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One integer will be randomly selected from the integers 11 to 60, incl

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Bunuel
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One integer will be randomly selected from the integers 11 to 60, incl [#permalink] New post   12 Jul 2018, 00:53
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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
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Re: One integer will be randomly selected from the integers 11 to 60, incl [#permalink] New post   12 Jul 2018, 00:57
5/48 = 0.10xx

Should be A
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Re: One integer will be randomly selected from the integers 11 to 60, incl [#permalink] New post   12 Jul 2018, 00:59
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Bunuel wrote:
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


1) check for squares :-
3^2=9 and 8^2=64
so 4^2, 5^2, 6^2 and 7^2 will all give us a perfect square

2) now for cubes
2^3=8, 3^3=27 and 4^3=64
so only one 3^3

total 4+1=5 out of 60-11+1=50 integers

probability = 5/50 = 1/10 =0.1

A
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Re: One integer will be randomly selected from the integers 11 to 60, incl [#permalink] New post   12 Jul 2018, 01:00
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vaibhav1221 wrote:
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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Re: One integer will be randomly selected from the integers 11 to 60, incl [#permalink] New post   12 Jul 2018, 01:12
chetan2u wrote:
vaibhav1221 wrote:
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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One integer will be randomly selected from the integers 11 to 60, incl [#permalink] New post   12 Jul 2018, 05:50
Bunuel wrote:
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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One integer will be randomly selected from the integers 11 to 60, incl [#permalink]
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