SOLUTION: the ratio of the ages of A:B=6:5, AND THE AGE of each C and D is 9/10 times that Of B . Age of F is less than A but greater then B . The ratio of age of B&E is 2:3 ,also age of A i

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Question 983538: the ratio of the ages of A:B=6:5, AND THE AGE of each C and D is 9/10 times that Of B . Age of F is less than A but greater then B . The ratio of age of B&E is 2:3 ,also age of A is 3 years less than E .What is the ratio of age of A and F ,if all the pages are in integers ?
Answer by KMST(5328) About Me  (Show Source):
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A= age of A,
B= age of B,
C= age of C,
D= age of D,
E= age of E, and
F= age of F.
We can use the ratios to calculate B , C , D , and E in terms of A :
The ratio of the ages of A:B=6:5. ---> A%2FB=6%2F5-->5A=6B-->%285%2F6%29A=B .
The age of each C and D is 9/10 times that Of B. ---> C=D=%289%2F10%29B=%289%2F10%29%285%2F6%29A=%283%2F4%29A .
The ratio of age of B&E is 2:3. ---> B%2FE=2%2F3-->2E=3B-->E=%283%2F2%29B-->E=%283%2F2%29%285%2F6%29A=%285%2F4%29A .
For all those ages to be integers, A must be a multiple of 6 and 4 ,
so it must be a multiple of 12 .
We can use "guess and check" from here, trying 12, 24, 36, etc as values for A .
We can also say that A=12N for some whole number N ,
and express everyone's age as a function of N :
A=12N ,
B=%285%2F6%29%2812N%29=10N,
C%2BD%2B%283%2F4%29%2812N%29=9N, and
E=%285%2F4%29%2812N%29=15N
The age of A is 3 years less than E. ---> 12N=15N-3-->15N-12N=3-->3N=3-->N=3%2F3-->N=1 .
So,
A=12N=12%2A1=12 and B=10N=10%2A1=10 .
Since the age of F is less than A but greater than B, F=11 ,
and the rati we need is
A%2FF=highlight%2812%2F11%29 .