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15 Jul 2020, 01:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (02:07) correct
22%
(02:11)
wrong
based on 88
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A candy dish contains only chocolates and caramels. There are three times as many chocolates as caramels. The chocolates are either vanilla-filled or nut-filled, and four times as many chocolates are nut-filled as vanilla-filled. If one piece of candy is selected randomly from the dish, what is the probability that it will be a nut-filled chocolate?
A. 3/20
B. 1/5
C. 1/4
D. 9/20
E. 3/5
A. 3/20
B. 1/5
C. 1/4
D. 9/20
E. 3/5
15 Jul 2020, 01:58
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Bunuel
Choco = 3*Caramels
Let, Vanila filled chocolates = x
Nut filled chocolates = 4*x
i..e. Total Chocolates = x+4x = 5x
Choco = 3*Caramels
i.e Caramels = 5x/3
Probability of Nut filled chocolate = (Nut filled chocolates) / Total Chco and Caramel \(= \frac{4x }{ (5x+5x/3)} = \frac{4*3}{20} = \frac{3}{5}\)
Answer: Option E
15 Jul 2020, 02:03
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Probability = Favorable Outcomes / Total Outcomes
Let the number of caramels = x
Therefore number of chocolates = 3x
Total Candies = Total Outcomes = x + 3x = 4x
Let the number of vanilla filled chocolates = y
Then number of nut filled chocolates = 4y
4y + y = 3x
5y = 3x
y = 3/5x = 0.6x
4y = Favorable Outcomes = 4 * 0.6x = 2.4x
The required probability = 2.4x / 4x = 24 / 40 = 3 / 5
Option E
Arun Kumar
Let the number of caramels = x
Therefore number of chocolates = 3x
Total Candies = Total Outcomes = x + 3x = 4x
Let the number of vanilla filled chocolates = y
Then number of nut filled chocolates = 4y
4y + y = 3x
5y = 3x
y = 3/5x = 0.6x
4y = Favorable Outcomes = 4 * 0.6x = 2.4x
The required probability = 2.4x / 4x = 24 / 40 = 3 / 5
Option E
Arun Kumar
