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Updated on: 24 Oct 2021, 02:09
Originally posted by mathbeginner on 23 Oct 2021, 22:46.
Last edited by Bunuel on 24 Oct 2021, 02:09, edited 2 times in total.
Last edited by Bunuel on 24 Oct 2021, 02:09, edited 2 times in total.
Renamed the topic and edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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67%
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In a box there are some numbered balls. The numbers on the ball are either the prime factor of 102021 or the roots of polynomial equation of \((x^4) - (31x^3) + (321x^2) - 1241x + 1430 = 0.\) If 3 balls are randomly taken out of the box, find the probability that the sum of three numbers is an even number.
(A) \(\frac{6}{7}\)
(B) \(\frac{4}{7}\)
(C) \(\frac{3}{7}\)
(D) \(\frac{1}{7}\)
(E) \(0\)
(A) \(\frac{6}{7}\)
(B) \(\frac{4}{7}\)
(C) \(\frac{3}{7}\)
(D) \(\frac{1}{7}\)
(E) \(0\)
24 Oct 2021, 00:54
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mathbeginner
PLease follow the rules and mention the source. This in no way is a GMAT type question.
I'll still give a solution as some part of it might help in other questions.
Although the question is not worded properly, I believe it means that the numbers on the balls are prime factors of 102021 and are roots of given polynomial.
Prime factor of 102021 - 3*31*1097
Solution of the polynomial: The degree 4 in \(x^4\) tells us that there are 4 roots, and 1430 is product of these 4 roots. 1430=2*5*11*13
We can also solve the polynomial but it will entail some complex steps.
So we have numbers: 2, 3, 5, 11, 13, 31 and 1097
Sum of three numbers will be EVEN when one is 2 and other two could be anything. Thus 1*6C2= \(\frac{6*5}{2}=15\)
Total ways = selecting 3 out of given 7 numbers = 7C3 = \(\frac{7*6*5}{3*2}=35\)
Probability(even) = \(\frac{15}{35}=\frac{3}{7}\)
24 Oct 2021, 12:03
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Given: In a box there are some numbered balls. The numbers on the ball are either the prime factor of 102021 or the roots of polynomial equation of \((x^4) - (31x^3) + (321x^2) - 1241x + 1430 = 0.\)
Asked: If 3 balls are randomly taken out of the box, find the probability that the sum of three numbers is an even number.
102021 = 3*31*1097
{3,31,1097} are prime factors of 102021.
\(x^4 - 31x^3 + 321x^2 - 1241x + 1430 = 0\)
\(x^4 - 31x^3 + 321x^2 - 1241x + 1430 = 0\)
x = {2,5,11,13} are the roots of the above polynomial.
We have {2,3,5,11,13,31,1097} as the set of numbers on the balls.
Since only 2 is an even number, for the sum of 3 numbers to be an even number, 1 number must be 2 and other 2 odd numbers.
Total ways of selecting 3 numbers out of {2,3,5,11,13,31,1097} = 7C3 = 35
Ways of selecting 3 numbers such that their sum is an even number = 6C2 = 15
The probability that the sum of three numbers is an even number = \(\frac{15}{35} = \frac{3}{7}\)
IMO C
Asked: If 3 balls are randomly taken out of the box, find the probability that the sum of three numbers is an even number.
102021 = 3*31*1097
{3,31,1097} are prime factors of 102021.
\(x^4 - 31x^3 + 321x^2 - 1241x + 1430 = 0\)
\(x^4 - 31x^3 + 321x^2 - 1241x + 1430 = 0\)
x = {2,5,11,13} are the roots of the above polynomial.
We have {2,3,5,11,13,31,1097} as the set of numbers on the balls.
Since only 2 is an even number, for the sum of 3 numbers to be an even number, 1 number must be 2 and other 2 odd numbers.
Total ways of selecting 3 numbers out of {2,3,5,11,13,31,1097} = 7C3 = 35
Ways of selecting 3 numbers such that their sum is an even number = 6C2 = 15
The probability that the sum of three numbers is an even number = \(\frac{15}{35} = \frac{3}{7}\)
IMO C
