014, 01:05
Best way is backsolving:

1) take C (14y.o.), so mean that S=14, J=28, their sum itself is 42+10 years from now is 52, it is over the 51,
so eliminate C,D,E
2) take B (10y.o.), meaning S=10,J=20, so 20+10+10=40 and for K=10-4/3=2+9=11, finally 40+11=51. It is correct

B

Originally posted by Temurkhon on 23 Nov 2014, 23:07.
Last edited by Temurkhon on 24 Nov 2014, 01:05, edited 1 time in total.
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   24 Nov 2014, 01:01
Jim ......... Stephanie ..................... Kate

.................. (a-4) ........................ \(\frac{1}{3} (a-4)\) ...................... (4 Years ago)


2a .............. a.............................................. (Current ages)


2a+5 ............ a+5 ........................ \(\frac{1}{3} (a-4) + 4 + 5\) .................. (Ages after 5 years)

Given that sum of ages post 5 years is 51

\(2a+5 + a+5 + \frac{1}{3} (a-4) + 9 = 51\)

a = 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   14 Mar 2019, 12:16
If you plug in the answers, you'll see that only B gives us an integer value for Kate's age 4 years ago (6/3 = 2).
Adding 15 to a non integer will still leave it a non integer. Thus, only B can be correct.
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   14 Mar 2019, 13:46
Is it normal for a GMAT question that Kate isn't born yet ?
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Re: Jim is twice as old as Stephanie, who, four years ago, was three times [#permalink] New post   06 Nov 2022, 03:35
Hello from the GMAT Club BumpBot!

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