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# A scientist is studying bacteria whose cell population

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A scientist is studying bacteria whose cell population [#permalink]

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16 Mar 2011, 06:55
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A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in
the population divides simultaneously. Four hours from now, immediately after the population doubles, the
scientist will destroy the entire sample. How many cells will the population contain when the bacteria is
destroyed?
(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.
(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the
sample.

Q))7. Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compounded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month? 333 383 402 433 483 Q. If ax + by + c = 0, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series? Q. If y = |x| – 100, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series? Q))Three fair coins are labeled with a zero (0) on one side and a one (1) on the other side. Jimmy flips all three coins at once and computes the sum of the numbers displayed. He does this over 1000 times, writing down the sums in a long list. What is the expected standard deviation of the sums on this list? (A) ½ (B) ¾ (C) Ö3/2 (D) Ö5/2 (E) 5/4 Q)) 7.51 8.22 7.86 8.36 8.09 7.83 8.30 8.01 7.73 8.25 7.96 8.53 A vending machine is designed to despense 8 ounces of coffee into a cup.After a test that recorded the number of ounces of coffee in each of 1,000 cups dispensed by the vending machine, the 12 listed amounts, in ounces, were selected from the data. If the 1,000 recorded amounts have a mean of 8.1 ounces and a standard standard deviation of 0.3 ounce, how many of the 12 listed amounts are within 1.5 standard deviations of the mean? Manager Joined: 03 Mar 2011 Posts: 90 Location: United States Schools: Erasmus (S) GMAT 1: 730 Q51 V37 GPA: 3.9 Followers: 2 Kudos [?]: 131 [2] , given: 12 Re: pls explain them frnds [#permalink] ### Show Tags 16 Mar 2011, 07:10 2 This post received KUDOS Last question. Since 8.1 is the mean of the sample and the standard deviation is 0.3, the interval which contained values within 1.5 standard deviations of the mean is (8.1-1.5*0.3;8.1+1.5*0.3)=(7.65;8.55) As we could see, the values 8.22, 7.86, 8.36, 8.09, 7.83, 8.30, 8.01, 7.73, 8.25, 7.96, 8.53 are within this interval, So, the answer is 11 _________________ If my post is useful for you not be ashamed to KUDO me! Let kudo each other! Manager Joined: 03 Mar 2011 Posts: 90 Location: United States Schools: Erasmus (S) GMAT 1: 730 Q51 V37 GPA: 3.9 Followers: 2 Kudos [?]: 131 [1] , given: 12 Re: pls explain them frnds [#permalink] ### Show Tags 16 Mar 2011, 07:15 1 This post received KUDOS 1 This post was BOOKMARKED ax + by + c = 0 y=-ax/b-c/b We know, that adding or substracting a real number to the random variable does not change its deviation. Multiplying by a real number, multiplies the standard deviation by the absolute of this number. So the answer is |a/b|*S If y = |x| – 100, then the answer could be different. If x>0 or x<0 for all observations, then the standard deviation does not change, while if there are both positive and negative values of x, then the standard deviation changes, and it could be calculated only knowing the actual values of x. _________________ If my post is useful for you not be ashamed to KUDO me! Let kudo each other! Math Forum Moderator Joined: 20 Dec 2010 Posts: 2021 Followers: 158 Kudos [?]: 1625 [3] , given: 376 Re: pls explain them frnds [#permalink] ### Show Tags 16 Mar 2011, 14:17 3 This post received KUDOS 2 This post was BOOKMARKED punyadeep wrote: Q)) A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed? (1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells. (2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample. Sol: Let the current time be 10:00AM. What will be the cell population at 2:00PM? 1. Let the cell population be x just before @ 10-2 = 8AM After doubling; population = 2x After that it has quadrupled; 2x*4 = 8x 10:00AM, the population is 8x And; 8x-2x=3750 6x = 3750 x = 625 We know the count was 625*2 = 1250 @8AM We know the count is 625*8 = 5000 @10AM We also know that the bacteria divided 2 times within last 2 hours; But, we don't know that constant interval yet after which bacteria divide. The cell could have divided at 8:50AM and 9:40AM with 50 minutes interval The cell could have divided at 8:41AM and 9:22AM with 41 minutes interval The cell could have divided at 8:59AM and 9:58AM with 59 minutes interval We wouldn't know exactly what will be the cell count after 4 hours unless we know that constant interval. Not Sufficient. (2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample. @1:00PM, the cell count = 40,000 Still, we don't know the interval. What if bacteria divides every minute or divides every hour. Not sufficient. Combing both; We know @10:00AM: count=1250 @1:00PM: count=40,000 Thus it must have divided 5 times in 5 hours 8:00AM: 1250 9:00AM: 2500 10:00AM:5000 11:00AM: 10000 12:00PM: 20000 1:00PM: 40000 @2:00PM count=80000 Sufficient. Ans: "C" punyadeep wrote: Q))7. Louie takes out a three-month loan of$1000. The lender charges him 10% interest per month compounded
monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the
nearest dollar, how much does Louie have to pay each month?
333 383 402 433 483

$$FV = PV(1+\frac{r}{n})^{(nt)}$$
FV = future value; Money to be returned after 3 months
PV = Present value; Money borrowed @ present = 1000
n = periods of application of the rate = 12 because it is compounded monthly. Thus there are total 12 periods in 1 year.
t = time in years = 3 month = 3/12 years.
r = 10% per month = 12*10 = 120% per year = 1.2

$$FV = 1000(1+\frac{1.2}{12})^{(12*\frac{3}{12})}$$
$$FV = 1000(1+0.1)^3$$
$$FV = 1000(1.1)^3 = 1000*1.331 = 1331$$

If 1331 has to be returned in 3 EMI(Equated monthly installments). Then every month, Louie must pay
1331/3 = $443 Well, the closest amount I see in the option is$10 away; \$433

Ans: "D"

Q. If ax + by + c = 0, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?

Q. If y = |x| – 100, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?

No idea what these convey and how to approach them.

Q))Three fair coins are labeled with a zero (0) on one side and a one (1) on the other side. Jimmy flips all three
coins at once and computes the sum of the numbers displayed. He does this over 1000 times, writing down
the sums in a long list. What is the expected standard deviation of the sums on this list?
(A) ½ (B) ¾ (C) Ö3/2 (D) Ö5/2 (E) 5/4

Sol:
With every toss of 3; we can get following possible results;

000 = 0
001 = 1
010 = 1
011 = 2
100 = 1
101 = 2
110 = 2
111 = 3

We will get the same results even if they are tossed 1000 times. If we just find the standard deviation for one subset, it will be the same for 1000 tosses.

Thus;
Sum of 0 - 1 time
Sum of 1 - 3 times
Sum of 2 - 3 times
Sum of 3 - 1 times

Thus;
$$Mean = \frac{1*0+3*1+3*2+1*3}{8} = \frac{12}{8} = 1.5$$

Standard deviation
$$\sigma^2 = \frac{1*(0-1.5)^2+3*(1.5-1)^2+3*(1.5-2)^2+1*(3-1.5)^2}{8}$$
$$\frac{(1.5)^2+3*(0.5)^2+3*(0.5)^2+(1.5)^2}{8}$$
$$\frac{2*(1.5)^2+6*(0.5)^2}{8}$$
$$\frac{4.5+1.5}{8} = \frac{6}{8} = \frac{3}{4}$$

$$\sigma^2 = \frac{3}{4}$$
$$\sigma = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$

Q)) 7.51 8.22 7.86 8.36 8.09 7.83 8.30 8.01 7.73 8.25 7.96 8.53
A vending machine is designed to despense 8 ounces of coffee into a cup.After a test that recorded the
number of ounces of coffee in each of 1,000 cups dispensed by the vending machine, the 12 listed amounts,
in ounces, were selected from the data. If the 1,000 recorded amounts have a mean of 8.1 ounces and a
standard standard deviation of 0.3 ounce, how many of the 12 listed amounts are within 1.5 standard
deviations of the mean?

Same as bagrettin's solution;
http://gmatclub.com/forum/pls-explain-them-frnds-110985.html#p893006
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Re: pls explain them frnds [#permalink]

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17 Mar 2011, 01:40
@ fluke....in the first problem..........bacteria doubles at constant intervals......so shdnt it be like this:
0800: x
doubles:2x
doubles:2*2x: 4x......at 1000
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Re: pls explain them frnds [#permalink]

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17 Mar 2011, 01:59
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@ fluke thnx so much for all the answers
regards
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Re: pls explain them frnds [#permalink]

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25 Mar 2011, 01:33
Q. If y = |x| – 100, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?

I guess answer should be 'S',as adding or subtracting from a series doesn't affect the SD.
Maybe Bunuel can help to confirm this answer.
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Re: pls explain them frnds [#permalink]

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25 Mar 2011, 04:36

In the above question x COULD be the set
x {1,2,3,4}
x {1,-2,3,-4}
x {1,2,3,-4} etc

y {-99,-98,-97,-96}
y {-99,-98,-97,-96}
y {-99,-98,-97,-96} etc

The SD of y is the same as the SD of {1,2,3,4} and NOT the SD of x. But the catch is I have oversimplified this problem. So the answer is definitive NO.
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Re: pls explain them frnds [#permalink]

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25 Mar 2011, 05:04
gmat1220 wrote:

In the above question x COULD be the set
x {1,2,3,4}
x {1,-2,3,-4}
x {1,2,3,-4} etc

y {-99,-98,-97,-96}
y {-99,-98,-97,-96}
y {-99,-98,-97,-96} etc

The SD of y is the same as the SD of {1,2,3,4} and NOT the SD of x. But the catch is I have oversimplified this problem. So the answer is definitive NO.

I was also apprehensive about the SD to be same as "S" because of |X|. Its now clear,but I think we can't have a unique answer in terms of "S".Isn't it?
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Re: pls explain them frnds [#permalink]

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25 Mar 2011, 14:05
I think the value of std. deviation for y series should be S^2 as we have multiplied with x.
-(a/b) *x1 -c/b
-(a/b) *x2 -c/b
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Re: pls explain them frnds [#permalink]

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25 Mar 2011, 14:15
If y = |x| – 100, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?
For this question, If the standard deviation of |x| series is 'S', the std deviation of Y series is also S.
However given standard deviation of x series is ‘S', we can't say std deviation of Y series is also S.
Because -ve values in X series is treated as positive value in Y, thereby impacting the std. deviation.
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Re: pls explain them frnds [#permalink]

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27 Mar 2011, 05:16
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@fluke, for the question 1, I think the answer is A, please opine.

At n intervals - P becomes 2P

Let t be the time now

From (1), t - 2 hrs back, population was x

x has become 4x

and 4x - x = 3750

=> 3x = 3750

=> x = 1250

So in 2 hrs population becomes 4 times

So it doubles every hour

1 hr from now - 2 * 5000

2hrs from now - 2 * 2 * 5000

Hence in four hours it will be (2)^4 * 5000 = 80000, sufficient

From (2)

t + 4, population will be doubled again

t + 3 hrs from now population will be doubled and = 40,000

So if the Population now is P

we don't know at what intervals the population doubles, it can double at 1/2 hr or 1 hr, so insufficient

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Re: pls explain them frnds [#permalink]

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27 Mar 2011, 05:26

End of 1 month - 1.1 * 1000

End of 2nd month - (1.1)^2 * 1000

End of 3rd month - (1.1)^3 * 1000

So monthly installment = (1.1)^3/3 * 1000

= 121 * 11/3

= 1331/3 = 443.66

The nearest answer is D - 433
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Re: pls explain them frnds [#permalink]

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27 Mar 2011, 05:47
@fluke, regarding the coins question, you've stated :

000 = 0

why can't be 000 = 1 ?
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Re: pls explain them frnds [#permalink]

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27 Mar 2011, 05:53
1 std dev above mean = 8.1 + 1.5 * 0.3

= 8.1 + 0.45 = 8.55

1 Std Dev below mean = 8.1 - 1.5 * 0.3

= 8.1 - 0.45 = 7.65

So out of the values - 7.51 8.22 7.86 8.36 8.09 7.83 8.30 8.01 7.73 8.25 7.96 8.53

11 values are within the range specified.
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Re: pls explain them frnds [#permalink]

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27 Mar 2011, 06:22
subhashghosh wrote:
@fluke, for the question 1, I think the answer is A, please opine.

At n intervals - P becomes 2P

Let t be the time now

From (1), t - 2 hrs back, population was x

x has become 4x

and 4x - x = 3750

=> 3x = 3750

=> x = 1250

So in 2 hrs population becomes 4 times

So it doubles every hour

1 hr from now - 2 * 5000

2hrs from now - 2 * 2 * 5000

Hence in four hours it will be (2)^4 * 5000 = 80000, sufficient

From (2)

t + 4, population will be doubled again

t + 3 hrs from now population will be doubled and = 40,000

So if the Population now is P

we don't know at what intervals the population doubles, it can double at 1/2 hr or 1 hr, so insufficient

See the attached image. Please raise concern, if any.
Attachments

Cell_Population.PNG [ 25.29 KiB | Viewed 8412 times ]

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Re: pls explain them frnds [#permalink]

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27 Mar 2011, 06:26
subhashghosh wrote:
@fluke, regarding the coins question, you've stated :

000 = 0

why can't be 000 = 1 ?

0+0+0 = 0
0+0+1 = 1
0+1+0 = 1
0+1+1 = 2
1+0+0 = 1
1+0+1 = 2
1+1+0 = 2
1+1+1 = 3
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Re: pls explain them frnds [#permalink]

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28 Mar 2011, 04:04
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Expert's post

Q. If ax + by + c = 0, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?

Q. If y = |x| – 100, and if the standard deviation of x series is ‘S’, what is the standard deviation of y series?

I received a PM asking for my input about these two questions. They don't make any sense, and you'll never see anything similar on the GMAT. It's anyone's guess what they mean by "y series" in these questions. A series in mathematics is a sum, and we don't take standard deviations of sums; we take standard deviations of sets.

If I were to guess what the question writer meant, I'd interpret the questions as bagrettin did above, and I agree with bagrettin's solutions. The first question seems to be trying to ask what happens to the standard deviation when you add a number to everything in a set, and when you multiply everything in a set by a constant. That is occasionally tested on the GMAT, and the important takeaways are the following:

* If you have a set S, and you *add* some constant k to every number in S, the standard deviation will *not* change, because none of the distances in your set will change;

* If you have a set S, and you *multiply* everything in that set by some constant m, then the standard deviation *will* change; it will be multiplied by m. Similarly, if you increase or decrease the value of everything in a set by x%, the standard deviation will increase or decrease by x%.

There is simply no way to answer the second question (the one which includes the absolute value). If you have the set {1, -1}, and you take the absolute value of everything in the set, the new standard deviation becomes zero. On the other hand, if you have the set {1, 3}, and take the absolute value of everything in the set, the standard deviation doesn't change. The question doesn't make any sense.

I don't know where these questions are from, but you should not waste your time studying them, or other questions from the same source. A few of the questions in the original post were from GMATPrep, but the two I've quoted in this post certainly are not official questions.
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Re: pls explain them frnds [#permalink]

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07 Apr 2011, 08:17
Q1

Q)) A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in
the population divides simultaneously. Four hours from now, immediately after the population doubles, the
scientist will destroy the entire sample. How many cells will the population contain when the bacteria is
destroyed?
(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.
(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the
sample

(1) Since the population quadrupled, we know that two intervals have passed in the last two hours (2^2 = 4, quadrupled), so we know the interval is an hour.

We also know that if 3750 is the increase after being quadrupled, then two intervals ago we had 1250 cells and now we have 5000.

Doubling four times for each hour leads to 5000 * 2^4 = 80000.

Sufficient

(2) The population doubles an hour before the cells are destroyed. We don't know how many more times it will double before the population is destroyed.

Insufficient

Ans: A

EDIT: Previoulsy thought it was D.
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