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DS Problem Set 02

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by gdk800 » Thu Dec 09, 2010 7:29 am
Q1) In sequence an = an-1 + k, where 2 ≤ n ≤ 15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10?
a. a1 = 24
b. a8 = 10

[spoiler]OA is B.
[/spoiler]



Q2) If 11 consecutive integers are listed from least to most, which is the mean of the 11 digits?
a. Average of the first 9 digits is 7
b. Average of last 9 digits is 9

OA is D.
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by shovan85 » Thu Dec 09, 2010 9:12 am
gdk800 wrote: Q2) If 11 consecutive integers are listed from least to most, which is the mean of the 11 digits?
a. Average of the first 9 digits is 7
b. Average of last 9 digits is 9

OA is D.
a. Average of the first 9 digits is 7

So the sum of first 9 digits = 63 and all are consecutive integers

Let a be the first intger then second will be a+1 third will a+2 up to 11th will be a+10
Thus sum of 11 digits = 11a+ (1+2+...+10) = 11a+ 55
Use Formula: Sum of n consecutive integers starting from 1 up to n = [n(n+1)]/2
Similarly first 9 integers sum = 9a+36 = 63

Now u know a so u can know 11a+55 and so u can know the mean by dividing the resulting sum by 11.

Thus sufficient.

b. Average of last 9 digits is 9

Try the above method. You can get this option also Sufficient.

IMO D
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by beat_gmat_09 » Thu Dec 09, 2010 9:42 am
gdk800 wrote:Q1) In sequence an = an-1 + k, where 2 ≤ n ≤ 15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10?
a. a1 = 24
b. a8 = 10
From the question stem it can be know that k can be +ve or -ve
A) a1 = 24, a2 can be incremented to > 24 if k is +ve , in this case all the terms will be > 10
if k is -ve a2 will be some value < 24 , k can be anything if k=-2 then a2=22 likewise k can be -20 in this case a2=4 <10
number of terms > 10 can be any value. Thus not sufficient.

B) a8=10, if k+ve all values from a9-a15 will be > 10, thus number of terms > 10 is 15-9 = 6
if k-ve, a7, a6,a5 ....a2 will all be <10 and a9,a10,a11.....a15 will be > 10, hence number of terms > 10 = 15-9 =6
In either case number of terms is 6. Sufficient.

Answer should be B.

Editing: This holds for non-integer values of k too.
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by fskilnik@GMATH » Thu Dec 09, 2010 10:24 am
If 11 consecutive integers are listed in increasing order, what is the average (arithmetic mean) of these integers?

(1) The average of the first 9 integers is 7
(2) The average of the last 9 integers is 9
Hi there!

Whenever you have a (finite) "arithmetic progression" you should remember that the mean and the median of the list (of the sequence terms) are the same, therefore the question may be rephrased as "what is the 6th term of this (ordered) list?"

(1) Therefore the 5th term is 7 (and from the fact that this is enough to know them all), this sttm is sufficient to answer the question asked.

(2) Therefore the 5th term BACKWARDS is 9, and from the fact that this is enough to know them all, this sttm is also sufficient to answer the question asked.

Regards,
Fabio.
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by fskilnik@GMATH » Thu Dec 09, 2010 10:34 am
beat_gmat_09 wrote:B) a8=10, if k+ve all values from a9-a15 will be > 10, thus number of terms > 10 is 15-9 = 6
if k-ve, a7, a6,a5 ....a2 will all be <10 and a9,a10,a11.....a15 will be > 10, hence number of terms > 10 = 15-9 =6
In either case number of terms is 6. Sufficient.
Just one small detail/correction, beat_gmat_09:

When the question stem says a_n = a_(n-1)+k, where 2 <= n <= 15, it does NOT mean that the sequence starts with a_2, because substituting n=2 (allowed by the examiner) in the definition given, we get a_2 = a_1 +k ... that means the examiner ADMITS, implicitly, that there is a_1 and, more than that, we implicitly understand that a_1 is the first term, otherwise we could use the same definition given to write a_1 in terms of a_0, but it is NOT the case. (The examiner did NOT tell us that we can apply the formula to n=1.)

The last paragraph understood, all beat_gmat_09´s arguments are kosher BUT we must say that in each case there are SEVEN possibilities, not six. That´s all.

Regards,
Fabio.
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by gdk800 » Fri Dec 10, 2010 12:40 pm
A sincere Thanks to all members for such detailed explanations.
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by fskilnik@GMATH » Fri Dec 10, 2010 1:11 pm
gdk800 wrote:A sincere Thanks to all members for such detailed explanations.
Beautiful (and unfortunately very rare) attitude, gdk800. I thank YOU for your kind gesture and I hope to be able to "find" you in other BTG posts.

Cheers,
Fabio.
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by goyalsau » Sat Dec 11, 2010 10:22 pm
shovan85 wrote:
gdk800 wrote: Q2) If 11 consecutive integers are listed from least to most, which is the mean of the 11 digits?
a. Average of the first 9 digits is 7
b. Average of last 9 digits is 9

OA is D.
Shovan, Is There any difference between Digits and integers,
According to me digits are from 0 to 9 , and integer can be any number that can be represented on the number line,

Now Question is the means of 11 digits , I thought that there is a series like 71 72 73 74 75 76 ..........

and we want to determine the mean of first 11 digits of the series

I) statement

The series i able to determine was 93 94 95 96 97 ...... to 104

we need to add first 11 digits that will be 9 + 3 +9 + 4 + 9 + 5 + 9 + 6 + 9 = 63 ,

This is the only series that will work

So sufficient,

II) statement

IF the average to digits are 9 then all digits are 9, Its not possible any other way,

For Nine 9 in at the last place. number must be 999999999.

We don't have any other information about the length of this integer, it can be 9 ( 20 times ) or 9 ( 9 times )
So i thought its insufficient.

Answer must be A,

Please tell me where i am wrong.
Saurabh Goyal
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by fskilnik@GMATH » Sun Dec 12, 2010 5:05 am
gdk800 wrote: If 11 consecutive integers are listed from least to most, which is the mean of the 11 digits?
The first sentence of the question stem seems (at least) odd, because we do not know which 11 digits the question stem is refering too... "the 11 digits" is not properly defined, unless the 11 given integers are in fact 11 digits, what would be a particular case without any special "merit".

That´s why (I guess) beat_gmat_09 and myself considered the words "digits" in the question stem typos and that´s why I rephrased the question stem (substituting digits by integers... have a look at my post). IMHO the question makes more sense like that (because I do not believe the examiner/creator of the problem had in mind something like goyalsau imagined).

As far as the difference between digits and integers are concerned, (non-negative) digits are the integers 0,1,2, ...,9 only, as goyalsau mentioned. (I´ve never seen GMAT using "negative digits", by the way.)

Regards,
Fabio.

P.S.: I know goyalsau question was addressed to Shovan, but I guess (and hope) neither him nor Shovan will be offended with my reply (both are some of my BTG friends, by the way). Anyway, goyalsau´s and Shovan´s feedback (not to mention beat_gmat_09´s) are more than welcomed, as usual, and I apologyse for any inconvenience my answer may have brought. (I am also a big fan of Mahatma, goyalsau.)
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by shovan85 » Sun Dec 12, 2010 6:20 am
fskilnik wrote:
gdk800 wrote: If 11 consecutive integers are listed from least to most, which is the mean of the 11 digits?
The first sentence of the question stem seems (at least) odd, because we do not know which 11 digits the question stem is refering too... "the 11 digits" is not properly defined, unless the 11 given integers are in fact 11 digits, what would be a particular case without any special "merit".

That´s why (I guess) beat_gmat_09 and myself considered the words "digits" in the question stem typos and that´s why I rephrased the question stem (substituting digits by integers... have a look at my post). IMHO the question makes more sense like that (because I do not believe the examiner/creator of the problem had in mind something like goyalsau imagined).

As far as the difference between digits and integers are concerned, (non-negative) digits are the integers 0,1,2, ...,9 only, as goyalsau mentioned. (I´ve never seen GMAT using "negative digits", by the way.)

Regards,
Fabio.
Great Explanation!! I think goyalsau your doubt is cleared. Let us know if you have still any concerns.
fskilnik wrote: P.S.: I know goyalsau question was addressed to Shovan, but I guess (and hope) neither him nor Shovan will be offended with my reply (both are some of my BTG friends, by the way). Anyway, goyalsau´s and Shovan´s feedback (not to mention beat_gmat_09´s) are more than welcomed, as usual, and I apologyse for any inconvenience my answer may have brought. (I am also a big fan of Mahatma, goyalsau.)
Hi Fabio, Nice explanation. I (nor any of BTG members) will never mind if someone tries to help other. Thanks for taking charge and explaining the same.

I was busy with something so could not respond to Gotalsau's Question. Thanks for helping:)
Last edited by shovan85 on Sun Dec 12, 2010 8:44 am, edited 1 time in total.
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by fskilnik@GMATH » Sun Dec 12, 2010 8:15 am
shovan85 wrote: Hi Fabio, Nice explanation. I (nor any of BTG members) will never mind if someone tries to help other. Thanks for taking charge and explaining the same.
It´s really my pleasure, Shovan! That´s why I really like to be part of this community! Not only some of the problems posed here (goyalsau´s exercises among them, for sure) and some of the comments/solutions presented are really VERY instructive and well-thought, but also the friendly and hard-working people/students we are in contact to are also a good motivation for all of us involved in the GMAT!

As some of the readers have already noticed, I have already adapted some problems I found here to put as special challenges to my brazilian students. I will keep doing that, whenever the opportunity rises, and I will leave here the version I created, for the interested readers.

Thank YOU all!

Cheers,
Fabio.
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by gdk800 » Mon Dec 13, 2010 10:12 am
fskilnik wrote:
gdk800 wrote:A sincere Thanks to all members for such detailed explanations.
Beautiful (and unfortunately very rare) attitude, gdk800. I thank YOU for your kind gesture and I hope to be able to "find" you in other BTG posts.

Cheers,
Fabio.
Thanks for your kind gesture Fabio. I too look forward to have some healthy discussions on various aspects of GMAT preparation in other BTG posts.

Cheers,
GDK800
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