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Bacteria

Expert replies
by GmatKiss » Mon Oct 31, 2011 11:52 pm
Data for a certain biology experiment are given in
the table above. If the amount of bacteria present
increased by the same factor during each of the two
3-hour periods shown, how many grams of bacteria
were present at 4:00 P.M. ?

(A) 12.0
(B) 12.1
(C) 12.2
(D) 12.3
(E) 12.4
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Last edited by GmatKiss on Tue Nov 01, 2011 12:08 am, edited 1 time in total.
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Source: — Problem Solving |

by smackmartine » Tue Nov 01, 2011 12:01 am
GmatKiss wrote:Data for a certain biology experiment are given in
the table above. If the amount of bacteria present
increased by the same factor during each of the two
3-hour periods shown, how many grams of bacteria
were present at 4:00 P.M. ?

(A) 12.0
(B) 12.1
(C) 12.2
(D) 12.3
(E) 12.4
I don't see any table.
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by GmatKiss » Tue Nov 01, 2011 12:09 am
smackmartine wrote:
GmatKiss wrote:Data for a certain biology experiment are given in
the table above. If the amount of bacteria present
increased by the same factor during each of the two
3-hour periods shown, how many grams of bacteria
were present at 4:00 P.M. ?

(A) 12.0
(B) 12.1
(C) 12.2
(D) 12.3
(E) 12.4
I don't see any table.
Sorry !! Attached it now..
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by shankar.ashwin » Tue Nov 01, 2011 12:13 am
10 to 12 (20% growth)
12 to 14.4 (20% growth)
A Think of it as a compound interest problem.
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by GMATGuruNY » Tue Nov 01, 2011 3:07 am
GmatKiss wrote:Data for a certain biology experiment are given in
the table above. If the amount of bacteria present
increased by the same factor during each of the two
3-hour periods shown, how many grams of bacteria
were present at 4:00 P.M. ?

(A) 12.0
(B) 12.1
(C) 12.2
(D) 12.3
(E) 12.4
We can plug in the answers, which represent the number of bacteria at 4pm.
We should start with A, the only integer value.

Answer choice A: 12
7pm/4pm = 14.4/12 = 144/120 = 6/5.
4pm/1pm = 12/10 = 6/5.
Success!

The correct answer is A.

We could also set up the following proportion:
7pm/4pm = 4pm/1pm
14.4/x = x/10
x² = 144
x = 12.
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by LalaB » Tue Nov 01, 2011 9:39 pm
if u need a formula, please see below (it is from mgmat guide book)-


(Original amount)*(factor)^t/s=future amount or

10*factor^(7-1)/3=14.4

factor=1.20

where (7-1) is time difference between 7.00 p.m. and 1.00 p.m. ; 3 expresses that every 3 hours the amount changes (4.00p.m.-1.00 p.m.=3 and 7.00 p.m.-4.00 p.m.=3).


so X = 10 *1.20=12

A is the answ
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