CAT 2018 Quant Questions

Question:
The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is

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Correct Answer: 8

putting n = 2, \({n^3} - 11{n^2} + 32n - 28 = 0\) i.e. (n-2) is a factor of \({n^3} - 11{n^2} + 32n - 28\)

\({n^3} - 11{n^2} + 32n - 28 = (n - 2)({n^2} - 9n + 14)\)

Also, \({n^2} - 9n + 14 = (n - 2)(n - 7)\).

\(\therefore {n^3} - 11{n^2} + 32n - 28 = {(n - 2)^2}(n - 7)\)

\( \Rightarrow \) \({n^3} - 11{n^2} + 32n - 28 > 0\)

\( \Rightarrow \) \({(n - 2)^2}(n - 7) > 0\)

Therefore, n-7>0

Or, the minimum value of n=8.

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CAT 2018 Quant Questions
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