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Seeking alternative answer method to OG Question

by crisprin117 » Mon Feb 24, 2014 7:52 am
OG Quant #90 Page 113 Problem Solving

The answer explanation provided is to set up a table and calculate each iteration of bacteria until it reaches 500,000. Can someone please provide me with a mathematical explanation. THANKS!

The population of a bacteria culture doubles every
2 minutes. Approximately how many minutes will
it take for the population to grow from 1,000 to
500,000 bacteria?
(A) 10
(8) 12
(C) 14
(D) 16
(E) 18
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by theCodeToGMAT » Mon Feb 24, 2014 8:49 am
1000 (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + ... ) = 500000

2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + .. = 500

(1)((2)^n - 1)/(2-1) = 500 [GP Formula]

2^n - 1 = 500
2^n = 501
n is nearly 9
So, total 9*2 = 18 minutes

[spoiler]{E}[/spoiler]?
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by Abhishek009 » Mon Feb 24, 2014 8:55 am
crisprin117 wrote:OG Quant #90 Page 113 Problem Solving

The answer explanation provided is to set up a table and calculate each iteration of bacteria until it reaches 500,000. Can someone please provide me with a mathematical explanation. THANKS!

The population of a bacteria culture doubles every
2 minutes. Approximately how many minutes will
it take for the population to grow from 1,000 to
500,000 bacteria?
(A) 10
(8) 12
(C) 14
(D) 16
(E) 18
2 ---> 4 ---> 8 ---> 16 .....

Sn = a + ar +ar^2 + ar^3 ...+ ar^(n-1)

Sn = a((r^n) - 1)/(r-1)

500000 = 1000((2^n) - 1)

=> 500 = 2^n - 1

=> 501 = 2^n

=> n ~ 9


Hence time required will be n *2 = > 18 minutes...
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by GMATGuruNY » Mon Feb 24, 2014 11:10 am
crisprin117 wrote:OG Quant #90 Page 113 Problem Solving

The answer explanation provided is to set up a table and calculate each iteration of bacteria until it reaches 500,000. Can someone please provide me with a mathematical explanation. THANKS!

The population of a bacteria culture doubles every
2 minutes. Approximately how many minutes will
it take for the population to grow from 1,000 to
500,000 bacteria?
(A) 10
(8) 12
(C) 14
(D) 16
(E) 18
We could use the formula for exponential growth:

Final Amount = Original Amount * (Multiplier)^(Number of changes)

In the problem above:
Final Amount = 500,000
Original Amount = 1000
Multiplier = 2 (since the population keeps doubling)
Number of changes = x.

Plugging these values into the formula, we get:
500,000 = 1000 * 2^x
2^x = 500
Since 2� = 512, x ≈ 9.

Since the number of changes = 9, and each change takes 2 minutes, the total number of minutes = 9*2 = 18.

The correct answer is E.

Similar problems:

https://www.beatthegmat.com/ps-exponent-t115416.html
https://www.beatthegmat.com/word-problem ... 14634.html
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by Brent@GMATPrepNow » Mon Feb 24, 2014 12:03 pm
crisprin117 wrote: The answer explanation provided is to set up a table and calculate each iteration of bacteria until it reaches 500,000. Can someone please provide me with a mathematical explanation. THANKS!

The population of a bacteria culture doubles every
2 minutes. Approximately how many minutes will
it take for the population to grow from 1,000 to
500,000 bacteria?
(A) 10
(8) 12
(C) 14
(D) 16
(E) 18
If you want the SUPER MATHEMATICAL solution (i.e., answer the question to several decimal places), you'll need to use logarithms to solve the equation 2^x = 500 [when we do so, we get: xlog2 = log500, which we can rewrite as x = log500/log2. When we evaluate, we get x = 8.97, so the number of minutes = 2(8.97)]


Of course, logs are NOT tested on the GMAT, so this point is somewhat moot :-)

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by Bill@VeritasPrep » Mon Feb 24, 2014 12:57 pm
Personally, I think I'd stick with the table method.
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by [email protected] » Mon Feb 24, 2014 5:52 pm
Hi crisprin117,

While it's understandable why you'd be interested in knowing the "math" behind a particular question, GMAT questions are written in such a way so that there is usually more than one approach to solving a given Quant (or Verbal) question. It's also worth noting that the Quant section of the GMAT is NOT a math Test; as such it takes more than just "math skills" to hit the highest scores in that section of the Test.

Your goal with any given question should be:
1) Get the question right, if possible
2) Do so in the fastest way possible

With this question, you can essentially "count doubles" and find the answer. This method is probably faster than any other approach that you might take. You can even speed up by ignoring the last 3 zeroes.

1k
becomes
2k
4k
8k
16k
32k
64k
128k
256k
512k

9 "doubles" at 2 minutes each = about 18 minutes

Final Answer: E

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by crisprin117 » Thu Mar 06, 2014 2:17 pm
Thanks everyone,all the feedback provides me with some perspective. I need to work on efficiency rather than accuracy.

I'm also rereading some sections of the Manhattan Gmat text and have an issue with a particular example in the Word Problems Guide.

Page 126:

Five identical pieces of wire are soldered together to form a longer wire, with the
pieces overlapping by 4 cm at each join. If the wire thus made is exactly 1 meter
long, how long is each of the identical pieces? (1 meter = 100 cm)

MG then states the total length of all the wires is 100 + 4(4) = 116 cm. I don't think takes into account that the wires are layered such that at each 4 cm layer occurs on two wires. I think it should be 100 + 4(4*2) = 132. Can anyone shed some light on this?
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by Matt@VeritasPrep » Thu Mar 06, 2014 5:19 pm
Crisprin, I tried it visually, and made x the length of each wire. Does this make sense? (Pic attached.)

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