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 Your Progress
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million
[#permalink]
Show Tags
15 Jun 2018, 23:20
2
00:00
A
B
C
D
E
Difficulty:
25% (medium)
Question Stats:
76% (01:32) correct 24% (01:23) wrong based on 93 sessions
HideShow timer Statistics
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?
Re: Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million
[#permalink]
Show Tags
16 Jun 2018, 00:44
Bunuel wrote:
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 just need to find how many 75 million (1million = \(10^6\)) make 5.5 billion (1billion = \(10^9\))
\(\frac{5.5*10^9}{75*10^6}\) ~73 Ans B
_________________
If this post helped you learn something pls give kudos
Re: Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 millio
[#permalink]
Show Tags
05 Jul 2018, 07:35
Bunuel wrote:
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
Re: Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 millio
[#permalink]
Show Tags
05 Jul 2018, 10:04
Bunuel wrote:
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) _________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
WE: Supply Chain Management (Energy and Utilities)
Re: Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 millio
[#permalink]
Show Tags
05 Jul 2018, 11:47
Bunuel wrote:
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)
_________________
Regards,
PKN Rise above the storm, you will find the sunshine
Re: Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 millio
[#permalink]
Show Tags
06 Jul 2018, 01:39
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
Re: Out of 5.5 billion of bacteria grown for an experiment, 1 in 75 millio
[#permalink]
Show Tags
07 Jul 2018, 02:50
1
Bunuel wrote:
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?
Re: Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million
[#permalink]
Show Tags
07 Jul 2018, 06:42
NandishSS wrote:
Bunuel wrote:
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?
Re: Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million
[#permalink]
Show Tags
07 Jul 2018, 17:26
Bunuel wrote:
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
_________________
Scott Woodbury-Stewart Founder and CEO
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions
Re: Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million
[#permalink]
Show Tags
07 Jul 2018, 22:34
\(2^0\) 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.
_________________
Regards,
Vaibhav
Sky is the limit.800 is the limit.
~GMAC
gmatclubot
Re: Out of 5.5 billion bacteria grown for an experiment, 1 in 75 million &nbs
[#permalink]
07 Jul 2018, 22:34
close
76% (01:32) correct 
