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B

5.5bn = 5500m

75m has 1 mutation
5500m has 5500/75 = 73.33
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Bunuel
Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 million has a particular mutation. Approximately how many of these bacteria have the mutation?


(A) 7
(B) 73
(C) 733
(D) 7,333
(E) 73,333



1 billion = 1000 million
5.5 billion = 5500 million
\(\frac{5500}{75}\) = 73.333

therefore answer is 73 option B
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Bunuel
Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 million has a particular mutation. Approximately how many of these bacteria have the mutation?


(A) 7
(B) 73
(C) 733
(D) 7,333
(E) 73,333

\(75\) million has \(1\) bacterial mutation
\(1\) million has \(\frac{1}{75}\) bacterial mutations
\(5500\) million has \(\frac{1}{75}*5500\) bacteria \(~\) \(73\) bacterial mutations , Answer must be (B)
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Bunuel
Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 million has a particular mutation. Approximately how many of these bacteria have the mutation?


(A) 7
(B) 73
(C) 733
(D) 7,333
(E) 73,333

Total no of bacteria having mutation=Total no of bacteria* rate of mutation=5.5 billion*\(\frac{1 nos}{75 million}\)=\(\frac{5.5*1000 million}{75million}\)=73(Approx.)
Ans. (B)
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Solution


Given:
    • Total number of bacteria grown for an experiment = 5.5 billion
    • 1 in every 75 million bacteria has a particular mutation

To find:
    • Approximately, how many of these 5.5 billion bacteria has the mutation

Approach and Working:
    • In 75 million bacteria, there is 1 bacteria with the mutation
    • In 5.5 billion bacteria grown for the experiment, number of bacteria which have the mutation = 1 * 5.5 billion/75 million = (5.5 * \(10^9\))/(75 * \(10^6\)) = 5500/75
    • The value is greater than 10 and less than 100

Hence, the correct answer is option B.

Answer: B
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Bunuel
Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 million has a particular mutation. Approximately how many of these bacteria have the mutation?


(A) 7
(B) 73
(C) 733
(D) 7,333
(E) 73,333

HI Bunuel ,

Is OA E? Can you please correct it? :-)
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NandishSS
Bunuel
Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 million has a particular mutation. Approximately how many of these bacteria have the mutation?


(A) 7
(B) 73
(C) 733
(D) 7,333
(E) 73,333

HI Bunuel ,

Is OA E? Can you please correct it? :-)

The OA is B. Edited. Thank you.
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Bunuel
Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million has a particular mutation. Approximately how many of the bacteria have the mutation?

(A) 7
(B) 73
(C) 733
(D) 7,333
(E) 73,333


We can create the proportion:

5,500,000,000/n = 75,000,000/1

n = 5,500,000,000/75,000,000 = 5500/75 = about 73

Answer: B
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\(2^0\) in 75 million
\(2^1\) in 150 million (number of bacteria with a mutation doubles every time the sample is doubled i.e 75-150-300-..)
\(2^2\) in 300 million
\(2^3\) in 600 million
\(2^4\) in 1200 million
\(2^5\) in 2400 million
\(2^6\) in 4800 million; \(2^6 = 64\)
\(2^7\) in 9600 million;\(2^7 = 128\)

An option between 64 and 128. Answer is B, 73.
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