BTGmoderatorDC wrote:Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?
A. 1/16
B. 5/42
C. 1/8
D. 3/16
E. 1/4
OA C
Source: Manhattan Prep
Let's first count the no. of factors of 210.
If a no. N is written as x^a*y^b*z^c, where x, y and z are prime factors of N, and a, b and c are positive integers, then
The no. of factors of N = (a+1)(b+1)(c+1)
210 = 2^1*3^1*5^1*7^1
Thus, the no. of factors of 210 = (1+1)(1+1)(1+1)(1+1) = 16;
Let's find out how many multiples of 42 are factors of 210. For that, we must divide 210 by 42.
210/42 = 5, a prime no.
Thus, 42 and 42*5 = 210 are the only two factors of 210 what are multiples of 42.
Thus, the probability that the ball is inscribed with a multiple of 42 = 2/16 = 1/8
The correct answer:
C
Hope this helps!
-Jay
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