If t > 0, is k < 8
(1) t + 6k = 48
(2) kt < 8
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
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