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19 Aug 2022, 01:38
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D
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Mbappé has two types of sneakers: red and blue. When packing for the World Cup, he puts equal number of sneakers in each of his suitcases. The first suitcase has 27% of all his red sneakers. Does the first suitcase have less than 25% of his blue sneakers ?
(1) No suitcase has more red sneakers than the first one.
(2) The first suitcase has between 23% and 27% of his blue sneakers.
(1) No suitcase has more red sneakers than the first one.
(2) The first suitcase has between 23% and 27% of his blue sneakers.
19 Aug 2022, 01:38
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Official Solution:
Mbappé has two types of sneakers: red and blue. When packing for the World Cup, he puts equal number of sneakers in each of his suitcases. The first suitcase has 27% of all his red sneakers. Does the first suitcase have less than 25% of his blue sneakers ?
Say Mbappé has x suitcases. Since he put equal number of sneakers in each of his suitcases, then each suitcase has 100/x% of all his sneakers. For example:
If Mbappé has 10 suitcases, then each will have 100/10 = 10% of all his sneakers.
Next, notice that, if a suitcase has LESS than \(\frac{100}{x}\%\) of Mbappé's red sneakers, it must have MORE than \(\frac{100}{x}\%\) of his blue sneakers (and vise-versa). For example:
If Mbappé has 10 suitcases, each having \(10\%\) of all his sneakers, and one specific suitcase has \(5\%\) of his red sneakers, then it must have MORE than \(10\%\) of his blue sneakers (if a suitcase has less than \(10\%\) of red sneakers and less than \(10\%\) of blue sneakers, it cannot have \(10\%\) of all sneakers).
(1) No suitcase has more red sneakers than the first one.
How many suitcases can Mbappé have? Can he have 3? No, because in this case at least one of his suitcases must have MORE than 100/3 ≈ 33% of his red sneakers (all of his 3 suitcases cannot have less than 33% of his read sneakers). So, Mbappé must have 4 or more suitcases. This means that each of them has LESS than 25% of all his sneakers. For example:
If he has 4 suitcases then each has 100/4 = 25% of all his sneakers;
If he has 5 suitcases then each has 100/5 = 20% of all his sneakers;
...
From the stem we know that the first suitcase has 27% (more than 25%) of all his red sneakers, thus it must have less than 25% of his blue sneakers. Sufficient.
(2) The first suitcase has between 23% and 27% of his blue sneakers.
From, the stem we now that the first suitcase has 27% of all his red sneakers and from this statement that the first suitcase has more than 23% of his blue sneakers:
\(0.23 \leq \frac{1}{x} \leq 0.27\)
Only integer which satisfies the above is \(x=4\). So, Mbappé has 4 suitcases and each has 25% of all his sneakers. Since the first suitcase has 27% (more than 25%) of all his red sneakers, then it must have less than 25% of his blue sneakers. Sufficient.
Answer: D
Mbappé has two types of sneakers: red and blue. When packing for the World Cup, he puts equal number of sneakers in each of his suitcases. The first suitcase has 27% of all his red sneakers. Does the first suitcase have less than 25% of his blue sneakers ?
Say Mbappé has x suitcases. Since he put equal number of sneakers in each of his suitcases, then each suitcase has 100/x% of all his sneakers. For example:
If Mbappé has 10 suitcases, then each will have 100/10 = 10% of all his sneakers.
Next, notice that, if a suitcase has LESS than \(\frac{100}{x}\%\) of Mbappé's red sneakers, it must have MORE than \(\frac{100}{x}\%\) of his blue sneakers (and vise-versa). For example:
If Mbappé has 10 suitcases, each having \(10\%\) of all his sneakers, and one specific suitcase has \(5\%\) of his red sneakers, then it must have MORE than \(10\%\) of his blue sneakers (if a suitcase has less than \(10\%\) of red sneakers and less than \(10\%\) of blue sneakers, it cannot have \(10\%\) of all sneakers).
(1) No suitcase has more red sneakers than the first one.
How many suitcases can Mbappé have? Can he have 3? No, because in this case at least one of his suitcases must have MORE than 100/3 ≈ 33% of his red sneakers (all of his 3 suitcases cannot have less than 33% of his read sneakers). So, Mbappé must have 4 or more suitcases. This means that each of them has LESS than 25% of all his sneakers. For example:
If he has 4 suitcases then each has 100/4 = 25% of all his sneakers;
If he has 5 suitcases then each has 100/5 = 20% of all his sneakers;
...
From the stem we know that the first suitcase has 27% (more than 25%) of all his red sneakers, thus it must have less than 25% of his blue sneakers. Sufficient.
(2) The first suitcase has between 23% and 27% of his blue sneakers.
From, the stem we now that the first suitcase has 27% of all his red sneakers and from this statement that the first suitcase has more than 23% of his blue sneakers:
\(0.23 \leq \frac{1}{x} \leq 0.27\)
Only integer which satisfies the above is \(x=4\). So, Mbappé has 4 suitcases and each has 25% of all his sneakers. Since the first suitcase has 27% (more than 25%) of all his red sneakers, then it must have less than 25% of his blue sneakers. Sufficient.
Answer: D
General Discussion
Krishnak13
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24 Oct 2022, 02:21
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Fantastic explanation. Could you please share some problems related to this concept
The learning that I can take from this question are:
1. If B1, B2 are two bags containing equal number of balls and each contain Red and Blue bags
Inference-1:
Number of total balls in each bag B1, B2 = 100/2 = 50%
Now, Let's say it is given as B1 has 25% red balls
Inference-2:
Since we know total balls in B1 = 50% and 25% red balls, the number of blue balls in bag-1 must be >50% for sure
2. If 100 coins are distributed among 'n' bags
At-least one bag will have coins greater than or equal to (100/n)
At-least one bag will have coins less than or equal to (100/n)
Application:
1. If say B1 has 27% of red balls in B1 and is the highest number of red balls in any bag.
=> If there are two bags, then B2 must have 73%
If there are 3 bags, B1 is given as 27% = <33% (=100/3) => the other bag must have more than 33%. But it contradicts the given condition, which is B1 has highest red balls.
If there are 4 bags, B1 is given as 27% = >25% (=100/4) => The other bag must have <25%. And 25% < % of red balls in B1. hence, min number of bags = 4.
Now once we know if there are blue balls => at-least one bag must have (<100/4 = < 25%) and have at-least one bag that has (>100/4 =>25%)
The learning that I can take from this question are:
1. If B1, B2 are two bags containing equal number of balls and each contain Red and Blue bags
Inference-1:
Number of total balls in each bag B1, B2 = 100/2 = 50%
Now, Let's say it is given as B1 has 25% red balls
Inference-2:
Since we know total balls in B1 = 50% and 25% red balls, the number of blue balls in bag-1 must be >50% for sure
2. If 100 coins are distributed among 'n' bags
At-least one bag will have coins greater than or equal to (100/n)
At-least one bag will have coins less than or equal to (100/n)
Application:
1. If say B1 has 27% of red balls in B1 and is the highest number of red balls in any bag.
=> If there are two bags, then B2 must have 73%
If there are 3 bags, B1 is given as 27% = <33% (=100/3) => the other bag must have more than 33%. But it contradicts the given condition, which is B1 has highest red balls.
If there are 4 bags, B1 is given as 27% = >25% (=100/4) => The other bag must have <25%. And 25% < % of red balls in B1. hence, min number of bags = 4.
Now once we know if there are blue balls => at-least one bag must have (<100/4 = < 25%) and have at-least one bag that has (>100/4 =>25%)
