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

 It is currently 03 May 2015, 03:07

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Baker's Dozen

Author Message
TAGS:
Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [25] , given: 5576

Baker's Dozen [#permalink]  08 Mar 2012, 12:27
25
KUDOS
Expert's post
51
This post was
BOOKMARKED
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!

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

Solution: baker-s-dozen-128782-20.html#p1057502

2. If $$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, 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

Solution: baker-s-dozen-128782-20.html#p1057503

3. For the past k days the average (arithmetic mean) cupcakes per day that Liv baked was 55. Today Bibi joined and together with Liv they baked 100 cupcakes, which raises the average to 60 cupcakes per day. What is the value of k?
A. 6
B. 8
C. 9
D. 10
E. 12

Solution: baker-s-dozen-128782-20.html#p1057504

4. What is the smallest positive integer $$k$$ such that $$126*\sqrt{k}$$ is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

Solution: baker-s-dozen-128782-20.html#p1057505

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: baker-s-dozen-128782-20.html#p1057507

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. $$\frac{yz}{yz+xz-xy}$$

C. $$\frac{yz}{yz+xz+xy}$$

D. $$\frac{xyz}{yz+xz-xy}$$

E. $$\frac{yz+xz-xy}{yz}$$

Solution: baker-s-dozen-128782-20.html#p1057508

7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A.$60,000
B. $90,000 C.$100,000
D. $120,000 E.$180,000

Solution: baker-s-dozen-128782-20.html#p1057509

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

Solution: baker-s-dozen-128782-20.html#p1057512

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. 1-y
C. -1-y
D. y-1
E. x-y

Solution: baker-s-dozen-128782-20.html#p1057514

10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

Solution: baker-s-dozen-128782-20.html#p1057515

11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12

Solution: baker-s-dozen-128782-40.html#p1057517

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

Solution: baker-s-dozen-128782-40.html#p1057519

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

Solution: baker-s-dozen-128782-40.html#p1057520
_________________
 Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes Veritas Prep GMAT Discount Codes
Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [0], given: 5576

Re: Baker's Dozen [#permalink]  11 Mar 2012, 00:50
Expert's post
SergeNew wrote:
Was it the wrong solution of #9. It seems to me that my post was removed

Even if it were wrong I wouldn't remove it, it's just on the second page: baker-s-dozen-128782-20.html#p1056538
_________________
Intern
Joined: 03 Sep 2010
Posts: 16
Followers: 2

Kudos [?]: 8 [0], given: 14

Re: Baker's Dozen [#permalink]  11 Mar 2012, 04:36
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well
Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [0], given: 5576

Re: Baker's Dozen [#permalink]  11 Mar 2012, 04:59
Expert's post
utkarshlavania wrote:
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well

If $$x<0$$ then $$|x|=-x$$ and $$\frac{|x|}{x}=\frac{-x}{x}=-1$$, ($$-\frac{x}{x}=-1$$ no matter whether $$x$$ is negative or positive).
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5449
Location: Pune, India
Followers: 1331

Kudos [?]: 6773 [0], given: 177

Re: Baker's Dozen [#permalink]  11 Mar 2012, 19:52
Expert's post
utkarshlavania wrote:
@karishma

then why is |x|/x =-1 , should not it be 1 as |x|=-x and denominator is negative as well

You are right that |x|=-x since x is negative but tell me, what is -x, negative or positive? Negative of negative gives you positive, right? So -x must be positive. Now, if x is negative,
-x/x must be positive/negative giving you -1.

You are confusing yourself too much with negatives and positives. Just think of it this way:

|x|/x = -x/x (By definition, since |x| = -x when x < 0)
-x/x = -1 ( x and x get canceled here leaving you with -1)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Senior Manager
Joined: 25 Feb 2010
Posts: 476
Followers: 4

Kudos [?]: 50 [0], given: 9

Re: Baker's Dozen [#permalink]  14 Mar 2012, 14:45
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. 1-y
C. -1-y
D. y-1
E. x-y

Note that $$\sqrt{a^2}=|a|$$. Next, since $$x<0$$ and $$y<0$$ then $$|x|=-x$$ and $$|y|=-y$$.

So, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.
_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [0], given: 5576

Re: Baker's Dozen [#permalink]  14 Mar 2012, 16:10
Expert's post
onedayill 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. 1-y
C. -1-y
D. y-1
E. x-y

Note that $$\sqrt{a^2}=|a|$$. Next, since $$x<0$$ and $$y<0$$ then $$|x|=-x$$ and $$|y|=-y$$.

So, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.

x and y are negative numbers, not odd numbers.

If $$x\geq{0}$$ then $$|x|=x$$;
If $$x<{0}$$ then $$|x|=-x$$. So if $$x$$ is negative then $$|x|=-x=-negative=positive$$. For example, if $$x=-2$$ then $$|x|=|-2|=2=-x$$.

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
_________________
Senior Manager
Joined: 12 Dec 2010
Posts: 282
Concentration: Strategy, General Management
GMAT 1: 680 Q49 V34
GMAT 2: 730 Q49 V41
GPA: 4
WE: Consulting (Other)
Followers: 8

Kudos [?]: 38 [0], given: 23

Re: Baker's Dozen [#permalink]  14 Mar 2012, 22:43
Got 9 correct with @ 2.5 mins/Q ... how about others ? (2 Q left without going to deep as killed more than 4 mins on those Q4,7- this proves that not to waste time on something you don't get idea in 1.5-2 mins time frame and missed 2 narrowly- Q9, 10: 10 I din't read 2y but only y so...)..will post in details some explanations, esp. those which are little different from already posted by the very own legend Mr B!!
_________________

My GMAT Journey 540->680->730!

~ When the going gets tough, the Tough gets going!

Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [0], given: 5576

Re: Baker's Dozen [#permalink]  16 Mar 2012, 02:47
Expert's post
chetan2u 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 fraction of the pool which pump A filled?
A. $$\frac{yz}{x+y+z}$$

B. $$\frac{yz}{yz+xz-xy}$$

C. $$\frac{yz}{yz+xz+xy}$$

D. $$\frac{xyz}{yz+xz-xy}$$

E. $$\frac{yz+xz-xy}{yz}$$

With pumps A and B both running and the drain unstopped the pool will be filled in a rate $$\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{yz+xz-zy}{xyz}$$ pool/hour. So, the pool will be filled in $$\frac{xyz}{yz+xz-xy}$$ hours (time is reciprocal of rate).

In $$\frac{xyz}{yz+xz-xy}$$ hours pump A will do $$\frac{1}{x}*\frac{xyz}{yz+xz-xy}=\frac{yz}{yz+xz-xy}$$ part of the job.

hi , i feel the ans to this Q is wrong. as we have to tell what fraction of the pool was filled up by pump A. it will not depend on the drain C as the filling was done only by pump A and B. ans shud be y/(x+y).
the given ans would be correct if we were to find "the amount of water pumped by A in that duration"

Actually that was the intended meaning of the question. I edited it so to avoid ambiguity: "which of the following represents the amount of water in terms of the fraction of the pool which pump A pumped into the pool?"

Hope now it's more precise.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [0], given: 5576

Re: Baker's Dozen [#permalink]  17 Mar 2012, 00:12
Expert's post
GMATD11 wrote:
Bunuel wrote:
2. If $$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}$$, 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

$$y=\frac{(3^5-3^2)^2}{(5^7-5^4)^{-2}}=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2=3^4*26^2*5^8*124^2=2^6*3^4*5^8*13^2*31^2$$.

Now, if you analyze each option you'll see that only $$52^4=2^4*13^4$$ is not a factor of $$y$$, since the power of 13 in it is higher than the power of 13 in $$y$$.

Pls correct the Typo 2^4 to 2^8

Typo corrected, thanks.
_________________
Intern
Joined: 09 Feb 2012
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 21

Re: Baker's Dozen [#permalink]  18 Mar 2012, 15:45
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.

$$P=\frac{favorable}{total}=\frac{810}{10^5}$$

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

Math Expert
Joined: 02 Sep 2009
Posts: 27168
Followers: 4223

Kudos [?]: 40895 [0], given: 5576

Re: Baker's Dozen [#permalink]  19 Mar 2012, 03:10
Expert's post
balas 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.

$$P=\frac{favorable}{total}=\frac{810}{10^5}$$

Hello Bunnel....

Can you kindly clarify this question and elaborate on the solution!!

Can you please specify which part didn't you understand? Thanks.
_________________
Manager
Joined: 22 Feb 2012
Posts: 93
Schools: HBS '16
GMAT 1: 740 Q49 V42
GMAT 2: 670 Q42 V40
GPA: 3.47
WE: Corporate Finance (Aerospace and Defense)
Followers: 2

Kudos [?]: 20 [0], given: 25

Re: Baker's Dozen [#permalink]  27 Mar 2012, 10:04
Awesome set of questions!
Intern
Joined: 14 Apr 2012
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 35

Re: Baker's Dozen [#permalink]  28 Apr 2012, 02:41
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)?
Why not : 9*9*1*1*1* 5!/2! ??

Last edited by mofasser08 on 28 Apr 2012, 03:00, edited 1 time in total.
Manager
Joined: 12 Feb 2012
Posts: 109
Followers: 1

Kudos [?]: 14 [0], given: 28

Re: Baker's Dozen [#permalink]  28 Apr 2012, 18:47
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.

$$P=\frac{favorable}{total}=\frac{810}{10^5}$$

Hey Bunuel,

Sorry to trouble you. If you could, I would appreciate if you could clarify more about 5C3. So this is how I approached the problem:

We have three - 6's. With two slots that can be filled by 9 numbers.

666 9 9 . Now the question is how many ways can we rearrange these codes of three 6's and 2 non 6's? Well, have 5 elements 3 identical (because we have 3 identical 6's and two non-6's) Hence we have 5!/(3!2!) ways of rearranging the numbers. Is this why we have 5C3 in the problem?

But the two non-6's can be identical or different. For example 66611 66622 66633 66644 66655 ... 66699... etc all satisfy the condition of only three 6's. The number of ways to rearrange the letters {aaabb} is 5!/(3!2!). Moreover, we can also have the two non-6's be different, 66612 66654 66624, etc . the number of ways to rearrange the letters {aaabc} is 5!/(3!).

What do we do in this situation? Do we look at this problem as (#6)(#6)(#6)(not #6)(not #6) therefore 5!/(3!2!). Three repeating elements of #6 and two repeating elements of not #6.

OR

Do we look at the problem as (#6)(#6)(#6)(x)(y), where x and y two numbers that are not 6, that may or may not be equal, therefore 5!/3! (3! for the 3 repeating 6's but we cannot say anything about whether x and y are equal to each other)

Bunuel, as always, thank you so much!! And again, sorry to trouble you.
Manager
Joined: 12 Feb 2012
Posts: 109
Followers: 1

Kudos [?]: 14 [0], given: 28

Re: Baker's Dozen [#permalink]  28 Apr 2012, 18:51
mofasser08 wrote:
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)?
Why not : 9*9*1*1*1* 5!/2! ??

Mofasser,

We pretty much have the same question. But I think the reasons it's 5C3 is because we have

(#6)(#6)(#6)(not #6)(not #6)=aaabb.

The number of ways to rearrange aaabb is 5!/(3!2!). But we need to check with the Oracle, Bunuel.
Intern
Joined: 01 Mar 2012
Posts: 32
Concentration: Operations, Finance
GMAT 1: 740 Q49 V41
GPA: 3.3
WE: Engineering (Manufacturing)
Followers: 1

Kudos [?]: 3 [0], given: 46

Re: Baker's Dozen [#permalink]  02 May 2012, 09:59
Question 1 3 number of 6's can be arranged among themselves in 1 way. Number of ways we can select three position from 5= 5C3. For the remaining two position we can choose any digit but 6. So number of possible patterns=(5C3)*9*9
Hence required probability=(5C3)*9*9/10^5
I hope those who have doubt can understand this explanation
Senior Manager
Joined: 21 Mar 2010
Posts: 316
Followers: 5

Kudos [?]: 22 [0], given: 33

Re: Baker's Dozen [#permalink]  08 May 2012, 17:46
Bunuel wrote:
7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A.$60,000
B. $90,000 C.$100,000
D. $120,000 E.$180,000

Fritz owns is $$\frac{2}{3}$$rd of the shares of the other three shareholders --> Fritz owns $$\frac{2}{2+3}=\frac{2}{5}$$th of all shares;
Luis owns is $$\frac{3}{7}$$th of the shares of the other three shareholders --> Luis owns $$\frac{3}{3+7}=\frac{3}{10}$$th of all shares;
Alfred owns is $$\frac{4}{11}$$th of the shares of the other three shareholders --> Alfred owns $$\frac{4}{4+11}=\frac{4}{15}$$th of all shares;

Together those three own $$\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}$$th of all shares, which means that Werner owns $$1-\frac{29}{30}=\frac{1}{30}$$. Hence from $3,600,000 Werner gets $$3,600,000*\frac{1}{30}=120,000$$. Answer: D. Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on? Math Expert Joined: 02 Sep 2009 Posts: 27168 Followers: 4223 Kudos [?]: 40895 [0], given: 5576 Re: Baker's Dozen [#permalink] 09 May 2012, 00:10 Expert's post mbafall2011 wrote: Bunuel wrote: 7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of$3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000 B.$90,000
C. $100,000 D.$120,000
E. $180,000 Fritz owns is $$\frac{2}{3}$$rd of the shares of the other three shareholders --> Fritz owns $$\frac{2}{2+3}=\frac{2}{5}$$th of all shares; Luis owns is $$\frac{3}{7}$$th of the shares of the other three shareholders --> Luis owns $$\frac{3}{3+7}=\frac{3}{10}$$th of all shares; Alfred owns is $$\frac{4}{11}$$th of the shares of the other three shareholders --> Alfred owns $$\frac{4}{4+11}=\frac{4}{15}$$th of all shares; Together those three own $$\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}$$th of all shares, which means that Werner owns $$1-\frac{29}{30}=\frac{1}{30}$$. Hence from$3,600,000 Werner gets $$3,600,000*\frac{1}{30}=120,000$$.

Could some one explain this to me how does one arrive at 2/2+3 = 2/5 and so on?

Explained here: baker-s-dozen-128782-40.html#p1059585
_________________
Manager
Joined: 04 Dec 2011
Posts: 81
Schools: Smith '16 (I)
Followers: 0

Kudos [?]: 15 [0], given: 13

Re: Baker's Dozen [#permalink]  04 Jun 2012, 10:16
Bunuel wrote:
mofasser08 wrote:
Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*5P3 ??

That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with $$C^3_5=10$$ we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: $$C^2_510$$ choosing which 2 places out of 5 will be occupied by 6's.

Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.

Total $$9*9*C^3_5=810$$.

Hope it's clear.

Hi Bunuel..can you tell me if I am thinking correct.

1/10 * 1/10 * 1/10 * 9/10 * 9/10 this gives me 3 chances of getting 6 and rest 2 for getting any other number.
now the above scenario can be arranged in 5C3 ways so

1/10 * 1/10 * 1/10 * 9/10 * 9/10 * 5C3..

I hope I am correct in this way..?

_________________

Life is very similar to a boxing ring.
Defeat is not final when you fall down…
It is final when you refuse to get up and fight back!

1 Kudos = 1 thanks
Nikhil

Intern
Joined: 15 May 2012
Posts: 41
Followers: 0

Kudos [?]: 5 [0], given: 94

Re: Baker's Dozen [#permalink]  12 Jun 2012, 17:39
Hi,
Thanks!
Re: Baker's Dozen   [#permalink] 12 Jun 2012, 17:39

Go to page   Previous    1   2   3   4   5   6   7   8    Next  [ 151 posts ]

Similar topics Replies Last post
Similar
Topics:
275 Devil's Dozen!!! 137 19 Mar 2012, 05:54
1 Turtles and Bakers beach 27 03 Feb 2009, 11:16
To Joshephine baker 4 13 Sep 2007, 18:41
Baker was perhaps not the most gifted soloist in the 12 19 Aug 2007, 04:56
Baker was perhaps not the most gifted soloist in the 14 29 May 2006, 07:41
Display posts from previous: Sort by