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Jim is twice as old as Stephanie, who, four years ago, was
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27 Apr 2012, 10:26
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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 I need an help to translate words into math. So. J= 2 S S4=3 (k4) 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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Re: Jim is twice as old as Stephanie, who, four years ago, was
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27 Apr 2012, 10:37
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
S4=3 (k4)
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 > S4=3(K4) > 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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Re: Jim is twice as old as Stephanie, who, four years ago, was
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27 Apr 2012, 16:14



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Re: Jim is twice as old as Stephanie, who, four years ago, was
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27 Apr 2012, 17:39
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



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Re: Jim is twice as old as Stephanie, who, four years ago, was
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28 Apr 2012, 07:43



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Re: Jim is twice as old as Stephanie, who, four years ago, was
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02 Jul 2013, 00:16



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Jim is twice as old as Stephanie, who, four years ago, was
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Updated on: 24 Nov 2014, 00:05
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=104/3=2+9=11, finally 40+11=51. It is correct
B
Originally posted by Temurkhon on 23 Nov 2014, 22:07.
Last edited by Temurkhon on 24 Nov 2014, 00:05, edited 1 time in total.



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Jim is twice as old as Stephanie, who, four years ago, was
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24 Nov 2014, 00:01
Jim ......... Stephanie ..................... Kate .................. (a4) ........................ \(\frac{1}{3} (a4)\) ...................... (4 Years ago) 2a .............. a.............................................. (Current ages)2a+5 ............ a+5 ........................ \(\frac{1}{3} (a4) + 4 + 5\) .................. (Ages after 5 years) Given that sum of ages post 5 years is 51 \(2a+5 + a+5 + \frac{1}{3} (a4) + 9 = 51\) a = 10 Answer = B
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Jim is twice as old as Stephanie, who, four years ago, was
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28 Nov 2015, 09:51
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
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29 Mar 2018, 15:53
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 StephanieSo 2x = James' present age. 4 years ago, Stephanie's was 3 times as old as KateIn other words, 4 years ago, Kate's age was 1/3 of Stephanie's age. 4 years ago, Stephanie'sage was x4, so Kate's age 4 years ago, was (x4)/3 So, Kate's present age = (x4)/3 + 4 In 5 years . . . Stephanie's age = x + 5 James' age = 2x + 5 Kate's age = (x4)/3 + 4 + 5 5 years from now, the sum of their ages will be 51So (x + 5) + (2x + 5) + (x4)/3 + 4 + 5 = 51 Simplify: 3x + (x4)/3 + 19 = 51 Subtract 19 from both sides: 3x + (x4)/3 = 32 Multiply both sides by 3: 9x + (x4) = 96 Solve . . . x = 10 Answer: B
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