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22 set 16

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Source: — Data Sufficiency |
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by kajcha » Thu Aug 30, 2007 12:09 pm
Is the ANS A?

t(n+1) = tn/2

Stmt 1 - t3 = 1/4 => t4 = 1/8 => t5 = 1/16 SUFF

Stmt 2 -> t1-t5 = 15/16 => does not give info for t1 => NOT SUFF
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Re: 22 set 16

by ratindasgupta » Thu Aug 30, 2007 12:15 pm
radhika1306 wrote:In the sequence of nonzero numbers t1, t2, t3, …, tn, …, tn+1 = tn / 2 for all positive integers
n. What is the value of t5?
(1) t3 = 1/4
(2) t1 - t5 = 15/16
1) If t3 = 1/4. we can work out t4 and then t5. So SUFF

2) t1 - t5 = 15/16. Assume t1 as x. So t5 is x/16. So x - x/16 = 15/16. So x = 1. So SUFF

Answer is D
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by kajcha » Thu Aug 30, 2007 12:18 pm
Ratin, for second stmt you calculated t5 = 15x/16. Why did you assume x = 1?
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by ratindasgupta » Thu Aug 30, 2007 12:31 pm
kajcha wrote:Ratin, for second stmt you calculated t5 = 15x/16. Why did you assume x = 1?
I have not calculated t5 as 15x/16. if t1 = x, the t2 = x/2, t3 = x/4, t4 = x/8 and t5 = x/16.

So the equation stands as x - x/16 = 15/16. By cross multiplying, you get the valueof x = 1.

Hope this clears it Kajcha.
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by kajcha » Thu Aug 30, 2007 12:36 pm
ratindasgupta wrote:
kajcha wrote:Ratin, for second stmt you calculated t5 = 15x/16. Why did you assume x = 1?
I have not calculated t5 as 15x/16. if t1 = x, the t2 = x/2, t3 = x/4, t4 = x/8 and t5 = x/16.

So the equation stands as x - x/16 = 15/16. By cross multiplying, you get the valueof x = 1.

Hope this clears it Kajcha.
Yup, it does.. got confused.. thanks
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by ratindasgupta » Thu Aug 30, 2007 12:41 pm
coolio.
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