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# I've been working through the PG 10th edition, and came

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I've been working through the PG 10th edition, and came [#permalink]  23 Feb 2007, 08:42
I've been working through the PG 10th edition, and came across 2 questions I had trouble with in a row. # 63 and 64 in problem solving.

If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?

A. 2/11
B. 1/3
C. 41/99
D. 2/3
E. 23/37

I knew 1/3 and 2/3 only had 1 digit repeating from memory, but how are you supposed to figure out the other 3 fractions without actually doing the division? I figured out the trick with fractions with a denominator of 11 after doing this problem, but I still wouldn't know how to easily compare these fractions.

Today Rose is twice as old as Sam and Sam is 3 years younger than Tina. If Rose, Sam, and Tina are all alive 4 years from today, which of the following must be true on that day?

1. Rose is twice as old as Sam
2. Sam is 3 years younger than Tina
3. Rose is older than Tina

(A) 1 only
(B) 2 only
(C) 3 only
(D) 1 and 2
(E) 2 and 3

1 is obviously not true, since you added 4 to the ages so it wouldn't stay doubles. 2 is obviously true, as the difference would stay the same.

The problem I have is with the third statment. If you R=2S and S=T-3, it comes out to R>T only if 2(T-3)>T or T>6. The OA says because of this restriction only statement 2 is true.

I got to this point, but I reasoned that since Sam is 3 years young than Tina, the youngest Tina could be is 3, since Sam couldn't have a negative age. 4 years later, the youngest Tina could be is 7, which is > 6 and thus makes statement 3 true. Am I right and is the OA wrong?
I've been working through the PG 10th edition, and came   [#permalink] 23 Feb 2007, 08:42
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# I've been working through the PG 10th edition, and came

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