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Jim is twice as old as Stephanie, who, four years ago, was [#permalink]
27 Apr 2012, 10:26

Expert's post

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

78% (03:29) correct
22% (02:33) wrong based on 100 sessions

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 ??

Re: Jim is twice as old as Stephanie, who, four years ago, was [#permalink]
27 Apr 2012, 10:37

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Jim is twice as old as Stephanie, who, four years ago, was [#permalink]
23 Nov 2014, 22:07

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

Last edited by Temurkhon on 24 Nov 2014, 00:05, edited 1 time in total.

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