1. Yachtchain has 75 stores. The sales at Store M represented the median for Yachtchains stores. How many yachts were sold by Store M?
1) At least 60% of the stores sold more than 50 yachts
2) More than 55% of the stores sold less than 52 yachts
2. Does the integer x have a factor p such that 1<p<x?
1) 17!+2 <= x <= 17!+ 18
2) x > 10! + 5
Could you explain the strategy and content to solve these two questions?
Thank you
M
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- JadranLee
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- dbernst
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1. Yachtchain has 75 stores. The sales at Store M represented the median for Yachtchains stores. How many yachts were sold by Store M?
1) At least 60% of the stores sold more than 50 yachts
2) More than 55% of the stores sold less than 52 yachts
In this problem, Yactchain has 75 stores. To represent the median sales, we would have to know how many sales were made at store 38 if the sales were arranged in ascending order (in an odd numbered set the median is the average of the first and last term, or (75 + 1)/2 in this case).
Statement (1): Tells us that at least 45 stores (60% of 75) sold more than 50 yachts. This is not sufficient to know exactly how many yachts were sold at the 38th store. Eliiminate AD from AD/BCE grid.
Statement (2): Tells us that at least 42 stores (55% of 75 = 41.25, so the next store would be 42) sold fewer than 52 yachts. This is not sufficient to know exactly how many yachts were sold at the 38th store. Eliiminate B from BCE grid.
Together: Statement (1) told us that at least 45 stores sold more than 50 yachts. In ascending order, that means stores number 31-75 sold more than 50 yachts. Statement (2) told us that at least 42 stores sold fewer than 52 yachts. In ascending order, that means stores number 1-42 sold fewer than 52 yachts. Thus, we know that stores number 31-42 (the overlap of the two sets) each sold more than 50 but fewer than 52 yachts. That means store number 38, which is included in this set, must have sold exactly 51 yachts.
The correct answer is C.
2. Does the integer x have a factor p such that 1<p<x?
1) 17!+2 <= x <= 17!+ 18
2) x > 10! + 5
In this problem, a good rephrase is virtually a necessity. The ONLY type of integer that DOES NOT have a factor between itself and one is a prime integer. Thus, the best rephrase for this problem is Is x prime?
Since Statement (2) looks easier to me, I am going to begin with a BD/ACE grid.
Statement (2): I know that x is greater than some huge number, but I have no idea whether it is prime (as an iteresting aside, the largest know prime number is more than 6 miillion digits long!). Since this statement does not definitely answer whether x is prime, eliminate BD from your BD/ACE grid.
Statement (1): This is actually a "trick" statement that is made to look more complex than it really is. I will use simpler number to explain. Suppose the statement had said 3!+2 < x < 3!+4. In this case 3! = (3)(2)(1), meaning the result - 6 - is a multiple of 3, 2. Therefore, this result CANNOT be prime because, by definition, it has factors smaller than itself but larger than 1. Furthermore, if I were to add 2 to the result (6+2), my new sum could not be prime since the new sum is simply one more multiple of 2. Similarly, if I were to add 3 to the result (6+3), my new sum could not be prime since the new sum is simply one more multiple of 3. Finally, if I were to add 4 to the result (6+4), my new sum could not be prime since the new sum must be even (as the original sum was even)
Using this logic for Statement (1), 17! cannot be prime because it contain all of the integers from 1-17 as factors. If I add 2 to 17!, the sum cannot be prime because 17! is even (2 is a factor) so 17!+2 is even. If I add 3 to 17!, the sum cannot be prime because 17! is a multiple of 3 (3 is a factor) so 17!+3 is a multiple of 3. This process repeats to all number up to 17! + 17. Finally, 17! + 18 cannot be prime because 17! is even and 18 is even, so 17!+18 must be even. Since none of the integers x from statement (1) can be prime, we have a definitive answer to our question.
The correct answer is A.
1) At least 60% of the stores sold more than 50 yachts
2) More than 55% of the stores sold less than 52 yachts
In this problem, Yactchain has 75 stores. To represent the median sales, we would have to know how many sales were made at store 38 if the sales were arranged in ascending order (in an odd numbered set the median is the average of the first and last term, or (75 + 1)/2 in this case).
Statement (1): Tells us that at least 45 stores (60% of 75) sold more than 50 yachts. This is not sufficient to know exactly how many yachts were sold at the 38th store. Eliiminate AD from AD/BCE grid.
Statement (2): Tells us that at least 42 stores (55% of 75 = 41.25, so the next store would be 42) sold fewer than 52 yachts. This is not sufficient to know exactly how many yachts were sold at the 38th store. Eliiminate B from BCE grid.
Together: Statement (1) told us that at least 45 stores sold more than 50 yachts. In ascending order, that means stores number 31-75 sold more than 50 yachts. Statement (2) told us that at least 42 stores sold fewer than 52 yachts. In ascending order, that means stores number 1-42 sold fewer than 52 yachts. Thus, we know that stores number 31-42 (the overlap of the two sets) each sold more than 50 but fewer than 52 yachts. That means store number 38, which is included in this set, must have sold exactly 51 yachts.
The correct answer is C.
2. Does the integer x have a factor p such that 1<p<x?
1) 17!+2 <= x <= 17!+ 18
2) x > 10! + 5
In this problem, a good rephrase is virtually a necessity. The ONLY type of integer that DOES NOT have a factor between itself and one is a prime integer. Thus, the best rephrase for this problem is Is x prime?
Since Statement (2) looks easier to me, I am going to begin with a BD/ACE grid.
Statement (2): I know that x is greater than some huge number, but I have no idea whether it is prime (as an iteresting aside, the largest know prime number is more than 6 miillion digits long!). Since this statement does not definitely answer whether x is prime, eliminate BD from your BD/ACE grid.
Statement (1): This is actually a "trick" statement that is made to look more complex than it really is. I will use simpler number to explain. Suppose the statement had said 3!+2 < x < 3!+4. In this case 3! = (3)(2)(1), meaning the result - 6 - is a multiple of 3, 2. Therefore, this result CANNOT be prime because, by definition, it has factors smaller than itself but larger than 1. Furthermore, if I were to add 2 to the result (6+2), my new sum could not be prime since the new sum is simply one more multiple of 2. Similarly, if I were to add 3 to the result (6+3), my new sum could not be prime since the new sum is simply one more multiple of 3. Finally, if I were to add 4 to the result (6+4), my new sum could not be prime since the new sum must be even (as the original sum was even)
Using this logic for Statement (1), 17! cannot be prime because it contain all of the integers from 1-17 as factors. If I add 2 to 17!, the sum cannot be prime because 17! is even (2 is a factor) so 17!+2 is even. If I add 3 to 17!, the sum cannot be prime because 17! is a multiple of 3 (3 is a factor) so 17!+3 is a multiple of 3. This process repeats to all number up to 17! + 17. Finally, 17! + 18 cannot be prime because 17! is even and 18 is even, so 17!+18 must be even. Since none of the integers x from statement (1) can be prime, we have a definitive answer to our question.
The correct answer is A.
1. Yachtchain has 75 stores. The sales at Store M represented the median for Yachtchains stores. How many yachts were sold by Store M?
1) At least 60% of the stores sold more than 50 yachts
2) More than 55% of the stores sold less than 52 yachts
2. Does the integer x have a factor p such that 1<p<x?
1) 17!+2 <= x <= 17!+ 18
2) x > 10! + 5
Could you explain the strategy and content to solve these two questions?
Thank you
Last edited by dbernst on Fri May 25, 2007 12:17 pm, edited 1 time in total.
- Jeff
Tough DS
For the second problem from the first poster.
2. Does the integer x have a factor p such that 1<p<x?
1) 17!+2 <= x <= 17!+ 18
2) x > 10! + 5
Let's rephrase the question. The question asks whether for some integer X, there is a factor, p, such that 1<p<x. This is equivalent to asking "Is X a prime number?" If so, then there will not be a p s.t. 1<p<x, otherwise if x is NOT prime, there will be a factor p s.t. 1<p<x.
Now that the question has been rephrased, look at the two facts. (2) looks easier to deal with so start there. Just knowing that x>10! + 5 does not answer the question of whether x is prime or not. Prime numbers occur periodically and without predictable pattern throughout the set of integers. So (2) is insufficient by itself and you can rule out B&D.
Let's look at (1). Since we know that x is an integer, 17!+2<=x<=17!+18 gives us 17 values to check. Remember that 17!= (17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1), so 17! has 17,16,15, etc. as factors. We also know that 17! is even since it has 2(several times over) as a factor. Recall that the sum of two even numbers is always an even number. So if 17! is even, then 17!+2, 17!+4, 17!+6...17!+18 all all even numbers and therefore have a factor p st 1<p<x. For even numbers one value of p would be 2.
So we just answered the question for 9 of the 17 possible values of x. What about 17!+3, 17!+5, 17!+7...17!+17? You can use similar reasoning. Since 17! is a multiple of 3, then 17!+3 must also be a multiple of 3. Likewise, since 17! is a multiple of 5, then 17!+5 must be a multiple of 5 and so on through 17!+17. So (1) is sufficient to answer the question and the answer is A.
cheers,
Jeff
2. Does the integer x have a factor p such that 1<p<x?
1) 17!+2 <= x <= 17!+ 18
2) x > 10! + 5
Let's rephrase the question. The question asks whether for some integer X, there is a factor, p, such that 1<p<x. This is equivalent to asking "Is X a prime number?" If so, then there will not be a p s.t. 1<p<x, otherwise if x is NOT prime, there will be a factor p s.t. 1<p<x.
Now that the question has been rephrased, look at the two facts. (2) looks easier to deal with so start there. Just knowing that x>10! + 5 does not answer the question of whether x is prime or not. Prime numbers occur periodically and without predictable pattern throughout the set of integers. So (2) is insufficient by itself and you can rule out B&D.
Let's look at (1). Since we know that x is an integer, 17!+2<=x<=17!+18 gives us 17 values to check. Remember that 17!= (17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1), so 17! has 17,16,15, etc. as factors. We also know that 17! is even since it has 2(several times over) as a factor. Recall that the sum of two even numbers is always an even number. So if 17! is even, then 17!+2, 17!+4, 17!+6...17!+18 all all even numbers and therefore have a factor p st 1<p<x. For even numbers one value of p would be 2.
So we just answered the question for 9 of the 17 possible values of x. What about 17!+3, 17!+5, 17!+7...17!+17? You can use similar reasoning. Since 17! is a multiple of 3, then 17!+3 must also be a multiple of 3. Likewise, since 17! is a multiple of 5, then 17!+5 must be a multiple of 5 and so on through 17!+17. So (1) is sufficient to answer the question and the answer is A.
cheers,
Jeff
- dbernst
- ManhattanGMAT Staff
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