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For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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27 Apr 2015, 04:20
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49% (02:49) correct 51% (02:48) wrong based on 146 sessions
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For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c? (1) min(max(a, b), c) = max(min(b, c), a) (2) max(max(a, b), c) – min(min(b, c), a) > c – a Kudos for a correct solution.
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Re: For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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27 Apr 2015, 06:26
Bunuel wrote: For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c?
(1) min(max(a, b), c) = max(min(b, c), a)
(2) max(max(a, b), c) – min(min(b, c), a) > c – a
Kudos for a correct solution. 1) two variants possible: \(a, b, c = 4, 5, 6\) \(min(max(a, b), c) = max(min(b, c), a)\) \(min(max(4, 5), 6) = max(min(5, 6), 4)\) \(5=5\) or \(a, b, c = 5, 4, 6\) \(min(max(a, b), c) = max(min(b, c), a)\) \(min(max(5, 4), 6) = max(min(4, 6), 5)\) \(5=5\) In first variant \(a<b\) in second \(a>b\) Insufficient 2) This statement can be simplified to \(max(a, b, c)  min (a, b, c) > c  a\) So \(a\) can't be minimum number because if \(a\) is minimum then \(max(a, b, c)  min (a, b, c) = c  a\) Sufficient Answer is B
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Re: For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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27 Apr 2015, 17:53
Bunuel wrote: For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c?
(1) min(max(a, b), c) = max(min(b, c), a)
(2) max(max(a, b), c) – min(min(b, c), a) > c – a
Kudos for a correct solution. The three numbers a, b , c can be arranged on the number line in 3 possible ways. 1. a<b<c, 2. a<c<b, 3. b<a<c Statement 1. All possibilities are possible true. 1. b = b, 2. c = c, 3. a = a Statement 2. a. ca = ca 2. ba > ca 3. cb > ca. So only 2 and 3 are possible. Therefore a<b<c is not possible. Hence Statement 2 is sufficient



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Re: For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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29 Apr 2015, 23:40
Bunuel wrote: For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c?
(1) min(max(a, b), c) = max(min(b, c), a)
(2) max(max(a, b), c) – min(min(b, c), a) > c – a Ans: B Known c>a, 1) if we consider the three scenarios: c>a>b, c>b>a, b>c>a after solving the condition from one we are getting three solutions: a=a,b=b,c=c we can not determine the relation between a,b and c from this. : INSUFFICIENT 2) same three scenarios as mentioned in (1) solutions: cb>ca, ca>ca and ba>ca: as (ca) can not be greater than itself so discarded. now from remaining two : we can easily get to the relation of a,b and c. : SUFFICIENT
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Re: For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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04 May 2015, 05:06
Bunuel wrote: For any numbers a and b, min(a, b) and max(a, b) represent the minimum and the maximum of a and b, respectively. If c > a, is a < b < c?
(1) min(max(a, b), c) = max(min(b, c), a)
(2) max(max(a, b), c) – min(min(b, c), a) > c – a
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:The question stem isn’t so bad—it just defines the min and max functions, and then asks whether b is between a and c, given that c is bigger than a. The statements, however, are another matter, because the functions are “nested.” That is, you have min’s and max’s inside each other. Here’s the general principle. Always work your way from the inside out, as you simplify complicated “nested” expressions. Statement 1: NOT SUFFICIENT. Let’s take the “Yes” case first, assuming that b is in fact between a and c, so that we have a < b < c. Take the left side first, and work your way from the inside out. min(max(a, b), c) = min(b, c) because b is bigger than a, so you can replace max(a, b) with b = b because b is smaller than c Right side: max(min(b, c), a) = max(b, a) because b is smaller than c, so you can replace min(b, c) with b = b because b is bigger than a However, what if the order is b < a < c, so that the answer to the question would be “No”? Left side: min(max(a, b), c) = min(a, c) because in this case, a is bigger than b, so you can replace max(a, b) with a = a because a is smaller than c Right side: max(min(b, c), a) = max(b, a) because b is smaller than c, so you can replace min(b, c) with b = a because a is bigger than b in this scenario. So again, the right side equals the left side. Trying to figure out this statement in the abstract is tough. A better way to deal with it is to assume an order of the variables (either “Yes” or “No” to the question), simplify the statement and determine whether it’s true. Normally, you want to assume that the statement is true and work to the question, but that’s hard to do in this situation. Also notice that any way you slice it, min(max(x, y), z) equals the middle number. So does max(min(y, z), x), even with any arrangement of variables inside. So statement (1) is always true, no matter whether b is the middle number, so the statement can’t be sufficient. Statement 2: SUFFICIENT The “max of the max” of the three variables will be the largest variable of the three. The “min of the min” of the three variables will be the smallest variable of the three. So if the largest variable minus the smallest variable is greater than c – a, then it can’t be true that c is the largest and that a is the smallest. The variable b must be either greater than c or smaller than a. If you need to, think through cases. The “Yes” answer to the question is a < b < c. So then the left side of the inequality would reduce to c – a, and the statement would be saying that c – a > c – a. That is impossible. So the answer to the question is a definite “No”—b is NOT between a and c. It must be either larger than c or smaller than a. A definite “No” answer to a Yes/No question on Data Sufficiency is sufficient. The correct answer is B.
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Re: For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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01 Jun 2019, 13:22
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Re: For any numbers a and b, min(a, b) and max(a, b) represent the minimum
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