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11 Jun 2025, 22:37
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
44% (03:02) correct
56%
(03:15)
wrong
based on 50
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A company got a batch of 300 single-colored jerseys printed, each of which is either blue, red, or black. Each jersey has a number from 1 to 300 printed on it and no two jerseys have the same number. The probability that a randomly chosen jersey is black and has a multiple of 5 written on it is 0.12, while the probability that the jersey selected is red or blue is 0.7. What is the maximum possible number of black jerseys that have an even number printed on them?
(A) 36
(B) 60
(C) 64
(D) 84
(E) 90
(A) 36
(B) 60
(C) 64
(D) 84
(E) 90
11 Jun 2025, 23:15
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Given:
To find: The maximum the number of black jerseys that have even numbers.
Solution:
Our task is to maximize the number of black jerseys that have even numbers. So, how do we do that?
Well, there are 150 even and 150 odd numbers from 1 to 300. If we want to maximize the count of even-numbered black jerseys, then we must minimize the number of even-numbered jerseys among the red and blue jerseys — meaning give as many odd numbers as possible to red and blue jerseys.
So, is it possible for all 90 black jerseys to have even numbers?
Yes, from a numbers point of view, there are enough even numbers left for the 90 black jerseys. But wait — the 36 black jerseys that are multiples of 5 must also fit this arrangement.
Let's check if these 36 can all be even:
Correct Answer: (D) 84
Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years
- Total jerseys = 300
- Each jersey is colored either red, blue, or black.
- Probability (Red or Blue) = 0.7
- Probability (Black and multiple of 5) = 0.12
To find: The maximum the number of black jerseys that have even numbers.
Solution:
- Since Probability (Red or Blue) = 0.7, we have:
- Total no. of red or blue jerseys = 0.7 × 300 = 210
- Therefore, no. of black jerseys = 300 – 210 = 90
- Now, Probability (Black and multiple of 5) = 0.12
- So, no. of black jerseys with multiples of 5 = 0.12 × 300 = 36
- Therefore, the other black jerseys are without multiples of 5 = 90 – 36 = 54
Our task is to maximize the number of black jerseys that have even numbers. So, how do we do that?
Well, there are 150 even and 150 odd numbers from 1 to 300. If we want to maximize the count of even-numbered black jerseys, then we must minimize the number of even-numbered jerseys among the red and blue jerseys — meaning give as many odd numbers as possible to red and blue jerseys.
- Red and blue total = 210 jerseys.
- The maximum no. of odd numbers they can take is 150 (because there are only 150 odd numbers available).
- So, these 210 jerseys will get:
- 150 odd numbers + 60 even numbers = 210 numbers.
- Therefore, at least 60 even numbers must go to red and blue jerseys, and the remaining 90 even numbers can go to black jerseys.
So, is it possible for all 90 black jerseys to have even numbers?
Yes, from a numbers point of view, there are enough even numbers left for the 90 black jerseys. But wait — the 36 black jerseys that are multiples of 5 must also fit this arrangement.
Let's check if these 36 can all be even:
- Total multiples of 5 from 1 to 300 = 60 (since 300 ÷ 5 = 60).
- Out of these 60 multiples of 5:
- Even multiples of 5 = multiples of 10 = 30 (since 300 ÷ 10 = 30).
- Odd multiples of 5 = remaining 30.
- Out of these 60 multiples of 5:
- But we need 36 black jerseys that are multiples of 5.
- So even if we take all 30 even multiples of 5, the remaining 6 black multiples of 5 must be odd.
- That means at least 6 black jerseys will have odd numbers — we cannot avoid this.
- So, the maximum number of black jerseys that can have even numbers = 90 – 6 = 84.
Correct Answer: (D) 84
Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years
11 Jun 2025, 23:30
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siddhantvarma
Quote:
The number of black jersery which are multiples of 5 = 0.12 * 300 = 36
Quote:
Probability that the jersey is neither red or blue = Probability that the jersey selected is black = 1 - 0.7 = 0.3
Hence, the number of black jerseys = 0.3 * 300 = 90
Among the 36 black jerseys which are a multiple of 5, we have some jerseys which are even and some which are odd.
The black jerseys which have an even number must be a multiple of 10.
Max number of black jerserys that are also a multiple of 10 = 300/10 = 30
Therefore, the number of black jerseys which are odd multiples of 5 = 36 - 30 = 6
As we are looking to maximize the number of black jerseys that have an even number printed on them, we assume that other than the 6 black jerseys which are odd multiple of 5, all other jerseys have an even number printed on them.
Therefore maximum number of black jerseys that have an even number printed on them = 90 - 6 = 84
Option D
