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# The cost C, in dollars of producing x units of a certain commodity is

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The cost C, in dollars of producing x units of a certain commodity is  [#permalink]

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Updated on: 24 Apr 2016, 11:36
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The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?

(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Originally posted by inakihernandez on 24 Apr 2016, 10:20. Last edited by Vyshak on 24 Apr 2016, 11:36, edited 1 time in total. Edited the url ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 6958 The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:58 11 4 inakihernandez wrote: The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a? (1) The cost of producing 500 units is$15.833.

(2) The cost of producing 1,000 units is twice the cost of producing 500 units.

Hi
the equation has got two constants a and b..

lets see the statements

(1) The cost of producing 500 units is $15.833. we have two constants a and b, and one equation Insuff (2) The cost of producing 1,000 units is twice the cost of producing 500 units here when you make two quation and equate them you will realize you can find answer.. $$C(1000) = 10.000 + b*1000 + a1000^2$$ .. $$C(500) = 10.000 + b*500 + a500^2$$ .. now $$C(1000) = 2* C(500)$$.. so $$10.000 + b*1000 + a1000^2 = 2( 10.000 + b*500 + a500^2 )$$.. $$10.000 + b*1000 + a1000^2 = 20.000 + b*1000 + 2*a*500^2$$ b will get cancelled and you can find value of a.. Suff Hope it helps _________________ 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 ##### General Discussion EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12649 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:47 2 Hi inakihernandez, You should post this question in the DS sub-Forum here: gmat-data-sufficiency-ds-141/ GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: The cost C, in dollars of producing x units of a certain commodity is  [#permalink]

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24 Apr 2016, 11:02
1
2
C(x) = 10.000 + bx + ax^2
a = ?

St1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833
We have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'
Not Sufficient

St2: C(1000) = 2*C(500)
10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)
a(10^6 - 5*10^5) = 10
Sufficient to determine the value of 'a'

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Re: The cost C, in dollars of producing x units of a certain commodity is  [#permalink]

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24 Apr 2016, 11:10
1
Vyshak wrote:
C(x) = 10.000 + bx + ax^2
a = ?

St1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833
We have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'
Not Sufficient

St2: C(1000) = 2*C(500)
10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)
a(10^6 - 5*10^5) = 10
Sufficient to determine the value of 'a'

how did you get this?
a(25*10^4 )
a(50*10^4)
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Re: The cost C, in dollars of producing x units of a certain commodity is  [#permalink]

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24 Apr 2016, 11:13
1
paidlukkha wrote:
Vyshak wrote:
C(x) = 10.000 + bx + ax^2
a = ?

St1: C(500) = 10 + 500b + a(25*10^4 ) = 15.833
We have 2 unknowns and hence value of 'a' cannot be determined without knowing the value of 'b'
Not Sufficient

St2: C(1000) = 2*C(500)
10 + 1000b + a(10^6) = 20 + 1000b + a(50*10^4)
a(10^6 - 5*10^5) = 10
Sufficient to determine the value of 'a'

how did you get this?
a(25*10^4 )
a(50*10^4)

Hi,

I have skipped some multiplication steps in my explanation.

x = 5*10^2 --> x^2 = 25*10^4

2*C(500) --> 2(10 + 500b + 25*10^4) --> 20 + 1000b + 2*25*10^4
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Re: The cost C, in dollars of producing x units of a certain commodity is  [#permalink]

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24 Apr 2016, 21:04
8
inakihernandez wrote:
The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?

(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Let me point out here that this is a (C) trap question. The first statement will give an equation in a and b. So we know that we cannot find the value of a. When we look at the second statement, we see another equation in a and b which is not a multiple of the first equation and suddenly, it seems that we will be able to solve both equations simultaneously and get the answer. So (C) seems to be correct. But that is just too easy. Beware of the (C) trap. This should make you think - can I solve the question using a single statement alone. Make the equation of statement 2 since it gives you some convoluted relative data. When you do, you see that b gets canceled and in fact, you can get the value of a. Note that you cannot find the value of b from the second statement alone but the question doesn't ask for the value of b. We only need the value of a and hence second statement is sufficient alone. _________________ 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 > GMAT self-study has never been more personalized or more fun. Try ORION Free! Current Student Joined: 28 Nov 2014 Posts: 867 Concentration: Strategy Schools: Fisher '19 (M) GPA: 3.71 Re: The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 09 Oct 2016, 04:29 The correct statement 1 is (1) The cost of producing 500 units of this commodity is$15833

It is NOT 15.833 although that won't affect the OA.
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Re: The cost C, in dollars of producing x units of a certain commodity is  [#permalink]

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09 Oct 2017, 06:44
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# Finding the matrix of a linear transformation from an upper triangular matrix to an upper triangular matrix.

The question that I am trying to solve is as follows:

Find the matrix $A$ of the linear transformation $T(M)= \begin{bmatrix} 7 & 3 \\ 0 & 1 \end{bmatrix} M$ from $U^{2×2}$ to $U^{2×2}$ (upper triangular matrices) with respect to the basis $\left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix}1&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$

$A=\begin{bmatrix} \cdot&\cdot&\cdot\\\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot\end{bmatrix}$

I don't know how to interpret this question. I would have thought that the matrix $A$ is $\begin{bmatrix} 7 & 3 \\ 0 & 1 \end{bmatrix}$ because that is the matrix which defines the linear transformation $T$. I am surprised that the required answer is $A\in M_{3\times 3}$. How can a basis with elements $\in M_{2\times 2}$ form a matrix $\in M_{3\times 3}$?

• Upper triangular matrices are a three dimensional vector space. But I don't understand your question. Usually $U$ indicates unitary matrices. – Emilio Novati Apr 11 '15 at 9:49
• Has three dimensions got anything to do with $A\in M_{3\times 3}$? – ahorn Apr 11 '15 at 11:21

Upper triangular matrices

$\begin{bmatrix} a&b\\ 0&c \end{bmatrix}$

form a vector space with canonical basis:

$e_1=\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} \quad e_2=\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix} \quad e_3=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}$

that is isomorphic to $\mathbb{R}^3$ by:

$e_1\rightarrow\vec e_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix} \quad e_2\rightarrow\vec e_2=\begin{bmatrix} 0\\1\\0 \end{bmatrix} \quad e_3\rightarrow \vec e_3=\begin{bmatrix} 0\\0\\1 \end{bmatrix}$

Your linear transformation $T$ is defined as a matrix multiplication:

$T(M)=\begin{bmatrix} 7&3\\ 0&1 \end{bmatrix} \begin{bmatrix} a&b\\ 0&c \end{bmatrix} = \begin{bmatrix} 7a&7b+3c\\ 0&c \end{bmatrix}$ so, in this canonical representation, it is given by the matrix $T_e$ such that:

$T_e\vec M=\begin{bmatrix} 7&0&0\\ 0&7&3\\ 0&0&1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix}= \begin{bmatrix} 7a\\ 7b+3c\\ c \end{bmatrix}$

Now you want a representation of the same transformation in a new basis:

$e'_1= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} \rightarrow\vec e'_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix} \quad e'_2=\begin{bmatrix} 1&1\\ 0&0 \end{bmatrix}\rightarrow\vec e'_2=\begin{bmatrix} 1\\1\\0 \end{bmatrix} \quad e'_3=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}\rightarrow \vec e'_3=\begin{bmatrix} 0\\0\\1 \end{bmatrix}$

This transformation of basis is represented by the matrices

$S= \begin{bmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \qquad S^{-1}= \begin{bmatrix} 1&-1&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$

So the matrix that represents the transformation $T$ in the new basis is:

$T_{e'}=S^{-1}T_eS= \begin{bmatrix} 7&0&-3\\ 0&7&3\\ 0&0&1 \end{bmatrix}$

• When you said "This transformation of basis is represented by the matricies", did you simply put the column vectors $\vec e'_1$, $\vec e'_2$ and $\vec e'_3$ next to each other to get $S$? Or was there a more nuanced way of getting $S$, such as by saying that $\vec e_j= (a_j+b_j+c_j)\vec e'_j$ where $j$ is the number of the column? – ahorn Apr 11 '15 at 15:49
• Symply I put the columns ( it's a general rule that you can easely verify by linearity). – Emilio Novati Apr 11 '15 at 16:31

Let $(X,Y,Z)$ be the basis of $U$, compute $T(X)$, $T(Y)$, $T(Z)$ and express them as a linear combination of $X$, $Y$ and $Z$.

• T(X)=7X; T(Y)=7Y; T(Z)=Z. How does that relate to A? – ahorn Apr 11 '15 at 11:20
• A={(7,0,0),(0,7,0),(0,0,1)} take the coefficients as elements of the matrix A. – 2ndYearFreshman Apr 11 '15 at 11:39
• I have tried to input all sorts of combinations, such as $\begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 7 & 0 & 0 \\ 7 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \text{and} \begin{bmatrix} 7 & 0 & 0 \\ 7 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ but these are still incorrect (I am answering an online test). Btw, welcome to math.stackexchange! What brought you to the site? – ahorn Apr 11 '15 at 14:25
• Try A={(7,0,0),(0,7,0),(-3,3,1)} Thank you,hopefully my answer will help.Your question brought me here. – 2ndYearFreshman Apr 11 '15 at 14:34


# How can a curve be one dimensional?

1. Apr 3, 2012

### rollcast

I've heard of curves being described as one dimensional but I don't understand how anything other than a straight line can be one dimensional as surely once the curve becomes, well, curved it is now in two dimensions?

I have illustrated this in the above diagram with the curve and straight line shown in red and the dimensions in green.

Thanks
A.

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1,706
2. Apr 3, 2012

### Office_Shredder

Staff Emeritus
The curve is embedded in a 2 dimensional plane, but the curve itself is one dimensional. For example, I can describe all the points using only one parameter in a very natural way (if the graph is y=f(x), then x is the parameter since it uniquely identifies y)

Really what you want is dimension to be an intrinsic property. If I took your curve and put it in 3 dimensions, is it a 3 dimensional object now? What about the fact that we're in spacetime, is it 4 dimensional? That's a bit silly, so instead we ask what properties does the curve have that is independent of the space that it is embedded in.

There are several possible definitions of dimension depending on context but the basic idea is that if it has dimension k you can describe every point using k parameters at least locally - by that I mean, for example the curve x2+y2=1 (a circle) you can describe the top half of it by the set of points of the form $(x,\sqrt{1-x^2})$ and the bottom half is of the form $(x,-\sqrt{1-x^2})$. So I wasn't able to describe the whole circle using the x-coordinate but I was able to describe each half of it using a single coordinate, which means the whole thing is 1 dimensional (in fact you can describe the whole circle using a single coordinate if you use trigonometric functions, which I encourage you to try if you haven't seen it)

3. Apr 3, 2012

### Bacle2

Would you think of the curve as having non-zero area? I don't mean a closed curve

that may enclose an area; I mean the curve itself.

4. Apr 3, 2012

### HallsofIvy

Staff Emeritus
Yes, such a curve is in two dimensions but it only "one dimensional" itself".

What is your definition of "dimension"? One common one is: a subset of Euclidean space is said to be "n-dimensional" if and only if any point can be identified using n real numbers.

Choosing one point on the curve, we can identify any point on the curve by its distance (positive in one direction, negative in the other) from the chosen point. Since that is a single number, the curve is one-dimensional.

5. Apr 3, 2012

### rollcast

Bit of a weird example here so the curve is like curved corridor in a building, each office on the corridor is numbered with respect to where it is positioned along the corridor and not where it is positioned relative to the 4 outer walls of the building.

6. Apr 3, 2012

### Matterwave

Perhaps a somewhat more intuitive way of thinking about it is, if you live "in" the curve (i.e. you're not allowed to go outside of it), you can only ever go forwards or backwards, never to the left or right or up or down.

7. Apr 3, 2012

### Bacle2

Altho you may want to consider separately the case of space-filling curves, like the Peano curve.

8. Apr 3, 2012

### Bacle2

I'm starting to doubt my answer, if we can define a curve as the continuous image of a connected interval. We can use the fact that, e.g., any compact metric space K is the continuous image of the Cantor set. Then take, e.g., K=[0,1]^n (mapping I into R^n), and extend continuously, using Tietze extension ( C is closed in I ; I is normal), then the image of f has non-empty interior in R^n.

Note/Edit: I'm referring to a more general definition of a curve, as the continuous image of an interval, and not the curves in the attached files.

Last edited: Apr 3, 2012
9. Apr 4, 2012

### chiro

Intuitively for any line, no whatever what its deformation or dimensionality for the embedded space is always a line. Similarly, a flat piece of paper no matter how its deformed or what space its embedded in is always a flat piece of paper.

If you are given a visualization of the object in question, you can get a very good idea of the number of parameters. If you can't visualize the object but are given a deterministic expression, then you have to use other methods.

If you are given a non-analytic version of a process where you only know the dimensionality of the embedded space and have to not only figure out the parameterization but also some sort of 'useful' definition for the process as a whole (implicit most likely but could be explicit), then you will have to use even different methods again which will be based on a combination of statistical theory and non-statistical mathematics.

One method for assessing parameterization is assessing density in various orientations for your function and statistical results used in data mining measure a kind of 'variation' density where you get principal components for your data set. This kind of idea could be applied to your analytic setting in the analytic perspective (not the statistical one).

10. Apr 4, 2012

### Matterwave

A curve has the same cardinality as an arbitrary n-dimensional hypercube, but that doesn't mean it has the same dimension. A space-filling curve defines a surjection, but not a homeomorphism.

11. Apr 4, 2012

### Bacle2

I was referring to a curve as the continuous, not homeomorphic image of an interval. I was also taking back my statement that a curve must have empty interior. As defined, I think the argument is correct, and a curve does not necessarily have an empty interior.

12. Apr 4, 2012

### Bacle2

Clearly, the image of an interval would depend on a single parameter. I was referring to the claim/belief that the continuous image of a curve would have no area, i.e., would have empty interior.

13. Apr 4, 2012

### Bacle2

Sorry, one last post on this, to try to avoid confusion:

0)Define a curve as the continuous image of an (connected)interval
Claim: A curve can have non-empty interior.

i)We have that every compact metric space is the continuous image of the
Cantor set C . So we map C subset I:=[0,1] into, say I^n, so there is an f with f(C)=I^n (n>1)

ii) By Tietze extension thm. ( C is closed in I normal) , we can extend f to f^, defined on the whole of I, continuously ( on each real variable). Then f^ is a continuous map such that f^(I) has non-empty interior in R^n.

14. Apr 4, 2012

### chiro

I'll have a look at your theorems later on, but the thing is if you can deform your object that you have in a way so that is linear, then apply the ideas of dimensionality to get not only the proper dimension but the parameterization of your object, then you are done. By finding the appropriate deformation, you have created a linear representation that can be dealt with in the linear context.

This is of course equivalent to finding the inverse for each part of your object with respect to the different 'branches' that exist. If you can find the inverse for each 'branch' of your object (might be multi-dimensional and multi-parameterized but the idea is the same), then you can find a linear representation which can be parameterized.

So the case with a line would result in basically an n dimensional system that represents a line with one parameter t where X(t) = At + (1-t)B in the deformed state to the linear space. If this parameterization wasn't only one dimensional, then the deformation wouldn't produce one but however many parameters for the object.

With the notion of topology, if you know the object is analytic continuous then there should exist a deformation to the linear representation which depends on the inverse within a given branch-volume that corresponds to where the derivatives are (which correspond to the ideas in the inverse function theorem which are calculated by finding zero-jacobians). With this information you can find the branches and thus deform that region to it's linear counterpart.

The reason I like to think of something in the deformed linear space is because linear spaces, decompositions, and parameterizations of linear systems are well understood both algebraically, algorithmically and also to an extent visually. It's very easy to parameterize something that is deformed to linear space than to try and analyze it in its non-deformed non-linear space with a function like say f(x,y,z,w,t,u,v,a) = blah for any analytic continuous blah.

You could also probably consider continuous representations that are not analytic over the whole domain and use the same argument based on topological reasoning but I will only speculate on this (intuitively it makes sense at least).

15. Apr 4, 2012

### Bacle2

I think the best we can say is that if f is C^1 (maybe Lipschitz) , then, with a bounded derivative, we can preserve the Hausdorff dimension.

16. Apr 4, 2012

### chiro

I'll go with that for finding the dimension (which is the OP's question), but in terms of understanding why (and where) dimension goes from say one thing to another it may be useful to see which 'parts' if they are isolated contribute to where the function requires more dimensions as opposed to less (it can happen but not always).

The only reason I am saying this (for the OP) is that visually you can see where dimensionality changes. I'm imagining some kind of implicit function that creates branches (like a bifurcation) and does this repeatedly which creates more dimensions at different parts.

Just out of curiosity, what would you say topologically is a way to do the above? In other words in the case of say a bifurcation like object, how would you isolate this kind of phenomenon in terms of the local characteristics (spatially) of the object?

17. Apr 4, 2012

### Bacle2

The way I see it, if the derivative is bounded, then the image of the balls in the covering will not expand by much.

Sorry, it is 4 a.m., here, I'm out for the night; will try to be back in case there are more posts.

18. Apr 4, 2012

### HallsofIvy

Staff Emeritus
Then any two-dimensional ball about a point on the curve will necessarily include some points not on the curve- a curve has no (2 dimensional) interior.

19. Apr 4, 2012

### Bacle2

Halls of Ivy:

I did a specific construction, with a specific definition of curve; what is it about my construction that you think is faulty?

It seems strange-enough that the hypercube ( as a compact metric space; a space with non-empty interior) is the continuous image of the Cantor set. Then what is wrong with the extension?

20. Apr 5, 2012

### TrickyDicky

I think the OP is simply reflecting on the fact that there is no such thing as intrinsic curvature in one dimension.


## cylinder with cone on top

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# If s is the sum of odd integers from 40 to 60, inclusive, and t is the

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If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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12 Oct 2017, 12:33
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?

A. 480
B. 490
C. 980
D. 990
E. 995

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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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12 Oct 2017, 12:39
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?

A. 480
B. 490
C. 980
D. 990
E. 995

Similar question: https://gmatclub.com/forum/if-x-is-equa ... 97544.html
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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12 Oct 2017, 12:48
There are total 10 odd numbers between 40 and 60.
hence, t=10

Now, the given sequence is 41,43,45,......,59
common difference is 2.
hence,
S= (41+59)*(10/2)
S=500.
S-t= 500-10= 490.

Kudos if it helps.
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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12 Oct 2017, 13:31
1
Are these formulas correct? I am having trouble with odd numbers for some reason.

N for odd numbers = (Last odd - First odd)/2)) +1

Sum of odd numbers = (First odd + Last odd)*N))/2

1. N = (59-41)/2))+1 = 10

2. Sum of odds = (41+59)*10))/2 = 500

3. 500 - 10 = 490 B

Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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12 Oct 2017, 13:36
1
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?

A. 480
B. 490
C. 980
D. 990
E. 995

The way I solve sums of arithmetic series, I need the number of terms no matter what. Start there.

Set t

Number of terms: number of odd integers from 40 to 60. First term is 41, last is 59.

Number of terms

$$\frac{LastTerm-FirstTerm}{interval} + 1$$

The "interval" for consecutive even and odds is always 2

$$\frac{59-41}{2} = \frac{18}{2} = (9 + 1)=$$ 10

Set s

Sum of arithmetic series
(Average)(Number of terms) =

$$\frac{FirstTerm+LastTerm}{2}*(10)$$ =

$$\frac{100}{2}(10) = (50)(10) = 500$$

s - t: 500 - 10 = 490

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If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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12 Oct 2017, 14:20
1
Are these formulas correct? I am having trouble with odd numbers for some reason.

N for odd numbers = (Last odd - First odd)/2)) +1

Sum of odd numbers = (First odd + Last odd)*N))/2

1. N = (59-41)/2))+1 = 10

2. Sum of odds = (41+59)*10))/2 = 500

3. 500 - 10 = 490 B

Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?

I think the reason you might be having trouble with odd numbers is that for consecutive odd integers, the interval (between the numbers), is two. Two is even; that could confuse. Or you might be forgetting to start with the first ODD term (e.g. here, first term is 41, not 40).

Bottom line: The "interval" between 1 and 3 is two. The interval between 2 and 4 is two. For consecutive odds and evens, in the formula, you divide by 2.

Take smaller numbers. Say, sum of odd integers from 0 to 6. So 1 + 3 + 5 = 9. There are three terms (1, 5, 9) whose sum is 9.

Number of terms in "odd integers from 0 to 6":

$$\frac{(Last - First)}{2}+ 1$$

$$\frac{(5-1)}{2} + 1 = (\frac{4}{2}+ 1) = (2 + 1) = 3$$ Correct

SUM: (Average)(# of terms)

$$\frac{(First + Last)}{2}(3)$$

$$\frac{(1 + 5)}{2} (3) = (3)(3) = 9$$
Bingo. Dividing by 2 for consecutive odd integers works.

When subtracting, you add one because subtraction doesn't include the first number. If in doubt: use small numbers that replicate your situation.

For example, # of terms from 1 to 4? The numerals are 1, 2, 3, 4. There are 4 terms. BUT (4 - 1) = 3.
We need one more. 3 + 1 = 4.

There's a mnemonic: "add one before you're done." It almost always applies. If in doubt, replicate your situation with small numbers. Rare in sequence sums but possible: If you see the word "exclusive" or the phrase "exclusive of," that is when you really need to check.

I have seen your other answers using this method. The ones I've seen are correct! (The one about -190 to 195? Correct, but long. -190 to +190 = 0; you are left with only 5 numbers to sum. STILL - correct.)

Here is a fantastically written, comprehensive post on all kinds of sequences by benjiboo , scroll down for arithmetic sequence

And here is a thread on sequences. VeritasPrepKarishma 's posts are great

Hope this barrage of information helps.
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the  [#permalink]

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13 Oct 2017, 11:23
N for odd numbers = (Last odd - First odd)/2)) +1

Sum of odd numbers = (First odd + Last odd)*N))/2

1. N = (59-41)/2))+1 = 10

2. Sum of odds = (41+59)*10))/2 = 500

3. 500 - 10 = 490 B
Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the &nbs [#permalink] 13 Oct 2017, 11:23
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# Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$

$$a + b = 2\\ a^2 + b^2 = 6$$

Find the value of $(a-b)^2$

My workings till I got stuck -

$$(a-b)^2 = a^2 - 2ab + b^2 \\ = a^2 + b^2 - 2ab\\ = 6 - 2 ab$$

I'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.

• $4= (a+b)^2=a^2+b^2+2ab$ – lulu Feb 26 '16 at 15:16
• Squaring the first equation then subtract the second equation from it. – Zhanxiong Feb 26 '16 at 15:17
• The above comments mean that you should consider the quantity $2a^2+2b^2-(a+b)^2$... – abiessu Feb 26 '16 at 15:19

Use $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$ Thus, here, $$(a-b)^2+4=12$$ $$(a-b)^2=8$$

We will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \iff (2-b)^2 + b^2 +6$

This means $2b^2-4b-2=0 \iff b^2 -2b -1 +0$. The solutions of this equation are $b=1-\sqrt{2}$ and $b=1+\sqrt{2}$. Then $a=1+\sqrt{2}$ or $a=1-\sqrt{2}$

Then, in any case $(a-b)^2=8$

• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42

$$a+b=2$$

$$\implies(a+b)^2=4$$

$$\implies a^2+b^2+2ab=4$$

$$\implies2ab=-2$$

Also,

$$(a-b)^2=a^2+b^2-2ab$$

$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)


# Probability that a random 13-card hand contains at least 3 cards of every suit?

A random 13-card hand is dealt from a standard deck of cards. What is the probability that the hand contains at least 3 cards of every suit? (Introduction to Probability, p.36)

My solution:

• There are $\binom{52}{13}$ possible hands.
• Because there are 13 cards for the hand, to obtain at least three cards of one suit per hand, we need to have exactly three cards of one suit per hand plus one additional card of any suit, thus $\binom{13}{3}^4 * 4 \binom{10}{1}$
• Result: $\frac{40*\binom{13}{3}^4}{\binom{52}{13}} = 0.4214$

However, simulating it in R yields:

deck <- rep(1:4, 13)
out <- replicate(1e5,{
hand <- sample(deck, size=13, replace=FALSE)
all(table(hand) >= 3)
})
mean(out)
> 0.14387

Can anybody tell me what is wrong?

EDIT

I'm afraid, the correct code should be.

deck <- rep(1:4, 13)
out <- replicate(1e5,{
hand <- sample(deck, size=13, replace=FALSE)
length(table(hand))==4 & all(table(hand) >= 3 )
})
mean(out)
> 0.10639

• The correct result is about $0.105$, yet the R simulation yields $0.14387$. This is the second R simulation you've posted today that gives a result significantly different from the theoretical result. There must be something wrong with the R code (up to and including the possibility that R's randomization is insufficiently random for your purpose) but I still don't know what the error is. You might want expert advice on this so that you aren't misled by an incorrect simulation in the future. Dec 7, 2014 at 17:17
• @DavidK R is completely fine, the idiot is me. As you mentioned, I posted two times simulations along with my questions and both of them were wrong, because I didn't give enought thought to the special cases. I'm sorry for the confusion I caused. Motivation for the code was that I didn't just want to ask for whether correct or not, but give some justification for my doubt and demonstrate previous work on the problem. Dec 7, 2014 at 17:32
• Good for you for finding that correction to the R code. The error was sufficiently subtle for me to miss it entirely, even knowing that something must be wrong. Dec 7, 2014 at 17:47

Dominik, your answer was off by a factor of 4. This happened because you counted a hand containing J,K,Q,A of spades (for example) 4 times: (JQK)(A), (QKA)(J), (KAJ)(Q), and (JAQ)(K)

• The answer above is a correct description of the basic issue that led to the overcount. Dec 7, 2014 at 15:40

We count the "favourables," the 4-3-3-3 hands. The suit in which we have $4$ cards can be chosen in $\binom{4}{1}$ ways. For each of these ways, the actual $4$ cards in that suit can be chosen in $\binom{13}{4}$ ways. For each of these ways, the cards in the other three suits can be chosen in $\binom{13}{3}^3$ ways, for a total of $\binom{4}{1}\binom{13}{4}\binom{13}{3}^3$.

Remark: Your counting procedure results in multiple counting. Think, for example, of the 4-3-3-3 hand that has the K, Q, J, 10 of hearts, and some specific cards for the rest of the hand. Your calculation views, for example, K, Q, J of hearts, and later 10 of hearts, as different from K, J, 10 of hearts, and then Q of hearts.

• As @judith pointed out above, it seems that $\binom{4}{1} \binom{13}{4} \binom{13}{3}^3 = \binom{13}{3}^4 \binom{10}{1}$. However, I still don't understand what I am overcounting. What do you mean by "the 4-3-3-3 hand that has the K, Q, J, 10 of hearts"? Could you rephrase/eleborate on your remark please? Dec 7, 2014 at 15:18
• You counted the $12$-card 3-3-3-3 hands and then added a card in one of the suits. Each 4-3-3-3 hand comes up in $4$ ways, so you are overcounting by a factor of $4$. Here is a simpler example. How many $3$ card hands have $2$ hearts and $1$ diamond? The obvious answer is $\binom{13}{2}\binom{13}{1}$. (Cont) Dec 7, 2014 at 15:33
• (Cont) Here is a wrong way of counting. Pick a heart and a diamond, $\binom{13}{1}^2$ ways. Then add a heart, The wrong way counts twice the hand that has K, Q of hearts, and 5 of diamonds, once as K of hearts, 5 of diamonds, then Q of hearts, and also as Q of hearts, 5 of diamonds, and then K of hearts. Dec 7, 2014 at 15:35
• Thank you for elaborating on it. I think I understand my error now. Dec 7, 2014 at 16:18
• You are welcome. This sort of inadvertent multiple counting happens fairly often, until one gets sensitized to it. Dec 7, 2014 at 16:21

I doubt, that it is still interesting for you, but I want to check myself.I think either you or I misunderstand the question. Because it asks to find the probability of the hand contains AT LEAST 3 cards of every suit and you calculated the probability of EXACTLY 3 cards of every suit. From my point of view answer must be:

1. for 0 cards of every suit there is no point to calculate, since nothing happened

2. for 1 cards of every suit (choose from 13-1)^3(choose from 13-remainig 10)(choose from 4-1) and divide all this to (choose from 52-13)

3. for 2 cards of every suit (choose from 13-2)^3(choose from 13-remainig 7)(choose from 4-1) and divide all this to (choose from 52-13)

Then just substract from 1 the calculated probability and you will get P(>=3)

• OP is definitely not calculating exactly three cards of every suit. Because there are 13 cards in a hand, the only distribution(s) that have at least three cards in every suit have three suits with three cards in them and one suit with four; this is what OP attempts to calculate (with the overcounting mentioned in other answers). The hand size is fixed, so the notion of '1 card of every suit' doesn't make sense; it's an impossible configuration. Jul 9, 2021 at 3:11

Another solution that's more mechanical and generalizable to other arrangements like two pair or full house:

We know we must have a 4-3-3-3 hand. You can pick any card for the first suit: 52. The next three cards must match the suit of the first so there are 12 x 11 x 10 possibilities for those. Now for the 4th card, we can pick any card not in the 1st suit: 39. Using the same reasoning as before, for the next two cards there are 12 x 11 possibilities. We continue until we have picked all cards:

$$X = (52 \times 12 \times 11 \times 10) \times (39 \times 12 \times 11) \times (26 \times 12 \times 11) \times (13 \times 12 \times 11)$$

$$X$$ gives us all arrangements of the pattern WWWW XXX YYY ZZZ, as such we overcount all the internal arrangements of W, X, Y, and Z ($$4!3!^3$$). We also overcount that X, Y, and Z can be interchanged ($$3!$$).

So the answer is $$X /(4!3!^4) / {52 \choose 13}$$.


# What is the integral of |x| ?

1. Aug 10, 2011

### agentredlum

The question is in the title.

This is not a homework question, just curiousity on my part.

2. Aug 10, 2011

### HallsofIvy

Staff Emeritus
If x< 0, then |x|= -x and its integral is $-x^2/2+ C$

If $x\ge 0$, then |x| x and its integral is $x^2/2+ C$

Last edited: Aug 10, 2011
3. Aug 10, 2011

### disregardthat

or |x|x/2 + C

4. Aug 10, 2011

### ReaverKS

How? I thought the whole purpose of the absolute value stemmed back to distances along a number line, therefore you couldn't get a negative value out of the absolute value of a number or a function?

5. Aug 10, 2011

### Citan Uzuki

If x<0, then -x>0.

6. Aug 10, 2011

### BloodyFrozen

7. Aug 10, 2011

### Bohrok

$$\int|x| dx = \frac{1}{2}x|x| + c$$If you write |x| as √(x2) and use integration by parts, you can find the integral.

8. Aug 10, 2011

### agentredlum

I know HallsofIvy figured it out because calculations were shown, but did the rest of you look it up?

HallsofIvy, can you combine your 2 solutions into a single solution to get wiki answer?

I know wiki mentions integration by parts, but did anyone use a simpler way to get the answer? It doesn't have to be rigorous, it just has to work. Don't worry about rigor, just functionality.

What is the integral of |x^3| ?

What is the integral of |x^n| for odd n?

EDIT* There is nothing wrong (IMHO) with looking up the answer, but there is nothing wrong with working it out for yourself.

9. Aug 10, 2011

### Bohrok

I never knew wikipedia showed how to integrate it; just noticed it right now. I personally did it once because I wanted to derive it myself, and for higher odd powers of x such as |x3| and there is a pattern.$$\int|x^3|dx = \frac{1}{4}x^3|x| + c, \int|x^5|dx = \frac{1}{6}x^5|x| + c, \text{etc.}$$So in general,$$\int|x^n|dx = \frac{1}{n + 1}x^n|x| + c \text{ for odd n}$$

10. Aug 10, 2011

### agentredlum

Oh, this is interesting. Is there any difference to the answer if you represent it as |x^n|x/(n+1) + c for odd n ?

Well done in deriving it for yourself.

11. Aug 10, 2011

### Mute

When you have an absolute value, you just split it into cases:

If $x > 0,~|x^n| = x^n$, so the integral is $x^{n+1}/(n+1)$.

If $x < 0,~|x^n| = - x^n$, so the integral is $-x^{n+1}/(n+1)$.

(for n odd, of course)
This can be written as a single results using the "sign function", $\mbox{sgn}(x)$, which returns the sign of x. Hence, the integral may be written

$$\mbox{sgn}(x) \frac{x^{n+1}}{n+1}.$$

For x not zero, the sign function is equivalent to |x|/x or x/|x| (as is hopefully obvious after thinking about it for a second), so you can insert the first form there to get the results people have been throwing around.

Note that since the sign function to an even power is always 1, you can always insert sign functions of even powers to shift around the exponents, so you can write the integral as

$$\frac{|x|^ax^b}{n+1},$$
where a + b = n+1, and a and b are odd numbers.

12. Aug 10, 2011

### Bohrok

No difference; it's the same way as how I have it. I should also add positive to the restriction on n: for positive odd n.

13. Aug 10, 2011

### agentredlum

Of course, right you are. I haven't explored what happens when n is negative. It is interesting to me that the integrals we are considering, |x^n|, for positive odd n, to get the answer you just write it down, multiply by x, divide by the total number of factors of x and add a 'c'. Integration by parts is not necessary. It makes me wonder if there are other functions out there that work simply like this, besides the normal x^n without absolute value symbol.

What is the integral of |x^3 + x| ?

Last edited: Aug 11, 2011
14. Aug 11, 2011

### Bohrok

I thought this would be difficult and involve integration by parts, but it turned out quite simple actually:
|x3 + x| = |x(x2 + 1)| = |x||x2 + 1| = |x|(x2 + 1) = |x|x2 + |x| = |x3| + |x|, using the facts that x2 + 1 = |x2 + 1| and x2 = |x2| since the former in each case is never negative, and |ab| = |a||b|, so
∫|x3 + x| dx = ∫|x3| dx + ∫|x| dx
Once you integrate those, you can combine them and make the result look more condensed.

If you had |x3 - x|, that would be a different story!

Last edited: Aug 11, 2011
15. Aug 11, 2011

### agentredlum

It's very nice how you turned this problem into something we have a shortcut for already!

So the integral of |x^3 + x| is |x^3|x/4 + |x|x/2 + C

I like your algebra manipulation of absolute value so let me use it too. Disregard the C for now...

(|x^3| + 2|x|)x/4 common denominator 4 and factor out x...now work inside the parenthesis...

|x|(|x^2| + 2)x/4 for the same reasons you gave above...

|x^2| + 2 = |x^2 + 2|

|x||x^2 + 2| = |x^3 + 2x|

|x^3 + 2x|x/4 again for same reasoning as you above... now 1/4 = 1/|4| so you can insert 1/4 into the parenthesis without fear, so you get finally...

|x^3/4 + x/2|x + C

Now if you compare this answer to the question...what is integral of |x^3 + x|, you write it down, divide x^3 by 4 inside absolute value (n + 1 for n = 3), divide x by 2 inside absolute value (n + 1 for n = 1), multiply the whole thing by x and add C

Putting it all together...

integral of |x^3 + x| = |x^3/4 + x/2|x + C

I think the shortcut is obvious now in this case and should work for any integral of |ax^3 + bx| where both a,b are non negative.

I'm still working on integral of |x^3 - x| and trying to make it co-operate with the shortcut, any ideas?

16. Aug 11, 2011

### disregardthat

Split the integral as such:

$$\int^t_{1} |x||x^2-1| dx = \int^t_{1} |x|(x^2-1) dx$$ for t >= 1,

$$\int^t_{-1} |x||x^2-1| dx = \int^t_{-1} |x|(x^2-1) dx$$ for t <= -1

$$\int^t_{0} |x||x^2-1| dx = \int^t_{0} |x|(1-x^2) dx$$ for -1 < t < 1

These three integrals are solved by the method above. For each domain, scale by a constant to get an answer on the form F(x) + C, but it is not necessary.

17. Aug 11, 2011

### agentredlum

Thank you. This is all very nice and important but right now i'm working on my own calculations using pencil and notebook. My PS3 browser does not decode LaTeX so its very hard sometimes to decode it using my brain. Can you solve all 3 integrals above and combine them into a single expression that gives the correct area of |x^3 - x| between the curve and the x-axis from x= -5 to x= 3 for example?

Let me give an simple example. The integral of x from x= -5 to x= 5 gives zero. I'm interested in the integral as AREA between 2 curves.

When you take the absolute value of a function it has 2 important graphical effects for my interests concerning area.

1) Portions below the x-axis are reflected by symmetry to portions above the x-axis with area unchanged.

2) Portions above the x-axis remain unchanged

This can be helpful when you have a function with portions below the x-axis. If one is interested in AREA then you don't have to find roots, split up the integral, use symmetry, etc. Just integrate it's absolute value. If you can find the integral then a single expression will give you the right answer for the area.

Now, I realise integrating |f(x)| is harder than integrating f(x) unless shortcuts can be found and we have shown in previous post's that at least some shortcuts exist. The question interesting to me is can the shortcuts be extended to include more general cases. Any ideas?

This can be helpful, for example, if you are writing a computer program designed to calculate areas. If the program recognises a particular expression as lending itself to the shortcut, then it doesn't have to do integration by parts or some other more exotic method.

Or consider it as a contest between bohrok, who knows the shortcut, and someone who doesn't know the shortcut. The question is find integral of |x^n| for positive odd n. Bohrok gets the answer in 1523 milliseconds, the other person not nearly as fast

Last edited: Aug 11, 2011
18. Aug 11, 2011

### Bohrok

19. Aug 11, 2011

### agentredlum

OH...MY...GOD you have no idea how much i want to thank you for that link. For months i've been hearing about wolfram alpha but i thought it was a computer program like Maple or Mathematica. I didn't know it was on the net, and my PS3 browser decodes it!!!!!

If i was in your neighborhood right now, i'd buy you a beer, i'd buy you 2 beers!.

As far as integral of |x^3 - x| my shortcuts are not working unless i made a mistake.

As for integral of |x^3| wolfram gave answer involving sgn(x) but you and i both know we don't need that. Haaaaaaaa.... Haaaa!

Humanity strikes back against the machines.

20. Aug 12, 2011

### agentredlum

According to wolframalpha...sqrt(abs(x)) has the same graph as abs(sqrt(x)) but that can't be right since domains are different

The first one has domain all real x

The second one has domain all rea NON-NEGATIVE x

What is going on here?

http://www.wolframalpha.com/input/?...,abs(x^(1/2))&asynchronous=false&equal=Submit

In the input interpretation it makes the correct interpretation but in the plots it doesn't.

Last edited: Aug 12, 2011


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