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17 Nov 2021, 04:27
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A jar contains \(x\) red marbles, \(y\) white marbles, \(z\) blue marbles, where \(x > y > z\), and no other marbles. How many red marbles are there in the jar?
(1) To ensure that at least one marble of each color is removed from the jar, a minimum of 55 marbles must be removed.
(2) To ensure that a red marble is removed from the jar, a minimum of 51 marbles must be removed.
(1) To ensure that at least one marble of each color is removed from the jar, a minimum of 55 marbles must be removed.
(2) To ensure that a red marble is removed from the jar, a minimum of 51 marbles must be removed.
17 Nov 2021, 04:27
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Official Solution:
A jar contains \(x\) red marbles, \(y\) white marbles, \(z\) blue marbles, where \(x > y > z\), and no other marbles. How many red marbles are there in the jar?
(1) To ensure that at least one marble of each color is removed from the jar, a minimum of 55 marbles must be removed.
How many marbles should we remove to ensure, to absolutely guarantee, that we have at least one marble of each color? Three? No, because all three might be, say, red, and in this case, we won't have at least one marble of each color. Ten? No, because all ten might be, say, red, and in this case, we won't have at least one marble of each color. So, how many? To answer this question, we should consider the worst-case scenario, the scenario in which we are absolutely certain to have at least one marble of each color.
The worst-case scenario would be if we first pick all \(x\) red marbles and then all \(y\) white marbles. Even at this stage, we still won't have one marble of each color. However, the next marble we pick will undoubtedly be blue, guaranteeing that we will have at least one marble of each color.
Hence, this statement implies that \(x + y + 1 = 55\). Not sufficient.
(2) To ensure that a red marble is removed from the jar, a minimum of 51 marbles must be removed.
Again, the worst-case scenario comes into play. In this case, we first remove all \(y\) white marbles and then all \(z\) blue marbles. After this point, we still do not have a red marble, but the next marble we remove will undeniably be red.
Hence, this statement implies that \(y + z + 1 = 51\). Not sufficient.
(1)+(2) We have \(x + y = 54\), \(y + z = 50\), and \(x > y > z\).
Since \(x + y = 54\) and \(x > y\), then \(y < 27\) (\(y\) must be less than half of 54).
Since \(y + z = 50\) and \(y > z\), then \(y > 25\) (\(y\) must be more than half of 50).
\(y < 27\) and \(y > 25\) leads to the conclusion that \(y = 26\).
Therefore, \(x = 28\), \(y = 26\), and \(z = 24\).
Sufficient.
Answer: C
A jar contains \(x\) red marbles, \(y\) white marbles, \(z\) blue marbles, where \(x > y > z\), and no other marbles. How many red marbles are there in the jar?
(1) To ensure that at least one marble of each color is removed from the jar, a minimum of 55 marbles must be removed.
How many marbles should we remove to ensure, to absolutely guarantee, that we have at least one marble of each color? Three? No, because all three might be, say, red, and in this case, we won't have at least one marble of each color. Ten? No, because all ten might be, say, red, and in this case, we won't have at least one marble of each color. So, how many? To answer this question, we should consider the worst-case scenario, the scenario in which we are absolutely certain to have at least one marble of each color.
The worst-case scenario would be if we first pick all \(x\) red marbles and then all \(y\) white marbles. Even at this stage, we still won't have one marble of each color. However, the next marble we pick will undoubtedly be blue, guaranteeing that we will have at least one marble of each color.
Hence, this statement implies that \(x + y + 1 = 55\). Not sufficient.
(2) To ensure that a red marble is removed from the jar, a minimum of 51 marbles must be removed.
Again, the worst-case scenario comes into play. In this case, we first remove all \(y\) white marbles and then all \(z\) blue marbles. After this point, we still do not have a red marble, but the next marble we remove will undeniably be red.
Hence, this statement implies that \(y + z + 1 = 51\). Not sufficient.
(1)+(2) We have \(x + y = 54\), \(y + z = 50\), and \(x > y > z\).
Since \(x + y = 54\) and \(x > y\), then \(y < 27\) (\(y\) must be less than half of 54).
Since \(y + z = 50\) and \(y > z\), then \(y > 25\) (\(y\) must be more than half of 50).
\(y < 27\) and \(y > 25\) leads to the conclusion that \(y = 26\).
Therefore, \(x = 28\), \(y = 26\), and \(z = 24\).
Sufficient.
Answer: C
01 Jun 2023, 03:54
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Bunuel please elaborate further
for statement (1), why do we assume that \(x + y + 1 = 55\)?
since the order is not given, it can also be \(z + y + 1 = 55\) meaning first we exhaust blue marbles, then white, then only then we get to the first red
by the same rationale, \(x + z + 1 = 55\) should also be valid, we exhaust the white then the red.
If those are true, we cannot infer which is which and the amount
for statement (1), why do we assume that \(x + y + 1 = 55\)?
since the order is not given, it can also be \(z + y + 1 = 55\) meaning first we exhaust blue marbles, then white, then only then we get to the first red
by the same rationale, \(x + z + 1 = 55\) should also be valid, we exhaust the white then the red.
If those are true, we cannot infer which is which and the amount
