Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 07 Jul 2015, 18:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m13 #9

Author Message
VP
Joined: 22 Nov 2007
Posts: 1104
Followers: 8

Kudos [?]: 216 [0], given: 0

A certain colony of bacteria doubles every morning while [#permalink]  17 Feb 2008, 05:42
A certain colony of bacteria doubles every morning while every evening 1000 bacteria die. What was the original size of the colony if the colony completely disappeared by the end of the third day?

 825
 850
 875
 910
 950
Manager
Joined: 02 Jan 2008
Posts: 159
Followers: 2

Kudos [?]: 71 [1] , given: 0

Re: Arithmetic [#permalink]  17 Feb 2008, 05:59
1
KUDOS
C
--
x = original size:
at the end of 1st day: 2x-1000
at the end of 2nd day: 2(2x-1000) -1000 = 4x-3000
at the end of 3rd day: 2(4x-3000)-1000
--
8x-7000=0; x= 875
Intern
Joined: 05 Jan 2009
Posts: 5
Location: Milan
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: m13 #9 [#permalink]  16 Jun 2009, 00:50
The first plausible number I found in reasonable time that causes 0 bacteria on Day 3 - evening is 800, but to have exactly 0 bacteria at the end of D3 the starting number for the colony must be 875:
Day 1 - morning: 1750
Day 1 - evening: 750
Day 2 - morning: 1500
Day 2 - evening: 500
Day 3 - morning: 1000
Day 3 - evening: 0
Manager
Joined: 14 Jun 2009
Posts: 96
Followers: 2

Kudos [?]: 7 [0], given: 9

Re: m13 #9 [#permalink]  16 Jun 2009, 01:08
I worked on this problem..backwards, as in i strtd from the end of day 3 and took 0, and worked their onwards.

Is tis right??

or culd it be 875, since 1750 is going to be the number on the morning of Day1.
Current Student
Joined: 13 Jan 2009
Posts: 372
Location: India
Followers: 19

Kudos [?]: 86 [0], given: 1

Re: m13 #9 [#permalink]  17 Jun 2009, 18:22

Day Morning Evening

I x x-1000
II 2x-2000 2x-3000
III 4x-6000 4x-7000

4x-7000=0

x=1750
SVP
Joined: 30 Apr 2008
Posts: 1889
Location: Oklahoma City
Schools: Hard Knocks
Followers: 35

Kudos [?]: 473 [0], given: 32

Re: m13 #9 [#permalink]  17 Jun 2009, 18:30
You can't do it this way because it's compounding on itself.

If you want to start backwards you have to do it like this (which I think was actually already posted in another method).

End of Day 3 = 0.....before losing 1000 = 1000, that morning before it doubled = 500

End of Day 2 = 500...before losing 1000 that night = 1500...that morning before doubled=750

End of Day 1 = 750...before losing 1000 that night = 1750...that morning before doubled = 875.

So 875 is the start, as said previously

My problem with this question though that it appears to asks for a certain number to start with, but in reality, there is a range which would still produce 0 at the end of day 3. if you start with 800 on day 1, you still end up with 0 at the end of day 3.

If you think of it as a range, then 875 > x > 750. If x has a value anywhere in that range, you will get 0 at the end of the 3rd day.

hemantsood wrote:

Day Morning Evening

I x x-1000
II 2x-2000 2x-3000
III 4x-6000 4x-7000

4x-7000=0

x=1750

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 18 Dec 2010
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Arithmetic [#permalink]  02 Jan 2011, 20:20
I think the best way to approach this question would be to backsolve. I picked D, then backsolved. That didnt work, so i checked B. B came out to be negative. Thus, the answer would be C.
Intern
Joined: 21 Sep 2010
Posts: 17
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Arithmetic [#permalink]  04 Jan 2011, 09:12
Day 0 Day 1 Day 2 Day3

Morning
Evening Must be 1000 before 1000 die and disappear

Solving backwards and filling in the table

Day 0 Day 1 Day 2 Day3

Morning 1750 1500 1000
Evening 875 750 500 Must be 1000 before 1000 die and disappear

Intern
Joined: 21 Sep 2010
Posts: 17
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Arithmetic [#permalink]  04 Jan 2011, 09:15
Day3
Evening: Must be 1000 before 1000 die and disappear.

Day3
Morning : 1000
Evening : Must be 1000 before 1000 die and disappear

Day2
Morning: 1500
Evening: 500

Day1
Morning: 1750
Evening: 750

Day 0: 1750/2 = 875

Senior Manager
Joined: 25 Nov 2011
Posts: 261
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 73 [0], given: 20

Re: Arithmetic [#permalink]  13 Feb 2012, 08:29
kamilshahid wrote:
I think the best way to approach this question would be to backsolve. I picked D, then backsolved. That didnt work, so i checked B. B came out to be negative. Thus, the answer would be C.

Yes, back solving should help us and I did in that way only. But there is no need to pick any answer and in fact that wastes time.

My back solving is like this:

third day night: input & output are 1000,0
third day morning: input & output are 500, 1000

second day night: input & output are 1500,500
second day morning: input & output are 750, 1500

first day night: input & output are 1750,750
first day morning: input & output are 875, 1750

But after seeing the first reply, I liked it very much and suggest to follow it because it is very methodical and probability of error is less.
_________________

-------------------------
-Aravind Chembeti

Senior Manager
Joined: 25 Nov 2011
Posts: 261
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 73 [0], given: 20

Re: Arithmetic [#permalink]  13 Feb 2012, 08:30
srp wrote:
C
--
x = original size:
at the end of 1st day: 2x-1000
at the end of 2nd day: 2(2x-1000) -1000 = 4x-3000
at the end of 3rd day: 2(4x-3000)-1000
--
8x-7000=0; x= 875

Thanks, I like this better than my method
_________________

-------------------------
-Aravind Chembeti

Re: Arithmetic   [#permalink] 13 Feb 2012, 08:30
Similar topics Replies Last post
Similar
Topics:
M13 #q 12 1 27 Nov 2008, 07:01
8 m13 q23 35 14 Sep 2008, 21:02
5 M13 Q30 13 03 Sep 2008, 18:02
13 M13 Q5 30 03 Sep 2008, 16:35
M13 System Issue 1 12 Aug 2008, 18:22
Display posts from previous: Sort by

# m13 #9

Moderators: WoundedTiger, Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.