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Re: Q is the set of all integers between A and B, inclusive. The average ( [#permalink]

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28 Oct 2014, 02:46

Can someone explain why the answer is B and not D?

Is it because, though the question states that it is a set of all integers between A & B, but it doesn't specifically state that integers can't be repeated. And hence, Option 1 isn't sufficient off its own?

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.

So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

This is a tricky question that baits you into making unjustified assumptions. Let's start with statement 2, as it is the easier of the 2 statements. If we know that the mean of m is 150, and that B is 190, it is pretty clear to see that we can see A will be 110 and thus statement 2 is sufficient. This is an evenly spaced set (each number has the same spacing between consecutive numbers, in this instance 1). In evenly spaced sets, the mean will always equal the median, helping us to realize quickly that A = 110 because 150 is 40 below 190, and 40 below 150 is 110.

Statement 1 is much more deceptive. At first glance it appears to give us the same information, and also tell us that A = 110. However, this is the case only if we are assuming that A and M are integers. If A is 111, for example, then M ends up being 150.5, and Q will still contain 40 terms that are greater than M (151-190, inclusive). This gives us the correct answer of B, with the trap answer being D.

Re: Q is the set of all integers between A and B, inclusive. The average ( [#permalink]

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25 May 2015, 01:47

regarding point 2 I agree with your explanation when we a and b are the only numbers in the set, although we do not know how many numbers there are in the set.

Consider the set has 6 numbers B = 190 x1, x2, x3, x4 are the other numbers

Statement 2 says: (A+B+X1+x2+x3+x4)/6 = 150

Substituting (190+A+X1+x2+x3+x4)/6 = 150

How do we find the value of A now???

Bunuel wrote:

minhphammr wrote:

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.

So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

regarding point 2 I agree with your explanation when we a and b are the only numbers in the set, although we do not know how many numbers there are in the set.

Consider the set has 6 numbers B = 190 x1, x2, x3, x4 are the other numbers

Statement 2 says: (A+B+X1+x2+x3+x4)/6 = 150

Substituting (190+A+X1+x2+x3+x4)/6 = 150

How do we find the value of A now???

Bunuel wrote:

minhphammr wrote:

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.

So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

Answer: B.

Hope it's clear.

Q is a set of consecutive integers, thus its average, m, equals to the average of the smallest and largest integers in the set.
_________________

Re: Q is the set of all integers between A and B, inclusive. The average ( [#permalink]

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25 May 2015, 01:56

Ok, that makes sense. I didn't notice the word consecutive in the question.

Thanks!!

minhphammr wrote:

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

[/quote][/quote]

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.

So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

Re: Q is the set of all integers between A and B, inclusive. The average ( [#permalink]

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30 May 2016, 04:17

Bunuel wrote:

minhphammr wrote:

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.

So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

Answer: B.

Hope it's clear.

Hi there! I think that its not clear that the integers in Q are consecutive. Where is that information derived from? It just states "integers between A and B, inclusive".

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

First of all, in order the answer to be B, it should be mentioned that A is an integer itself. Suppose we are told that.

So, we are told that the average of integers between integer A and 190 is m. Notice that Q is a set of consecutive integers, thus its average, m, equals to the median.

(1) Q contains 40 terms that are greater than m.

If A is odd, then the number of integers between odd A and 190 would be even, so in this case m would be the average of two middle terms, so not an integer. Thus, in this case we would have that there are 80 integers in the set, 40 less than m and 40 greater than m. This gives the value of A as 111.

If A is eve, then the number of integers between even A and 190 would be odd, so in this case m would be the middle terms, so an integer. Thus, in this case we would have that there are 81 integers in the set, 40 less than m, m itself and 40 greater than m. This gives the value of A as 110.

(2) m=150 --> m = (A + B)/2 --> 150 = (A + 190)/2 --> A = 110. Sufficient.

Answer: B.

Hope it's clear.

Hi there! I think that its not clear that the integers in Q are consecutive. Where is that information derived from? It just states "integers between A and B, inclusive".

Integers between A and B, inclusive are all integers between A and B, inclusive in order. For example, integers between 2 and 7, inclusive are 2, 3, 4, 5, 6, and 7.
_________________

Re: Q is the set of all integers between A and B, inclusive. The average ( [#permalink]

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29 Jul 2017, 03:45

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Re: Q is the set of all integers between A and B, inclusive. The average ( [#permalink]

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07 Aug 2017, 12:52

minhphammr wrote:

Q is the set of all integers between A and B, inclusive. The average (arithmetic mean) of Q is m. If A<B, and B=190, what is the value of A?

(1) Q contains 40 terms that are greater than m.

(2) m=150

Official Solution

The issue is consecutive integers. Since Q contains a set of consecutive integers, the average m is also the middle of the set - the number with an equal number of terms greater and smaller than it. Recall that the average of a set of consecutive integers can also be easily calculated as the average of its first and last terms.

Stat. (1): Since m is the middle of the set, it follows that there are 40 members smaller than m. This statement wants to lead you to the conclusion that there are 81 members in set Q: 40 members greater than m, 40 members smaller than m, and m itself. Thus, A can be calculated by counting down 81 consecutive integers from B=190 (A in this case would be 110, but you don't really have to know that).

But is that the only possible value for A? The problem lies with the assumption that m itself is a member of set Q, but that fact is not stated anywhere. In fact, there is even no guarantee that m is an integer. If set Q has an even number of members (i.e. only 80 terms), then m, as the median of the set, is calculated by averaging the pair of middle terms; m would still have 40 members greater and 40 members smaller than it, but will not itself be a member of set Q. However, since in this case set Q has only 80 members, this will determine a different value for A (i.e. A=111). No single value can be determined for A, so Stat.(1)->IS->BCE.

Stat. (2): If m is an integer, it follows that m is also a member of set Q. Since the average is also the median, then number of terms smaller than m is equal to the number of terms greater than m, until B=190. It is therefore possible to count down the same number of consecutive integers from 150 and find a single value for A.

Answer: B

Alternative explanation for stat. (2): since the average of a set of consecutive integers is the average of the first and last members of the set, (A + B)/2 = m. Since B=190 and m=150, this leads to a single equation with A: (A+190)/2 = 150. From here you can determine a single value for A.

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