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A set of numbers contains 7 integers and has an average
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15 May 2012, 01:36
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A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set? A. 33 B. 35 C. 38 D. 48 E. 75
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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 01:59
Joy111 wrote: A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?
A. 33 B. 35 C. 38 D. 48 E. 75 The average of 7 numbers is 23 > the sum of these numbers is 7*23=161; The median of 7 numbers is 23 > 23 is the middle term: {*, *, *, 23, *, *, *}; The largest value is equal to 15 more than 4 times the smallest number > say the smallest number is x then the largest number would be 4x+15, so our set is {x, *, *, 23, *, *, 4x+15}; Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. Since the sum is 161 then x+x+x+23+23+23+4x+15=161 > x=11. The range is (4x+15)x=3x+15=48. Answer: D.
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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 03:36
gmihir wrote: Also, can you please explain this part: "Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. "  Even if 2nd, 3rd, numbers had been anything between x & 23, vice versa, if 4th and 5th numbers had been any numbers between 23 and 4x+15, the range (4x + 15  x) would have been same. So, the question is how did we guess 2nd, 3rd, 5th and 6th numbers of this set ? Thanks! for a set of 7 numbers if the median is 23 then we can have 2 cases 1) all numbers are 23 { 23,23,23,23,23,23,23}, in this case too average is 23 2) {# <= 23, # <= 23, # <= 23, 23( center digit equal to 23) , # >= 23, >= 23, >= 23}, median of 7 numbers is 23 and average is 23, means that half of the numbers are below or equal to the median and half of the numbers are equal to or above the median . Now the given condition in the question is that the maximum = 4X + 15 , where x is the smallest one .So we know that all numbers are not equal . now in order to maximize the range we have to keep the other 6 as small as possible . { x,x,x,23,23,23,4X+15} now if the smallest is x then largest is 4x+15, this is given, so we cannot change these two. the center has to be 23 as this is the median. so the smallest value for the 2, and 3, digit has to be equal to the smallest one which is equal to x , we cannot take anything larger then x as that could make the range smaller , but want to find the largest range , so we have to take the smallest numbers. similarly: the smallest value for 4th and 5th digit cannot be smaller than 23, as 23 is the median , and cannot be larger the 23 as again we are looking for the largest range, so we have the keep the 4th and 5th digits as small as possible , and the smallest we can get for the 4th and 5th digit is 23. hence Bunuel's Statement Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. Hope it is clear to you now .Please ask if any doubt remains .




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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 02:14
Bunnel,
Isn't it true that when mean & median are same for the given set of integers, the nubmers have to be equally spaced ? i had read this somewhere so tried solving the problem in that direction. Can you please clarify? Thanks!



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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 02:22
gmihir wrote: Bunnel,
Isn't it true that when mean & median are same for the given set of integers, the nubmers have to be equally spaced ? i had read this somewhere so tried solving the problem in that direction. Can you please clarify? Thanks! In any evenly spaced set the arithmetic mean (average) is equal to the median. For example for a set {1, 3, 5, 7, 9} mean=median=5. But the reverse of this property is not always true. For example a set {0, 1, 1, 1, 2} has the median as well as mean equal to 1, but this set is not evenly spaced. Hope it's clear.
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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 02:23
Also, can you please explain this part: "Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. "  Even if 2nd, 3rd, numbers had been anything between x & 23, vice versa, if 4th and 5th numbers had been any numbers between 23 and 4x+15, the range (4x + 15  x) would have been same. So, the question is how did we guess 2nd, 3rd, 5th and 6th numbers of this set ? Thanks!



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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 03:55
Hi ,
I have same query.
Lets say we have a series ( x, x, x , 23 , 23 , 23 , 4x+15) the range in this case 4x+15  x = 3x+15
The range is the same even for ( x , x+1 , x+ 3 , 23 , y , y+ 1 , 4x + 15) where 23 < y < 4x because from my understanding range is the difference between the largest and the smallest number in a set . The values in between do not have any significance on the range.
Is this a correct understanding , because from you answer range seems to mean something else.
Thanks in advance.
Regards bharath



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Re: A set of numbers contains 7 integers and has an average
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Updated on: 09 Nov 2012, 06:37
mazerath wrote: Hi ,
I have same query.
Lets say we have a series ( x, x, x , 23 , 23 , 23 , 4x+15) the range in this case 4x+15  x = 3x+15
The range is the same even for ( x , x+1 , x+ 3 , 23 , y , y+ 1 , 4x + 15) where 23 < y < 4x because from my understanding range is the difference between the largest and the smallest number in a set . The values in between do not have any significance on the range.
Is this a correct understanding , because from you answer range seems to mean something else.
Thanks in advance.
Regards bharath Hi Remember that the total has to 161 , and we need one variable to sort this out , so by taking x and y , it would be difficult to find the range. To find the answer the most efficient way is to take smallest one as x and largest one as 4x+15 and , and keep the others as small as possible . Technically { 11, 11, 12, 23, 23, 24, 59 }this set is also possible , but to arrive at 11 and 59 , we have to proceed as shown by bunuel, one cannot solve the sum assuming the set to be { x, x, x+1, 23, y, y+1,4x+15}. solving this way will be complicated and time consuming . Hope it helps .
Originally posted by Joy111 on 15 May 2012, 04:14.
Last edited by Joy111 on 09 Nov 2012, 06:37, edited 2 times in total.



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Re: A set of numbers contains 7 integers and has an average
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15 May 2012, 04:36
I guess the point is to maximize range , the values in between had to be as low as possible , because they are bounds by a fixed sum. Keeping the values in between to a minimum and adding that in x will yield a higher range as the range is 3x+15..
Thanks for the help.
Bharath



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Re: A set of numbers contains 7 integers and has an average
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17 May 2012, 23:39
Isn't the average of first and last term supposed to equal the average hence (4x+15+x)/2 should equal 23 but then we get some other result for x??



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Re: A set of numbers contains 7 integers and has an average
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17 May 2012, 23:48
kaka1989 wrote: Isn't the average of first and last term supposed to equal the average hence (4x+15+x)/2 should equal 23 but then we get some other result for x?? The average of a set equals to the average of the first and the last terms if a set is evenly spaced, but if a set is not evenly spaced (like we have in original question) then this relationship is not necessary to be true.
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Re: A set of numbers contains 7 integers and has an average
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22 May 2012, 09:28
I have a question here: now in order to maximize the range we have to keep the other 6 as small as possible
Range is to do with first and last term in a series, how do we assume the other terms. Am I missing something?



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Re: A set of numbers contains 7 integers and has an average
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24 May 2012, 10:34
pavanpuneet wrote: I have a question here: now in order to maximize the range we have to keep the other 6 as small as possible
Range is to do with first and last term in a series, how do we assume the other terms. Am I missing something? It is true that the Range has to do with the first and the last terms in a series. In this question, the sum of the numbers is fixed. If we do not keep the other 6 as small as possible, we will not be able to maximise the difference between the largest and the smallest no.(= the Range) In effect, we are distributing the fixed sum in such a manner that we minimise the other nos, so that we can maximise the largest no (and so maximise the Range). Hope that makes some sense



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Re: A set of numbers contains 7 integers and has an average
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25 May 2012, 08:11
hey really nice question! I must admit couldn't fig out a way ahead for over 3mins! i totally missed the point that median has been provided and hence all 6 nos can be 23!



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Re: A set of numbers contains 7 integers and has an average
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04 Aug 2012, 03:59
mazerath wrote: Hi ,
I have same query.
Lets say we have a series ( x, x, x , 23 , 23 , 23 , 4x+15) the range in this case 4x+15  x = 3x+15
The range is the same even for ( x , x+1 , x+ 3 , 23 , y , y+ 1 , 4x + 15) where 23 < y < 4x because from my understanding range is the difference between the largest and the smallest number in a set . The values in between do not have any significance on the range.
Is this a correct understanding , because from you answer range seems to mean something else.
Thanks in advance.
Regards bharath I had a similar problem while solving the question. A way to look at this is that we need to maximize 3x+15, which is the range. Hence we need to maximize x. While maximize x, the set ( x , x+1 , x+ 3 , 23 , y , y+ 1 , 4x + 15) will do worse than ( x, x, x , 23 , 23 , 23 , 4x+15) when the total is fixed.(161) . Makes sense?



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Re: A set of numbers contains 7 integers and has an average
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28 Oct 2013, 00:41
Bunuel wrote: Joy111 wrote: A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?
A. 33 B. 35 C. 38 D. 48 E. 75 The average of 7 numbers is 23 > the sum of these numbers is 7*23=161; The median of 7 numbers is 23 > 23 is the middle term: {*, *, *, 23, *, *, *}; The largest value is equal to 15 more than 4 times the smallest number > say the smallest number is x then the largest number would be 4x+15, so our set is {x, *, *, 23, *, *, 4x+15}; Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. Since the sum is 161 then x+x+x+23+23+23+4x+15=161 > x=11. The range is (4x+15)x=3x+15=48. Answer: D. Hi Bunuel, Can we take median ~= range/2. Then median(23) ~= 46/2 So nearest value is 48 in the answers. Thanks, Sahil Bansal



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Re: A set of numbers contains 7 integers and has an average
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28 Oct 2013, 00:53
bsahil wrote: Bunuel wrote: Joy111 wrote: A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?
A. 33 B. 35 C. 38 D. 48 E. 75 The average of 7 numbers is 23 > the sum of these numbers is 7*23=161; The median of 7 numbers is 23 > 23 is the middle term: {*, *, *, 23, *, *, *}; The largest value is equal to 15 more than 4 times the smallest number > say the smallest number is x then the largest number would be 4x+15, so our set is {x, *, *, 23, *, *, 4x+15}; Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. Since the sum is 161 then x+x+x+23+23+23+4x+15=161 > x=11. The range is (4x+15)x=3x+15=48. Answer: D. Hi Bunuel, Can we take median ~= range/2. Then median(23) ~= 46/2 So nearest value is 48 in the answers. Thanks, Sahil Bansal No. What would be your answer if 46 were among the options?
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Re: A set of numbers contains 7 integers and has an average
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28 Oct 2013, 11:22
bsahil wrote: Bunuel wrote: Joy111 wrote: A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?
A. 33 B. 35 C. 38 D. 48 E. 75 The average of 7 numbers is 23 > the sum of these numbers is 7*23=161; The median of 7 numbers is 23 > 23 is the middle term: {*, *, *, 23, *, *, *}; The largest value is equal to 15 more than 4 times the smallest number > say the smallest number is x then the largest number would be 4x+15, so our set is {x, *, *, 23, *, *, 4x+15}; Now, in order to maximize the range we need to make the second and the third numbers equal to x and the fifth and sixth numbers equal to 23, so the set should be {x, x, x, 23, 23, 23, 4x+15}. Since the sum is 161 then x+x+x+23+23+23+4x+15=161 > x=11. The range is (4x+15)x=3x+15=48. Answer: D. Hi Bunuel, Can we take median ~= range/2. Then median(23) ~= 46/2 So nearest value is 48 in the answers. Thanks, Sahil Bansal The best way to answer this query is to visualize this problem as having to balance 6 members of equal weights on a seesaw such that 3 members are on either side of the number 23, which acts as the fulcrum. The farther the members on one side are from the fulcrum the farther the the members on the other side have to be to counter the weight of these members. Hence, if you push all 3 weights on the left side to the lowest possible value (x), and keep 2 weights on the right at the fulcrum, then the 1 remaining weight on the right would need to be placed at the maximum possible distance from the fulcrum; thus maximizing the range. Attachment:
fulcrum.png [ 5.79 KiB  Viewed 34422 times ]



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Re: A set of numbers contains 7 integers and has an average
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15 Mar 2015, 09:10
Joy111 wrote: A set of numbers contains 7 integers and has an average (arithmetic mean) value as well as a median value of 23. If the largest value is equal to 15 more than 4 times the smallest number, what is the largest possible range for the numbers in the set?
A. 33 B. 35 C. 38 D. 48 E. 75 let the numbers be a,b,c,d,e,f,g (d=23) since mean is also 23, sum of all 7 integers would be 161 now, g = 4a+15 range = ga = 3a+15 substitute this in the options, only a and d give an integer value of a. 3a+15 = 33 => a = 6 and g = 39 set becomes something like > 6 b c 23 e f 39 b+c+e+f = 93 if all are 23 then sum becomes 92, so one feasible set is b=c=e=23 and f=24. but we need to find the largest range 3a+15 = 48 a = 11 and g = 59 set becomes > 11 b c 23 e f 59 b+c+e+f = 68 => b=c=11 and e=f=23 hence 48 is the greatest range.
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Re: A set of numbers contains 7 integers and has an average
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25 May 2015, 23:14
Alternate Way : Smallest number =x ,Largest number =4x+15 .Range =(4x+15x) > 3x+15 . Now look at the answer choices and solve the equation 3x+15 = any one of 33,35,38,48,75 Only Option A and D result in a integer value of X . Since we are asked the maximum possible range ,hence D is the answer . Press Kudos if you like the solution .
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