Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 May 2013, 10:33

MGMAT Challenge: Decimals on Deposit

Author Message
TAGS:
SVP
Affiliations: HEC
Joined: 28 Sep 2009
Posts: 1525
Concentration: Economics, Finance
GMAT 1: 730 Q48 V44
Followers: 64

Kudos [?]: 404 [0], given: 390

MGMAT Challenge: Decimals on Deposit [#permalink]  18 Jul 2010, 10:32
00:00

Question Stats:

54% (21:46) correct 45% (16:19) wrong based on 0 sessions
Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30
[Reveal] Spoiler: OA

_________________
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11610
Followers: 1800

Kudos [?]: 9593 [6] , given: 828

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  18 Jul 2010, 11:42
6
KUDOS
bmillan01 wrote:
Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30

Theory:
Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only b (denominator) is of the form 2^n5^m, where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028, as 250 (denominator) equals to 2*5^3. Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \frac{x}{2^n5^m}, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.)

BACK TOT THE ORIGINAL QUESTION:

Question: does average=\frac{p}{d} has less than 5 decimal places? Where p=prime>100 and d is the chosen day.

If the chosen day, d, is NOT of a type 2^n5^m (where n and m are nonnegative integers) then average=\frac{p}{d} will not be a terminating decimal and thus will have more than 5 decimal places.

How many such days are there of a type 2^n5^m: 1, 2, 4, 5, 8, 10, 16, 20, 25 (1=2^0*5^0, 2=2^2, 4=2^2, 5, 8=2^3, 10=2*5, 16=2^4, 20=2^2*5, 25=5^2), total of 9 such days (1st of June, 4th of June, ...).

Now, does p divided by any of these d's have fewer than 5 decimal places? Yes, as \frac{p}{d}*10,000=integer for any such d (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: P=\frac{9}{30}=\frac{3}{10} .

Hope it's clear.
_________________
Intern
Joined: 09 Dec 2008
Posts: 29
Location: Vietnam
Schools: Somewhere
Followers: 0

Kudos [?]: 0 [0], given: 2

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  20 Jul 2010, 00:37
Bunuel wrote:
bmillan01 wrote:
Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30

Theory:
Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only b (denominator) is of the form 2^n5^m, where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028, as 250 (denominator) equals to 2*5^2. Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \frac{x}{2^n5^m}, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.)

BACK TOT THE ORIGINAL QUESTION:

Question: does average=\frac{p}{d} has less than 5 decimal places? Where p=prime>100 and d is the chosen day.

If the chosen day, d, is NOT of a type 2^n5^m (where n and m are nonnegative integers) then average=\frac{p}{d} will not be a terminating decimal and thus will have more than 5 decimal places.

How many such days are there of a type 2^n5^m: 1, 2, 4, 5, 8, 10, 16, 20, 25 (1=2^0*5^0, 2=2^2, 4=2^2, 5, 8=2^3, 10=2*5, 16=2^4, 20=2^2*5, 25=5^2), total of 9 such days (1st of June, 4th of June, ...).

Now, does p divided by any of these d's have fewer than 5 decimal places? Yes, as \frac{p}{d}*10,000=integer for any such d (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: P=\frac{9}{30}=\frac{3}{10} .

Hope it's clear.

Great explanation!!! I learn a lot from this
Manager
Joined: 11 Jul 2010
Posts: 233
Followers: 1

Kudos [?]: 30 [0], given: 20

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  21 Jul 2010, 11:56
bunuel... this is great. thanks...

the way you piece it together is sometimes scary... just curious - what was your gmat score
Manager
Joined: 11 Jul 2010
Posts: 233
Followers: 1

Kudos [?]: 30 [0], given: 20

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  21 Jul 2010, 12:21
Actually wanted to seek 1 clarification to better understand this:

p/d *10,000=integer

p is a prime integer greater than 100

d can be one of the 9 numbers

To test the tendency to leave a certain desired number of decimal places, upon division of p by d why is it ok to multiply p by a common multiple (10k here) of the 9 numbers in the denominator?

very crude general example (which I am hoping is an analogy):
37 divided by 7 leaves R of 2 and certain decimal places; 37 * 14 divided by 7 leaves no remainder ---> how can the later scenario be used to test whether a certain desired number of decimal places are left by the first scenario...
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11610
Followers: 1800

Kudos [?]: 9593 [0], given: 828

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  21 Jul 2010, 13:13
gmat1011 wrote:
Actually wanted to seek 1 clarification to better understand this:

p/d *10,000=integer

p is a prime integer greater than 100

d can be one of the 9 numbers

To test the tendency to leave a certain desired number of decimal places, upon division of p by d why is it ok to multiply p by a common multiple (10k here) of the 9 numbers in the denominator?

very crude general example (which I am hoping is an analogy):
37 divided by 7 leaves R of 2 and certain decimal places; 37 * 14 divided by 7 leaves no remainder ---> how can the later scenario be used to test whether a certain desired number of decimal places are left by the first scenario...

Your example is not good as \frac{37}{7} will be recurring decimal (will have infinite number of decimal places).

How many decimal places will terminating decimal \frac{p}{2^n*5^m} have? (p is prime number)

Consider following examples:
0.2 has 1 decimal place --> 1.2*10=12=integer (multiplying by 10 with 1 zero);
0.25 has 2 decimal places --> 1.25*10^2=125=integer (multiplying by 100 with 2 zeros);
0.257 has 3 decimal places --> 1.257*10^3=1257=integer (multiplying by 100 with 3 zeros);
0.2571 has 4 decimal places --> 1.2571*10^4=12571=integer (multiplying by 100 with 4 zeros);
...

So, terminating decimal, \frac{p}{2^n*5^m} (where p is prime number), will have k decimal places, where k is the least value in 10^k for which \frac{p}{2^n*5^m}*10^k=integer.

In our original question least value of k for which \frac{p}{d}*10^k=integer, for all 9 d's, is 4 or when 10^k=10,000 (k=4 is needed when d=16).

Hope it's clear.
_________________
Manager
Joined: 11 Jul 2010
Posts: 233
Followers: 1

Kudos [?]: 30 [0], given: 20

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  22 Jul 2010, 04:22
Great - many thanks.
Manager
Joined: 22 Jun 2010
Posts: 58
Followers: 1

Kudos [?]: 3 [0], given: 10

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  23 Jul 2010, 05:03
awesome explanation!
Manager
Joined: 06 Jul 2010
Posts: 113
Followers: 1

Kudos [?]: 2 [0], given: 9

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  23 Jul 2010, 06:36
Brilliant explanation. Never knew that dividing by (2^m)(5^n) gives a terminating decimal .. Thanks for the info
Manager
Joined: 24 Jan 2010
Posts: 168
Location: India
Schools: ISB
Followers: 2

Kudos [?]: 10 [0], given: 14

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  23 Jul 2010, 09:09
Thanks for the explanation Bunuel
_________________

_________________
If you like my post, consider giving me a kudos. THANKS!

Intern
Joined: 15 May 2010
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  23 Jul 2010, 20:54
Thanks for the explanation....
Intern
Joined: 27 Aug 2010
Posts: 24
Followers: 0

Kudos [?]: 3 [0], given: 2

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  19 Sep 2010, 05:40
This is definitely a 800+ question
Good explanation and thnx for the theory, very useful!
Manager
Joined: 06 Aug 2010
Posts: 229
Location: Boston
Followers: 2

Kudos [?]: 50 [0], given: 5

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  06 Oct 2010, 17:18
One question about this problem. The problem doesn't ask if the decimal will be terminating, but rather if the decimal will have less than 5 places. Your solution checks for termination, but how do you check for the number of decimal places? Couldn't some of the possibilities result in termination with more than 5 decimal places?
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11610
Followers: 1800

Kudos [?]: 9593 [1] , given: 828

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  07 Oct 2010, 02:10
1
KUDOS
TehJay wrote:
One question about this problem. The problem doesn't ask if the decimal will be terminating, but rather if the decimal will have less than 5 places. Your solution checks for termination, but how do you check for the number of decimal places? Couldn't some of the possibilities result in termination with more than 5 decimal places?

You should read the last part:
"Now, does p divided by any of these d's have fewer than 5 decimal places? Yes, as \frac{p}{d}*10,000=integer for any such d (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: P=\frac{9}{30}=\frac{3}{10} .

This issue is also discussed in the posts following the one with solution.
_________________
Manager
Joined: 03 Aug 2010
Posts: 111
GMAT Date: 08-08-2011
Followers: 1

Kudos [?]: 12 [0], given: 63

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  08 Oct 2010, 15:39
Bunuel, is it important to know that the numerator in this problem is greater than 100?
Senior Manager
Joined: 30 Nov 2010
Posts: 267
Schools: UC Berkley, UCLA
Followers: 1

Kudos [?]: 48 [0], given: 66

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  28 Jan 2011, 08:29
Bunuel wrote:
bmillan01 wrote:
Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30

Theory:
Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only b (denominator) is of the form 2^n5^m, where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028, as 250 (denominator) equals to 2*5^2. Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \frac{x}{2^n5^m}, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.)

BACK TOT THE ORIGINAL QUESTION:

Question: does average=\frac{p}{d} has less than 5 decimal places? Where p=prime>100 and d is the chosen day.

If the chosen day, d, is NOT of a type 2^n5^m (where n and m are nonnegative integers) then average=\frac{p}{d} will not be a terminating decimal and thus will have more than 5 decimal places.

How many such days are there of a type 2^n5^m: 1, 2, 4, 5, 8, 10, 16, 20, 25 (1=2^0*5^0, 2=2^2, 4=2^2, 5, 8=2^3, 10=2*5, 16=2^4, 20=2^2*5, 25=5^2), total of 9 such days (1st of June, 4th of June, ...).

Now, does p divided by any of these d's have fewer than 5 decimal places? Yes, as \frac{p}{d}*10,000=integer for any such d (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: P=\frac{9}{30}=\frac{3}{10} .

Hope it's clear.

So you're saying that by controlling the terminating decimal using d=2^m*5^n and, and you can make it an integer by multiplying it by 10000 (if the number must be five decimal places to the left.
Therefore, you can choose of days 1, 2, 4, 5, 8, 10, 16, 20, or 25. (making that 9 days). - But they're not prime??? What am I missing here... I'm the slow one of the lot please help me out
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11610
Followers: 1800

Kudos [?]: 9593 [0], given: 828

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  28 Jan 2011, 08:44
mariyea wrote:
Bunuel wrote:
bmillan01 wrote:
Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30

Theory:
Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only b (denominator) is of the form 2^n5^m, where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028, as 250 (denominator) equals to 2*5^2. Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \frac{x}{2^n5^m}, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.)

BACK TOT THE ORIGINAL QUESTION:

Question: does average=\frac{p}{d} has less than 5 decimal places? Where p=prime>100 and d is the chosen day.

If the chosen day, d, is NOT of a type 2^n5^m (where n and m are nonnegative integers) then average=\frac{p}{d} will not be a terminating decimal and thus will have more than 5 decimal places.

How many such days are there of a type 2^n5^m: 1, 2, 4, 5, 8, 10, 16, 20, 25 (1=2^0*5^0, 2=2^2, 4=2^2, 5, 8=2^3, 10=2*5, 16=2^4, 20=2^2*5, 25=5^2), total of 9 such days (1st of June, 4th of June, ...).

Now, does p divided by any of these d's have fewer than 5 decimal places? Yes, as \frac{p}{d}*10,000=integer for any such d (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: P=\frac{9}{30}=\frac{3}{10} .

Hope it's clear.

So you're saying that by controlling the terminating decimal using d=2^m*5^n and, and you can make it an integer by multiplying it by 10000 (if the number must be five decimal places to the left.
Therefore, you can choose of days 1, 2, 4, 5, 8, 10, 16, 20, or 25. (making that 9 days). - But they're not prime??? What am I missing here... I'm the slow one of the lot please help me out

Nominator is a prime number>100 (on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100) and denominator is that day: {the average daily deposit}={sum of deposits up to and including that day}/{# of days}=p/d (so denominator d must not be a prime it should be of a type d=2^m*5^n. It's nominator p which is given to be a prime>100).
_________________
Senior Manager
Joined: 30 Nov 2010
Posts: 267
Schools: UC Berkley, UCLA
Followers: 1

Kudos [?]: 48 [0], given: 66

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  28 Jan 2011, 08:54
Bunuel wrote:
Nominator is a prime number>100 (on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100) and denominator is that day: {the average daily deposit}={sum of deposits up to and including that day}/{# of days}=p/d.

Oh so since we are limited to having five decimal places we were able to control that by choosing a using a denominator that gives us a terminating decimal and we multiply that by 10000 to make that number greater than 100. And the denominator is the 30 days out of which we can choose the 9 days.

But if the Nominator is supposed to be a prime number then we are either supposed to choose from 2 and 5 and nothingelse. Please have patience with me. I just need to understand.
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela

GMAT Club team member
Joined: 02 Sep 2009
Posts: 11610
Followers: 1800

Kudos [?]: 9593 [1] , given: 828

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  28 Jan 2011, 09:11
1
KUDOS
mariyea wrote:
Bunuel wrote:
Nominator is a prime number>100 (on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100) and denominator is that day: {the average daily deposit}={sum of deposits up to and including that day}/{# of days}=p/d.

Oh so since we are limited to having five decimal places we were able to control that by choosing a using a denominator that gives us a terminating decimal and we multiply that by 10000 to make that number greater than 100. And the denominator is the 30 days out of which we can choose the 9 days.

But if the Nominator is supposed to be a prime number then we are either supposed to choose from 2 and 5 and nothingelse. Please have patience with me. I just need to understand.

I'm not sure understood your question.

Anyway: average=\frac{prime}{day}. In order this value (p/d) to be terminating decimal d must be of a type 2^n5^m. Because if the chosen day, d, is NOT of a type 2^n5^m then average=\frac{p}{d} will not be a terminating decimal and thus will have more than 5 decimal places.

There are 9 such days: 1, 2, 4, 5, 8, 10, 16, 20, 25. For example average=\frac{prime}{1} or average=\frac{prime}{2} or average=\frac{prime}{4}, ..., average=\frac{prime}{25} all will be terminating decimals (and for ALL other values of d: 3, 6, 7, ... p/d will be recurring decimal thus will have infinite number of decimal places so more than 5). Also for all these 9 values of d average=p/d not only be terminating decimal but also will have fewer than 5 decimal places (average=\frac{prime}{16} will have max # of decimal places which is 4).

Hope it's clear.
_________________
Senior Manager
Joined: 30 Nov 2010
Posts: 267
Schools: UC Berkley, UCLA
Followers: 1

Kudos [?]: 48 [0], given: 66

Re: MGMAT Challenge: Decimals on Deposit [#permalink]  28 Jan 2011, 12:38
Bunuel wrote:
mariyea wrote:
Bunuel wrote:
Nominator is a prime number>100 (on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100) and denominator is that day: {the average daily deposit}={sum of deposits up to and including that day}/{# of days}=p/d.

Oh so since we are limited to having five decimal places we were able to control that by choosing a using a denominator that gives us a terminating decimal and we multiply that by 10000 to make that number greater than 100. And the denominator is the 30 days out of which we can choose the 9 days.

But if the Nominator is supposed to be a prime number then we are either supposed to choose from 2 and 5 and nothingelse. Please have patience with me. I just need to understand.

I'm not sure understood your question.

Anyway: average=\frac{prime}{day}. In order this value (p/d) to be terminating decimal d must be of a type 2^n5^m. Because if the chosen day, d, is NOT of a type 2^n5^m then average=\frac{p}{d} will not be a terminating decimal and thus will have more than 5 decimal places.

There are 9 such days: 1, 2, 4, 5, 8, 10, 16, 20, 25. For example average=\frac{prime}{1} or average=\frac{prime}{2} or average=\frac{prime}{4}, ..., average=\frac{prime}{25} all will be terminating decimals (and for ALL other values of d: 3, 6, 7, ... p/d will be recurring decimal thus will have infinite number of decimal places so more than 5). Also for all these 9 values of d average=p/d not only be terminating decimal but also will have fewer than 5 decimal places (average=\frac{prime}{16} will have max # of decimal places which is 4).

Hope it's clear.

I understand now Thank you so much!
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
-Nelson Mandela

Re: MGMAT Challenge: Decimals on Deposit   [#permalink] 28 Jan 2011, 12:38
Similar topics Replies Last post
Similar
Topics:
[#29] 2 Min. Challenge : Decimal 2 20 Mar 2004, 07:32
PS: MGMAT Challenge of the week 12 03 Nov 2005, 22:55
ps: football - mgmat challenge 1 24 Sep 2007, 18:34
2 MGMAT Challenge: Zenzizenzizenzic Zurprise 5 13 Jul 2010, 18:42
mgmat challenge set overlap with mgmat cat tests ? 3 10 Oct 2011, 09:14
Display posts from previous: Sort by