A set consists of n consecutive integers in which the smalle : Data Sufficiency (DS)

vibhav

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21 Jul 2013, 07:20

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guys any algebraic approach to solve this problem?

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mau5

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

From F. S 1, we know that the initial average = \(\frac{n+1}{2}\). Now, let the element removed be e. Thus, \(15*(n-1)+e = \frac{n*(n+1)}{2}\)

Also, the element e can take values only between [1,n], \(\to 1\leq{e}\leq{n} \to \\
\\
1\leq{{ \frac{n*(n+1)}{2}}-{15(n-1)}}\leq{n}\) \(\to\) \(2\leq{n^2-29n+30}\leq{2n}\).

On solving for the inequalities individually, we get

\(n^2-29n+30\leq{2n} \to\) \(n^2-31n+30\leq{0}\) \(\to 1\leq{n}\leq{30}\)

AND

\(2\leq{n^2-29n+30} \to\) \(0\leq{n^2-29n+28}\) \(\to n\leq{1}\)OR \(n\geq{28}\), the former value is not possible.

Thus, \(28\leq{n}\leq{30}\) , Insufficient.

Similarly, from F.S 2, we get that \(16*(n-1)+e^' = \frac{n*(n+1)}{2}\) and just as above, we get \(30\leq{n}\leq{32}\) ,Insufficient.

Taking both fact statements together, we have n = 30 as the only possible solution.

C.

Wow tough one and a awesome solution !
Is there any alternate algebraic solution to this one , solving 4 inequalities will certainly take a lot of time?

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Here is how I am looking at it (heuristic, at best):

The average of the first n consecutive integers is [n+1][/2]. So based on statement 1, I can choose to pick
1) the first 29 numbers and take the number 15 (middle value) off, or
2) pick the first 30 numbers and knock 30 (end value) off, or
3) pick the first 28 numbers and knock 1 (start value) off. So statement 1 is insufficient. We can have a similar set of options for statement 2.

With both statements, however, the choice of numbers to choose reduces to 1( The first 30 numbers)

stne

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VeritasPrepKarishma

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

This question was made to help you understand how averages work in consecutive integers.

The takeaway here is this: When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5. The reason is that when you add a number at the extreme, the average moves by 0.5. So when you remove a number from the extreme, it will move by 0.5 only. If you remove a number from the middle somewhere, the average will move by less than 0.5.

Say, average of 1, 2, 3, 4 is 2.5.
Average of 1, 2, 3, 4, 5 is 3 (when you added a number at the end, the average increased by 0.5)
So if you remove a number from this list, the maximum the average can move is 0.5 (in case you remove 5 again or in case you remove 1 - the numbers at the extreme)

Let me give the OE provided by this question for people who come across this question:

Notice that the average of n consecutive positive integers is either the integer in the middle (if there are odd number of integers) or average of 2 consecutive integers in the middle (if there are even number of integers). So average of n integers could be 15, 15.5, 16, 16.5 etc.
When one of the numbers is erased, the average goes down/up depending on whether the number was higher/lower than the average. When a number is removed, the maximum change cannot be more than 0.5. Take an example to understand this:
1, 2, 3, 4, 5
Average = 3. There are numbers on either side of 3 that average out to 3 e.g. 2 and 4, 1 and 5. When you remove one of these numbers, an imbalance is created. Say, if you remove 1, there is nothing to balance out 5 which will be 2 more than the average. 2 will be distributed among the remaining 4 numbers and hence the average will increase by 2/4 = 0.5. This is the maximum change in average.
Statement 1: The original average would be 14.5/15/15.5. Value of n would be different for each average.
If average is 14.5, n = 28
If average is 15, n = 29
If average is 15.5, n = 30
Not sufficient.
Statement 2: The original average would be 15.5/16/16.5. Value of n would be different for each average. Not sufficient.
Taking both statements together, original average must be 15.5 and n must be 30. Sufficient.
Answer (C)

Yes very nice solution,
More GMAT type, now it looks doable within the time frame.
Thank you for the new concept .

KarishmaB

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stne

Yes very nice solution,
More GMAT type, now it looks doable within the time frame.
Thank you for the new concept .

Actually the concept is not new. You knew that average of 1, 2, 3, 4, 5 is 3 while average of 1, 2, 3, 4, 5, 6 is 3.5 i.e. a difference of 0.5.
This question is an advanced application of the same thing. When you add a number at the extreme, the average moves by 0.5. When you remove a number from the extreme, the average will move by 0.5. This is how GMAT frames its 700+ level questions. The concept will be the same, the application will be a little more obscure. That is why it is very important to really understand the concept.

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stne

Wow tough one and a awesome solution !
Is there any alternate algebraic solution to this one , solving 4 inequalities will certainly take a lot of time?

So this might not be helping anyone as it is outside the scope of GMAT, yet just for knowing the origin of why things happen, here is the proof for what Karishma said : "When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5"

For any evenly spaced series, let the initial average be \(A_i\) and the common difference be d and the no of terms be n.

\(A_f - A_i = \frac{d}{2} + \frac{1-t_n}{n-1}\); where , \(A_f\) is the final average after removing the\(t_n\)term.

For example 3,4,5,6,7. Here, \(A_i = 5\), n = 5 and d = 1.

Thus, the final average after removing say the \(2^{nd}\) term, \(t_n\) = 2 is :

\(A_f - A_i = \frac{1}{2} + \frac{1-2}{5-1}\)

\(A_f = A_i + \frac{1}{4} = 5+0.25 = 5.25\)

To verify, the average of 3,5,6,7 = \(\frac{21}{4} = 5.25\)

One can fix d=1, fix\(t_n\) = first term and last term individually and get the required proof.

IMO it's a good problem, but I wouldn't expect such a question on GMAT. If you know this concept good enough, else it's too difficult to think of it on the spot.

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21 Dec 2013, 22:50

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stne

VeritasPrepKarishma

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

This question was made to help you understand how averages work in consecutive integers.

The takeaway here is this: When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5. The reason is that when you add a number at the extreme, the average moves by 0.5. So when you remove a number from the extreme, it will move by 0.5 only. If you remove a number from the middle somewhere, the average will move by less than 0.5.

Say, average of 1, 2, 3, 4 is 2.5.
Average of 1, 2, 3, 4, 5 is 3 (when you added a number at the end, the average increased by 0.5)
So if you remove a number from this list, the maximum the average can move is 0.5 (in case you remove 5 again or in case you remove 1 - the numbers at the extreme)

Let me give the OE provided by this question for people who come across this question:

Notice that the average of n consecutive positive integers is either the integer in the middle (if there are odd number of integers) or average of 2 consecutive integers in the middle (if there are even number of integers). So average of n integers could be 15, 15.5, 16, 16.5 etc.
When one of the numbers is erased, the average goes down/up depending on whether the number was higher/lower than the average. When a number is removed, the maximum change cannot be more than 0.5. Take an example to understand this:
1, 2, 3, 4, 5
Average = 3. There are numbers on either side of 3 that average out to 3 e.g. 2 and 4, 1 and 5. When you remove one of these numbers, an imbalance is created. Say, if you remove 1, there is nothing to balance out 5 which will be 2 more than the average. 2 will be distributed among the remaining 4 numbers and hence the average will increase by 2/4 = 0.5. This is the maximum change in average.
Statement 1: The original average would be 14.5/15/15.5. Value of n would be different for each average.
If average is 14.5, n = 28
If average is 15, n = 29
If average is 15.5, n = 30
Not sufficient.
Statement 2: The original average would be 15.5/16/16.5. Value of n would be different for each average. Not sufficient.
Taking both statements together, original average must be 15.5 and n must be 30. Sufficient.
Answer (C)

Yes very nice solution,
More GMAT type, now it looks doable within the time frame.
Thank you for the new concept .

Ans should be 'D' :-D

as per the solution below:

From statement 1: 15(n-1)=n/2*(1+n)-(1+(n-1))
solving this,we get n=1 or n=30 => n=30 because it can't be n=1
From statement 2:16(n-1)=n/2*(1+n)-(1+(n-1))
solving this,we get n=1 or n=32 => n=32 because it can't be n=1

Correct me if i am wrong. :oops:

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"A set consists of n consecutive integers in which the smallest term is \(1\)" also means \(n\) is the largest number of the set.

If the first number to be removed from the set is \(A\), and the second number is \(B\). Then from both the statements together, we can draw that:

\(A = B + (n-1)\)

Since A is a number of the set, must \(A <= n\) => \(B <= 1\)

Since \(B\) is also a number of the set and the smallest number of the set is \(1\) => \(B = 1\)

As \(\frac{n(n+1)}{2} - B = 16n -16\)

=> \(n=30\)

KarishmaB

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30 Dec 2013, 07:17

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danjay

Ans should be 'D' :-D

as per the solution below:

From statement 1: 15(n-1)=n/2*(1+n)-(1+(n-1))
solving this,we get n=1 or n=30 => n=30 because it can't be n=1
From statement 2:16(n-1)=n/2*(1+n)-(1+(n-1))
solving this,we get n=1 or n=32 => n=32 because it can't be n=1

Correct me if i am wrong. :oops:

I don't know what you have done here but notice that each statement alone is not sufficient.

1, 2, 3 .... n

Take statement 1 alone:

Say, n = 28
Average = 14.5
If you remove 1, you get average = 15

Say, n = 30
Average = 15.5
If you remove 30, you get average = 15

So n could be 28 or 30.

Similarly statement 2 alone is also not sufficient.

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

For consecutive integers in the original condition we have 2 variables (the starting value and the total number of values) and since the value starts from 1, we just need 1 more equation. Since there is 1 each in 1) and 2), D is likely the answer.
in case of 1), the answer is n=29(removed value=15), n=30(removed value=30). The answer is not unique, therefore the condition is not sufficient
in case of 2), the answer is n=31(removed value=16), n=30(removed value=1). The answer is not unique, therefore the condition is not sufficient.
Using both 1) & 2) together, the answer is n=30 and the conditions are sufficient. Therefore the answer is C

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VeritasPrepKarishma

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

This question was made to help you understand how averages work in consecutive integers.

The takeaway here is this: When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5. The reason is that when you add a number at the extreme, the average moves by 0.5. So when you remove a number from the extreme, it will move by 0.5 only. If you remove a number from the middle somewhere, the average will move by less than 0.5.

Say, average of 1, 2, 3, 4 is 2.5.
Average of 1, 2, 3, 4, 5 is 3 (when you added a number at the end, the average increased by 0.5)
So if you remove a number from this list, the maximum the average can move is 0.5 (in case you remove 5 again or in case you remove 1 - the numbers at the extreme)

Let me give the OE provided by this question for people who come across this question:

Notice that the average of n consecutive positive integers is either the integer in the middle (if there are odd number of integers) or average of 2 consecutive integers in the middle (if there are even number of integers). So average of n integers could be 15, 15.5, 16, 16.5 etc.
When one of the numbers is erased, the average goes down/up depending on whether the number was higher/lower than the average. When a number is removed, the maximum change cannot be more than 0.5. Take an example to understand this:
1, 2, 3, 4, 5
Average = 3. There are numbers on either side of 3 that average out to 3 e.g. 2 and 4, 1 and 5. When you remove one of these numbers, an imbalance is created. Say, if you remove 1, there is nothing to balance out 5 which will be 2 more than the average. 2 will be distributed among the remaining 4 numbers and hence the average will increase by 2/4 = 0.5. This is the maximum change in average.
Statement 1: The original average would be 14.5/15/15.5. Value of n would be different for each average.
If average is 14.5, n = 28
If average is 15, n = 29
If average is 15.5, n = 30
Not sufficient.
Statement 2: The original average would be 15.5/16/16.5. Value of n would be different for each average. Not sufficient.
Taking both statements together, original average must be 15.5 and n must be 30. Sufficient.
Answer (C)

Hello Karishma,

How can average be 15. In understand n-1 is 15 as per S-1, so actual average for n could be 14.5 (increase of 0.5 removing lowest number) or 15.5 (decrease of 0.5 removing highest number). But I don't get when it will be 15, so n = 29 ?

KarishmaB

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coolkl

VeritasPrepKarishma

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

This question was made to help you understand how averages work in consecutive integers.

The takeaway here is this: When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5. The reason is that when you add a number at the extreme, the average moves by 0.5. So when you remove a number from the extreme, it will move by 0.5 only. If you remove a number from the middle somewhere, the average will move by less than 0.5.

Say, average of 1, 2, 3, 4 is 2.5.
Average of 1, 2, 3, 4, 5 is 3 (when you added a number at the end, the average increased by 0.5)
So if you remove a number from this list, the maximum the average can move is 0.5 (in case you remove 5 again or in case you remove 1 - the numbers at the extreme)

Let me give the OE provided by this question for people who come across this question:

Notice that the average of n consecutive positive integers is either the integer in the middle (if there are odd number of integers) or average of 2 consecutive integers in the middle (if there are even number of integers). So average of n integers could be 15, 15.5, 16, 16.5 etc.
When one of the numbers is erased, the average goes down/up depending on whether the number was higher/lower than the average. When a number is removed, the maximum change cannot be more than 0.5. Take an example to understand this:
1, 2, 3, 4, 5
Average = 3. There are numbers on either side of 3 that average out to 3 e.g. 2 and 4, 1 and 5. When you remove one of these numbers, an imbalance is created. Say, if you remove 1, there is nothing to balance out 5 which will be 2 more than the average. 2 will be distributed among the remaining 4 numbers and hence the average will increase by 2/4 = 0.5. This is the maximum change in average.
Statement 1: The original average would be 14.5/15/15.5. Value of n would be different for each average.
If average is 14.5, n = 28
If average is 15, n = 29
If average is 15.5, n = 30
Not sufficient.
Statement 2: The original average would be 15.5/16/16.5. Value of n would be different for each average. Not sufficient.
Taking both statements together, original average must be 15.5 and n must be 30. Sufficient.
Answer (C)

Hello Karishma,

How can average be 15. In understand n-1 is 15 as per S-1, so actual average for n could be 14.5 (increase of 0.5 removing lowest number) or 15.5 (decrease of 0.5 removing highest number). But I don't get when it will be 15, so n = 29 ?

If the new average is 15, it is possible that the old average is 15 too.
If n is 29, the numbers would be
1, 2, 3, 4, 5, ... 13, 14, 15, 16, 17, ... 27, 28, 29

Now if the number removed is 15, the new average will still be 15.

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VeritasPrepKarishma

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

This question was made to help you understand how averages work in consecutive integers.

The takeaway here is this: When a number is removed from a list of consecutive integers, the maximum change in the average cannot be more than 0.5. The reason is that when you add a number at the extreme, the average moves by 0.5. So when you remove a number from the extreme, it will move by 0.5 only. If you remove a number from the middle somewhere, the average will move by less than 0.5.

Say, average of 1, 2, 3, 4 is 2.5.
Average of 1, 2, 3, 4, 5 is 3 (when you added a number at the end, the average increased by 0.5)
So if you remove a number from this list, the maximum the average can move is 0.5 (in case you remove 5 again or in case you remove 1 - the numbers at the extreme)

Let me give the OE provided by this question for people who come across this question:

Notice that the average of n consecutive positive integers is either the integer in the middle (if there are odd number of integers) or average of 2 consecutive integers in the middle (if there are even number of integers). So average of n integers could be 15, 15.5, 16, 16.5 etc.
When one of the numbers is erased, the average goes down/up depending on whether the number was higher/lower than the average. When a number is removed, the maximum change cannot be more than 0.5. Take an example to understand this:
1, 2, 3, 4, 5
Average = 3. There are numbers on either side of 3 that average out to 3 e.g. 2 and 4, 1 and 5. When you remove one of these numbers, an imbalance is created. Say, if you remove 1, there is nothing to balance out 5 which will be 2 more than the average. 2 will be distributed among the remaining 4 numbers and hence the average will increase by 2/4 = 0.5. This is the maximum change in average.
Statement 1: The original average would be 14.5/15/15.5. Value of n would be different for each average.
If average is 14.5, n = 28
If average is 15, n = 29
If average is 15.5, n = 30
Not sufficient.
Statement 2: The original average would be 15.5/16/16.5. Value of n would be different for each average. Not sufficient.
Taking both statements together, original average must be 15.5 and n must be 30. Sufficient.
Answer (C)

Hi Karishma,

Here n is number of consecutive numbers in the set. How did you get n=29 directly? I mean n can be 2 (15+16/2=31.5, if we remove 16 , the avg is 15) and n can be 4. Like wise if we consider both statements:
statement 1+2,
14.5< sum/n< 15.5--from statement 1
15.5< Sum/n<16.5 ----from statement 2

if n=2,
29<sum<31
31<sum<33

if n=4,
58<sum<62
62<sum<66

we can clearly see that n can be 2 and n can be 4. How did you directly infer n=30? I am not getting this part. Pls help me understand the logic

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vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

VeritasKarishma and MathRevolution Request you to please let me know if my approach of solving this question is correct

initial sum of AP would be sum=[n*(1+n)]/2

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

Now lets say x was removed from the series then the sum would be {[n*(1+n)]/2} - x
and the new avg of the series would be [{[n*(1+n)]/2} - x]/(n-1) =15
if we solve it further we get n^2 - 29n - x = -30 -----eq1

Clearly Not Sufficient.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

the new avg of the series would be [{[n*(1+n)]/2} - y]/(n-1) =16
if we solve it further we get the final equation as n^2 - 31n - y =-32 -----eq2 [here y represents the new number that was removed]

Clearly Not Sufficient.

Combining both (1) and (2) we get

n^2 - 29n - x = -30
n^2 - 31n - y =-32

Now solving further we get the Final Equation as below:
n= [(x-y)/2]+1

I know both x and y are integers and also "n" has to be an integer. Hence there would exist values for x, y that will satisfy the this equation to give an integer value.

Hence C is the answer.

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DhruvS

vibhav

A set consists of n consecutive integers in which the smallest term is 1. What is the value of n?

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

VeritasKarishma and MathRevolution Request you to please let me know if my approach of solving this question is correct

initial sum of AP would be sum=[n*(1+n)]/2

(1) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 15.

Now lets say x was removed from the series then the sum would be {[n*(1+n)]/2} - x
and the new avg of the series would be [{[n*(1+n)]/2} - x]/(n-1) =15
if we solve it further we get n^2 - 29n - x = -30 -----eq1

Clearly Not Sufficient.

(2) When one of the numbers is removed from the set, the average of the remaining numbers in the set is 16.

the new avg of the series would be [{[n*(1+n)]/2} - y]/(n-1) =16
if we solve it further we get the final equation as n^2 - 31n - y =-32 -----eq2 [here y represents the new number that was removed]

Clearly Not Sufficient.

Combining both (1) and (2) we get

n^2 - 29n - x = -30
n^2 - 31n - y =-32

Now solving further we get the Final Equation as below:
n= [(x-y)/2]+1

I know both x and y are integers and also "n" has to be an integer. Hence there would exist values for x, y that will satisfy the this equation to give an integer value.

Hence C is the answer.

There are infinite integer values that x and y can take to give you integer values for n. Hence, you cannot arrive at (C) using that logic. A lot more thought needs to go into the concept of mean to arrive at the answer here as discussed in this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... -the-gmat/

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21 Nov 2020, 13:10

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VeritasKarishma

Hello VeritasKarishma,
A prompt question that occurred to me while I was trying to solve this problem during the practice test,
Here is says "A set consists of n consecutive integers" so does it mean that we look for sets like 1,2,3,4,5...
or that we can also include even/odd consecutive sets such as 2,4,6,8... / 1,3,5,7... So generally speaking when a question states about consecutive integers does it restricts them to the first example I mentioned 1,2,3,4,5... or are all of the above (even-odd consecutives) also taken into consideration?

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VeritasKarishma

Hello VeritasKarishma,
A prompt question that occurred to me while I was trying to solve this problem during the practice test,
Here is says "A set consists of n consecutive integers" so does it mean that we look for sets like 1,2,3,4,5...
or that we can also include even/odd consecutive sets such as 2,4,6,8... / 1,3,5,7... So generally speaking when a question states about consecutive integers does it restricts them to the first example I mentioned 1,2,3,4,5... or are all of the above (even-odd consecutives) also taken into consideration?

n consecutive integers means sets such as 1, 2,3 ,4 5... only.
If you have 2, 4, 6, 8..., this is a set containing n consecutive even integers.
1,3 ,5 ,7 ..., is a set containing n consecutive odd integers.

A set consists of n consecutive integers in which the smalle

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A set consists of n consecutive integers in which the smalle

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