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13 Sep 2007, 09:24
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A
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85% (01:11) correct
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A certain culture of bacteria quadruples every hour. If a container with these bacteria was half full at 10:00 a.m., at what time was it one-eighth full?
(A) 9:00 a.m.
(B) 7:00 a.m.
(C) 6:00 a.m.
(D) 4:00 a.m.
(E) 2:00 a.m.
(A) 9:00 a.m.
(B) 7:00 a.m.
(C) 6:00 a.m.
(D) 4:00 a.m.
(E) 2:00 a.m.
25 Feb 2011, 02:37
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beatgmat
You certainly don't need that much thinking on this one.
To go from one-eighth (1/8) full to half (1/2) full culture of bacteria should quadruple: 1/8*4=1/2, as it quadruples every hour then container was one-eighth full at 10:00 a.m -1 hour = 9:00 a.m.
Answer: A.
25 Feb 2011, 00:56
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Sol:
Method1:
*******
We know the bacteria quadruples every hour: i.e. if at x hrs, the count is 100, at time x+1 hrs, the count will be 400, x+2: 1600 and so on;
Reverse is also true; if at x hours the count is 1600, x-1 hrs, it would have been 1600/4=400; x-2:400/4=100 and so on...
Same way:
@ 10:00 am: The count was 1/2(Half)
@ 9:00 am: The count would have been (1/2)/4 = 1/8
Thus the bacteria count was 1/8 @ 9:00 AM.
******************************************
Method2:
*******
A more mathematical and little cumbersome approach would be the Geometric Progression:
To find the nth term in a GP;
\(A_n=a*r^{(n-1)}\)
We know
\(a=1/2\); This is the first term and represents the count of bacteria at time 10:00AM
\(r=1/4\)
\(A_n=1/8\); This is the \(n^{th}\) term and represents the count of bacteria (n-1) hours before 10
\(n=?\)
\(A_n=a*r^{(n-1)}\)
\(\frac{1}{8}=\frac{1}{2}*(\frac{1}{4})^{(n-1)}\)
\(\frac{1}{4}=\frac{1}{4^{(n-1)}}\)
Equating the denominator:
n-1=1
n=2;
So; we know the bacteria reached 1/8 @ n=2; Time = 2-1=1 hour before 10:00AM=9:00AM
The latter method may be little subtle but is a generalized way to deal with such questions. The former should be your first choice to solve such questions and latter may be used only as a fall back method.
Ans: "A"
Method1:
*******
We know the bacteria quadruples every hour: i.e. if at x hrs, the count is 100, at time x+1 hrs, the count will be 400, x+2: 1600 and so on;
Reverse is also true; if at x hours the count is 1600, x-1 hrs, it would have been 1600/4=400; x-2:400/4=100 and so on...
Same way:
@ 10:00 am: The count was 1/2(Half)
@ 9:00 am: The count would have been (1/2)/4 = 1/8
Thus the bacteria count was 1/8 @ 9:00 AM.
******************************************
Method2:
*******
A more mathematical and little cumbersome approach would be the Geometric Progression:
To find the nth term in a GP;
\(A_n=a*r^{(n-1)}\)
We know
\(a=1/2\); This is the first term and represents the count of bacteria at time 10:00AM
\(r=1/4\)
\(A_n=1/8\); This is the \(n^{th}\) term and represents the count of bacteria (n-1) hours before 10
\(n=?\)
\(A_n=a*r^{(n-1)}\)
\(\frac{1}{8}=\frac{1}{2}*(\frac{1}{4})^{(n-1)}\)
\(\frac{1}{4}=\frac{1}{4^{(n-1)}}\)
Equating the denominator:
n-1=1
n=2;
So; we know the bacteria reached 1/8 @ n=2; Time = 2-1=1 hour before 10:00AM=9:00AM
The latter method may be little subtle but is a generalized way to deal with such questions. The former should be your first choice to solve such questions and latter may be used only as a fall back method.
Ans: "A"
