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# During a 10-week summer vacation, was the average

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Manager
Joined: 12 Mar 2009
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During a 10-week summer vacation, was the average  [#permalink]

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Updated on: 22 Oct 2013, 03:42
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Difficulty:

55% (hard)

Question Stats:

62% (02:11) correct 38% (02:02) wrong based on 251 sessions

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During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob read per week?

(1) Twice the average number of books that Carolyn read per week was greater than 5 less than twice the average number of books that Jacob read per week.

(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more than Jacob.

Originally posted by vaivish1723 on 22 Jan 2010, 23:27.
Last edited by Bunuel on 22 Oct 2013, 03:42, edited 2 times in total.
OA edited.
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Posts: 64099

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23 Jan 2010, 00:27
4
5
vaivish1723 wrote:
During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob
(1) Twice the average number of books that Carolyn read per week was greater
than 5 less than twice the average number of books that Jacob read per week.
(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more
than Jacob.

Oa is .

Let $$c$$ be the average # of books that Carolyn read per week;
Let $$j$$ be the average # of books that Jacob read per week;

Question is c>j?

(1) $$2c>2j-5$$, if $$c=10<j=11$$ --> $$2*10=20>2*11-5=17$$ BUT if $$c=10>j=5$$ --> $$2*10=20>2*5-5=5$$, two different answers. Not sufficient.

(2) Clearly insufficient. In the second half of the 10 week period, Carolyn read 3 books more than Jacob. So her average for the second half will be greater than Jacob's, but we know nothing about the first half.

(1)+(2) Combined two statements also not sufficient to determine whether c>j.

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Senior Manager
Joined: 06 Aug 2011
Posts: 309
Re: During a 10-week summer vacation, was the average  [#permalink]

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30 Oct 2012, 03:56
1
Bunuel..it is necessary to try with num..

i have done lyk this..take C as the average of carolyn and J as the avg of jacob..

2c>2j-5
2j-2c<5

(J-C)<5/2..

so we dont knw its positive or negative..if its positive that means J avg is more than carolyn and vice versa..

Statement 2 is insufficient..

both are still insufficient to give the ans ..

So e..
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Re: During a 10-week summer vacation, was the average  [#permalink]

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30 Oct 2012, 04:47
sanjoo wrote:
Bunuel..it is necessary to try with num..

i have done lyk this..take C as the average of carolyn and J as the avg of jacob..

2c>2j-5
2j-2c<5

(J-C)<5/2..

so we dont knw its positive or negative..if its positive that means J avg is more than carolyn and vice versa..

Statement 2 is insufficient..

both are still insufficient to give the ans ..

So e..

Yes, you can do this way as well.
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22 Oct 2013, 03:36
Bunuel wrote:
vaivish1723 wrote:
During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob
(1) Twice the average number of books that Carolyn read per week was greater
than 5 less than twice the average number of books that Jacob read per week.
(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more
than Jacob.

Oa is .

Let $$c$$ be the average # of books that Carolyn read per week;
Let $$j$$ be the average # of books that Jacob read per week;

Question is c>j?

(1) $$2c>2j-5$$, if $$c=10<j=11$$ --> $$2*10=20>2*11-5=17$$ BUT if $$c=10>j=5$$ --> $$2*10=20>2*5-5=5$$, two different answers. Not sufficient.

(2) Clearly insufficient. In the second half of the 10 week period, Carolyn read 3 books more than Jacob. So her average for the second half will be greater than Jacob's, but we know nothing about the first half.

(1)+(2) Combined two statements also not sufficient to determine whether c>j.

Bunuel, answer is seen in "edit" in the first post, kind of destroys the purpose of 'spoiler', could you please edit it out?
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Re: During a 10-week summer vacation, was the average  [#permalink]

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07 Apr 2016, 17:37
vaivish1723 wrote:
During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob read per week?

(1) Twice the average number of books that Carolyn read per week was greater than 5 less than twice the average number of books that Jacob read per week.

(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more than Jacob.

my approach to prove that A, B, and C are not sufficient.

1. suppose average for C is 4, and so is for J.
2x4=8
2x4 - 5 = 3
so A for C > than A for J.

but it can be that A read 5/week and J read 4/week
10>3 true again, but values differ.

A and D are out

2. clearly insufficient.
B is out.

1+2
40 total books
C in first 5 weeks read 40/2 -3 = 17 books, and 23 in last 5 weeks
J on the contrary read 20 books in the first 5 weeks, and 20 in the last 5 weeks.
satisfies both 1 and 2, but averages are equal

now.
suppose C=5books/week = 50 books total
J = 4books/week = 40 total
C during first 5 weeks read 46 books, and 4 in the last 5 weeks
J read 39 in the first 5 weeks, and 1 in the last 5 weeks.
again, satisfies both conditions from 1 and 2..
2 outcomes - C is out
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Re: During a 10-week summer vacation, was the average  [#permalink]

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20 Aug 2019, 04:44
Bunuel wrote:
vaivish1723 wrote:
During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob
(1) Twice the average number of books that Carolyn read per week was greater
than 5 less than twice the average number of books that Jacob read per week.
(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more
than Jacob.

Oa is .

Let $$c$$ be the average # of books that Carolyn read per week;
Let $$j$$ be the average # of books that Jacob read per week;

Question is c>j?

(1) $$2c>2j-5$$, if $$c=10<j=11$$ --> $$2*10=20>2*11-5=17$$ BUT if $$c=10>j=5$$ --> $$2*10=20>2*5-5=5$$, two different answers. Not sufficient.

(2) Clearly insufficient. In the second half of the 10 week period, Carolyn read 3 books more than Jacob. So her average for the second half will be greater than Jacob's, but we know nothing about the first half.

(1)+(2) Combined two statements also not sufficient to determine whether c>j.

Bunuel hello, how can we prove algebraically that both statements 1 and 2 are not sufficient? I tried by testing some numbers but this approach is way too time consuming.
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Re: During a 10-week summer vacation, was the average  [#permalink]

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20 Aug 2019, 05:48
vaivish1723 wrote:
During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob read per week?

(1) Twice the average number of books that Carolyn read per week was greater than 5 less than twice the average number of books that Jacob read per week.

(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more than Jacob.

Asked: During a 10-week summer vacation, was the average (arithmetic mean) number of books that Carolyn read per week greater than the average number of books that Jacob read per week?

Let the average (arithmetic mean) number of books that Carolyn read per week be c
Let the average number of books that Jacob read per week be j

(1) Twice the average number of books that Carolyn read per week was greater than 5 less than twice the average number of books that Jacob read per week.
2c > 2j-5 => c>j may or may not be true
NOT SUFFICIENT

(2) During the last 5 weeks of the vacation, Carolyn read a total of 3 books more than Jacob.
No data is provided for first 5 weeks
NOT SUFFICIENT

IMO E
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Re: During a 10-week summer vacation, was the average   [#permalink] 20 Aug 2019, 05:48