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If t > 0, is k < 8

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If t > 0, is k < 8 [#permalink] New post 25 Nov 2012, 23:27
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If t > 0, is k < 8

(1) t + 6k = 48
(2) kt < 8

Last edited by Bunuel on 26 Nov 2012, 01:19, edited 1 time in total.
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Re: If t > 0, is k < 8 [#permalink] New post 25 Nov 2012, 23:58
(1) k = (48 - t)/6. For t>0, k<8. Hence (1) is sufficient.
(2) kt<8 when k is very large and t is a very small fraction OR when k is a very small fraction and t is very large. Hence (2) is insufficient.

Therefore A.
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Re: If t > 0, is k < 8 [#permalink] New post 26 Nov 2012, 04:29
ritumaheshwari02 wrote:
If t > 0, is k < 8

(1) t + 6k = 48
(2) kt < 8



Hello,

St 1--> t+6K=48, Since t>0, value of K will always be less than 8. If t=0 or negative only then value of K will be equal to 8 or > than 8

St 2 Kt<8, t>0, there will be range of value and K<8 will depend upon value of t. If t is large for ex 48 then k<1/6 and therefore less than 8
but if t is 1/2 then k<16 but may be greater than or less than 8.

Therefore Ans should be A

Thanks
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Re: If t > 0, is k < 8 [#permalink] New post 27 Nov 2012, 02:40
If t > 0, is k < 8

(1) t + 6k = 48
(2) kt < 8

if t is +ve is k less than 8 , no info about t and k of being intigers or not.


from 1
k = (48-t)/6 = 8 - t/6 , as long as t >0 thus k<8....sufficient

from 2
obviously not sufficient

A
Re: If t > 0, is k < 8   [#permalink] 27 Nov 2012, 02:40
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