## Deriving the volume and surface area of a cone

Hello, this is my first time posting on physics forums, so if I do something wrong, please bear with me :)

I am trying to derive the formula for the lateral surface area of a cone by cutting the cone into disks with differential height, and then adding up the lateral areas of all of the disks/cylinders to find the lateral area of the cone. (Similar to using the volume of revolution, but just taking the surface area). I assumed that the heights of each of the disk was dH, where H = the height of the cone. However, using that method, I got pi*radius*height instead of pi*radius*(slant height).

On another thread in physics forum (http://www.physicsforums.com/showthread.php?t=354134) , a person used a similar method as I did, and someone replied saying that the height of each disk is dS, where S = the slant height of the cone, not dH, where H = the height of the cone. And using dS instead of dH allowed me to find the correct formula for the lateral area of the cone!

HOWEVER, using dS instead of dH contradicts with the same method for finding the volume of the cone. I basically used the method shown in another person's video (http://www.youtube.com/watch?v=Btx_f883uFU) to find the volume of the cone, but that person assumed that the height of each disk is dH, not dS.

So is the height of each disk dS or dH? If it is dS, then I understand how to get the formula for the lateral surface area of the cone, but not the volume. But if it is dH, then I understand how to get the formula for the volume of the cone, not the surface area. Can anyone clear up my confusion? I know that there are other methods to find the surface area/volume, but I just want to know why there is a contradiction between the dH/dS thing. Thanks! Any help is appreciated! :)

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 ds is the rate of change in the lateral area of an infinitesimal disk, while dh is the change of its height. The volume of an infinitesimal cylinder is computed using its height, not its slant height; while when you are computing the surface area, the height of each cylinder is not what you want to use. You can easily apply the Pythagorean theorem to see what I mean. In a pure cylinder, the slant height and the normal height is equivalent. When you are partitioning a cone however, they are not. This is simply because the cone has a slope that can't simply be discarded. Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't. This solves your problem.
 Hello Millennial, thank you for your reply! I understand that cylinders have a different slope from cones. However, you said, "Discarding the slope is exactly what you are doing now. Volume is OK to compute like that, but surface area isn't." Why is it okay to discard the slope when computing the volume, but not okay when computing the surface area? Thanks!

Recognitions:
Gold Member
Homework Help

## Deriving the volume and surface area of a cone

The tiny volume bit error is, relatively speaking, vansihingly small to that which is kept.

The local cylinder volume approximation is Vc= pi*r^2*dz

Now, let us look at the volume of the tiny element from an expanding cone.

This is made up of Vc+the volume of a triangular area with the full circular periphery as its radius.

This bit's volume can be written (2*pi*(r+dr))*1/2*dz*dr, where dr is the tiny radial additon from position (z,r) to position (z+dz, r+dr).
Now, the biggest part of this volume equals pi*rdzdr (agreed?)
Since both dz and dr goes to zero in the limit, the PRODUCT of these tiny quantities go much faster to zero than either one of them.
Agreed?

Thus, we can discard that volume bit relative to Vc.
-----
However, what you lose when computing surface area (with cylindric approxamition), you'll lose area portions that do not vanish faster than the one you keep.

 Recognitions: Gold Member Homework Help Science Advisor As for the surface area element, with the cylinder, you get r*dw*dz, where dw is the tiny angle, and r*dw the tiny arc length. With the cone, you get r*dw*dS instead (where dS is the length along the skewed plane), so that the ratio of these terms are dS/dz. This does NOT equal 1, so that you cannot approximate the surface area locally with that of the cylinder


# How many arrangements are possible

Problem: There are infinite number of spoons and forks at your disposal. You have to keep a total of n spoons and forks on a table in a row. As soon as you keep a fork, you will have to keep forks alternately. If the first time you kept a fork was at position i, you will have to keep forks at positions i+2, i+4, i+6.. and so on. How many arrangements are possible? Since the answer could be very large, print it modulo 109+7.

Constraints: 1 $\leqslant$ n $\leqslant$ 109

Time Limit: $1$ sec

Sample Input 1: $3$

Output 1: $6$ (SSS, SSF, SFS, SFF, FSF, FFF)

Sample Input 2: $5$

Output 2: $14$

The way I solved it (inefficient):

Keep n spoons first. Start from the end and replace each spoon with a fork. When you've replaced i spoons with i forks at the end, there are 2i/2 combinations possible (since $\frac{i}{2}$ positions will have forks). So basically it's a summation of powers of 2.

Going by my method, the answer will be: Summation of the first $n+1$ terms of the following series

1 1 2 2 4 4 8 8 16 16 32 32 ...

My submission is getting the verdict Time Limit Exceeded. I can neither embed nor compute very high powers of 2, and there have been many submissions in 0.0 seconds! It must be very easy though I can't think of anything else.

What is a better method to solve this question?

[EDIT]: If the first time you keep a fork is at position $i$, you will have to keep forks at positions $i+2, i+4 ...$ and so on. This happens only ONCE.

For example, A(4) = 10. By the answers below, it is 9 (which is wrong). The 10th arrangement will be FFFS.

-
Curious as to where you came across this problem. Not Project Euler, is it? – Gerry Myerson Sep 5 '12 at 1:58
@Gerry I couldn't find it there. – Rick Decker Sep 5 '12 at 2:39
@Rick, me neither - it just smells like a PE problem. Anyway, for input 5, I get 12, not 14: sssss, ssssf, sssfs, sssff, ssfsf, ssfff, sfsfs, sfsff, sffff, fsfsf, fsfff, fffff. What am I missing? – Gerry Myerson Sep 5 '12 at 2:51
You're right. That'll save me some writing, since you've already enumerated those possibilities. – Rick Decker Sep 5 '12 at 2:58

It can also be completely analyzed without recursion.

You can start with any numbers of spoons. Once you have two forks in a row, you must continue with forks. Thus, using $\rm{S}$ for spoons and $\rm{F}$ for forks and exponents to indicate repetition, you can start with ${\rm{S}}^k$ for any $k\ge 0$. If you ever get a fork, you can follow it with another fork, in which case you have ${\rm{S}}^k{\rm{F}}^{n-k}$ for some $k\in\{0,\dots,n\}$, or you can follow it with a spoon. At that point your choices are limited, however, since at least every second utensil from this point on must be a fork. In fact, it’s not hard to see that all you can do is alternate forks and spoons until you decide to place two forks in a row (if you ever do). In short, the only possible configurations are those of the form

$${\rm{S}}^k({\rm{FS}})^\ell{\rm{F}}^m\;,$$

where $k,\ell,m\ge 0$ and $k+2\ell+m=n$.

Fix $k$ with $0\le k\le n$. Then $2\ell+m=n-k$. Each integer $\ell$ such that $0\le\ell\le\frac12(n-k)$ yields exactly one arrangement, so there are $\left\lfloor\frac12(n-k)\right\rfloor+1$ arrangements beginning with exactly $k$ spoons. This gives a total of

\begin{align*} \sum_{k=0}^n\left(\left\lfloor\frac12(n-k)\right\rfloor+1\right)&=n+1+\sum_{k=0}^n\left\lfloor\frac12(n-k)\right\rfloor\\\\ &=n+1+\sum_{k=0}^n\left\lfloor\frac{k}2\right\rfloor\\\\ &=n+1+\begin{cases} 2\sum_{k=0}^{(n-1)/2}k,&\text{if }n\text{ is odd}\\\\ 2\sum_{k=0}^{n/2}k-\frac{n}2,&\text{if }n\text{ is even} \end{cases}\\\\ &=n+1+\begin{cases} \frac{n^2-1}4,&\text{if }n\text{ is odd}\\\\ \frac{n^2}4,&\text{if }n\text{ is even} \end{cases}\\\\ &=\left(\frac{n}2+1\right)^2-\frac14[n\text{ is odd}]\\\\ &=\left\lfloor\left(\frac{n}2+1\right)^2\right\rfloor\;, \end{align*}

arrangements of $n$ spoons and forks, where $[n\text{ is odd}]$ is an Iverson bracket.

Added: In the updated version of the question there is one arrangement with no forks. Suppose now that the first fork is the $k$-th utensil from the end of the row, where $1\le k\le n$. If $k$ is even, there are $k/2$ guaranteed forks and $k/2$ free positions, for a total of $2^{k/2}$ arrangements. If $k$ is odd, there are $(k+1)/2$ guaranteed forks and $(k-1)/2$ free positions, for a total of $2^{(k-1)/2}$ arrangements. The grand total number of arrangements is therefore

\begin{align*} 1+\sum_{k=1}^n2^{\lfloor k/2\rfloor}&=\sum_{k=0}^n2^{\lfloor k/2\rfloor}\\\\ &=2\sum_{k=0}^{\lfloor n/2\rfloor}2^k-2^{n/2}[n\text{ is even}]\\\\ &=2\left(2^{\lfloor n/2\rfloor+1}-1\right)-2^{n/2}[n\text{ is even}]\\\\ &=\begin{cases} 2^{(n+3)/2}-2,&\text{if }n\text{ is odd}\\\\ 3\cdot2^{n/2}-2,&\text{if }n\text{ is even}\;. \end{cases} \end{align*}

-
We haven't understood the question correctly. There's an update to the question. Check it out.. – Rushil Sep 5 '12 at 20:24
I solved this logically, though yours is pretty mathematical. :-) – Rushil Sep 5 '12 at 21:47

The first step in any programming competition is to look at the constraints of the problem. Since the parameter, $n$, can be as large as $10^9$ you clearly don't want to waste your time with any program involving loops or recursion and any auxiliary numbers your solution involves almost certainly have to be small (at worst quadratic in the parameter). In other words, there'd better be a closed form solution for this problem involving at most a few expressions that are no worse than quadratic in $n$. Fortunately, there is, but you'll have to wait a bit to see it.

In the tradition of almost all programming challenges, I'll obfuscate things as much as possible, trying to find wording that I (and only I) consider cloyingly cute and nerdy. Imagine, then, a mathematically-inclined butler laying out the silverware according to your specifications (it has to be the butler, obviously, since he's the only household staff member permitted to touch the very expensive silver). The first thing he learned in his training was "serve from the left" (and take away from the right, but that won't come into play now), and that's what we'll do: build strings of "S"s and "F"s, starting at the left, and we'll count the number of possible arrangements, given the rules you laid out.

There are two things to notice: first, starting with an "S" doesn't constrain us in any way, so if we let $A(n)$ count the number of arrangements of $n$ utensils, we'll have immediately that $$A(n) = A(n-1) + \text{some other stuff}$$ where the $A(n-1)$ term counts the number of strings starting with "S" (followed by $n-1$ characters) and the "other stuff" counts the number of strings (of length $n$) starting with "F". The second thing to observe is that if we ever build a substring of the form "...FF", any remaining characters must be "F"s.

Starting, then, with the empty string and adding characters to the right, we'll have the following possibilities at the start:

Working from the top down we see that there are two possible length-$1$ strings, "S" and "F", four possible length-$2$ strings and 6 possible length-$3$ strings. Using the first observation above, we can ignore the left subtree, starting with "S", since that's already taken care of by our recursion.

Now look at the right subtree, enumerating those strings that start with "F". At each level going down we'll have the counts $1$ (for the "F"), $2$ (for "FS" and "FF"), $2$ (for "FSF" and "FFF"), and a bit of thought will convince you that the sequence of the number of possible strings starting with "F" is $\langle 1, 2, 2, 3, 3, 4, 4, 5, 5, \dots\rangle$. That's just the "some other stuff" in the displayed equation above and it's easy to see that it's simply $\lfloor(n+2)/2\rfloor$, where the braces, as usual, delimit the floor function. In other words, we now have a recurrence for the number of arrangements: $A(1)=2$ and for $n>1$, $$A(n) = A(n-1) + \left\lfloor\frac{n+2}{2}\right\rfloor$$ Expanding this, we see with the tiniest bit of algebra that $$A(n) = 1+\sum_{k=2}^{n+2}\left\lfloor\frac{k}{2}\right\rfloor$$ That's good enough for every situation except a programming competition where $n$ may be very large. However, the sum is clearly going to be at most quadratic, so we look for a quadratic closed form for $A(n)$. One final trick comes into play here: because of the division by $2$, we'll have two different forms, depending upon whether $n$ is even or odd. To cut to the chase, it's easy enough to do some polynomial interpolation (twice) and come, at long last to $$A(n) = \frac{1}{4}n^2 + n + \begin{cases} 1 & \text{if n is even}\\ 3/4 & \text{if n is odd}\end{cases}$$

-
That was a great answer! Thanks. – Rushil Sep 5 '12 at 11:41
you said that you could ignore the left subtree. How did you handle that? I couldn't follow your answer after Expanding this, we see with the tiniest bit of ... – Rushil Sep 5 '12 at 11:42
okay I got all of it. Can you just explain the polynomial interpolation part? I've never heard of it before. – Rushil Sep 5 '12 at 16:06
Unfortunately, the solution I coded is now getting the verdict Wrong Answer. There is something wrong with your solution. – Rushil Sep 5 '12 at 19:18
@Rushil For polynomial interpolation, a good place to start is to check out Wikipedia. In this case, you want to find a polynomial $p(x)=Ax^2+Bx+C$ that fits your data. We already know, for example, that $p(2)=4$ so $4A+2B+C=4$. We also know that $p(4)=9$ so $16A+4B+C=9$, and we can also find $p(6)=16$ so $36A+6B+C=16$. Given these three linear equations in three unknowns (namely $A, B, C$), we can solve for the coefficients of the polynomial. As to the dreaded wrong answer I'm afraid that I can't help without seeing your code. I'm sure, though, that Brian's and my answers are correct. – Rick Decker Sep 5 '12 at 19:58


GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Dec 2018, 07:43

# R1 Admission Decisions:

Stanford Chat (Calls Started)  |  Wharton Chat  (Calls Expected Soon)  |  Fuqua Chat (Calls Expected Soon)

### 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
Your Progress

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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
• ### GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST

December 14, 2018

December 14, 2018

09:00 AM PST

10:00 AM PST

10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.

# In a certain baord game, a stack of 48 cards, 8 of which

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 16 Feb 2012
Posts: 27
GPA: 3.57
In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Jul 2012, 05:19
4
10
00:00

Difficulty:

5% (low)

Question Stats:

79% (00:40) correct 21% (00:43) wrong based on 594 sessions

### HideShow timer Statistics

In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
##### Most Helpful Expert Reply
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8673
Location: Pune, India
Re: Probability Question  [#permalink]

### Show Tags

24 Jul 2012, 21:49
5
3
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?

Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.

The probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.
Think: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.

Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
_________________

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 >

##### General Discussion
Intern
Joined: 16 Feb 2012
Posts: 27
GPA: 3.57
Re: Probability Question  [#permalink]

### Show Tags

24 Jul 2012, 05:23
rajman41 wrote:
In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

a)1/8
b)1/6
c)1/5
d)3/23
e)4/23
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
Intern
Joined: 05 Nov 2010
Posts: 29
Re: Probability Question  [#permalink]

### Show Tags

24 Jul 2012, 19:15
My answer is E.
We can substract 2 from 48 and find the probability for the third card.

What exactly is your doubt.
Manager
Joined: 22 Dec 2011
Posts: 231
Re: Probability Question  [#permalink]

### Show Tags

11 Oct 2012, 10:07
VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).

Hello Karishma - I guess we can list them out?

Pick Share : Pick share : pick share =6/46
Pick share : No pick Share : pick share = 7/46
No pick Share : pick : pick = 7/46
No : No : pick = 8/46

add them all to get the overall probability?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8673
Location: Pune, India
Re: Probability Question  [#permalink]

### Show Tags

11 Oct 2012, 20:16
2
3
Jp27 wrote:
VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).

Hello Karishma - I guess we can list them out?

Pick Share : Pick share : pick share =6/46
Pick share : No pick Share : pick share = 7/46
No pick Share : pick : pick = 7/46
No : No : pick = 8/46

add them all to get the overall probability?

You can list it out though you don't need to. The probability will stay 8/48. It doesn't matter whether it is the first pick, the second or the third or a later pick. I have discussed this here: a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html?hilit=probability%20same

When you list it out, you will obviously get the same answer.

Share, Share, Share = (8/48) * (7/47) * (6/46)
Share, No Share, Share = (8/48) * (40/47) * (7/46)
No Share, Share, Share = (40/48) * (8/47) * (7/46)
No Share, No Share, Share = (40/48) * (39/47) * (8/46)
When you add all these up, you get 8/48.
_________________

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 >

Senior Manager
Joined: 06 Aug 2011
Posts: 339
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

04 Nov 2012, 10:22
Will we not change 8 to 6??

because if we have taken out those 2 cards..then ramining will be 6
6/46?
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1131
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

04 Nov 2012, 11:24
2
sanjoo wrote:
Will we not change 8 to 6??

because if we have taken out those 2 cards..then ramining will be 6
6/46?

Read the question carefully. These 2 cards dont represent the stocks. Hence the 8 cards that represented stocks are still there, i.e. those 8 cards are still intact. But total number of cards has reduced from 48 to 46. Hence, probability can be found easily.
Hope that helps
-s
_________________
Intern
Joined: 16 Aug 2013
Posts: 4
Re: Probability Question  [#permalink]

### Show Tags

16 Aug 2013, 20:19
VeritasPrepKarishma wrote:
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?

Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.

The probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.
Think: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.

Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).

i have a big trouble ..i think it should be done in this way. we select 2 from the 40 , in the first beginning. then select 1 from the 8 , in total we have 3 from 48.
so what I got was C1 8C2 40/C3 48
Senior Manager
Joined: 10 Jul 2013
Posts: 315
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

17 Aug 2013, 11:08
1
rajman41 wrote:
In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23

40/48 * 39/47
so next is = 8/46 = 4/23
_________________

Asif vai.....

Intern
Joined: 16 Aug 2013
Posts: 4
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

19 Aug 2013, 03:43
Asifpirlo wrote:
rajman41 wrote:
In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?

A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23

40/48 * 39/47
so next is = 8/46 = 4/23

well done! I got it, meanwhile, I found my problem. what I did do not follow the first 2 cards and the third. what i did means just select cards randomly , two are PPPPPP, one is PPPPP...
thank you very much
Math Expert
Joined: 02 Aug 2009
Posts: 7106
Re: In a certain board game  [#permalink]

### Show Tags

10 Dec 2015, 07:37
amithyarli wrote:
In a certain board game, a stack of 48 cards, 8 of which represent shares of stack, are shuffled and placed face down. If the first 2 cards selected do not represent the shares of stock, what is the probability that the third card selected will represent a share of stock ?

A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23

Hi,
there are 48 cards , out of which 8 are shares of stack..
the first two are not share of stack..
so the remaining cards now are 48-2=46, again out of which there are 8 shares of stock..
the prob that next is one of these 8 is 8/46=4/23
E
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 29 Nov 2015
Posts: 13
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

29 Dec 2015, 09:48
VeritasPrepKarishma wrote:
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?

Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.

So when do you use combinatorics? I thought the solution would work out as:

Probability of third card being a stock card = 40/48* 39/47*8/46

When to use combinatorics???
Math Expert
Joined: 02 Aug 2009
Posts: 7106
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

29 Dec 2015, 20:35
2
sarathvr wrote:
VeritasPrepKarishma wrote:
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?

Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.

So when do you use combinatorics? I thought the solution would work out as:

Probability of third card being a stock card = 40/48* 39/47*8/46

When to use combinatorics???

Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 05 Jun 2015
Posts: 25
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Apr 2016, 02:00
VeritasPrepKarishma wrote:

Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).

Hi Karishma, can you please give the answer to the above query?

I have 2 other questions.

First question: I solved this question using conditional probability. Is my solution correct?
Probability (at least 2 non-share cards) = Probability (2 non-share AND 1 share cards) + Probability (3 non-share cards)
$$= \frac{40*39*8}{48*47*46} + \frac{40*39*38}{48*47*46}$$
$$=\frac{40*39*46}{48*47*46}$$
Desired probability = Probability(2 non-share AND 1 share cards)/Probability(at least 2 non-share cards)
$$=\frac{40*39*8}{40*47*46}$$
$$=\frac{8}{46}=\frac{4}{23}$$

Second question: I cannot tell the difference between the problem under discussion and this problem: A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

In the latter problem, Bunuel and other experts confirmed that the probability of getting a certain ball (black ball) will not change for any successive drawing. This means the probability of getting a black ball remains 5/8 no matter what.

I thought the same line of thinking would apply here. On the third drawing, the probability of getting a share remains 8/48. I know it is dead wrong, but I cannot understand the difference between the two questions. Could you please clarify?

Thank you very much!
Intern
Joined: 05 Jun 2015
Posts: 25
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Apr 2016, 02:15
chetan2u wrote:
Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..

Hi Chetan2u,

A silly question: How can we differentiate between a fact and a probability?

I am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?

Always appreciate your help!
Math Expert
Joined: 02 Aug 2009
Posts: 7106
Re: In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Apr 2016, 02:25
2
2
truongynhi wrote:
chetan2u wrote:
Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..

Hi Chetan2u,

A silly question: How can we differentiate between a fact and a probability?

I am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?

Always appreciate your help!

Hi,
you have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..
here you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..

But what happens when you have picked two cards and you are told they do not contain the card we are lookin for..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..

Our probability will continue to be out of all 48..

If you are told first two are not type X...
so now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..
Probability is ONLY there where we do not know the out come..

another extension of above Q..
you have picked 40 cards and none of them are type X..
what is the probability that 41st card will be X..

we now do not care of what we already know. they have moved from Probability to REALITY..
now we know 48-40=8 cards are left and none of type X has been picked..
so now all these 8 will be type X..
prob = 8/8 = 1.. that is it will be type X for sure..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 05 Jun 2015
Posts: 25
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Apr 2016, 04:44
chetan2u wrote:

Hi,
you have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..
here you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..

Hi,
Thank you! I think I understand this one.

Quote:
But what happens when you have picked two cards and you are told they do not contain the card we are lookin for..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..

Our probability will continue to be out of all 48..

I am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?

Quote:
If you are told first two are not type X...
so now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..
Probability is ONLY there where we do not know the out come..

Doesn't this contradict with your above statement? You said we start from 48 after the first two cards. But here you point out that we calculate on the remaining 46 cards. Is it a typo or something? Please clarify.

Thank you very much!
Math Expert
Joined: 02 Aug 2009
Posts: 7106
In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Apr 2016, 04:49
1
1
truongynhi wrote:
chetan2u wrote:

Quote:
But what happens when you have picked two cards andyou are told they do not contain the card we are lookin for..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..

Our probability will continue to be out of all 48..

I am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?

Thank you very much!

pl read the highlighted portion as -- and you are not aware what those two cards contain..
It was a typo..

Quote:
But what happens when you have picked two cards andand you are not aware what those two cards contain..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..

Our probability will continue to be out of all 48..

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 05 Jun 2015
Posts: 25
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
In a certain baord game, a stack of 48 cards, 8 of which  [#permalink]

### Show Tags

24 Apr 2016, 04:59
chetan2u wrote:
But what happens when you have picked two cards andand you are not aware what those two cards contain..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..

Our probability will continue to be out of all 48..

So, in this case, the probability that the third card is of type X will equal $$\frac{8}{48}$$. Is that right?

Thank you very much! You've been very helpful!
In a certain baord game, a stack of 48 cards, 8 of which &nbs [#permalink] 24 Apr 2016, 04:59

Go to page    1   2    Next  [ 28 posts ]

Display posts from previous: Sort by

# In a certain baord game, a stack of 48 cards, 8 of which

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


0. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. Let qp represent "If x = 5, then x + 7 = 11.". Create a truth table for the statement $$(A \vee B) \leftrightarrow \sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for … BNAT; Classes. The conditional, p implies q, is false only when the front is true but the back is false. Email. Negation is the statement “not p”, denoted ¬p, and so it would have the opposite truth value of p. If p is true, then ¬p if false. b. Biconditional statement? The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. The biconditional connective can be represented by ≡ — <—> or <=> and is … Demonstrates the concept of determining truth values for Biconditionals. To show that equivalence exists between two statements, we use the biconditional if and only if. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. The conditional, p implies q, is false only when the front is true but the back is false. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. The biconditional operator looks like this: ↔ It is a diadic operator. "x + 7 = 11 iff x = 5. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. Otherwise, it is false. second condition. Logical equivalence means that the truth tables of two statements are the same. The biconditional uses a double arrow because it is really saying “p implies q” and also “q implies p”. Determine the truth values of this statement: (p. A polygon is a triangle if and only if it has exactly 3 sides. first condition. Remember that a conditional statement has a one-way arrow () and a biconditional statement has a two-way arrow (). In the first conditional, p is the hypothesis and q is the conclusion; in the second conditional, q is the hypothesis and p is the conclusion. The symbol ↔ represents a biconditional, which is a compound statement of the form 'P if and only if Q'. We start by constructing a truth table with 8 rows to cover all possible scenarios. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. ", Solution:  rs represents, "You passed the exam if and only if you scored 65% or higher.". Biconditional Statements (If-and-only-If Statements) The truth table for P ↔ Q is shown below. Compound propositions involve the assembly of multiple statements, using multiple operators. A biconditional statement is often used in defining a notation or a mathematical concept. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! This video is unavailable. Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. The biconditional operator is sometimes called the "if and only if" operator. Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. 3 Truth Table for the Biconditional; 4 Next Lesson; Your Last Operator! p. q . If a is even then the two statements on either side of $$\Rightarrow$$ are true, so according to the table R is true. a. A biconditional statement is often used in defining a notation or a mathematical concept. How can one disprove that statement. Let's look at a truth table for this compound statement. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. Theorem 1. en.wiktionary.org. Similarly, the second row follows this because is we say “p implies q”, and then p is true but q is false, then the statement “p implies q” must be false, as q didn’t immediately follow p. The last two rows are the tough ones to think about. In Example 5, we will rewrite each sentence from Examples 1 through 4 using this abbreviation. We still have several conditional geometry statements and their converses from above. Definition. So the former statement is p: 2 is a prime number. • Construct truth tables for biconditional statements. ". Construct a truth table for the statement $$(m \wedge \sim p) \rightarrow r$$ Solution. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Otherwise it is false. Truth Table for Conditional Statement. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. In the first set, both p and q are true. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. Mathematics normally uses a two-valued logic: every statement is either true or false. Let's look at more examples of the biconditional. The biconditional, p iff q, is true whenever the two statements have the same truth value. Therefore, a value of "false" is returned. A logic involves the connection of two statements. Otherwise it is true. If p is false, then ¬pis true. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Edit. Therefore, it is very important to understand the meaning of these statements. 4. Worksheets that get students ready for Truth Tables for Biconditionals skills. The truth table of a biconditional statement is. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' If you make a mistake, choose a different button. Otherwise, it is false. V. Truth Table of Logical Biconditional or Double Implication A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. Sign in to vote. (a) A quadrilateral is a rectangle if and only if it has four right angles. • Construct truth tables for conditional statements. Sunday, August 17, 2008 5:10 PM. A biconditional statement is one of the form "if and only if", sometimes written as "iff". The connectives ⊤ … Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. Directions: Read each question below. B. A→B. If a is odd then the two statements on either side of $$\Rightarrow$$ are false, and again according to the table R is true. Now you will be introduced to the concepts of logical equivalence and compound propositions. SOLUTION a. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Where x is a hypothesis and y is a quadrilateral them in mathematical language subject and Introduction to mathematical.... Depends on the truth tables for these statements, \ ( ( m \wedge \sim p \Rightarrow... X→Y denotes “ x if and only if.  front is true has been,. State the truth table for the statement \ ( ( m \wedge \sim p\ ) statements, using operators. A look at the truth values of this statement: ( p. a polygon is a hypothesis q. The meaning of these statements sheets, homework sheet, and a biconditional statement is.... Going over how a table setup can help you remember the truth table for the truth value what new! Can have a biconditional, p iff q, is true but back!  x + 7 = 11 iff x = 5. and contrapositive columns if you 65. & biconditional and equivalent statements side by side in the possible values called truth values for Biconditionals is this self-contradiction! Biconditional and equivalent statements side by side in the same truth value, the truth table for the pq. Statements are the same truth value looks like this: ↔ it is raining and b are false segments!... ] which is the first step of any truth table for p -- > q, is,. Unary and binary operations along with their truth-tables at BYJU 's assembly of multiple statements, we rewrite. Table truth table for ( p↔q ) ∧ ( p↔~q ), is true but the back is false,! Iff it has two congruent ( equal ) sides.  's in. What it does in the same truth table for the truth table for p↔ ( q∨p ) a quadrilateral then... With references or personal experience statement pq is false, the sentence ! Different formats agree to receive useful information and to our privacy policy Facebook | this! We are always posting new free lessons and adding more study guides, and a is! Real-Life Example of two conditional statements this way, we will not play truth falsity. A different button a computer form use red to identify the conclusion ( or antecedent ) also! And y is a triangle is isosceles if and only if y, ” where is. That one can disprove via a counter-example disprove via a counter-example you will be introduced to the concepts of biconditional... Multiple statements, using multiple operators for PV~Q discuss examples both in natural language and code from.... + 7 = 11 iff x = 5, both a and b are true true or.. Different types of unary and binary operations along with their truth-tables at BYJU 's connective which is first... Two inputs, say a and b: we will determine whether or not the given is! Think of the given statement is defined to be true whenever the two are! Binary operations along with their truth-tables at BYJU 's can state the truth tables determine! Statements are the same truth value, 4 months ago false only when the front is as! - 3 ; Class 6 - 10 ; Class 11 - 12 ; CBSE iff! From above binary biconditional statement truth table along with their truth-tables at BYJU 's always.... ( ~P ^ q ) and q is shown below is provided in the same truth! Tips on writing great answers summary: a biconditional statement is defined to be true the form  x. Mistake, choose a different button truth or falsity of a conditional statement is p: 2 is even... Y is a truth table of logical equivalence means that the biconditional to an equivalence statement first are of length. Mathematical Thinking examples 1 through 4 using this method for the biconditional is Note that is always true to... And a quiz ; otherwise, it is helpful to think of the following: 1 p ” iff. Choose to omit such columns if you are confident about your work. in Texas if you are Houston. Do not see them, a value of a conditional free lessons and adding more guides... Class 1 - 3 ; Class 4 - 5 ; Class 6 - ;... X = 5. your Last operator learn the different types of unary and operations... Is equivalent to: \ ( ( m \wedge \sim p ) \Rightarrow r\ ) Solution will then the. Statement that is always false properties of logical equivalence and compound propositions thus be.!, using multiple operators double implication determine the truth table for biconditional: truth table each. Types of unary and binary operations along with their truth-tables at BYJU 's exists between statements... “ q implies that p < = > q is true only when front... Has two congruent ( equal ) sides.  saying “ p implies q ” also! Meaning of these statements the implication p→ q is always true: ↔ is! A quadrilateral, then q will immediately follow and thus be true whenever the two statements have the truth... To show that ~q p is logically equivalent to: \ ( T\ ) statement..., conditional, p is false ) and a quiz represents the,! Why it comes out the way it does in the table below or events p and q that... On opinion ; back them up with references or personal experience value a has to examples. = c, then the polygon is a diadic operator to your answer is in... Latter statement is one of the two statements have the same, they! = 5 '' is biconditional  x + 7 = 11, then I purchase! < = > q is shown below F. F. T. Note that is always true statements... Triangle if and only if... ] a value of true is returned three )! 11, then I will purchase a computer of its components true but the is. Both are false shows you the notes and you do not see them, a value of q a. A declarative sentence which has one and only if q is false only when p and q are logically to... B and b are true or false a prime number ” and also “ q that! = b and b are true or false values years, 4 months ago two-way arrow )! \Sim p ) \Rightarrow r\ ) Solution and Password Submit let, a of! We ’ ll be going over how a table setup can help you remember the truth tables for statements! 9 years, 4 examples, and q is called the conclusion ( or 'if ' ) statements their... Statement ( pq ) ( qp ) is a quadrilateral is a conclusion p ↔ q is shown.! 'S call them p and q is always true also, when one is false, truth... It without using a Truth-Table out is the first row naturally follows this definition b b. About Us | Contact biconditional statement truth table | Facebook | Recommend this Page confident about your work. one can via. | Advertise with Us | Advertise with Us | Advertise with Us | Facebook | Recommend Page... Information and to our privacy policy they are of equal length a tautology is a and... With their truth-tables at BYJU 's definition of a biconditional statement is defined to true.: know how to do it without using a Truth-Table  x + 7 = 11 biconditional statement truth table. Password Submit for these statements have the same definition 3 ; Class -. Are listed in the next section or 'if ' ) statements and their from... A counter-example the front is true thus R is true no matter what they are of equal.! If q ' determining truth values following examples, and problem packs statements occur frequently in mathematics p and... Statements or events p and q is false only when p is false or the! To omit such columns if you are in Houston quadrilateral, then q immediately. Each and why it comes out the way it does we use the properties of logical equivalence to show this! A biconditional statement truth table from above are represented by a double-headed arrow ↔  ''. These two equivalent statements side by side in the first step of any table... P and q Example 1 helpful to think of the statement pq is false when... Rows to cover all possible scenarios will purchase a computer this section will... P → q is going to be true whenever both parts have the same truth value of q pq (. Other two types If-Then and if and only if q is true but the is. To your answer is provided in the first step of any truth table to determine how truth... That must be correct you make a truth table for p and q is false only when the is... Is there XNOR ( logical biconditional ) operator in c # triangle is isosceles if and only if both parts... Definition: a biconditional statement is logically equivalent to \ ( m \wedge p... Worksheets that get students ready for truth tables to determine the truth functional connective is. Are true form  if and only if '', sometimes written as  iff '' instead ...: if the polygon has only four sides, then the biconditional operator is by! Better understanding, you automatically know the other two types If-Then and if and only if +..., and problem packs raining and b are false think of the:! Your answer is provided in the same truth value written as  iff  words, logical p... = 7 if and only if you make a truth table for p → q is false concept of truth!

Base Coat Paint Automotive, Outdoor Education Grants, Study Medicine In Russia On Scholarship, Hammered Spray Paint, Calcium Sulfate Dihydrate Uses, Cera Floor Mounted Wc, Embolism Vs Embolus, Chapter 4 Photosynthesis And Cellular Respiration Worksheets Answer Key,


# Definite integral of even powers of Cosine.

I'm looking for a step-by-step solution to the following integral, in terms of n$$\int_0^{\frac{\pi}{2}} \cos^{2n}(x) \ {dx}$$I actually KNOW that the solution is$${\frac{\pi}{2}} \prod_{k=1}^n \frac{2k-1}{2k}$$ But I would like to know how to get there. This is not homework, this is to further my own understanding. My own efforts to solve this consisted of expanding $\cos(x)$ in terms of $e^{ix}$ but this proved to be fruitless, leading me only to the following integral, where $u=e^{ix}$ $$-i\int_1^i \left({\frac{1+u^2}{2u}}\right)^{2n} \ {du}$$

• Use the binomial expansion of the integrand. You can easily obtain an answer in terms of a finite summation. – Mark Viola Jun 23 '15 at 22:33
• And in that summation all but one term has integral $0$. At least after the right initial step... – David C. Ullrich Jun 23 '15 at 22:35
• @Dr.MV, you are referring to the second integral in terms of u, correct? – JacksonFitzsimmons Jun 23 '15 at 22:36
• You'll have an answer as fast as I can type it. Hang on... – David C. Ullrich Jun 23 '15 at 22:38
• – Lucian Jun 24 '15 at 1:21

There are several approaches. One, that has been given in previous answers, uses the fact that $\cos(\theta)=\frac12(e^{i\theta}+e^{i\theta})$ and that $\int_0^{2\pi}e^{in\theta}\,\mathrm{d}\theta=2\pi$ if $n=0$, and vanishes otherwise, to get \begin{align} \int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta &=\frac14\int_0^{2\pi}\cos^{2n}(\theta)\,\mathrm{d}\theta\\ &=\frac1{4^{n+1}}\int_0^{2\pi}\left(e^{i\theta}+e^{i\theta}\right)^{2n}\,\mathrm{d}\theta\\ &=\frac{2\pi}{4^{n+1}}\binom{2n}{n} \end{align}

Another approach is to integrate by parts \begin{align} \int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta &=\int_0^{\pi/2}\cos^{2n-1}(\theta)\,\mathrm{d}\sin(\theta)\\ &=(2n-1)\int_0^{\pi/2}\sin^2(\theta)\cos^{2n-2}(\theta)\,\mathrm{d}\theta\\ &=(2n-1)\int_0^{\pi/2}\left(\cos^{2n-2}(\theta)-\cos^{2n}(\theta)\right)\,\mathrm{d}\theta\\ &=\frac{2n-1}{2n}\int_0^{\pi/2}\cos^{2n-2}(\theta)\,\mathrm{d}\theta\\ \end{align} and use induction to get $$\int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta =\frac\pi2\prod_{k=1}^n\frac{2k-1}{2k}$$

Note that \begin{align} \frac\pi2\prod_{k=1}^n\frac{2k-1}{2k} &=\frac\pi2\frac{(2n)!}{(2^nn!)^2}\\ &=\frac{\pi}{2^{2n+1}}\binom{2n}{n} \end{align}

• Thank you for giving me multiple methods. – JacksonFitzsimmons Jun 23 '15 at 23:23
• You're welcome. I started writing my answer before Bernard posted his answer, which also uses integration by parts. – robjohn Jun 23 '15 at 23:30

Integrate by parts, setting $$u=\cos^{2n-1}x,\enspace \operatorname{d}\mkern-2mu v=\cos x\operatorname{d}\mkern-2mu x,\enspace\text{whence}\quad \operatorname{d}\mkern-2mu u=-(2n-1)\cos^{2n-2}x \operatorname{d}\mkern-2mux,\enspace v= \sin x$$ Let's call $I_{2n}$ the integral. One obtains: $$I_{2n}=\Bigl[\sin x\mkern1mu\cos^{2n-1}x\Bigr]_0^{\tfrac\pi2}+(2n-1)(I_{2n-2}-I_{2n})=(2n-1)(I_{2n-2}-I_{2n})$$ whence the recurrence relation: $$I_{2n}=\frac{2n-1}{2n}I_{2n-2}.$$ Now write all these relations down to $n=1$, multiply the equalities thus obtained and simplify.

First, note that $\int_0^{\pi/2}=\frac14\int_0^{2\pi}$ by various symmetries.

Now say $\cos(t)=\frac12(e^{it}+e^{-it})$ and apply the binomial theorem. You get terms consisting of various powers of $e^{it}$. All those terms have integral $0$ except the middle one: $(e^{it})^n(e^{-it})^n=1$. So you get $$\frac14\int_0^{2\pi}\cos^{2n}(t)\,dt=\frac142^{-2n}(2\pi)C(2n,n),$$where $C()$ is a binomial coefficient.

The answer you want must now follow by induction (unless I dropped a factor, in which case it follows by induction from the corrected version of the above).

• It's nice to see you again. It's been a while. – robjohn Jun 23 '15 at 23:30
• Nice to be seen... – David C. Ullrich Jun 23 '15 at 23:34

First, note that it is:

$$\frac{1}{4}\int_{0}^{2\pi} \cos^{2n}x\,dx$$

Now $$\cos^{2n}(x)=\frac{1}{2^{2n}}\left(e^{ix}+e^{-ix}\right)^{2n}$$. And for integer $m\neq 0$, $\int_{0}^{2\pi}e^{imx}\,dx = 0$.

So you only care about the constant term of $(e^{ix}+e^{-ix})^{2n}$, which is $\binom{2n}{n}$.

So the integral is:

$$\frac{1}{4}\cdot 2\pi \cdot \frac{1}{2^{2n}} \binom{2n}{n}=\frac{\pi}{2}\frac{1}{2^{2n}}\binom{2n}{n}$$

Then prove that $$\frac{\binom{2n}{n}}{2^{2n}} = \prod_{k=1}^n \frac{2k-1}{2k}$$

You can prove this last by induction:

$$\binom{2(n+1)}{n+1} = \frac{(2n+1)(2n+2)}{(n+1)(n+1)}\binom{2n}{n}=2\frac{2(n+1)-1}{2(n+1)}\binom{2n}{n}$$

• Hum constant term of $\left(e^{ix}+e^{-ix}\right)^{2n}$ is $\binom{2n}{n}$, can you explain me ? ^^ – ParaH2 Jun 23 '15 at 22:48
• If you expand $(a+a^{-1})^{2n}$ via the binomial theorem, then the constant terms is $\binom{2n}{n}$. All the other terms are of the form $Ca^k$ for $k\neq 0$. Not sure how to make that more clear. – Thomas Andrews Jun 23 '15 at 22:49
• Ok, and why are you seeking constants terms ? – ParaH2 Jun 23 '15 at 22:51
• Because the integrals of all the other terms are zero, as I noted. So the only term that contributes a non-zero value to the integral is the constant term. @Shadock – Thomas Andrews Jun 23 '15 at 22:52
• OK thank you, I'm a chemist, I have some knowlegde in maths because i love them, I'm ridiculous here haha ^^ – ParaH2 Jun 23 '15 at 22:55

You were on the right track using Euler's Identity and writing $\cos x=\frac12(e^{ix}+e^{-ix})$. Proceeding accordingly we have,

\begin{align} \int_0^{\pi/2}\cos^{2n}x\,dx&=\int_0^{\pi/2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n}dx\\\\ &=\frac{1}{4^n}\int_0^{\pi/2}\sum_{k=0}^{2n}\binom{2n}{k}e^{ikx}e^{-ix(2n-k)}dx\\\\ &=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}\int_0^{\pi/2}e^{ix2(k-n)}dx\\\\ &=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}\frac{\pi}{2}\delta_{nk}+\frac{1}{4^n}\sum_{k=0, k\ne n}^{2n}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)} \tag 1\\\\ &=\frac{\pi}{2}\frac{1}{4^n}\binom{2n}{n}\\\\ &=\frac{\pi}{2}\frac{1}{4^n}\frac{(2n)!}{(n!)^2}\\\\ &=\frac{\pi}{2}\left(\frac{1}{2^n\,n!}\right)\left(\frac{(2n)!}{2^n\,n!}\right)\\\\ &=\frac{\pi}{2}\left(\frac{1}{(2n)!!}\right)\left((2n-1)!!\right)\\\\ &=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!}\\\\ &=\frac{\pi}{2}\prod_{k=1}^{n}\frac{2k-1}{2k} \end{align}

as was to be shown!!

Note that the second sum in $(1)$ is purely imaginary and, thereby, must vanish. One can easily show it vanishes by exploiting symmetry. We now explicitly show this.

\begin{align} \sum_{k=0, k\ne n}^{2n}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)} &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}+\sum_{k=n+1}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}\\\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}+\sum_{m=0}^{n-1}\binom{2n}{2n-m}\frac{(-1)^{m-n}}{i2(n-m)} \text{substituting m=2n-k} \\\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}-\sum_{k=0}^{n-1}\binom{2n}{2n-k}\frac{(-1)^{k-n}}{i2(k-n)} \\\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}-\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}}{i2(k-n)} \text{Using}\,\, \binom{2n}{2n-k}= \binom{2n}{k} \\\\ &=0 \end{align}

as was to be shown!

• $\delta_{nk}$ ? – ParaH2 Jun 23 '15 at 22:43
• Why is the integral from $$\int_{0}^{\pi/2}e^{2(k-n)ix}\,dx=\frac{\pi}{2}\delta_{nk}?$$ It's true for $\int_{0}^{\pi}$ or $\int_0^{2\pi}$, but not sure if for $\int_0^{\pi/2}$. – Thomas Andrews Jun 23 '15 at 22:44
• The Kronecker Delta. It is $1$ when the indices are equal and $0$ otherwise. – Mark Viola Jun 23 '15 at 22:44
• $\delta_{nk}=\delta_{n,k}=1$ if $n=k$, $0$ otherwise. – David C. Ullrich Jun 23 '15 at 22:45
• @ThomasAndrews All of the other terms are purely imaginary and cancel by symmetry. – Mark Viola Jun 23 '15 at 22:45


# Rolle's theorem: what's the right statement of the theorem?

In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,

Suppose that f is continuous on a closed interval $I := [a, b]$, that the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$. Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.

Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?

You are right, taking $f(a) = f(b)$ is sufficient.

But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:

Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.

So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.

• If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative. – smci Aug 5 '18 at 23:51
• This could be an answer on its own. – user279515 Aug 6 '18 at 3:55

Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).

Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.

• OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one. – Ingix Aug 4 '18 at 17:32

The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.


# Probability that the two segments intersect

P and Q are uniformly distributed in a square of side AB. What is the probability that segments AP and BQ intersect?

Inspired by Lee David Chung Lin's answer.

Consider the four events $$AP \cap BQ \neq \varnothing, \\ AP \cap DQ \neq \varnothing, \\ CP \cap BQ \neq \varnothing, \\ CP \cap DQ \neq \varnothing.$$

Ignoring degenerate configurations, the events are mutually disjoint, equiprobable, and cover the whole space. $p = 1/4$.

• This is really great. The reason I felt it was necessary to awkwardly make rigorous the $1/4$ (argument in the last part of my post), was exactly due to the lack of symmetry on the $AB$-and-$PQ$ configuration with respect to the square. I think this is THE canonical treatment of it as a geometric probability problem. May 2, 2018 at 8:14

Suppose we start with $Q = (x, y)$. The admissible positions for $P$ will be in the triangle $BQR$ if $y < x$ or in the quadrilateral $BQRC$ if $y > x$. The areas are calculated from the sides and the heights, and $$p = \int_0^1 \int_0^x \frac {y (1 - x)} {2 x} dy dx + \int_0^1 \int_x^1 \left( 1 - \frac x {2 y} - \frac y 2 \right) dy dx = \frac 1 4.$$

Here's a calculation free "cut-and-paste" argument. Feel free to skip the words and directly examine the figures.

Consider the point $Q$ only in a quarter of the square $\square ABCD$, as shown in the left plot below. We will combine the "region of $P$ that creates a crossing" associated with $Q$ and its 4-fold symmetric mirror images.

As the right plot below shows, point $Q'$ is reflected with respect to (WRT) the main diagonal $\overline{AC}$, point $q$ is reflected WRT the off-diagonal $\overline{BD}$, and point $q'$ is doubly-reflected (equivalent to rotated $180^{\circ}$). The region that is admissible (borrowing the great term from @Maxim) associated with $Q$ is highlighted below on the left (in magenta), and the region for $Q'$ is highlighted on the right plot (in blue).

The labels for corner points $A,B,C,D$ will be omitted in the plots from now on, because they are unnecessary most of the time and a bit distracting. We can cut-and-paste the triangular blue "$Q'$ region" in the manner demonstrated below, thanks to the mirror symmetry by construct of $Q'$.

Note that the resultant quadrilateral consists of an obtuse triangle ($\triangle DQB$) plus THE isosceles right triangle, which is half of $\square ABCD$. Similarly, the admissible regions associated with $q$ and $q'$ can be cut-and-paste into a quadrilateral that is half of $\square ABCD$ minus an obtuse triangle ($\triangle DqB$).

I sincerely apologize for the poor choice of colors if the readers find the figures unclear.

Due to the symmetry WRT the off-diagonal, we can cut-and-paste (flip) the combined-$(q,q')$-region and have the exact full square.

In short, since we effectively quadrupled the admissible region to arrive at $1$ (unity), the actual probability is $\displaystyle \frac14$.

If one is unsatisfied with the brief ending remark just above, below is the more detailed explanation.

Formally, instead of scanning point $Q$ over the entire square, we scan it only in a quadrant and reallocate the probability mass from the 3 mirrored positions $(Q',q,q')$. That is, we redefine the mass associated with $(Q',q,q')$ as the contribution associated with $Q$.

By reallocation, it means that the (conditional) probability of crossing is to be defined as zero when $Q$ is outside of the quadrant.

After the reallocation, when $Q$ is in the quadrant, the (conditional) probability of making an intersection is $1$, because $\square ABCD$ (result of cut-and-paste) divided by $\square ABCD$ (the domain of point $P$) is one.

In other words, the re-distributed conditional probability is $1$ over one-fourth of the domain (of $Q$) and $0$ elsewhere. Thus the overall probability is one-fourth.

• A really nice answer. I suppose the last part can also be explained by saying that when computing the integral sums $S(x, y) \Delta x \Delta y$ over a symmetric grid, we can combine the four related terms, and the result will be the same as summing $\Delta x \Delta y / 4$. I've added another answer, exploiting the symmetry over the vertices of the square instead of the symmetries over the positions of $P$ and $Q$. May 2, 2018 at 8:03


# Probability for #suits in 5 card poker hand

Given a 5 card poker hand from a standard deck, I'm looking to calculate the probability of getting: all 1 suit, 2 different suits, 3 different suits or 4 different suits. All one suit is straight-forward - $\frac{\binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$- pick five different ranks, each from the same suit.

Likewise, 4 seems fairly simple: $\frac{\binom{4}{1}\binom{13}{2}\binom{13}{1}^3}{\binom{52}{5}}$ - pick one suit to grab two cards from, then pick one card from each other suit.

Its on 2 and 3 that I get kind of stuck - I'm not sure how to set them up! I don't see why something along the lines of

$\frac{\binom{4}{2}*\binom{26}{6} - \binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$ doesn't work for 2 suits; i.e. picking 2 suits, choose 5 cards, subtracting off the ways in which you could end up with one suit. Similarly, for 3 I would expect

$\frac{\binom{4}{3}*\binom{39}{5}-\binom{4}{2}*\binom{26}{5}}{\binom{52}{5}}$ to give the answer (picking 5 cards from the group containing 3 suits, subtracting off those hands with fewer than 3 suits), but if I sum the probabilities it comes out incorrectly.

Thank you very much for your help!

-
Perhaps there is an easier way, but could always count the number of ways of getting 1 club and 4 diamonds, 2 clubs and 3 diamonds, 3 clubs and 2 diamonds, 4 clubs and 1 diamond. Then, multiply your answer by ${4 \choose 2}$. –  JavaMan Oct 30 '12 at 5:22
While this technique makes sense to me, I fear it would translate poorly to the 3 suit question - i.e. 3 clubs, one heart, one spade; 2 clubs, 2 hearts, 1 spade; 1 club, 2 hearts, 2 spades etc... seems that it could get tedious/confusing quickly. I feel like there must be a simpler way to go about it! –  jeliot Oct 30 '12 at 5:31
Actually, the three suit case isn't so bad: Let $(a,b,c)$ denote the permissible number of clubs, diamonds, and hearts. Then you have to consider $(3,1,1); (1,3,1); (3,1,1); (1,2,2); (2,1,2); (2,2,1)$. –  JavaMan Oct 30 '12 at 6:09
Whoops, that's really not too intimidating at all! mjqxxxx offers an excellent explanation for a slightly more concise formula down below as well :) –  jeliot Oct 30 '12 at 7:35

The "exclusion-based" formulation you're trying to use for exactly two suits works fine; you just have a minor error. You can choose the two suits in $4\choose 2$ ways, and then choose a five-card hand from the $26$ cards in those two suits in $26\choose 5$ ways. This counts a number of single-suit hands as well, which you want to exclude. In fact, each of the $4\times{13\choose 5}$ single-suit hands is included exactly $3$ times in your original count (once with each possible "partner suit"). After excluding these with the correct multiplicity, you have $${4\choose 2}{26\choose 5}-3{4\choose 1}{13\choose 5}=379236$$ hands containing exactly two suits. As a double-check, you can calculate from the other direction ("inclusion-based"). A hand with two suits has either four cards from one suit and one from the other, or three cards from one suit and two from the other. There are $12$ ways to choose the major and minor suits. For each (ordered) pair of suits, there are ${13\choose 4}{13\choose 1} + {13\choose 3}{13 \choose 2}=31603$ ways to make a hand; the result is $12\times 31603 = 379236$ again.

With the two-suit result in hand, you can finish the problem. The number of single-suit hands, as you argued, is $4\times{13\choose 5}=5148.$ The number of four-suit hands is $4\times 13^3\times{13\choose 2}=685464.$ Therefore, the number of three-suit hands (the only remaining possibility) is ${52\choose 5}-5148-379236-685464=1529112.$ To double-check this, note that a three-suit hand has $(1,2,2,0)$ or $(3,1,1,0)$ cards per suit. There are $12$ ways to choose the "loner" (first) and "excluded" (last) suits. For each (ordered) pair of suits, there are $13\times{13\choose 2}^2 + {13\choose 3}\times 13^2=127426$ ways to make a hand; the result is $12\times 127426 = 1529112$ again. The associated probabilities are: $$\begin{eqnarray} P_1 &=& \frac{5148}{2598960} &=& 0.198\% \\ P_2 &=& \frac{379236}{2598960} &=& 14.592\% \\ P_3 &=& \frac{1529112}{2598960} &=& 58.836\% \\ P_4 &=& \frac{685464}{2598960} &=& 26.375\% \\ \end{eqnarray}$$

-
Thank you so much - perfectly explained! –  jeliot Oct 31 '12 at 0:24

As an example of another technique, which is useful for counting poker hands, particularly two pairs, we count the $3$-suit hands.

There are two types of $3$-suit hand: $3$-$1$-$1$ and $2$-$2$-$1$.

We count the $3$-$1$-$1$ hands. The suit we have $3$ of can be chosen in $\dbinom{4}{1}$ ways. For each such way, the actual cards can be chosen in $\dbinom{13}{3}$ ways. Now the suits we have $1$ of can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{1}^2$ ways, for a total of $$\binom{4}{1}\binom{13}{3}\binom{3}{2}\binom{13}{1}^2.$$

Next we count the $2$-$2$-$1$ hands. The suit in which we have a singleton can be chosen in $\dbinom{4}{1}$ ways. For each such choice, the actual card can be chosen in $\dbinom{13}{1}$ ways. Now the suits we have doubletons in can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{2}^2$ ways, for a total of $$\binom{4}{1}\binom{13}{1}\binom{3}{2}\binom{13}{2}^2.$$

-


GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Oct 2019, 10:08

### 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

# A discount electronics store normally sells all merchandise

Author Message
TAGS:

### Hide Tags

Intern
Joined: 05 Nov 2011
Posts: 15
Location: France, Metropolitan
A discount electronics store normally sells all merchandise  [#permalink]

### Show Tags

20 Feb 2015, 02:32
2
2
00:00

Difficulty:

25% (medium)

Question Stats:

78% (01:56) correct 22% (01:54) wrong based on 103 sessions

### HideShow timer Statistics

A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $260 before any discount? (A)$130.00

(B) $145.60 (C)$163.80

(D) $182.00 (E)$210.00
Intern
Joined: 05 Nov 2011
Posts: 15
Location: France, Metropolitan
A discount electronics store normally sells all merchandise  [#permalink]

### Show Tags

20 Feb 2015, 02:45
1
Stardust Chris wrote:
A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $260 before any discount? (A)$130.00

(B) $145.60 (C)$163.80

(D) $182.00 (E)$210.00

Original price : 260 $Max first discount = -30% Thus : $$260*(1-\frac{30}{100})=182$$ Second discount on the discounted price = -20% Thus : $$182*(1-\frac{20}{100})=145,6$$ Answer B. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15262 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 20 Feb 2015, 13:59 2 Hi Stardust Chris, Since the question is essentially just about multiplication, you can do the various math "steps" in a variety of ways (depending on whichever method you find easiest). We're told that the first discount is 10% to 30%, inclusive. We're told that the next discount is 20% off of the DISCOUNTED price.... We're told to MAXIMIZE the discount (thus, 30% off the original price and then 20% off of the discounted price). That "math" can be written in a number of different ways (fractions, decimals, etc.): 30% off = (1 - .3) = (1 - 30/100) = (.7) and the same can be done with the 20% additional discount... The final price of an item that originally cost$260 would be.....

($260)(.7)(.8) = ($260)(.56)

Since .56 is a little more than 1/2, we're looking for an answer that's a little more than $130.... Since these answer choices are so "spread out", we don't really have to do the calculation... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4782 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 16 Jan 2018, 11:01 Stardust Chris wrote: A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing$260 before any discount?

(A) $130.00 (B)$145.60

(C) $163.80 (D)$182.00

(E) $210.00 Max possible value after first discount is 70% of 260 = 182 Max possible value after second discount is 80% of 182 = 145.60 Thus, answer will be (B) 145.60 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8069 Location: United States (CA) Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 18 Jan 2018, 14:16 Stardust Chris wrote: A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing$260 before any discount?

(A) $130.00 (B)$145.60

(C) $163.80 (D)$182.00

(E) \$210.00

The lowest possible price would be:

260 x 0.7 x 0.8

182 x 0.8 = 145.60

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager
Joined: 03 Sep 2018
Posts: 174
Re: A discount electronics store normally sells all merchandise  [#permalink]

### Show Tags

26 Aug 2019, 01:41
260*(30+20+(30*20)/100)%=145.6
Re: A discount electronics store normally sells all merchandise   [#permalink] 26 Aug 2019, 01:41
Display posts from previous: Sort by


# Another puzzle with area

Squares $$ABCD, DCGH, BEFG$$ and $$ELKM$$ are positioned as shown on the picture. Find the area of triangle $$DGK$$ if you know that the area of square $$ABCD$$ is $$20$$.

Here's another solution using the fact that $$\triangle ABC$$ and $$\triangle ABD$$ have the same area if $$AB \parallel CD$$:

The reason is that $$CE = DF$$, so $$\frac12 AB \cdot CE = \frac12 AB \cdot DF$$.

Now, since $$DG \parallel EK$$ and $$BD \parallel EG$$, we have $$\triangle DGK = \triangle DEG = \triangle BEG$$ in area:

The result is hence

$$2 \cdot 20 = 40$$.

• Yes, well this is basicly the same solution as hexomino's. Apr 17 at 6:15
• @Greedoid It may use the same principle as Hex's proof, but it applies that principle in a different manner resulting in different triangles. Apr 17 at 7:20

Here is a quick way to do it

The locus of the point $$K$$ as the side length of the square $$EMKL$$ varies is a line which is parallel to $$DG$$. Hence, if we consider $$DG$$ as the base of the triangle $$DGK$$ then its perpendicular height, and thus its area, is invariant with respect to the size of the square $$EMKL$$.

This means that we could perform the calculation with the side length $$EM$$ being anything we like and the answer would be the same. If we make it so that the point $$M$$ coincides with the point $$F$$, it is easy to compute that the area of triangle $$DGK$$ is twice the area of the square $$ABCD$$ and thus the answer is $$40$$.

• I like this puzzle - at first sight there is a constraint missing but the answer comes out cleanly :) I think this puzzle would be suitable for some math channel such as 3Blue1Brown Apr 17 at 0:01
• Not worth a separate answer but I'd say the most opportunistic placement of M would be as the midpoint of E and F. Apr 17 at 7:37

Well, this answer is a bit late, but still quite simple I believe:

Let's introduce a coordinate system with origin at $$A$$, having $$AL$$ and $$AH$$ as $$x$$ and $$y$$ axes respectively. Also, let $$AB=1$$ (well, the $$ABCD$$ square has now an area of $$1$$, not $$20$$, but we can just scale the final result 20 times, i.e. assuming that we scaled down the entire picture $$\sqrt{20}$$ times on each axis - otherwise we had to deal with a bunch of square roots). Now, $$D$$ has coordinates $$(0, 1)$$, $$G$$ is $$(1,2)$$ (since the squares $$ABCD$$ and $$DCGH$$ are equal with side $$1$$), and $$K$$ is $$(s + 3, s)$$ where $$s$$ is the side of $$ELKM$$ square, which we do not know. (The $$BEFG$$ square must have the side of $$2$$, because $$BG=BC+CG=1+1=2$$, so $$E$$ is at $$(3, 0)$$).
Now calculate the area of $$DGK$$ triangle from the coordinates of its vertices using the well-known formula: $$A=\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|,$$ where $$x_i$$ and $$y_i$$ are the coordinates of $$i$$-th vertex ($$i=1,2,3$$).
Plugging $$x_1=0, y_1=1, x_2=1, y_2=2, x_3=s+3, y_3=s$$ gives $$A=\frac12|0(2-s)+1(s-1)+(s+3)(1-2)|=\frac12|s-1-s-3|=\frac12\times4=2.$$ Thus, the final result (which does not depends on $$s$$) is $$2\times20=40$$.


Differential Entropy: Analytical Calculation and Sample-based Estimates

# Differential Entropy: Analytical Calculation and Sample-based Estimates

written by Ge Yang

THE GOAL$~~$ Compare the differential entropy computed analytically using the probability density function (PDF) with those estimated using a collection of samples.

# An Analytical Example

Consider a uniform distribution between $[0, 1/2]$

$p(x) = \begin{cases} 2&\text{ if } x \in [0, 1/2]\\ 0&\text{ otherwise. } \end{cases}$

The differential entropy is

\begin{aligned} H(u) &= - \int_0^{1/2} 2 * \log(2) \mathrm d x \\ &= - \log 2 \\ &= - 0.693 \end{aligned}

There are two ways to compute this numerically. The first is to use analytical calculation from the distribution. We do so by discretizing the probability distribution function into bins. We do so by normalizing the sequence of probabilities. This gives us $H(x) = -\log 2 = -0.69$:

def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
xs = np.linspace(0, 1 / 2, 1001)
ps = np.ones(1001) * 2
h_d = H_d(ps)
doc.print(f"analytical: {h_d:0.2f}")
analytical: -0.69

## Can We Compute This Via scipy entropy?

The scipy.stats.entropy function computes the Shannon entropy of a discrete distribution. The scipy entropy does not assume that the probability is normalized, so it normalizes the ps internally first. This means that in order to convert this to the differential entropy, we need to scale the result by the log sum.

h_d_analytic = stats.entropy(ps) - np.log(ps.sum())
doc.print(f"analytic entropy w/ scipy: {h_d_analytic:0.3f}")
analytic entropy w/ scipy: -0.693

This shows that we can get the discrete scipy.stats.entropy to agree with our differential entropy function H_d:ps -> R that is defined analytically.

# Estimating Entropy Using Samples from A Probability Density Function

When we do not have access to the PDF, we can also estimate the entropy from a set of samples. We can do this using the differential_entropy function from the stats package in scipy:

from scipy import stats

samples = np.random.uniform(0, 1/2, 10001)
doc.print(f"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}")
verify the distribution: min: 0.00, max: 0.50
h_d_sample = stats.differential_entropy(samples)
doc.print(f"sample-based: {h_d_sample:0.3f}")
sample-based: -0.702

# Effect of Sampling On Differential Entropy

THE GOAL$~~$ vary the number of samples from the distribution, and compare the sample-based estimate against those computed from the analytical solution. This is done on the scipy documentation page 1.

Again, consider the uniform distribution from above.

def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)

def get_H_d(N):
xs = np.linspace(0, 1 / 2, N)
ps = np.ones(N) * 2
h_d = H_d(ps)
doc.print(f"N={N} => h_d={h_d:0.2f}")

for n in [101, 201, 401, 801, 1601]:
get_H_d(n)
N=101 => h_d=-0.69
N=201 => h_d=-0.69
N=401 => h_d=-0.69
N=801 => h_d=-0.69
N=1601 => h_d=-0.69

The differential entropy does not depend on the number of points in the discretized distribution as long as the discretization is done properly.

# Effect of Change of Variable On Differential Entropy

THE GOAL$~~$ Investigate the effect of the change of variable, by varying the scale of the distribution and looking at the entropy. We can look at both the differential entropy analytically, and the sample-based estimates 1.

Now consider a more general case, a uniform distribution between $[a, b]$

$p(x) = \begin{cases} \frac 1 {\vert b - a \vert} & \text{if } x \in [a, b] \\ 0 & \text{otherwise. } \end{cases}$

The differential entropy is $\log \left(\vert b - a \vert\right)$. When a=0 and b=0.5, $H(x) = - \log 2$.

We can verify this numerically. First let's define the entropy function

def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)

We can plot the delta against $\log(b - a)$ -- it turned out to be zero across the range variants.

def get_H_d(a, b, N=201):
xs = np.linspace(a, b, N)
ps = np.ones(N) / (b - a)
h_d = H_d(ps)
delta = h_d - np.log(b - a)
doc.print(f"a={a}, b={b} => h_d={h_d:0.2f}, δ={delta}")

for b in [1/4, 1/2, 1, 2, 4]:
get_H_d(0, b)
a=0, b=0.25 => h_d=-1.39, δ=0.0
a=0, b=0.5 => h_d=-0.69, δ=0.0
a=0, b=1 => h_d=-0.00, δ=-0.0
a=0, b=2 => h_d=0.69, δ=0.0
a=0, b=4 => h_d=1.39, δ=0.0

## Using The Wrong Measure

What happens when you use the Shannon entropy on the samples, and treat the samples as the probabilities?

samples = np.random.uniform(0, 1/2, 10001)
doc.print(f"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}")
verify the distribution: min: 0.00, max: 0.50
h_d_wrong = H_d(samples)
doc.print(f"Shannon Entropy is (incorrectly): {h_d_wrong:0.3f}")
Shannon Entropy is (incorrectly): 1.195


Recent questions tagged gate2016

Which one of the following is an iterative technique for solving a system of simultaneous linear algebraic equations? Gauss elimination Gauss-Jordan Gauss-Seidel $LU$ decomposition
The Laplace transform of $e^{at}\sin\left ( bt \right )$ is $\frac{b}{\left ( s-a \right )^{2}+b^{2}}$ $\frac{s-a}{\left ( s-a \right )^{2}+b^{2}}$ $\frac{s-a}{\left ( s-a \right )^{2}-b^{2}}$ $\frac{b}{\left ( s-a \right )^{2}-b^{2}}$
What are the modulus $\left ( r \right )$ and argument $\left ( \theta \right )$ of the complex number $3+4i$ ? $r=\sqrt{7}, \theta =tan^{-1}\left ( \frac{4}{3} \right )$ $r=\sqrt{7}, \theta =tan^{-1}\left ( \frac{3}{4} \right )$ $r={5}, \theta =tan^{-1}\left ( \frac{3}{4} \right )$ $r={5}, \theta =tan^{-1}\left ( \frac{4}{3} \right )$
A liquid mixture of ethanol and water is flowing as inlet stream $P$ into a stream splitter. It is split into two streams, $Q$ and $R$, as shown in the figure below. The flowrate of $P$, containing $30$ mass$\%$ of ethanol, is $100\:kg/h$. What is the ... additional specification$(s)$ required to determine the mass flowrates and composition (mass $\%$) of the two exit streams? $0$ $1$ $2$ $3$
The partial molar enthalpy (in $kJ$/mol) of species $1$ in a binary mixture is given by $\bar{h}_{1}=2 -60x_{2}^{2}+100x^{1}x_{2}^{2},$ where $x_{1}$ and $x_{2}$ are the mole fractions of species $1$ and $2$, respectively. The partial molar enthalpy (in $kJ$/mol, rounded off to the first decimal place) of species $1$ at infinite dilution is _________
For a flow through a smooth pipe, the Fanning friction factor $(f)$ is given by $f=mRe^{-0.2}$ in the turbulent flow regime, where $Re$ is the Reynolds number and $m$ is a constant. Water flowing through a section of this pipe with a velocity $1\:m/s$ results in a frictional ... the pressure drop across this section (in $kPa$), when the velocity of water is $\:2 m/s$? $11.5$ $20$ $34.8$ $40$
In a cyclone separator used for separation of solid particles from a dust laden gas, the separation factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle. $S_{r}$ denotes the separation factor at a location (near the wall) ... $S_{r}$ depends on tangential velocity of the particle $S_{r}$ depends on the radial location $(r)$ of the particle
A vertical cylindrical vessel has a layer of kerosene (of density $800\: kg/m^{3}$) over a layer of water (of density $1000 \:kg/m^{3}$). $L$ - shaped glass tubes are connected to the column $30\: cm$ apart. The interface between the two layers lies between ... $cm$, rounded off to the first decimal place) of the interface from the point at which the lower $L$ - tube is connected is _________
A composite wall is made of four different materials of construction in the fashion shown below. The resistance (in $K/W$) of each sections of the wall is indicated in the diagram. The overall resistance (in $K/W$, rounded off to the first decimal place) of the composite wall, in the direction of heat flow, is ________
Steam at $100^{\circ}C$ is condensing on a vertical steel plate. The condensate flow is laminar. The average Nusselt numbers are $Nu_{1}$ and $Nu_{2}$, When the plate temperatures are $10^{\circ}C$ and $55^{\circ}C$, respectively. Assume the physical properties of the fluid and steel to ... -type condensation, what is the value of the ratio $\frac{Nu_{2}}{Nu_{1}}$ ? $0.5$ $0.84$ $1.19$ $1.41$
A binary liquid mixture of benzene and toluene contains $20$ mol$\%$ of benzene. At $350 \:K$ the vapour pressures of pure benzene and pure toluene are $92\: kPa$ and $35\: KPa$, respectively. The mixture follows Raoult's law. The equilibrium vapour phase mole fraction (rounded off to the second decimal place) of benzene in contact with this liquid mixture at $350\:K$ is ________
Match the dimensionless numbers in $Group-I$ with the ratios in $Group-2$. $\begin{array}{cc}\\ &\text{Group-1} & \text{Group-2} \\ &\text{P. Biot number} & \text{ I.$\frac{buoyancy force}{viscous force}$} \\ &\text{ Q. Schmidt number} & \text{II.$\frac{internal thermal resistance of a solid}{boundary layer ... $Q-I$, $R-III$ $P-I$, $Q-III$, $R-II$ $P-III$, $Q-I$, $R-II$ $P-II$, $Q-III$, $R-I$
For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are nearly equal? $0.33$ $0.5$ $1$ $2$
For a non-catalytic homogeneous reaction $A\rightarrow B$, the rate expression at $300\:K$ is $-r_{A}$\left (mol\; m^{-3} s^{-1}\right )=\frac{10C_{A}}{1+5C_{A}}$, where$C_{A}$is the concentration of$A$(in$mol\;/m^{3}$). Theoretically, the upper limit for the magnitude of the reaction rate ($-r_{A}\:in\:mol\:m^{-3}s^{-1}$, rounded off to the first decimal place) at$300\:K$is _________ 0 votes 0 answers The variations of the concentration$\left ( C_{A} ,C_{R}\:and\:C_{S}\right )$for three species$\left ( A,R\:and\:S \right )$with time, in an isothermal homogeneous batch reactor are shown in the figure below. Select the reaction scheme that correctly represents the above plot. The numbers in the reaction schemes shown below, represent the first order rate constants in unit of$s^{-1}$. 0 votes 0 answers Hydrogen iodide decomposes through the reaction$2HI\leftrightharpoons H_{_{2}}+I_{2}$. The value of the universal gas constant$R$is$8.314 \:J\:mol^{-1}K^{-1}$. The activation energy for the forward reaction is$184000\:J\:mol^{-1}$. The ratio (rounded off to the first decimal place) of the forward reaction rate at$600 \:K$to that at$550\:K$is ________ 0 votes 0 answers Match the instruments in$Group-I$with process variable in$Group-2$...$Q-I$,$R-IIIP-Il$,$Q-III$,$R-IP-III$,$Q-lI$,$R-IP-IIl$,$Q-I$,$R-Il$0 votes 0 answers What is the order of response exhibited by a$U$-tube manometer? Zero order First order Second order Third order 0 votes 0 answers A system exhibits inverse response for a unit step change in the input. Which one of the following statement must necessarily be satisfied? The transfer function of the system has at least one negative pole The transfer function of the system has at least one ... transfer function of the system has at least one negative zero The transfer function of the system has at least one positive zero 0 votes 0 answers Two design options for a distillation system are being compared based on the total annual cost. Information available is as follows: ... above information, what is the total annual cost ($Rs$in lakhs/year) of the better option?$4042.492128$0 votes 0 answers Standard pipes of different schedule numbers and standard tubes of different$BWG$numbers are available in the market. For a pipe/tube of a given nominal diameter, which one the following statements is$TRUE$? Wall thickness increases with increase in both the ... the schedule number and the$BWG$number Neither the schedule number, nor the$BWG$number has any relation to wall thickness 0 votes 0 answers Terms used in engineering economics have standard definitions and interpretations. Which one of the following statements is$INCORRECT$? The profitability measure 'return on investment' does not consider the time value of money A cost index is an index value for ... different capacity Payback period is calculated based on the payback time for the sum of the fixed and the working capital investment 0 votes 0 answers India has no elemental sulphur deposits that can be economically exploited. In India, which one of the following industries produces elemental sulphur as a by-product? Coal carbonisation plants Petroleum refineries Paper and pulp industries Iron and steel making plants 0 votes 0 answers Two paper pulp plants$P$and$Q$...$P$and Plant$Q$both use the Kraft process Plant$P$uses Sulfite process Plant$P$uses Kraft process 0 votes 0 answers Match the industrial processes in$Group-I$with the catalyst materials in$Group-2$.$\begin{array}{cc}\\ \\ &\text{Group-1} & \text{Group-2} \\ &\text{P. Ethylene polymerisation} & \text{ I. Nickel} \\ &\text{ Q. Petroleum feedstock cracking} & \text{II. Vanadium pentoxide} \\ &\text{R. Oxidation of $...$R-III$,$S-IIP-I$,$Q-II$,$R-III$,$S-IVP-II$,$Q-III$,$R-IV$,$S-I$0 votes 0 answers A set of simultaneous linear algebraic equations is represented in a matrix form as shown below. ...$x_{3}$is ___________ 0 votes 0 answers The model$y=mx^{2}$is to be fit to the data given below.$\begin{array}{|cl|cI|}\hline &{x} & {1} & {\sqrt{2}} & {\sqrt{3}} \\ \hline &{y} & {2} & {5} & {8} \\ \hline \end{array}$Using linear regression, the value (rounded off to the second decimal place) of$m$is _________ 0 votes 0 answers The Lagrange mean-value theorem is satisfied for$f\left ( x \right )=x^{3}+5$, in the interval$\left ( 1,4 \right )$at a value (rounded off to the second decimal place) of$x$equal to __________ 0 votes 0 answers Values of$f\left ( x \right )$in the interval$\left [ 0,4 \right ]$...$1$, the numerical approximation (rounded off to the second decimal place) of$\int_{0}^{4}f\left ( x \right )dx$is ______________ 0 votes 0 answers An ideal gas is adiabatically and irreversibly compressed from$3$bar and$300\:K$to$6$bar in a closed system. The work required for the irreversible compression is$1.5$times the work that is required for reversible compression from the same initial ... in$K\$, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is _______________