May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th. Jun 01 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 27 Jul 2011
Posts: 138

Re: Baker's Dozen
[#permalink]
Show Tags
19 Apr 2013, 00:20
Bunuel wrote: I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?A. 460 B. 490 C. 493 D. 455 E. 445 Solution: bakersdozen12878220.html#p1057507 Actually need help from the moderators to know where my approach is going wrong: it is it is told at least 1 red , at least one blue will be there. So i am leaving 1 red , 1 blue=> Now i have 6 Red , 4 Blue ..I have to select 8 marbels... So it means i have to select 8 marbels out of 10 marbels (colors don't matter) = \(C^10_8\) = \(C^10_2\) = 45 I know i am not getting the ans...but what's wrong with this I m not able to spot...
_________________
If u can't jump the 700 wall , drill a big hole and cross it .. I can and I WILL DO IT ...need some encouragement and inspirations from U ALL



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India

Re: Baker's Dozen
[#permalink]
Show Tags
21 Apr 2013, 08:19
sujit2k7 wrote: Bunuel wrote: I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar?A. 460 B. 490 C. 493 D. 455 E. 445 Solution: bakersdozen12878220.html#p1057507 Actually need help from the moderators to know where my approach is going wrong: it is it is told at least 1 red , at least one blue will be there. So i am leaving 1 red , 1 blue=> Now i have 6 Red , 4 Blue ..I have to select 8 marbels... So it means i have to select 8 marbels out of 10 marbels (colors don't matter) = \(C^10_8\) = \(C^10_2\) = 45 I know i am not getting the ans...but what's wrong with this I m not able to spot... Responding to a pm: From the options, it is obvious that the marbles are considered unique i.e. the red marbles are all different, say they are numbered, and the blue marbles are all different. The problem with your approach is that when you leave 1 red and 1 blue, you don't provide for this. Also, once you account for the uniqueness, there will be double counting  when you leave Red1 and Blue1 and then choose 8 out of the remaining 10 and say you are left with Red2 and Red3 too, this will be counted as different from when you leave Red2 and Blue1 and then choose 8 out of the remaining 10 and you are left with Red1 and Red3. These two cases are actually the same. Rule of thumb is that 'at least one' cases are best done using the complement approach which is 'Total  None' (as shown by Bunuel in the solution).
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 12 Dec 2012
Posts: 217
Concentration: Leadership, Marketing
GMAT 1: 540 Q36 V28 GMAT 2: 550 Q39 V27 GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)

Re: Baker's Dozen
[#permalink]
Show Tags
11 May 2013, 19:25
Bunuel wrote: 12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z? A. 12 B. 20 C. 24 D. 29 E. 33
Given \(x=qy+3\), where \(q\) is a quotient, an integer \(\geq0\). Which means that the least value of \(x\) is when \(q=0\), in that case \(x=3\). This basically means that \(x\) is less than \(y\). For example 3 divided by 4 yields remainder of 3.
Thus we have that: \(x\) is divided by \(y\) the remainder is 3 > minimum value of \(x\) is 3; \(y\) is divided by \(z\) the remainder is 8 > minimum value of \(y\) is 8 and minimum value of \(z\) is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);
So, the smallest possible value of \(x+y+z\) is 3+8+9=20.
Answer: B. why z is minimum= 9?
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India

Re: Baker's Dozen
[#permalink]
Show Tags
11 May 2013, 21:27
TheNona wrote: Bunuel wrote: 12. x, y and z are positive integers such that when x is divided by y the remainder is 3 and when y is divided by z the remainder is 8. What is the smallest possible value of x+y+z? A. 12 B. 20 C. 24 D. 29 E. 33
Given \(x=qy+3\), where \(q\) is a quotient, an integer \(\geq0\). Which means that the least value of \(x\) is when \(q=0\), in that case \(x=3\). This basically means that \(x\) is less than \(y\). For example 3 divided by 4 yields remainder of 3.
Thus we have that: \(x\) is divided by \(y\) the remainder is 3 > minimum value of \(x\) is 3; \(y\) is divided by \(z\) the remainder is 8 > minimum value of \(y\) is 8 and minimum value of \(z\) is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);
So, the smallest possible value of \(x+y+z\) is 3+8+9=20.
Answer: B. why z is minimum= 9? When m is divided by n, the remainder can be anything from 0 to (n1). Take examples: When m is divided by 8, the remainder can 0 or 1 or 2 till 7. Can the remainder be 8? No because that means the number is divisible by 8. Say you divide 20 by 8, you get remainder 4. You divide 21 by 8, you get remainder 5. You divide 22 by 8, you get remainder 6. You divide 23 by 8, you get remainder 7. You divide 24 by 8, you get remainder 0, isn't it? The remainder will not be 8  it will be 0. Hence if you know that the remainder when you divide m by n is 7, you can say that n must be at least 8. Similarly, when a number is divided by z and remainder is 8, z must be at least 9.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 17 Oct 2012
Posts: 49

Re: Baker's Dozen
[#permalink]
Show Tags
04 Jul 2013, 03:03
Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=[color=#ff0000]3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).[/color]
Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).
Answer: E. Bunuel, could you please kindly explain this part? How did you come up with this factorization? Many thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
04 Jul 2013, 03:24
BankerRUS wrote: Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=[color=#ff0000]3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).[/color]
Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).
Answer: E. Bunuel, could you please kindly explain this part? How did you come up with this factorization? Many thanks! Consider this: \((abac)^2=(a(bc))^2=a^2*(bc)^2\), so: \((3^53^2)^2=(3^2*(3^31))^2=(3^2)^2*(3^31)^2=3^4*(3^31)^2=3^4*26=3^4*2^2*13^2\). \((5^75^4)^2=(5^4*(5^31))^2=(5^4)^2*(5^31)^2=5^8*(5^31)^2=5^8*124^2=5^8*2^4*31^2\) \((3^4*2^2*13^2)(5^8*2^4*31^2)=2^6*3^4*5^8*13^2*31^2\). Hope it's clear.
_________________



Senior Manager
Joined: 08 Apr 2012
Posts: 346

Re: Baker's Dozen
[#permalink]
Show Tags
18 Sep 2013, 06:00
Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).
Answer: E. Can someone please explain how the move from the second part of the equation to the third part was?



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
18 Sep 2013, 07:19
ronr34 wrote: Bunuel wrote: 2. If \(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}\), then y is NOT divisible by which of the following? A. 6^4 B. 62^2 C. 65^2 D. 15^4 E. 52^4
\(y=\frac{(3^53^2)^2}{(5^75^4)^{2}}=(3^53^2)^2*(5^75^4)^2=3^4*(3^31)^2*5^8*(5^31)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2\).
Now, if you analyze each option you'll see that only \(52^4=2^8*13^4\) is not a factor of \(y\), since the power of 13 in it is higher than the power of 13 in \(y\).
Answer: E. Can someone please explain how the move from the second part of the equation to the third part was? Consider this: \((abac)^2=(a(bc))^2=a^2*(bc)^2\), so \((3^53^2)^2=(3^2*(3^31))^2=(3^2)^2*(3^31)^2=3^4*(3^31)^2\). Hope it helps.
_________________



Intern
Joined: 23 May 2012
Posts: 1

Re: Baker's Dozen
[#permalink]
Show Tags
08 Oct 2013, 07:32
Hi Bunuel,
On Mr. Wallace's Briefcase problem, I was wondering if you can clarify the step where you multiple by 5 choose 3? Can I think of it as SSSNN ["S" representing the number 6 and "N" representing a non6 digit] and the number of ways to arrange this would be 5!/[3!*2!] since there are 3 repeats for S and 2 repeats for N. Is my reasoning the a logical way to approach this problem as well? Thnks!



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
17 Oct 2013, 03:56
nyknicksah20 wrote: Hi Bunuel,
On Mr. Wallace's Briefcase problem, I was wondering if you can clarify the step where you multiple by 5 choose 3? Can I think of it as SSSNN ["S" representing the number 6 and "N" representing a non6 digit] and the number of ways to arrange this would be 5!/[3!*2!] since there are 3 repeats for S and 2 repeats for N. Is my reasoning the a logical way to approach this problem as well? Thnks! Yes, that is correct: \(C^3_5=\frac{5!}{3!2!}\).
_________________



Intern
Joined: 23 Nov 2013
Posts: 1
Concentration: Finance

Re: Baker's Dozen
[#permalink]
Show Tags
01 Dec 2013, 03:32
Hello Bunuel, Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest? \(x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = 185\) \(x = 41\) Then we take 5 smallest integers, which are the latter in our set \((x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40\) \(x=  41\) \(41*5 + 40=165\) answer A? Please correct me if I'm wrong. Bunuel wrote: 8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is 185 what is the sum of the 5 smallest integers of set A? A. 165 B. 175 C. 195 D. 205 E. 215
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)
Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=185\) > \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=185\) > \((5x+20)+20=185\) > \(5x+20=205\)
Answer: D.



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
01 Dec 2013, 06:54
trailrunner wrote: Hello Bunuel, Since we have all negative numbers in the set, shouldn't we add up the first 5 numbers as the largest? \(x + (x+2) + (x+4) + (x+6) + (x+8) = 5x +20 = 185\) \(x = 41\) Then we take 5 smallest integers, which are the latter in our set \((x+4) + (x+6) + (x+8) + (x+10) + (x+12) = 5x +40\) \(x=  41\) \(41*5 + 40=165\) answer A? Please correct me if I'm wrong. Bunuel wrote: 8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is 185 what is the sum of the 5 smallest integers of set A? A. 165 B. 175 C. 195 D. 205 E. 215
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)
Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=185\) > \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=185\) > \((5x+20)+20=185\) > \(5x+20=205\)
Answer: D. Five largest integers from {\(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\)} are {\(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\)} no matter whether x is negative or positive.
_________________



Intern
Joined: 13 Feb 2014
Posts: 1
Concentration: Technology, Healthcare
WE: Business Development (Computer Software)

Re: Baker's Dozen
[#permalink]
Show Tags
14 Feb 2014, 01:39
Bunuel wrote: 9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)? A. 1+y B. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. Shouldn't 1y be 1y? since y will be positive always?



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
17 Feb 2014, 07:50
dnawap123 wrote: Bunuel wrote: 9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)? A. 1+y B. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. Shouldn't 1y be 1y? since y will be positive always? No. When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). Since given that y is a negative number (y<0), then \(y = y\), hence \(1y=1(y)=1+y\). Hope it helps.
_________________



Manager
Joined: 10 Mar 2014
Posts: 186

Re: Baker's Dozen
[#permalink]
Show Tags
03 Jul 2014, 00:52
Bunuel wrote: 13. If \(x=\frac{(8!)^{10}(8!)^6}{(8!)^{5}(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}39\)? A. 0 B. 6 C. 7 D. 12 E. 14
Apply \(a^2b^2=(ab)(a+b)\): \(x=\frac{(8!)^{10}(8!)^6}{(8!)^{5}(8!)^3}=\frac{((8!)^{5}(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}39=(8!)^2+139=(8!)^238\).
Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^238\) will have 0038=62 as the last digits: 6*2=12.
Answer: D. Hi Bunnel, I have a doubt here I did it in a following way ((8!)^6 ((8!)^41))/((8!)^3 ((8!)^21)) then I will get (8!)^5 as i can ignore 1. (8!)^41 = (8!)^4. I have seen this kind of solution on some of the problems where you ignore 1 . I did the same here but getting different result. Please suggest.



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
03 Jul 2014, 06:09
PathFinder007 wrote: Bunuel wrote: 13. If \(x=\frac{(8!)^{10}(8!)^6}{(8!)^{5}(8!)^3}\), what is the product of the tens and the units digits of \(\frac{x}{(8!)^3}39\)? A. 0 B. 6 C. 7 D. 12 E. 14
Apply \(a^2b^2=(ab)(a+b)\): \(x=\frac{(8!)^{10}(8!)^6}{(8!)^{5}(8!)^3}=\frac{((8!)^{5}(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}(8!)^3}=(8!)^{5}+(8!)^3\).
Next, \(\frac{x}{(8!)^3}39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}39=(8!)^2+139=(8!)^238\).
Now, since \(8!\) has 2 and 5 as its multiples, then it will have 0 as the units digit, so \((8!)^2\) will have two zeros in the end, which means that \((8!)^238\) will have 0038=62 as the last digits: 6*2=12.
Answer: D. Hi Bunnel, I have a doubt here I did it in a following way ((8!)^6 ((8!)^41))/((8!)^3 ((8!)^21)) then I will get (8!)^5 as i can ignore 1. (8!)^41 = (8!)^4. I have seen this kind of solution on some of the problems where you ignore 1 . I did the same here but getting different result. Please suggest. That's totally wrong. We need to find the tens and units digits of some expression, ignoring anything there will change the result. Where did I say that we can ignore something when we want to find units/tens digit of an expression???
_________________



Intern
Joined: 11 Jul 2013
Posts: 31

Baker's Dozen
[#permalink]
Show Tags
29 Jul 2014, 05:40
Bunuel wrote: 6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xzxy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xzxy}\)
E. \(\frac{yz+xzxy}{yz}\)
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}\frac{1}{z}=\frac{yz+xzzy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xzxy}\) hours (time is reciprocal of rate).
In \(\frac{xyz}{yz+xzxy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xzxy}=\frac{yz}{yz+xzxy}\) amount of the water into the pool.
Answer: B. can you please tell me if there is any difeerence in these two statements amount of water in terms of the fraction of the pool which pump A pumped into the pool? amount of water which pump A pumped into the pool? and if there is a difference can you please explain? i had solved the question but was stumped by the language, is it simply asking me the amount of water pumped by A thanks



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
29 Jul 2014, 09:16
tyagigar wrote: Bunuel wrote: 6. A pool has two water pumps A and B and one drain C. Pump A alone can fill the whole pool in x hours, and pump B alone can fill the whole pool in y hours. The drain can empty the whole pool in z hours, where z>x. With pumps A and B both running and the drain C unstopped till the pool is filled, which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool? A. \(\frac{yz}{x+y+z}\)
B. \(\frac{yz}{yz+xzxy}\)
C. \(\frac{yz}{yz+xz+xy}\)
D. \(\frac{xyz}{yz+xzxy}\)
E. \(\frac{yz+xzxy}{yz}\)
With pumps A and B both running and the drain unstopped the pool will be filled in a rate \(\frac{1}{x}+\frac{1}{y}\frac{1}{z}=\frac{yz+xzzy}{xyz}\) pool/hour. So, the pool will be filled in \(\frac{xyz}{yz+xzxy}\) hours (time is reciprocal of rate).
In \(\frac{xyz}{yz+xzxy}\) hours A will pump \(\frac{1}{x}*\frac{xyz}{yz+xzxy}=\frac{yz}{yz+xzxy}\) amount of the water into the pool.
Answer: B. can you please tell me if there is any difeerence in these two statements amount of water in terms of the fraction of the pool which pump A pumped into the pool? amount of water which pump A pumped into the pool? and if there is a difference can you please explain? i had solved the question but was stumped by the language, is it simply asking me the amount of water pumped by A thanks The first question asks about the fraction of the water pumped by A. For example, if the capacity of the pool is 100 gallons and A pumped 50, then the fraction would be 50/100. The second question is about the amount. In this case the answer would be 50 gallons.
_________________



Manager
Joined: 23 Jan 2012
Posts: 60

Re: Baker's Dozen
[#permalink]
Show Tags
07 Sep 2014, 09:30
Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hello Bunuel. My P&C area is kind of weak, and I am working on it. In the solution why is it important to pay attention to no. of ways three 6's can be used in the password? Can you please elaborate on why have you used the Combination formula at all in the solution? Thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: Baker's Dozen
[#permalink]
Show Tags
07 Sep 2014, 09:39
p2bhokie wrote: Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. Hello Bunuel. My P&C area is kind of weak, and I am working on it. In the solution why is it important to pay attention to no. of ways three 6's can be used in the password? Can you please elaborate on why have you used the Combination formula at all in the solution? Thanks. We have 5digit password: ABCDE, out of which there are three 6's. These 6's can take any place: ABC, ABD, ABE, ... \(C^3_5\) gives the number of ways to choose which 3 digits (which 3 letters out of 5) will be 6's: 666DE 66C6E ...
_________________







Go to page
Previous
1 2 3 4 5 6
Next
[ 106 posts ]



