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Bunuel
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vaibhav1221
5/48 = 0.10xx

Should be A

5/48 is not 0.1...
since 11 and 60 are inclusive it will be 5/50
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vaibhav1221
5/48 = 0.10xx

Should be A

5/48 is not 0.1...
since 11 and 60 are inclusive it will be 5/50

My bad, made a silly calculation error to get max outcomes as 48. Plus i did not divide 5/48 any further to get 0.104. Thanks for pointing it out.

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Bunuel
One integer will be randomly selected from the integers 11 to 60, inclusive. What is the probability that the selected integer will be a perfect square or a perfect cube?

(A) 0.1
(B) 0.125
(C) 0.16
(D) 0.5
(E) 0.9


Integer Range : 11 -60. Total no. of Integers = 60 - 11 + 1 = 50

Case 1: Perfect Square:

\(4^2\) =16
\(5^2\) = 25
\(6^2\) = 36
\(7^2\) = 49

Case 2 : Perfect Cube:

\(3^3\) = 27

We know that we need to add if it's a OR probability.

\(\frac{4}{50}\) + \(\frac{1}{50}\)

= \(\frac{5}{50}\)
= \(\frac{1}{10}\)
= 0.1

Thus the best answer is A.
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