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79 laboratories raise the bacterium, the laboratory have 48

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79 laboratories raise the bacterium, the laboratory have 48

by Max@Math Revolution » Tue Mar 08, 2016 4:09 pm
79 laboratories raise the bacterium, the laboratory have 48 culture dishes on average, which has about 25,075 bacteria each. How many bacteria are there approximately?

A. 10^7
B. 10^8
C. 10^9
D. 10^10
E. 10^11


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by Brent@GMATPrepNow » Tue Mar 08, 2016 8:00 pm
Max@Math Revolution wrote:79 laboratories raise the bacterium, the laboratory have 48 culture dishes on average, which has about 25,075 bacteria each. How many bacteria are there approximately?

A. 10^7
B. 10^8
C. 10^9
D. 10^10
E. 10^11
So, we want to find the approximate value of (25,075)(48)(79)
Here's how I would do this.

First deal with (25,075)(48)
25,075 is very close to 25,000, so let's go with that.
Now, (25,000)(4) = 100,000, so (25,000)(40) = 1,000,000, which means (25,000)(48) is a BIT BIGGER than 1,000,000, which I will denote as 1,000,000+

So, (25,075)(48)(79) ≈ (1,000,000+)(79)
≈ 80,000,000+
≈ 100,000,000
≈ 10^8 [I know this because 100,000,00 has 8 zeros]

Answer: B

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by Max@Math Revolution » Sat Mar 12, 2016 5:29 am
79 laboratories raise the bacterium, the laboratory have 48 culture dishes on average, which has about 25,075 bacteria each. How many bacteria are there approximately?

A. 10^7
B. 10^8
C. 10^9
D. 10^10
E. 10^11


-> Due to approximately, 79=80, 48=50, 25,075=25,000 are derived, which makes (79)(48)(25,075)=(80)(50)(25,000)=10^8. The answer is B.
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