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In a certain game, a large container is filled with red,

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In a certain game, a large container is filled with red, [#permalink] New post 08 Jul 2006, 19:13
"In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? "

You may recognize this question from the combination lesson, however since it is one of the practice questions, there are no solutions. Can you help, anyone? 8-)
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 [#permalink] New post 08 Jul 2006, 19:49
I broke down the Prime factorization of 147,000 to get ---> 7^2*5^3*3^1*2^3

7^2= 49 red beads.

Alternatively, notice the pattern that 147,000 is divisble by 49. 49 is equal to 7^2.
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 [#permalink] New post 08 Jul 2006, 19:56
GMATT73 wrote:
I broke down the Prime factorization of 147,000 to get ---> 7^2*5^3*3^1*2^3

7^2= 49 red beads.

Alternatively, notice the pattern that 147,000 is divisble by 49. 49 is equal to 7^2.


GMATT73, stupid question but can we just take prime factor of 147?
7x7x3
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 [#permalink] New post 08 Jul 2006, 20:15
First of all, thank you for the speedy reply.

My dilemma now lies in the fact that your answer is not in the list of possible solutions which leads me to think that a typo exists in the question or that perhaps we need to reevaluate the question.

Here's the list of choices.
(A) 5
(B) 4
(C) 3
(D) 2
(E) 0
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 [#permalink] New post 08 Jul 2006, 20:31
shampoo wrote:
GMATT73 wrote:
I broke down the Prime factorization of 147,000 to get ---> 7^2*5^3*3^1*2^3

7^2= 49 red beads.

Alternatively, notice the pattern that 147,000 is divisble by 49. 49 is equal to 7^2.


GMATT73, stupid question but can we just take prime factor of 147?
7x7x3


No, the one who is stupid here is me. You are absolutely right.

It`s the exponent we are searching for.

7^2= 49

2 red beads

(D)
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 [#permalink] New post 08 Jul 2006, 20:46
2 red beads were removed. Looks like we will be here for a while unless you can explain to me why in the world we are considering the exponent here?

The question asks, "how many red beads were removed?" and like you correctly put it, it must be 49 or the question is supposed to read 147 not 147,000 in which case 2 would be the perfect answer.

GMAT please shed some more light! :?:
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 [#permalink] New post 08 Jul 2006, 20:57
Yes Ryan, 2 red beads were removed. The question states that each red bead has a "point value" of 7 points, and the product of the point values equals 147,000 points.

7*7*5*5*5*3*2*2*2 = 147,000 (points, not beads)

7*7= 2 red beads.

I hope that clarified things.

(probably would have misread the question on G-Day and chose 49 as a trap answer :oops: )
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 [#permalink] New post 08 Jul 2006, 21:05
You are definately the man!

Thanks for sparking my intellect on this one. So its really all about Prime Factorization this one, where is the testing on combination theory in this one? :-D


Thanks again.

But, if you're not too busy Mr. CEO. Here's another one of my hurdles.

The probability that it will rain in NYC on any given day in July is 30%. what is the probability that it will rain on exactly 3 days from July 5 to July 10 ? There are 4 Fashion magazines and 4 Car magazines. Four magazines are drawn at random, what is the probability that all fashion magazines will be drawn?
a. 1
b. ?
c. 1/3
d. 1/8
e. 1/70
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 [#permalink] New post 08 Jul 2006, 21:51
These are two different problems, please seperate the two with relevant answers. Thanx!
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 [#permalink] New post 08 Jul 2006, 22:00
Rayn wrote:
But, if you're not too busy Mr. CEO. Here's another one of my hurdles.

The probability that it will rain in NYC on any given day in July is 30%. what is the probability that it will rain on exactly 3 days from July 5 to July 10 ?
There are 4 Fashion magazines and 4 Car magazines. Four magazines are drawn at random, what is the probability that all fashion magazines will be drawn?
a. 1
b. ?
c. 1/3
d. 1/8
e. 1/70


Allow me to answer.

I think there are two question in just one. :-D :-D :-D
Q1:
Prob of rain on any give day = 3/10
Prob of NO rain on any give day = 7/10
July 5 to July 10 (Both Inclusive) = 6 days
Prob that it rains on three days and not on other 3 days = (6!/(3!*3!)) * (3/10)^3 * (7/10)^3

= 20 * (27/1000) * (343/1000)

Q2:

Total cases of drawing 4 magzines out of 8 = 8C4 = 70
Cases when all 4 will be fashion magzines = 1
Prob = 1/70
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 [#permalink] New post 08 Jul 2006, 22:18
PS Dashiya, your math seems to be correct if the magazines are not replaced. Or dependent probability.

4/8*3/7*2/6*1/5 ---> 24/1680 ---> 1/70

If the magazines are replaced, then the answer is 1/2 (which might be the ? answer choice ;)
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 [#permalink] New post 09 Jul 2006, 06:30
Quote:
Allow me to answer.

I think there are two question in just one. :-D :-D :-D
Q1:
Prob of rain on any give day = 3/10
Prob of NO rain on any give day = 7/10
July 5 to July 10 (Both Inclusive) = 6 days
Prob that it rains on three days and not on other 3 days = (6!/(3!*3!)) * (3/10)^3 * (7/10)^3
= 20 * (27/1000) * (343/1000)


Thanks for the explanation. I like your approach but can you quickly explain to me why we are not multiplying the "total cases of rain on three of the six days" by 30% alone.

Why must we include the 70% Is it because the question says "exactly" or is there another hint in the question. And how would the question need to be framed in order to consider only the 30%?

Sorry for being long-winded on this...
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 [#permalink] New post 09 Jul 2006, 13:50
Rayn wrote:
Quote:
Allow me to answer.

I think there are two question in just one. :-D :-D :-D
Q1:
Prob of rain on any give day = 3/10
Prob of NO rain on any give day = 7/10
July 5 to July 10 (Both Inclusive) = 6 days
Prob that it rains on three days and not on other 3 days = (6!/(3!*3!)) * (3/10)^3 * (7/10)^3
= 20 * (27/1000) * (343/1000)


Thanks for the explanation. I like your approach but can you quickly explain to me why we are not multiplying the "total cases of rain on three of the six days" by 30% alone.

Why must we include the 70% Is it because the question says "exactly" or is there another hint in the question. And how would the question need to be framed in order to consider only the 30%?

Sorry for being long-winded on this...


Out of six days we exactly three days of rain and three days of no rain. You cannot take just three days only because you have to make sure than on the other three days it did not rain.

For example: First 3 days rain. Then to calculate the correct probability you should make sure that it don't rain other wise it will not be exactly 3 days.

Total cases of three days rain and no rain of other three are

6!/(3!*3!) like RRRNNN, RNRNRN etc..... but in all cases there are three rain terms and three non-rain terms. So for one combination we have (3/10)^3 * (7/10)^3 and for all we have

6!/(3!*3!) * (3/10)^3 * (7/10)^3

Hope this helps.
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 [#permalink] New post 09 Jul 2006, 14:39
Thanks... :lol:

Makes perfect sense.
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 [#permalink] New post 10 Jul 2006, 16:28
Question Source: Challenge 25

Is plugging in the best on this type or is there an easier way to manipulate the variables? I dislike the time I waste in trying to find relationships.

Gold depreciated at a rate of X% per year between 2000 and 2005. If the gold cost S dollars in 2001 and T dollars in 2003, what was the price of gold in 2002?



a) T*S/2
b) T*sqrt(T/S)
c) T*sqrt( S )
d) T*S/sqrt( T )
e) sqrt(S*T
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 [#permalink] New post 10 Jul 2006, 16:34
e) sqrt(S*T)

By the way, the best way to get a hit on your questions with a wide ranging explanation would be to post it as a different thread...
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 [#permalink] New post 10 Jul 2006, 16:47
Thanks for the Blog Tip...

Can you give me a brief explanation of how you arrived at your answer?
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P&C keep it simple [#permalink] New post 21 Nov 2006, 09:27
====================
Rayn wrote:

But, if you're not too busy Mr. CEO. Here's another one of my hurdles.

The probability that it will rain in NYC on any given day in July is 30%. what is the probability that it will rain on exactly 3 days from July 5 to July 10 ? There are 4 Fashion magazines and 4 Car magazines. Four magazines are drawn at random, what is the probability that all fashion magazines will be drawn?
a. 1
b. ?
c. 1/3
d. 1/8
e. 1/70
=========================

BLISSFUL replied: Check this expln below:
1) July 5 to july 10 -> 5,6,7,8,9,10: 6 days
Out of 6 days, it should rain EXACTLY on 3 days .
-> It can be in 6C3 ways --(A)
Now, Prob of Rain on any day = 0.3
P of NO Rain on any day = 0.7

P of Rain on exactly 3 days = (0.3)^3 * (0.7)^(6-3) -- (B)

But it can rain 3 days on any different combinations of 3 days out of 6 days.

Therefore, Combinging A) & B): The right answer is: 6C3 * (0.3)^3 * (0.7)^3

******

If its for 7 total days, it will be like
7C3 * (0.3)^3 * (0.7)^(7-3)

*******

You can clearly see it resembles the formula:
nCr * p^r * (1-p)^(n-r)

= Number of ways in which a thing can happen exactly r times( ouf of n total times) when p is the probability if that thing happening.


If you are CRAZY like me: you can expand this problem like..

In a leap year February in DC beltway, if Traffic Jam is likely 3/4 of the time on any given day, whats the Probability that Traffic JAm happens on exactly 5 days of this month ?

Easy to answer: 29C5 * (3/4)^5 * (1/4)^(29-5)

Got it ?..Now Kick this Problem in its a..
--------------------------------------

2) 4 Fashion Mags , 4 Car Mags
when 4 mags chosen at random,
P of all are Fasion Mgs =

No of fav outcomes/ Total No of outcomes =

4C4 / (4+4)C4= 1/8C4 = 4*3*2*1 / 8*7*6*5 = 1/70

****
3 Fashion Mags, 4 Car Mags

P of all fashion mags in a random single pick of 4 Mags =
fav/total = 3C3/(3+4)C3= 3C3/7C3

And so on..

Hope this helps !
***
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 [#permalink] New post 21 Nov 2006, 14:23
Someone wrote this:
--------------------------------------------------------------------------------
PS Dashiya, your math seems to be correct if the magazines are not replaced. Or dependent probability.

4/8*3/7*2/6*1/5 ---> 24/1680 ---> 1/70

If the magazines are replaced, then the answer is 1/2 (which might be the ? answer choice
-------------------------------------------------------------------------------
Am I crazy or is that wrong? If they were replaced and you drew 4 times you would have 1/2 odds of drawing a fashion magazine each time, meaning that the prob for all for draws would be 1/2 *1/2 *1/2 *1/2 = 1/16?
Am I missing something or is this correct?
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 [#permalink] New post 22 Nov 2006, 07:29
Rayn wrote:
Thanks for the Blog Tip...

Can you give me a brief explanation of how you arrived at your answer?


Allow me to explain:

Gold depreciated at a rate of X% per year between 2000 and 2005. If the gold cost S dollars in 2001 and T dollars in 2003, what was the price of gold in 2002?

Let the price in 2000 be Y

2001: Y - XY/100 = S
2002: (Y - XY/100) - (Y- XY/100)(X) = (Y- XY/100)(1-X)
2003: (Y- XY/100)(1-X) - (Y- XY/100)(1-X)(X) = (Y- XY/100)(1-X)(1-X) = T

So to sum it up, from 2003, (1-X)^2 = T/(Y- XY/100) = T/S.
So (1-X)=sqrt(T/S).

Price of 2002 = (Y - XY/100)*(1-X)
(subtitue the bold equations above in this one)
= S * sqrt(T/S) = sqrt(T*S).

Hope it's clear enough :roll:
  [#permalink] 22 Nov 2006, 07:29
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