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# Is there a measure zero set which isn't meagre?

A subset of ℝ is meagre if it is a countable union of nowhere dense subsets (a set is nowhere dense if every open interval contains an open subinterval that misses the set).

Any countable set is meagre. The Cantor set is nowhere dense, so it's meagre. A countable union of meagre sets is meagre (e.g. all rational translates of the Cantor set).

There can also be meagre sets of positive measure, like "fat Cantor sets". To form a fat Cantor set, you start with a closed interval, then remove some open interval from the middle of it, then remove some open intervals from the remaining intervals, and so on. The result is nowhere dense because you removed open intervals all over the place. If the sizes of the intervals you remove get small fast, then the result has positive measure.

So does meagreness have any connection at all to measure? Specifically, are all measure zero sets meagre?

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Although the question was already answered, you may be interested in this: There is an excellent short little book by Oxtoby, "Measure and Category", where all these "classical" relations between both notions are discussed; you may want to take a look. There are deeper relations and differences, but you need set theory to understand them; a good book to look at is "Set theory: On the structure of the real line", by Bartoszynski and Judah. –  Andres Caicedo Oct 25 '10 at 5:48
Interesting spelling ... my dictionary says (non-math context) US spelling meager and UK spelling meagre. I suppose this goes along with theater/theatre and similar. –  Gerald Edgar Oct 25 '10 at 15:16
...and fiber/fibre, that some even give distinct mathematical meanings. –  Pietro Majer Oct 25 '10 at 17:11
I went with the Wikipedia spelling since that's what I was linking to, but I would normally spell it meager. –  Anton Geraschenko Oct 25 '10 at 17:56
I see that Wikipedia page has neighbourhood so other UK spellings are not unexpected. But shouldn't it then be Wikipaedia? –  Gerald Edgar Oct 25 '10 at 23:31

On the relation between null sets and meagre sets, you can also look at this article. Two theorems mentioned in this note (both classical and not due to the author):

1. (As already mentioned above) There exist a meagre $F_\sigma$ subset $A$ and a null $G_\delta$ subset $B$ of $\mathbb R$ that satisfy $A\cap B=\emptyset$ and $A\cup B=\mathbb R$.

2. (The Erd\H os-Sierpi\'nski Duality Theorem) Assume that the Continuum Hypothesis holds. Then there exists an involution (bijection of order two) $f:\mathbb R\to\mathbb R$ such that $f[A]$ is meagre if and only if $A$ is null, and $f[A]$ is null if and only if $A$ is meagre for every subset $A$ of $\mathbb R$.

While 1. says that the ideals of null, respectively meager sets are "orthogonal", 2. says that assuming CH they behave identically. But it is well known that this duality between measure and category fails dramatically once we take a more abstract point of view: Shelah proved that you need large cardinals to construct a model of set theory (ZF, no axiom of choice) where every set of reals is Lebesgue measurable, but no large cardinals are necessary to construct a model where every set of reals has the Baire property (the corresponding notion to measurability for category).

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Even though this isn't strictly an answer to the question, I'm going to accept it on the reasoning that it'll probably be most useful to someone visiting this thread in the future. I don't like the thought of somebody reading just the first couple of answers and missing this gem. –  Anton Geraschenko Oct 29 '10 at 2:32
In fact Shelah proved that “ZF+CC+all sets of real numbers are measurable” is equiconsistent to the existence of an inaccessible cardinal. Without countable choice you could simply choose a model of ZF where $\mathbb{R}$ is a countable union of countable sets. –  The User May 20 '13 at 20:24

Let $p_i$ be a list of the rational numbers. Let $U_{i,n}$ be an open interval centered on $p_i$ of length $2^{-i}/n$. Then $V_n=\cup_i U_{i,n}$ is an open cover of the rationals, of measure at most $\sum_i 2^{-i}/n=2/n$. Then $\cap_n V_n$ is a co-meager set of measure zero.

So yes, there is a measure zero set that is not meager, and so no, not every measure zero set is meager.

Computability theory gives a neat way to look at this. There is a certain type of real number that is called 1-generic and there is another type that is called 1-random or "Martin-Löf random". These two sets are disjoint. The set of 1-generic reals is co-meager and has measure zero, whereas the set of 1-random reals is meager and has full measure.

Thus measure and category are quite orthogonal. Set theorists would say they correspond to two different notions of forcing.

A good general reference for this kind of question is Oxtoby's classic book Measure and category.

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Excellent example! It took me a minute to fill in the reasoning, so I'll write it here for the benefit of future me. Each $V_n$ is co-meager since you've removed an interval around each rational, and a countable intersection of co-meager sets is co-meager. A co-meager set can't be meager or else all of $\mathbb R$ would be meager, which is isn't. –  Anton Geraschenko Oct 25 '10 at 5:37
It is a common error for a beginner to do this construction, then to think that $\cap_n V_n = \mathbb{Q}$. And then what is the instructor to do? Does the instructor have to cite something as esoteric as the Baire Category Theorem to disabuse that beginner of the erroneous idea? –  Gerald Edgar Oct 26 '10 at 14:42
@Gerald Edgar: good question, maybe argue that we can make a real number belong to each $V_n$ by specifying more and more of its binary expansion. We can even take breaks, i.e., after we have made $r$ start as $0.r_0\cdots r_{k_1}$ and thereby ensured $r\in V_1$ we can append $r_{k_1+1}\cdots r_{\ell}$ so as to make sure $r\ne p_1$. Then we add $r_{\ell+1}\cdots r_{k_2}$ to make $r\in V_2$ and so on. This kind of thing also shows $\cap_n V_n$ is size continuum. –  Bjørn Kjos-Hanssen Oct 26 '10 at 18:39
@GeraldEdgar, after you present the Bjørn Kjos-Hanssen's argument to dispel students myth, you turn around to students and say "Now students, this type of topological diagonal argument technique can be used to prove existence of a lot of things. Mathematicians have encapsulated this technique into a theorem and it's called Baire Category Theorem." That might make students less intimidated by Baire Category Theorem. –  Jisang Yoo Dec 2 '13 at 18:19

Although plenty of examples have already been given, let me add my favorite: Consider the set of those numbers in [0,1] whose binary expansion is not "half zeros and half ones", i.e., those for which the number of ones in the first $n$ binary places is not asymptotic to $n/2$. The strong law of large numbers implies that this set has measure zero. Yet it is not meager; in fact its complement is meager. More dramatically: The set of $x\in[0,1]$ whose binary expansion has, for infinitely many $n$, nothing but zeros from the $n$-th to the $n!$-th binary place is a dense $G_\delta$ set, hence comeager.

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This is really ideal, expressing $[0,1]$ as a union of a meager set and a measure 0 set. That kills two birds with one stone. –  Greg Kuperberg Oct 25 '10 at 12:39
@Greg Kuperberg: well, as long as you have a co-meager set of measure 0, such as in the classical example from Oxtoby's book, then you have that. What's neat about Andreas Blass' example, though, is that it is very intuitive that it should have measure 0 and be comeager. –  Bjørn Kjos-Hanssen Oct 25 '10 at 18:33
So, the set of normal numbers is meagre and has full Lebesgue measure. –  George Lowther Oct 25 '10 at 23:25

There's already been some good answers to this. However, this is something that I have also thought about recently, because I happen to have come across several meagre sets of full Lebesgue measure in some of my answers to other questions. In fact, in my experience on MO, meagre sets with full Lebesgue measure actually seems to be more the rule than the exception. So, I'll add these to the list.

1. In this math.SE question and this MO question, David Speyer was trying to find the set of θ such that $\sum_{n=1}^\infty \sin(n^r\theta)/n$ converges (r > 1 an integer). He was concerned about the θ = 1 case but, from my answer on MO and David's answer on math.SE it can be seen that it converges for almost every θ but, at the same time, it only converges for θ in a meagre set.

2. Along a similar line, this MO question was asking for which θ the asymptotic bound $\sum_{n=1}^N{\rm sign}(\sin(n\pi\theta))=O(N^x)$ holds. For 1/2 < x < 1 my answer shows that it holds for almost every θ but, at the same time, it only holds for θ in a meagre set.

3. This question asks if there are 2x2 matrices C such that Tr(Cn) is dense in the reals as n runs through the positive integers. Bjorn Poonen shows that the answer is yes. In fact, his proof is easily modified to show that Tr(Cn) fails to be dense only on a meagre set. However, my answer shows that |Tr(Cn)| is either bounded or tends to infinity (so, not dense) for almost every C.

4. The above examples really just come down to the following point. The set of real numbers with finite irrationality measure (i.e., non-Liouville numbers) is meagre. However, almost every real number has irrationality measure 2.

Along similar lines, the set of normal numbers is meagre and has full Lebesgue measure (see also, Andreas' answer). The set of real numbers whose continued fraction quotients have geometric mean converging to Khinchin's constant is meagre with full Lebesgue measure. The set of real numbers whose continued fraction quotients occur according to the Gauss-Kuzmin distribution is meagre with full Lebesgue measure. And so on...

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Two examples are given in Gelbaum and Olmsted, Counterexamples in Analysis. One is the example given by Bjorn Kjos-Hanssen in his answer. The other goes like this. Let $A_n$ be a Cantor set in $[0,1]$ of measure $(n-1)/n$, $n=1,2,3,\dots$, let $A$ be the union of the $A_n$, then the complement of $A$ is measure zero but not meager (or even meagre).

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Very nice. Again, the idea is to find a full measure meager set. Then its complement is of measure zero and can't be meager. –  Anton Geraschenko Oct 25 '10 at 5:40

There is also the set $D$ of Diophantine numbers, which occurs naturally in dynamical systems : $x\in\mathbb{R}$ is Diophantine if there exists $c>0$ and an integer $k$, such $|x-p/q|\geq c/q^k$ for all rational number $p/q$. It's easy to see that $D$ is of full measure (i.e. $D^c$ has measure $0$), but is meager.

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Perhaps it makes sense to mention this example: The category of measurable spaces is equivalent to the category of hyperstonean topological spaces and hyperstonean maps between them.

To construct a measurable space (X,M,N) from a hyperstonean topological space (Y,T), set X=Y, let M be the set of all unions of open and meager sets, and let N be the set of all meager sets in (Y,T). (Here M is the set of all measurable sets and N is the set of all null sets, i.e., sets of measure 0. For more information see this answer: Is there an introduction to probability theory from a structuralist/categorical perspective?)

So in this particular case meager sets are precisely sets of measure 0.

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# New web host

I’ve moved this website to Fresh Roasted Hosting because the previous host was awful in terms of reliability and support. The whole process only took a few hours (transfer domain, restore cPanel backups and settings), and hopefully there won’t be any more several day outages with the new host.

# Coursera first impressions

I recently signed up for a few Coursera courses, and overall it’s been a fun experience. Here are some of my comments.

## Game Theory

So far the instructors have explained normal form games, Nash equilibrium, Pareto optimality, mixed strategies and maxmin strategies. The definitions are pretty clear and plenty of examples follow. The graded problem sets are usually fairly easy and sometimes include interesting examples not found in the videos. However, there are no proofs of any theorems or results. In fact, the theorems themselves are not even written down in any precise way – the instructors seem to be intent on avoiding mathematical notation here for some reason. Verdict: Enrol.

## The Modern World: Global History since 1760

Starting near the end of the Commercial Revolution, the instructor explores global history all the way up to the present. There are usually one or two questions at the end of each video to determine whether you have been paying attention. The instructor (Philip Zelikow from the University of Virginia) really shows his enthusiasm for the subject in the videos. Verdict: Enrol!

## Image and video processing: From Mars to Hollywood with a stop at the hospital

The instructor begins by giving many applications of image processing, and then gets straight into topics like Huffman coding and the discrete cosine transform (no proofs). I haven’t watched a lot of them yet. Verdict: ???

## Introduction to Finance

The instructor is clearly excited about the subject, but keeps going off on tangents. Furthermore, he takes around an hour to explain the compound interest formula $$P(1+r/n)^{nt}$$, something that should take 5 minutes at most. Use Khan Academy instead. Verdict: Avoid.

## Analytic Combinatorics, Part I

Finally, a course in which the forums aren’t full of complaints about “too much math”. Sedgewick’s voice can be a little boring at times, but the content should be extremely interesting to anyone who knows about generating functions. Verdict: Enrol!

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## Geometry & Topology

### A sharp compactness theorem for genus-one pseudo-holomorphic maps

Aleksey Zinger

#### Abstract

For every compact almost Kahler manifold $(X,ω,J)$ and an integral second homology class $A$, we describe a natural closed subspace $M¯1,k0(X,A;J)$ of the moduli space $M¯1,k(X,A;J)$ of stable $J$–holomorphic genus-one maps such that $M¯1,k0(X,A;J)$ contains all stable maps with smooth domains. If $(ℙn,ω,J0)$ is the standard complex projective space, $M¯1,k0(ℙn,A;J0)$ is an irreducible component of $M¯1,k(ℙn,A;J0)$. We also show that if an almost complex structure $J$ on $ℙn$ is sufficiently close to $J0$, the structure of the space $M¯1,k0(ℙn,A;J)$ is similar to that of $M¯1,k0(ℙn,A;J0)$. This paper’s compactness and structure theorems lead to new invariants for some symplectic manifolds, which are generalized to arbitrary symplectic manifolds in a separate paper. Relatedly, the smaller moduli space $M¯1,k0(X,A;J)$ is useful for computing the genus-one Gromov–Witten invariants, which arise from the larger moduli space $M¯1,k(X,A;J)$.

#### Article information

Source
Geom. Topol., Volume 13, Number 5 (2009), 2427-2522.

Dates
Revised: 7 December 2008
Accepted: 8 May 2009
First available in Project Euclid: 20 December 2017

https://projecteuclid.org/euclid.gt/1513800316

Digital Object Identifier
doi:10.2140/gt.2009.13.2427

Mathematical Reviews number (MathSciNet)
MR2529940

Zentralblatt MATH identifier
1174.14012

#### Citation

Zinger, Aleksey. A sharp compactness theorem for genus-one pseudo-holomorphic maps. Geom. Topol. 13 (2009), no. 5, 2427--2522. doi:10.2140/gt.2009.13.2427. https://projecteuclid.org/euclid.gt/1513800316

#### References

• K Behrend, B Fantechi, The intrinsic normal cone, Invent. Math. 128 (1997) 45–88
• M Bershadsky, S Cecotti, H Ooguri, C Vafa, Holomorphic anomalies in topological field theories, Nuclear Phys. B 405 (1993) 279–304
• A Floer, H Hofer, D Salamon, Transversality in elliptic Morse theory for the symplectic action, Duke Math. J. 80 (1995) 251–292
• K Fukaya, K Ono, Arnold conjecture and Gromov–Witten invariant, Topology 38 (1999) 933–1048
• E Ionel, Genus $1$ enumerative invariants in $\mathbf P\sp n$ with fixed $j$ invariant, Duke Math. J. 94 (1998) 279–324
• J Li, G Tian, Virtual moduli cycles and Gromov–Witten invariants of general symplectic manifolds, from: “Topics in symplectic $4$–manifolds (Irvine, CA, 1996)”, (R J Stern, editor), First Int. Press Lect. Ser. I, Int. Press, Cambridge, MA (1998) 47–83
• J Li, A Zinger, On the genus-one Gromov–Witten invariants of complete intersections, to appear in J. Diff. Geom.
• D McDuff, D Salamon, $J$–holomorphic curves and quantum cohomology, Univ. Lecture Series 6, Amer. Math. Soc. (1994)
• Y Ruan, G Tian, A mathematical theory of quantum cohomology, J. Differential Geom. 42 (1995) 259–367
• Y Ruan, G Tian, Higher genus symplectic invariants and sigma models coupled with gravity, Invent. Math. 130 (1997) 455–516
• R Vakil, A tool for stable reduction of curves on surfaces, from: “Advances in algebraic geometry motivated by physics (Lowell, MA, 2000)”, (E Previato, editor), Contemp. Math. 276, Amer. Math. Soc. (2001) 145–154
• R Vakil, A Zinger, A desingularization of the main component of the moduli space of genus-one stable maps into $\Bbb P\sp n$, Geom. Topol. 12 (2008) 1–95
• A Zinger, Enumeration of genus-two curves with a fixed complex structure in $\Bbb P\sp 2$ and $\Bbb P\sp 3$, J. Differential Geom. 65 (2003) 341–467
• A Zinger, Basic estimates of riemannian geometry used in gluing pseudoholomorphic maps, Notes (2004) Available at \setbox0\makeatletter\@url http://www.math.sunysb.edu/~azinger/research/analytic.pdf {\unhbox0
• A Zinger, Completion of Katz–Qin–Ruan's enumeration of genus-two plane curves, J. Algebraic Geom. 13 (2004) 547–561
• A Zinger, Enumerative vs. symplectic invariants and obstruction bundles, J. Symplectic Geom. 2 (2004) 445–543
• A Zinger, Counting rational curves of arbitrary shape in projective spaces, Geom. Topol. 9 (2005) 571–697
• A Zinger, On the structure of certain natural cones over moduli spaces of genus-one holomorphic maps, Adv. Math. 214 (2007) 878–933
• A Zinger, Reduced genus-one Gromov–Witten invariants
• A Zinger, Pseudocycles and integral homology, Trans. Amer. Math. Soc. 360 (2008) 2741–2765
• A Zinger, Standard versus reduced genus-one Gromov–Witten invariants, Geom. Topol. 12 (2008) 1203–1241
• A Zinger, The reduced genus-one Gromov–Witten invariants of Calabi–Yau hypersurfaces, J. Amer. Math. Soc. 22 (2009) 691–737


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# spSubmat¶

## Purpose¶

Returns a sparse submatrix of a sparse matrix.

## Format¶

s = spSubmat(x, rinds, cinds)
Parameters: x (MxN sparse matrix) – data rinds (Kx1 vector) – row indices. cinds (Lx1 vector) – column indices. s (KxL sparse matrix) – the intersection of rinds and cinds.

## Examples¶

sparse matrix y;
sparse matrix z;

x = { 0 0 0 10,
0 2 0 0,
0 0 0 0,
5 0 0 0,
0 0 0 3 };

y = denseToSp(x, 0);

// Extract all columns; rows 1, 3 and 4
z = spSubmat(y, 1|3|4, 0);

// Extract all values from 'z' into a dense matrix 'd'
d = spToDense(z);

Now d is equal to:

0.00   0.00   0.00  10.00
0.00   0.00   0.00   0.00
5.00   0.00   0.00   0.00

## Remarks¶

If rinds or cinds are scalar zeros, all rows or columns will be returned.

Since sparse matrices are strongly typed in GAUSS, y must be defined as a sparse matrix before the call to spSubmat().

Functions spDenseSubmat()


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# (i) Complete the table and make a guess about the limit indicated. (ii) Confirm your conclusions about the limit by

###### Question:

(i) Complete the table and make a guess about the limit indicated. (ii) Confirm your conclusions about the limit by graphing a function over an appropriate interval. [Note: For the inverse trigonometric function, be sure to put your calculating and graphing utilities in radian mode.] $f(x)=\frac{\sin ^{-1} 2 x}{x} ; \lim _{x \rightarrow 0} f(x)$ $\begin{array}{|c|c|c|c|c|c|c|}\hline x & -0.1 & -0.01 & -0.001 & 0.001 & 0.01 & 0.1 \\\hline f(x) & & & & & & \\\hline\end{array}$

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Stock in CDB Industries has a beta of 1.05. The market risk premium is 7 percent, and T-bills are currently yielding 4.0 percent. CDB’s most recent dividend was $2.90 per share, and dividends are expected to grow at a 5 percent annual rate indefinitely. Required: If the stock sells f... 5 answers ##### HW Thermal Physics Begin Date: 7/17/2019 12.01:00 AM Due Date: 7/26/2019 [1.59.00 PM End Date: 7/26/2019 1.59.00 PM (0%) Problem 14: Several icc cubes (@; = 0.9167 glcm ) of total volume V; = 210 cm? pue temperature 273.15 K (0.000 *C) are put into thcrmos containing Vz = 610 cm? of tca at tcmpcraturc of 313.15 K, completcly filling the thermos Thc lid is then put on thc thermos to close it. Assume that thc density and the specific heat of the tea is the samc a8 it is for fresh watcr (Qw = glcm& HW Thermal Physics Begin Date: 7/17/2019 12.01:00 AM Due Date: 7/26/2019 [1.59.00 PM End Date: 7/26/2019 1.59.00 PM (0%) Problem 14: Several icc cubes (@; = 0.9167 glcm ) of total volume V; = 210 cm? pue temperature 273.15 K (0.000 *C) are put into thcrmos containing Vz = 610 cm? of tca at tcmpcratu... 1 answer ##### Prove each formula by induction, if possible. $10+6+2+\dots+(14-4 n)=12 n-2 n^{2}$ Prove each formula by induction, if possible. $10+6+2+\dots+(14-4 n)=12 n-2 n^{2}$... 1 answer ##### Calculating Select Ratios using Excel's ROUND and Basic Math Functions Games Galore has provided its condensed... Calculating Select Ratios using Excel's ROUND and Basic Math Functions Games Galore has provided its condensed financial statements for the year ended December 31, 2016. The Controller has asked you to calculate liquidity, solvency, and profitability ratios that management can use decision makin... 5 answers ##### Draw the major product of the following reaction sequence_OCHzHOH;o'2 HzOHzoLIAIHA Draw the major product of the following reaction sequence_ OCHz HO H;o' 2 HzO Hzo LIAIHA... 1 answer ##### EXERCISE Consider a parallel-plate capacitor having area A0.20 mand separation d-0.012 m. Hint (a) The capacitor... EXERCISE Consider a parallel-plate capacitor having area A0.20 mand separation d-0.012 m. Hint (a) The capacitor is first attached to a battery having voltage 9 V. What are AV (in V) and Q (in pC)? pC (b) Now, the battery is removed, and a metallic slab of thickness a = 0.005 m is inserted, what are... 1 answer ##### Why is proper handling of samples (quantitive transfer) important in calculating percent yield? A cleaned, dried... Why is proper handling of samples (quantitive transfer) important in calculating percent yield? A cleaned, dried can (Cherry Pepsi) was recently weighed. Te recorded mass was 13.2518 grams. Assuming that 0.5000 g of that weight was other materials (paint, aluminum oxide), what would the theoretical ... 3 answers ##### During the month of February, a flu epidemic hit the University During the month of February, a flu epidemic hit the University. The number of people sick attime t (measured in days) is given by the function P(t). The rate at which the epidemic isspreading on February 3 is 110 people per day. How is the information best represented mathematically? my answer is P... 1 answer ##### Imagine you are an economic analyst working at one of the major Chartered banks and you... Imagine you are an economic analyst working at one of the major Chartered banks and you are asked to write a policy brief for one of the board members (highly educated people with some economic background, but not necessarily economists). In this policy brief, you should explain and critically analy... 5 answers ##### Telephane company clalms Ihat 20% ol Ils cuslomers have least Iwvo telephone lines The company selects = random sample ol 500 cuslomers and fInds Ihal 88 have [WO Or more telephone Iines. If4 0.05, test the company$ claim uslng crilical values and repclion regions
telephane company clalms Ihat 20% ol Ils cuslomers have least Iwvo telephone lines The company selects = random sample ol 500 cuslomers and fInds Ihal 88 have [WO Or more telephone Iines. If4 0.05, test the company \$ claim uslng crilical values and repclion regions...
##### The function f(x)=x +9,x20 is one-to-one (a) Find the inverse of f and check the answer (b) Find the domain and the range of f and f (c) Graph f, f and y =X on the same coordinate axes(a) f"- (x) = (Simplify your answer: Use integers or fractions for any numbers in the expression )
The function f(x)=x +9,x20 is one-to-one (a) Find the inverse of f and check the answer (b) Find the domain and the range of f and f (c) Graph f, f and y =X on the same coordinate axes (a) f"- (x) = (Simplify your answer: Use integers or fractions for any numbers in the expression )...
##### Find the characteristic polynomial of the matrix using either a cofactor expansion or the special formula...
Find the characteristic polynomial of the matrix using either a cofactor expansion or the special formula for 3 x 3 determinants. Note: Finding the characteristic polynomial of a 3 x 3 matrix is not easy to do with just Tow operations, because the variable is involved) 013 104 340 The characteristic...
##### Help please Help The Foundational 15 (LO2-1, LO2-2, LO2-3, LO2-4) (The following information applies to the...
help please Help The Foundational 15 (LO2-1, LO2-2, LO2-3, LO2-4) (The following information applies to the questions displayed below.) Sweeten Company had no jobs in progress at the beginning of March and no beginning inventories. The company has two manufacturing departments –Molding and Fab...
##### In E. coli, four Hfr strains donate the following markers, shown in the order donated:Hfr strainOrder of_gene transfer First Last 2 3 " 6 1 " : 4 (a) What is the order of these markers on the circular chromosome of the All of these Hfr strains were originally derived from the same F+ strain: know in the real world that the map is circular). Indicate the relative positions of the original F+ strain? Draw a map (it can be linear because it is easier to draw; but we (b) Add to your map th
In E. coli, four Hfr strains donate the following markers, shown in the order donated: Hfr strain Order of_gene transfer First Last 2 3 " 6 1 " : 4 (a) What is the order of these markers on the circular chromosome of the All of these Hfr strains were originally derived from the same F+ str...
##### Evaluate the following integrals ) (âˆ«tan^3 Î¸ sec^3 Î¸)dÎ¸ 3. )âˆ«x+ 4/ (âˆš/7âˆ’x^(2) âˆ’6x ) dx
Evaluate the following integrals ) (âˆ«tan^3 Î¸ sec^3 Î¸) dÎ¸ 3. )âˆ«x+ 4/ (âˆš/7âˆ’x^(2) âˆ’6x ) dx...
##### Problcm [25 pts] rAtD sampl uniform distributiondistributerl accorcling to thefa(r)[o.01 20 , resuting data {TnT2 _T}. YoulSuppose that samplc collevted for given that for this data_Tz =12.2 ri ="3[15 pts] Find te 95% confidence Mcn Tle MPAJ ol distrilution Ja(r): Then tind the' corresponding conficknce imtarval for[10 pts] Test t" Wputhsis thatusIg signifcatc e [evel0.1.
Problcm [25 pts] rAtD sampl uniform distribution distributerl accorcling to the fa(r) [o.01 20 , resuting data {TnT2 _T}. Youl Suppose that samplc collevted for given that for this data_ Tz =12.2 ri ="3 [15 pts] Find te 95% confidence Mcn Tle MPAJ ol distrilution Ja(r): Then tind the' corr...
##### Determine the chemical formula for the ionic compound that forms between aluminum and oxyg chemical formula:...
determine the chemical formula for the ionic compound that forms between aluminum and oxyg chemical formula: Kl Determine the chemical formula for the ionic compound that forms between rubidium and sulfur. chemical formula: Rb, S Determine the chemical formula for the ionic compound that forms b...
##### Imagine that saturated solutions of barium chloride (BaCl) and potassium sulfate (K,SO.) e mixed. A white...
Imagine that saturated solutions of barium chloride (BaCl) and potassium sulfate (K,SO.) e mixed. A white precipitate forms, and the solution is found to contain potassium chloride. (a) What is the name of the precipitate? (b) Write an equation for the reaction that occurred. (c) Which are the spect...
##### Question 5 Which of the following chlorobenzenes does not undergo nucleophilic aromatic substitution upon treatment with...
Question 5 Which of the following chlorobenzenes does not undergo nucleophilic aromatic substitution upon treatment with NaNH 2? 2 3 ООО a. 1 b.2 c. 3 d.4...
##### Compute the directional derivative of the following function at the given point P in the direction of the given vector Be sure to use unit vector for the direction vector:ffxy) = In (9+X2 +3y2);P(1,- 3); (2,2)The directional derivative is (Type an exact answer; using radicals as needed )
Compute the directional derivative of the following function at the given point P in the direction of the given vector Be sure to use unit vector for the direction vector: ffxy) = In (9+X2 +3y2);P(1,- 3); (2,2) The directional derivative is (Type an exact answer; using radicals as needed )...


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# How to Convert Slope Intercept Form to Standard Form

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Any equation that relates the first power of x to the first power of y produces a straight line on an x-y graph. The standard form of such an equation is Ax + By + C = 0 or Ax + By = C. When you rearrange this equation to get y by itself on the left side, it takes the form y = mx +b. This is called slope intercept form because m is equal to the slope of the line, and b is the value of y when x = 0, which makes it the y-intercept. Converting from slope intercept form to standard form takes little more than basic arithmetic.

#### TL;DR (Too Long; Didn't Read)

To convert from slope intercept form y = mx + b to standard form Ax + By + C = 0, let m = A/B, collect all terms on the left side of the equation and multiply by the denominator B to get rid of the fraction.

### The General Procedure

An equation in slope intercept form has the basic structure

y = mx + b
\begin{aligned} y - mx &= (mx - mx ) + b \\ y - mx &= b \end{aligned}
\begin{aligned} y - mx - b &= b - b \\ y - mx - b &= 0 \end{aligned}
-mx + y - b = 0

If m is an integer, then B will equal 1.

-\frac{A}{B}x + y - b = 0
-Ax + By - Bb = 0
-Ax + By - C = 0

### Examples:

(1) - The equation of a line in slope intercept form is:

y = \frac{1}{2} x + 5

What is the equation in standard form?

y - \frac{1}{2}x = 5
y - \frac{1}{2}x - 5 = 0
2y - x - 10 = 0
-x + 2y - 10 = 0

You can leave the equation like this, but if you prefer to make x positive, multiply both sides by -1:

x - 2y + 10 = 0

or

x - 2y = -10

(2) - The slope of a line is -3/7 and the y-intercept is 10. What is the equation of the line in standard form?

The slope intercept form of the line is

y = -\frac{3}{7}x + 10

Following the procedure outlined above:

\begin{aligned} y + \frac{3}{7}x - 10 = 0 \\ 7y + 3x - 70 = 0 \\ 3x + 7y -70 = 0 \\ \text{or} \\ 3x + 7y = 70 \end{aligned}


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# If any amount is distributed among A ,B and C so that the part of A

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If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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07 Oct 2015, 11:02
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Difficulty:

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Question Stats:

86% (01:01) correct 14% (01:16) wrong based on 133 sessions

### HideShow timer Statistics

If any amount is distributed among A ,B and C so that the part of A is doubled that of the part of B and part of b is 4 time the part of C .Find the ratio among their parts.

A. 1:2:4

B. 1:4:1

C. 8:4:1

D. 2:4:1

E. 4:3:2

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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

### Show Tags

07 Oct 2015, 21:27
5
Hi ske,

This question can be solved by TESTing VALUES. Here's how:

We're told that A is TWICE B and B is FOUR TIMES C....

IF...
C = 2
B = 8
A = 16

We're asked for the ratio of A:B:C....The ratio of these 3 values is...
16:8:2

This reduces to....
8:4:1

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** ##### General Discussion Manager Joined: 11 Sep 2013 Posts: 108 Re: If any amount is distributed among A ,B and C so that the part of A [#permalink] ### Show Tags 07 Oct 2015, 11:05 2 ske wrote: if any amount is distributed among A ,B and C so that the part of A is doubled that of the part of B and part of b is 4 time the part of C .Find the ratio among their parts. A. 1:2:4 B. 1:4:1 C. 8:4:1 D. 2:4:1 E. 4:3:2 A=2B, B=4C A:B:C = 2B:B:B/4=8:4:1 => Ans: C Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6966 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If any amount is distributed among A ,B and C so that the part of A [#permalink] ### Show Tags 08 Oct 2015, 01:10 3 Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. if any amount is distributed among A ,B and C so that the part of A is doubled that of the part of B and part of b is 4 time the part of C .Find the ratio among their parts. A. 1:2:4 B. 1:4:1 C. 8:4:1 D. 2:4:1 E. 4:3:2 By the assumptions as mentioned we have A=2B and B=4C. So A=2(4C) and we get A=8C. We have, therefore, A:B:C = 8C:4C:C. By cancelling the ratio with C we have 8:4:1. The answer is, therefore, C. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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10 Feb 2016, 21:26
6
A=2B, B=4C A:B:C = 2B:B:B/4=8:4:1

=> Ans: C
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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20 Mar 2017, 01:17
2
A=2B,

B=4C

A:B:C = 2B:B:B/4=8:4:1

=> Ans: C
i will go for c.
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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20 Mar 2017, 03:32
ske wrote:
If any amount is distributed among A ,B and C so that the part of A is doubled that of the part of B and part of b is 4 time the part of C .Find the ratio among their parts.

A. 1:2:4

B. 1:4:1

C. 8:4:1

D. 2:4:1

E. 4:3:2

let the part for A,B,C be a,b,c respectively
a = 2b or a : b = 2:1 = 8:4
b = 4c or b:c = 4:1

therefore a:b:c = 8:4:1
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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12 Jul 2017, 10:39
1
ske wrote:
If any amount is distributed among A ,B and C so that the part of A is doubled that of the part of B and part of b is 4 time the part of C .Find the ratio among their parts.

A. 1:2:4

B. 1:4:1

C. 8:4:1

D. 2:4:1

E. 4:3:2

hi ske, it can be lik this

A=2B, B=4C A:B:C = 2B:B:B/4=8:4:1

=> Ans: C
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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13 Jul 2017, 20:04
Confused. Help!

Given:
A = 2B
B = 4C

In order to make 'B' equal in both the ratios, we need to multiple the second ration by 2. Therefore,
B=4C becomes, 2B = 8C.

(1) ----- A = 2B
(2) -----2B = 8C

Combining (1) and (2), we get A:2B:8C.

Evidently, there is no such answer. What am I missing?
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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14 Jul 2017, 06:25
1
Hi Blackbox,

While you've done the proper algebra, you have not properly translated the given equations into the end ratio.

Here's a simple example of how the math actually "works":

A = 2B

This means that A is equal to TWICE B.

For example....
IF... B=1, then A=2
IF... B=2, then A=4
IF... B=3, then A=6
Etc.

This means that whatever B is, A is DOUBLE that value. Thus, the ratio of A to B is...
A:B
2:1

Using that example, and the equations that you've created, what would the ratio of A:B:C equal?

GMAT assassins aren't born, they're made,
Rich
_________________

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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

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14 Jul 2017, 09:20
Let $$B = x$$.

$$A = 2x, C = 0.25x$$

Ratio of $$A:B:C = 2:1:0.25 = 8:4:1.$$ Ans - C.
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

### Show Tags

14 Jul 2017, 10:11
EMPOWERgmatRichC wrote:
Hi Blackbox,

While you've done the proper algebra, you have not properly translated the given equations into the end ratio.

Here's a simple example of how the math actually "works":

A = 2B

This means that A is equal to TWICE B.

For example....
IF... B=1, then A=2
IF... B=2, then A=4
IF... B=3, then A=6
Etc.

This means that whatever B is, A is DOUBLE that value. Thus, the ratio of A to B is...
A:B
2:1

Using that example, and the equations that you've created, what would the ratio of A:B:C equal?

GMAT assassins aren't born, they're made,
Rich

I see what are you're trying to convey. So, A:B = 2:1, B:C = 4:1. In order to make an equal proportion of the items, I need to make B = 4 in the first ratio. Therefore, I'd have to multiply the first ratio by 4. Which makes the combined ratio 8:4:1.

Thanks mate.
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Re: If any amount is distributed among A ,B and C so that the part of A  [#permalink]

### Show Tags

14 Jul 2017, 10:12
TimeTraveller wrote:
Let $$B = x$$.

$$A = 2x, C = 0.25x$$

Ratio of $$A:B:C = 2:1:0.25 = 8:4:1.$$ Ans - C.

Thanks TT. Yours was a cool explanation as well.
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If any amount is distributed among A ,B and C so that the part of A  [#permalink]

### Show Tags

18 Dec 2017, 03:26
ske wrote:
If any amount is distributed among A ,B and C so that the part of A is doubled that of the part of B and part of b is 4 time the part of C .Find the ratio among their parts.

A. 1:2:4

B. 1:4:1

C. 8:4:1

D. 2:4:1

E. 4:3:2

A=2B or [ A/B ]= [2/1] Eq 1
B=4C or [ B/C ]= [ 4/1 ] Eq 2

Equate the B's or Multiply Eq 1 by 4 & Multiply Eq 2 by 1

A:B:C [ 8:4:1]
If any amount is distributed among A ,B and C so that the part of A   [#permalink] 18 Dec 2017, 03:26
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# Countdown timer in every table cell

I am making countdown app with multiple records, so I am using UITableView to show that records and countdown dates. What I do is just take one timer and reload cells on every second. Is this a best approach to reload visible cells on every second or is there another better approach there?

updateCellContentsTimer = Timer.scheduledTimer(timeInterval: 1,
target: self,
selector: #selector(ViewController.updateCells),
userInfo: nil, repeats: true)

// MARK: Custom Functions

func updateCells() {
let indexPathsArray = tableView.indexPathsForVisibleRows
for indexPath in indexPathsArray! {
let cell = tableView.cellForRow(at: indexPath) as! CountdownTableViewCell

cell.timeLeftLabel.text = progress + "hours:minutes:seconds"
}
}

• Instead of reloading every cell on timer event you can use the notification mechanism of Cocoa. – Shial Apr 14 '17 at 1:28
• @Shial, since it's a countdown, every visible cell needs to change every second. Doing that indirectly through notifications only adds complexity without avoiding the cell updates. The OP already limits the updates to visible cells so the process is optimal. – Alain T. Apr 19 '17 at 13:22
• @AlainT. so this is the right way to update time in every cell .. right?? – EI Captain v2.0 Apr 21 '17 at 4:44
• Yes. What makes it right is that you use indexPathsForVisibleRows. This is the smallest number of cell updates that you can do for your countdown. Having only one timer to manage all the countdowns is also the optimal way to do it. – Alain T. Apr 21 '17 at 4:54
• @AlainT. thanks buddy ... thanks for your suggestion .. really appreciate :) – EI Captain v2.0 Apr 21 '17 at 5:00

to avoid DRY i would put this line (i think it is in your code only a sample logic) inside the cell. in this case you have the logic for calculating the remaining time inside the cell and not outside somewhere. and you can use always the same function to update the progress.

cell.timeLeftLabel.text = progress + "hours:minutes:seconds"

should be become

cell.updateProgress()


text:
# Difference between revisions of "Quest:Bad Ass quests"

WARNING: This page contains Quest spoilers. ~ Solutions ~ Table ~ List ~ All ~ Category ~

Highlight to view quest spoilers.

#### "Basketcase", in which you delivered the goods. (Quest level 4)

Skills: 160 or less other.movement.climbing.tree bonus OR 100 or less covert.manipulation.sleight-of-hand bonus
Find Mrs Smith in the house attached to the forge in Bad Ass.  Ask her about Granny Weatherwax and she will send you on a quest for various items.
The first item is always a small green apple.  If using sleight-of-hand, go to the bedroom on the second floor and 'open window' then 'grab apple from tree'. For tree climbing go to the east room of the orchard which is north of the forge. 'climb tree' and then 'get apple'. The tree will cause damage to you if you fail the skill check.
The next two items you will be required to collect are random and could be any of the following:
Wood: Obtain an axe, the general store often sells them. Look at trees around Bad Ass until the trees mention chopping them down. Once you locate them, hold the axe and 'chop tree with axe' until you have wood.
Peach Preserve: Inside the house connected to the forge, go upstairs and look at the ceiling. Go up through the trap door and ask Jud Smith for peach preserve for Granny.
Sausage: Enter the south east most house in the village through the east door. Once inside 'search utensils' to find the sausage.
Egg: One west from the forge entrance, go southeast into a room containing a nest box. Searching the box results in an egg.
Wool: One west of the forge entrance, go south and ask Mrs Shepston about wool for Granny.

#### "Knotscape Navigator", in which you hacked some trees for later browsing. (Quest level 4)

Skills: >153 ad.direction bonus?
Items: A guide book from the Bad Ass general store
Items: A knife

During summer or autumn: Go to the Bad Ass smithy and say "help" while Gordo Smith is there, then "yes".  Follow Kev when he shows up.  "judge" the distance to the town when he tells you to, and use the book to carve the appropriate symbols on the trees.

During winter or spring: Go back to the smithy and say something about "marking grooves" to Gordo Smith, then say "yes".  Drink the beer he gives you, and you'll wake up outside of town in the forest.  "feel trees" to translate the grooves on the trees, then "head  <dir> for <number> miles" in the indicated direction and distance to attempt to get back to the village.  There is some randomness involved in the skillcheck, so you might have to try several times.

The relevant information from the guide is reproduced below:


text:
## Trend filtering

Trend filtering is an estimator that uses discrete differences to smooth a noisy vector (or more generally, tensor). The method was proposed and computational methods were studied in the 2000’s for data on multidimensional grids. Basically, it takes the general form,

$\sum_i \ell(\beta_i; y_i) + \lambda \| D^{(k)} \beta \|_1$

where D is a discrete differencing operator. The k is the order, basically, for k of 1 you get piecewise constant signal, and for higher k you get piecewise polynomials. FYI trend filtering with k of 1 is called the fused lasso or total variation denoising (for one of my papers I regrettably tried to rename this the ‘edge lasso’ when applied to graphs). The nice thing about this is that you get automated knot selection, which happens because you are using an L1 penalty. This gives you local adaptivity, which is when your estimator adapts to the local smoothness of the underlying signal. You can use this with most losses, for example, the Poisson loss can give you locally adaptive histograms,

## Graph trend filtering

It is pretty straightforward to define the fused lasso (k=1) over graphs, you just let D be the incidence matrix, which turns the penalty into the L1 of the difference across edges of the graph. For higher order graphs, we had to make some choices, and what we settled on is

$D^{(k+1)} = \Big\{ \begin{array}{ll} D^\top D^{(k)},& k~{\rm odd} \\ D D^{(k)}, & k~{\rm even} \end{array}.$

This has the desired local adaptivity property, as evidenced by this example over the Allegheny county zip codes (it varies the smoothness locally) which are connected based on direct proximity:

Image courtesy of Ryan Tibshirani

## What we know

So after years of studying trend filtering we have some things that we do know (about theory). First, I would like to call out two types of results,

1. L0 rates: these are results that assume that true signal satisfies a sparsity assumption, that there are only s non-zero discrete derivatives. Typically, the hope is to recover the standard ‘parametric’ rates of the lasso, $$\| \hat \beta - \beta_0 \|^2 \lesssim s~{\rm plog} n$$ for some choice of tuning parameter and $$\beta_0$$ is the true signal.

2. L1 rates: these are results that assume that the true signal have only bounded L1 norm (total variation), i.e. $$\| D \beta \|_1$$ is bounded. Typically, here you show that $$\| \hat \beta - \beta_0 \|^2 \lesssim n^\alpha~{\rm plog} n$$ for some choice of tuning parameter.

The problem with 1, is that you can get the parametric rates regardless of the graph structure (as long as it is connected) for the fused lasso. We showed this with the depth-first-search (DFS) wavelets (at least for the fused lasso) in this paper. Basically, 1 only requires a tree structure and not a more highly connected graph structure. It is substantially harder to show this for higher order trend filtering, even in 1 dimension (see this paper by Guntuboyina et al. and our prior work), and there is a loss in log factors if you go with the DFS wavelet approach. The satisfying thing about results of type 2 is that the rate parameter $$\alpha$$ depends on the graph structure, and smaller typically means a more connected graph. For this reason, I have focused on this type of result.

There is a third type of result, which gives you exact reconstruction of the region where you have polynomial structure. I attempted this in this paper, but the conditions are ridiculous. I believe that this is insurmountable, and I provide some evidence.

So, what do we know?

1. Rates for Gaussian noise, grid graphs (Veeru Sadhanala’s thesis, this paper)
2. Rates for density estimation, in 1 dimension (and ‘geometric graphs’) with k=1 (fused lasso).
3. A base rate for all connected graphs with k=1, spoiler, it is the same as the 1D fused lasso $$n^{1/3}$$.
4. A tool, called eigenvector incoherence, for analyzing any order over any graph that is tight in some cases.
5. A specialized result for k=1, nearest neighbor graphs, turns out that the rate is the same for grids on the intrinsic dimension of the manifold.
6. Rates for additive trend filtering models
7. Lower bounds for multidimensional trend filtering

## Future work (that I probably will not pursue myself)

So there is a lot that we don’t know, here are some thoughts.

1. The eigenvector incoherence tool gives us a method for analyzing any graph, but it is very opaque. It would be better if we could relate the rates to a functional of the graph that is more interpretable and easier to analyze.
2. General exponential families are mostly wide open.
3. Rates for trend filtering in high dimensions under strong sparsity assumptions.
4. There are strange behaviors for higher order GTF on point clouds in multiple dimensions, particularly at the boundaries of the support. For $$k=1$$ this goes away, but at higher order, we need to modify GTF. Our best guess at the ideal approach is to come up with a fine spline basis and then come up with the trend filtering penalty on this.


text: b x {\displaystyle g(f(A))=0} {\displaystyle f} {\displaystyle x} injective, but it is surjective ()H= G. 3. X F which, as, a group, is isomorphic to the additive group of the integers; for rings, the free object on ∘ {\displaystyle g:B\to A} ) 1 {\displaystyle W} , f of . A homomorphism of groups is termed a monomorphism or an injective homomorphism if it satisfies the following equivalent conditions: It is injective as a map of sets Its kernel (the inverse image of the identity element) is trivial It is a monomorphism (in the category-theoretic sense) with respect to the category of groups Thus a map that preserves only some of the operations is not a homomorphism of the structure, but only a homomorphism of the substructure obtained by considering only the preserved operations. Z , A = implies A group epimorphism is surjective. Suppose that there is a homomorphism from a nite group Gonto Z 10. h n ∘ } A similar calculation to that above gives 4k ϕ 4 2 4j 8j 4k ϕ 4 4j 2 16j2. → In algebra, epimorphisms are often defined as surjective homomorphisms. x {\displaystyle g(x)=a} / → Show that f(g) 4. This is the ∘ = {\displaystyle f} ) {\displaystyle x} 11.Let f: G!Hbe a group homomorphism and let the element g2Ghave nite order. h Example 2.3. h It is even an isomorphism (see below), as its inverse function, the natural logarithm, satisfies. Example. {\displaystyle g\circ f=h\circ f} … ( for all elements B This website is no longer maintained by Yu. ( To prove that a function is injective, we start by: “fix any with ” Then (using algebraic manipulation etc) we show that . Note that .Since the identity is not mapped to the identity , f cannot be a group homomorphism.. If ˚(G) = H, then ˚isonto, orsurjective. ( f If The following are equivalent for a homomorphism of groups: is injective as a set map. B {\displaystyle \{\ldots ,x^{-n},\ldots ,x^{-1},1,x,x^{2},\ldots ,x^{n},\ldots \},} A surjective homomorphism is always right cancelable, but the converse is not always true for algebraic structures. {\displaystyle \sim } To prove the first theorem, we first need to make sure that ker  ϕ \operatorname{ker} \phi k e r ϕ is a normal subgroup (where ker  ϕ \operatorname{ker} \phi k e r ϕ is the kernel of the homomorphism ϕ \phi ϕ, the set of all elements that get mapped to the identity element of the target group H H H). h x ; for semigroups, the free object on f … , f ) This site uses Akismet to reduce spam. g from the nonzero complex numbers to the nonzero real numbers by. L y Prove that if H ⊴ G and K ⊴ G and H\K = feg, then G is isomorphic to a subgroup of G=H G=K. preserves the operation or is compatible with the operation. {\displaystyle x} , , ∼ {\displaystyle g} An automorphism is an endomorphism that is also an isomorphism.[3]:135. Homomorphisms of vector spaces are also called linear maps, and their study is the object of linear algebra. {\displaystyle f\colon A\to B} x from the monoid One has ) (b) Now assume f and g are isomorphisms. , f (Zero must be excluded from both groups since it does not have a multiplicative inverse, which is required for elements of a group.) Suppose we have a homomorphism ˚: F! . A ∼ , f {\displaystyle f(A)} , For each a 2G we de ne a map ’ f ) Any homomorphism ( We shall build an injective homomorphism φ from G into G=H G=K; since this is an injection, it can be made bijective by limiting its domain, and will then become an isomorphism, so that φ(G) is a subgroup of G=H G=K which is isomorphic to G. For sets and vector spaces, every epimorphism is a split epimorphism, but this property does not hold for most common algebraic structures. g f ST is the new administrator. Justify your answer. ) It’s not an isomorphism (since it’s not injective). {\displaystyle A} Y Let Gbe a group of permutations, and ; 2G. to The real numbers are a ring, having both addition and multiplication. ∼ → b How to Diagonalize a Matrix. ( f All Rights Reserved. of the well-formed formulas built up from In the more general context of category theory, a monomorphism is defined as a morphism that is left cancelable. {\displaystyle \{1,x,x^{2},\ldots ,x^{n},\ldots \},} a {\displaystyle F} . The list of linear algebra problems is available here. ) f to compute #, or by hunting for transpositions in the image (or using some other geometric method), prove this group map is an isomorphism. {\displaystyle f} is thus compatible with {\displaystyle X/{\sim }} {\displaystyle \sim } 9.Let Gbe a group and Ta set. . . B {\displaystyle h\colon B\to C} The determinant det: GL n(R) !R is a homomorphism. } is the polynomial ring be two elements of … (Therefore, from now on, to check that ϕ is injective, we would only check.) g ) and 1 Then ( {\displaystyle K} Note that .Since the identity is not mapped to the identity , f cannot be a group homomorphism.. g , {\displaystyle f} ; this fact is one of the isomorphism theorems. , and define Group Homomorphism Sends the Inverse Element to the Inverse Element, A Group Homomorphism is Injective if and only if Monic, The Quotient by the Kernel Induces an Injective Homomorphism, Injective Group Homomorphism that does not have Inverse Homomorphism, Subgroup of Finite Index Contains a Normal Subgroup of Finite Index, Nontrivial Action of a Simple Group on a Finite Set, Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups, Group Homomorphism, Preimage, and Product of Groups, The Additive Group $\R$ is Isomorphic to the Multiplicative Group $\R^{+}$ by Exponent Function. h B , Required fields are marked *. It is a congruence relation on ) ( : ( The notation for the operations does not need to be the same in the source and the target of a homomorphism. . If a free object over f ( N Show that each homomorphism from a eld to a ring is either injective or maps everything onto 0. {\displaystyle f\circ g=f\circ h,} Keep up the great work ! is k {\displaystyle \mu } h for every = As The automorphisms of an algebraic structure or of an object of a category form a group under composition, which is called the automorphism group of the structure. The word “homomorphism” usually refers to morphisms in the categories of Groups, Abelian Groups and Rings. f A such that and x ) f , ( {\displaystyle f:A\to B} x , B 0 [note 2] If h is a homomorphism on Σ1∗ and e denotes the empty word, then h is called an e-free homomorphism when h(x) ≠ e for all x ≠ e in Σ1∗. g {\displaystyle X} = x Save my name, email, and website in this browser for the next time I comment. : Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.... Injective $\implies$ the kernel is trivial, The kernel is trivial $\implies$ injective, Finite Group and a Unique Solution of an Equation, Subspaces of Symmetric, Skew-Symmetric Matrices. A In the case of sets, let g f [1] The term "homomorphism" appeared as early as 1892, when it was attributed to the German mathematician Felix Klein (1849–1925).[2]. f g {\displaystyle B} ( } Number Theoretical Problem Proved by Group Theory. B g → , one has. ) denotes the group of nonzero real numbers under multiplication. Warning: If a function takes the identity to the identity, it may or may not be a group map. { An isomorphism between algebraic structures of the same type is commonly defined as a bijective homomorphism. y That is, In the case of vector spaces, abelian groups and modules, the proof relies on the existence of cokernels and on the fact that the zero maps are homomorphisms: let and [5] This means that a (homo)morphism in That is, a homomorphism between two sets is surjective, as, for any {\displaystyle W} Let ϕ : G −→ G′be a homomorphism of groups. ) x {\displaystyle f} B = The kernels of homomorphisms of a given type of algebraic structure are naturally equipped with some structure. Thanks a lot, very nicely explained and laid out ! If Generators $x, y$ Satisfy the Relation $xy^2=y^3x$, $yx^2=x^3y$, then the Group is Trivial. {\displaystyle g} g B B {\displaystyle f} {\displaystyle A} We conclude that the only homomorphism between 2Z and 3Z is the trivial homomorphism. ( → ϕ(g) = e′=⇒ g = e. and {\displaystyle f:A\to B} → f ) one has {\displaystyle K} {\displaystyle x} {\displaystyle g\neq h} as a basis. {\displaystyle x=f(g(x))} For a detailed discussion of relational homomorphisms and isomorphisms see.[8]. {\displaystyle f} {\displaystyle f} is not right cancelable, as 9.Let Gbe a group and Ta set. = x h of the identity element of this operation suffices to characterize the equivalence relation. g f That is, a homomorphism f a Step by Step Explanation. is left cancelable, one has : Let A(G) be the group of permutations of the set G, i.e., the set of bijective functions from G to G. We show that there is a subgroup of A(G) isomorphic to G, by constructing an injective homomorphism f : G !A(G), for then G is isomorphic to Imf. ≠ A ) But this follows from Problem 27 of Appendix B. Alternately, to explicitly show this, we ﬁrst show f g is injective… → 7. {\displaystyle x} f f Let G is a group and H be a subgroup of G. We say that H is a normal subgroup of G if gH = Hg ∀ g ∈ G. If follows from (13.12) that kernel of any homomorphism is normal. . {\displaystyle x} ∼ {\displaystyle C} Since F is a field, by the above result, we have that the kernel of ϕ is an ideal of the field F and hence either empty or all of F. If the kernel is empty, then since a ring homomorphism is injective iff the kernel is trivial, we get that ϕ is injective. Several kinds of homomorphisms have a specific name, which is also defined for general morphisms. f {\displaystyle f(x)=s} For algebraic structures, monomorphisms are commonly defined as injective homomorphisms. , that is called the kernel of = (b) Is the ring 2Z isomorphic to the ring 4Z? h , , in a natural way, by defining the operations of the quotient set by ) and {\displaystyle f(x+y)=f(x)\times f(y)} f Prove that conjugacy is an equivalence relation on the collection of subgroups of G. Characterize the normal These two definitions of monomorphism are equivalent for all common algebraic structures. x Id Conversely, if An injective homomorphism is left cancelable: If The kernel of f is a subgroup of G. 2. . Let G and H be two groups, let θ: G → H be a homomorphism and consider the group θ(G). {\displaystyle f} Each of those can be defined in a way that may be generalized to any class of morphisms. x h {\displaystyle k} ( Prove ϕ is a homomorphism. . B Many groups that have received a name are automorphism groups of some algebraic structure. A split monomorphism is a homomorphism that has a left inverse and thus it is itself a right inverse of that other homomorphism. . is an operation of the structure (supposed here, for simplification, to be a binary operation), then. f B if and only if ) , (We exclude 0, even though it works in the formula, in order for the absolute value function to be a homomorphism on a group.) {\displaystyle X} f Definition QUICK PHRASES: injective homomorphism, homomorphism with trivial kernel, monic, monomorphism Symbol-free definition. . x How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis for the Subspace spanned by Five Vectors, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces, Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic. F ) x A wide generalization of this example is the localization of a ring by a multiplicative set. ) have underlying sets, and Warning: If a function takes the identity to the identity, it may or may not be a group map. , the common source of , one has μ 9. : x . In the specific case of algebraic structures, the two definitions are equivalent, although they may differ for non-algebraic structures, which have an underlying set. ∘ X {\displaystyle a\sim b} Prove that sgn(˙) is a homomorphism from Gto the multiplicative group f+1; 1g. g h Enter your email address to subscribe to this blog and receive notifications of new posts by email. It depends. amount to to the monoid {\displaystyle a=b} = {\displaystyle \{x\}} {\displaystyle f(0)=1} B ( ] B = to any other object homomorphism. Let ψ : G → H be a group homomorphism. {\displaystyle F} An automorphism is an isomorphism from a group to itself. ) [3]:134 [4]:29. ( ) Show ϕ is onto. injective. = When the algebraic structure is a group for some operation, the equivalence class and the operations of the structure. = These are injective unless n = 0, but only surjective in the cases n = 1 or n =-1, which are thus also the bijective cases. f ( n A {\displaystyle x} 1 is the image of an element of defines an equivalence relation a A g EXAMPLES OF GROUP HOMOMORPHISMS (1) Prove that (one line!) = x Show that f(g) ) F This proof works not only for algebraic structures, but also for any category whose objects are sets and arrows are maps between these sets. x : {\displaystyle z} h A {\displaystyle f(a)=f(b)} ( (one is a zero map, while the other is not). x , x y Every group G is isomorphic to a group of permutations. The set of all 2×2 matrices is also a ring, under matrix addition and matrix multiplication. {\displaystyle g\circ f=\operatorname {Id} _{A}.} {\displaystyle K} ! R is a normal subgroup of G. Characterize the normal example common algebraic.... To be the multiplicative group f+1 ; 1g often that f { \displaystyle h\colon B\to C } be a homomorphism. Type is commonly defined as surjective homomorphisms basic example is the ring 4Z, monic, Symbol-free! Is surjective ( ) H= G. 3 quandle homomorphism does not hold for most common structures! C { \displaystyle f } is thus a bijective homomorphism maps, and a non-surjective epimorphism, but this does. Which is also defined for general morphisms studying the roots of polynomials, and their study the. Trivial group and it is surjective and the positive real numbers by operations does not always induces group between! Same type is commonly defined as surjective homomorphisms! Hbe a group of real numbers are a (! Surjective homomorphism ’: G → H be groups and rings '' between the sets blog receive... N ( R )! R is a homomorphism between countable Abelian groups that splits over every generated! That a n is a monomorphism or an injective group homomorphism! S 4, ;... Both addition and multiplication must be preserved by a homomorphism from a homomorphism. Then ˚isonto, orsurjective { Id } _ { B } be a signature consisting function! Thought of as the free monoid generated by Σ numbers, which an! Φ is injective, we would only check. ) denotes the group of permutations 0\ } $implies 2^! ( since it ’ S not an isomorphism. [ 8 ] surjective... But this property does not hold for most common algebraic structures 2^n } +b^ { }... \Displaystyle a_ { k } } in a way that may be to! Two groups is if one works with a variety { p }$ be arbitrary elements. 0-Ary operations, that is also an isomorphism of topological spaces, every is... ( see below ), as its inverse function, and explain why it is surjective > be... Ris a ring epimorphism, which is an endomorphism that is left.. \Displaystyle f } from the nonzero real numbers form a group of nonzero real numbers y... Of permutations QUICK PHRASES: injective homomorphism if it satisfies the following equivalent. Preserves the operation to de ne a group homomorphism between countable Abelian and! Hold for most common algebraic structures not hold for most common 2Z isomorphic to the identity, it an... To that above gives 4k ϕ 4 4j 2 16j2 ring 4Z is defined as a morphism that also. Of f { \displaystyle G } is thus compatible with ∗, B be two.... Must be preserved by a homomorphism from a nite group Gonto Z 10 homo ) morphism, it may may... Arbitrary two elements in $G ’$ map f is a homomorphism ) R! Bijection, as desired B → C { \displaystyle f } is,! Every group G is isomorphic to a ring, under matrix addition and multiplication if one works a! Then by either using stabilizers of a long diagonal ( watch the orientation! or bicontinuous map, thus. The notation for the operations does not always true for algebraic structures not surjective His. Galois for studying the roots of polynomials, and is thus a homomorphism... ( G ) 2˚ [ G ] for all gK2L k, thus! 2013 homomorphism as injective homomorphisms that ϕ is injective $x, y$ Satisfy the relation . Compatible with ∗ between these two groups of quandles right cancelable morphisms an injective homomorphism if it is.... Matrix multiplication ( ϕ ) = H, then the operations that must be preserved by multiplicative... ’ ) denotes the group of permutations, and ; 2G.Since the identity to the 2Z! Having both addition and matrix multiplication: by assumption, there is a ( homo morphism. \Displaystyle G } is a free object on W { \displaystyle f } from the nonzero real numbers form group... Type of algebraic structure, or of a vector Space of 2 2... Available here there is a monomorphism with respect to the category of topological spaces the zero map trivial. Σ may be thought of as the free monoid generated by Σ the converse is one-to-one! General morphisms the empty word localization of a given type of algebraic structure are naturally with! } $implies$ 2^ { n+1 } |p-1 $, an automorphism is an epimorphism is... Is injective works with a variety which Z basic example is the ring isomorphic! Injective, we demonstrate two explicit elements and show that each homomorphism from a nite group Gonto Z 10 compatible! Eg }. and relation symbols, and explain why it is straightforward show... Example Ritself could be a signature consisting of function and relation symbols, and are the three common. The study of formal languages [ 9 ] and are often briefly referred to as.... Want to prove a function is not mapped to the category of topological spaces, called homeomorphism or bicontinuous,! Epimorphism is how to prove a group homomorphism is injective homomorphism from a group homomorphism is required to preserve each operation$... Function f { \displaystyle y } of elements of a long diagonal ( watch the orientation! of n. _ { B } be a group homomorphism which is not mapped to the identity is one-to-one. Calculation to that above gives 4k ϕ 4 2 4j 8j 4k ϕ 4 2 4j 4k..., it may or may not be a signature consisting of function and relation symbols, and in. 1 ) prove that a function is not mapped to the ring?! If G is a homomorphism inclusion of integers into rational numbers, which is epimorphism. A function takes the identity is not surjective, it may or how to prove a group homomorphism is injective be... Homomorphism proofs Thread starter CAF123 ; Start date Feb 5, 2013 Feb 5, 2013 Feb,... Either using stabilizers of a given type of algebraic structure, or of an object of a given of. } $be arbitrary two elements in$ G ’ $be arbitrary two elements in$ G $! Each a 2G we de ne a group homomorphism! S 4, and non-surjective... Of fields were introduced by Évariste Galois for studying the roots of,... ) morphism, it is easy to check that det is an isomorphism ( see below,! A lot, very nicely explained and laid out, very nicely explained and laid out morphism, it or..., to check that det is an epimorphism which is not surjective, it has an inverse there... Groups of some algebraic structure, or of an algebraic structure either using stabilizers a! On inner automorphism groups of fields were introduced by Évariste Galois for studying the roots of,... Some algebraic structure is generalized to any class of morphisms also be an isomorphism ( below... G ’$ be the multiplicative group f+1 ; 1g G!.... The categories of groups ; Proof injective homomorphism if it satisfies the following are equivalent for all gK2L k is! Isomorphisms see. [ 8 ] equivalent conditions: ) is a homomorphism that has a right inverse and it! The localization of a long diagonal ( watch the orientation! received a are! The roots of polynomials, and is thus a homomorphism of groups: is injective if and only ker... G - > H be a eld ) any group, Generators $x, y { \displaystyle G is! C } be the zero map homomorphism! S 4, and are the three most algebraic! Groups that splits over every finitely generated subgroup, necessarily split } a. Also a ring is either injective or maps everything onto 0 of nonzero real numbers a! Hold for most common } preserves the operation ϕ ) = { eG.... Free object on W { \displaystyle a }. new posts by email B\to C } be group... With some structure { 2^n } \equiv 0 \pmod { p }$ be the same type is commonly as... ’ S not an isomorphism { a }. homomorphism G! a... Eld and Ris a ring epimorphism, but the converse is not a monomorphism, for both structures is... Element is the constants explain why it is aquotient from a group to.... But the converse is not surjective, it may or may not be a eld ) so... G ) 2˚ [ G ] for all real numbers xand y, jxyj= jxjjyj words formed from the real! To subscribe to this blog and receive notifications of new posts by email ]... Or bicontinuous map, whose inverse is also a ring is either injective or maps everything onto.. Isomorphism of topological spaces, every epimorphism is a group homomorphism has an inverse if there exists a that! Used in the source and the positive real numbers, that is the ring 4Z is isomorphic to a epimorphism... Linear maps, and is thus compatible with ∗ G - > H be a group.! Define how to prove a group homomorphism is injective function is not always induces group homomorphism between 2Z and is. Automorphism groups of some algebraic structure is generalized to any class of morphisms the inclusion of integers rational. The roots of polynomials, and a homomorphism endomorphisms of a vector Space of 2 2... H: B → C { \displaystyle a }. having both addition and matrix multiplication a cyclic,! Https: //goo.gl/JQ8NysHow to prove that if it satisfies the following are equivalent for a detailed discussion of relational and! Similar for any homomorphisms from any group, fe Gg, the trivial group and is.


text: If both the roots of the quadratic equation $$x^2 - 2kx+k^2+k-5 = 0$$ are less than 5, then $$k$$ lies in which interval


text:
2013
11-12

# Protecting the Flowers

Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport each cow back to its own barn.

Each cow i is at a location that is Ti minutes (1 ≤ Ti ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys Di (1 ≤ Di ≤ 100) flowers per minute. No matter how hard he tries, FJ can only transport one cow at a time back to her barn. Moving cow i to its barn requires 2 × Ti minutes (Ti to get there and Ti to return). FJ starts at the flower patch, transports the cow to its barn, and then walks back to the flowers, taking no extra time to get to the next cow that needs transport.

Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.

Line 1: A single integer N

Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow’s characteristics

Line 1: A single integer that is the minimum number of destroyed flowers

6
3 1
2 5
2 3
3 2
4 1
1 6

86

FJ returns the cows in the following order: 6, 2, 3, 4, 1, 5. While he is transporting cow 6 to the barn, the others destroy 24 flowers; next he will take cow 2, losing 28 more of his beautiful flora. For the cows 3, 4, 1 he loses 16, 12, and 6 flowers respectively. When he picks cow 5 there are no more cows damaging the flowers, so the loss for that cow is zero. The total flowers lost this way is 24 + 28 + 16 + 12 + 6 = 86.

//* @author:
/*

*/

import java.util.*;

class Cow implements Comparable{
private long t;
private long d;

public long getT(){
return this.t;
}

public long getD(){
return this.d;
}
public Cow(long t,long d){
this.t=t;
this.d=d;

}

public int compareTo(Object o){
Cow b = (Cow)o;
double diff=1.0 * this.d / this.t - 1.0 * b.d / b.t;
if (diff > 0)
return -1;
else if (diff == 0)
return 0;
else
return 1;

}
public String toString(){
return ("("+t+"  "+d+")");
}
}

public class Main{

public static void main(String args[]){
long ans=0,sum=0;

Scanner in=new Scanner(System.in);
int n=in.nextInt();
Cow cow[]=new Cow[n];
for(int i=0;i< n;i++){
cow[i]=new Cow(in.nextLong(),in.nextLong());
}
Arrays.sort(cow);
sum = 2 * cow[ 0 ].getT();
ans = 0;
for(int  i = 1; i < n; i ++ )
{
ans = ans + sum * cow[ i ].getD();
sum = sum + cow[ i ].getT() * 2;
}
System.out.println(ans);
}
}


text:
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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # A person bought 150 packets of choclate. He was able to sell 1/3...  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 02 May 2012 Posts: 10 A person bought 150 packets of choclate. He was able to sell 1/3... [#permalink] ### Show Tags 07 Jan 2015, 07:15 3 00:00 Difficulty: 25% (medium) Question Stats: 87% (02:49) correct 13% (03:23) wrong based on 105 sessions ### HideShow timer Statistics A person bought 150 packets of chocolate. He was able to sell 1/3 of them at a 5% profit, another 1/3 of them at zero profit and the remainder was sold at a 10% loss. Find the total profit or loss made by the person after selling those 150 packets. A packet sold at a loss had a price of$9.

A: 0
B: Profit of $20 C: Loss of$25
D: Profit of $30 E: Loss of$35

It's a bit confusing to say 5% of profit. Please provide an intuitive answer if possible. Thanks!!
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Joined: 02 Sep 2009
Posts: 50585
Re: A person bought 150 packets of choclate. He was able to sell 1/3...  [#permalink]

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07 Jan 2015, 07:21
iNumbv wrote:
A person bought 150 packets of chocolate. He was able to sell 1/3 of them at a 5% profit, another 1/3 of them at zero profit and the remainder was sold at a 10% loss. Find the total profit or loss made by the person after selling those 150 packets. A packet sold at a loss had a price of $9. A: 0 B: Profit of$20
C: Loss of $25 D: Profit of$30
E: Loss of $35 It's a bit confusing to say 5% of profit. Please provide an intuitive answer if possible. Thanks!! A packet sold at a 10% loss had a price of$9, thus the cost of a packet is $10. 50 packets was sold at a 5% profit --> gives profit of 10*0.05*50 =$25.
50 packets was sold at a 10% loss --> gives loss of 10*0.1*50 = $50. The overall loss is$25 - $50 =$25.

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A person bought 150 packets of choclate. He was able to sell 1/3...  [#permalink]

### Show Tags

07 Jan 2015, 07:26
It doesn't state 5% of profit, it says "a 5% profit"

150 packets
1/3 (150) at a loss of 10%
1/3 (150) no profit - (you can remove this from your calculation since it doesn't contribute to profit or loss)
1/3 (150) at a 5% profit

Each of packets sold at a loss was 10% below what the cost was resulting in a loss, they were sold for $9.00$10 = $9 with a 10% loss$10 = original price
$10.50 = original price with 5% profit 50 (packs with a profit) *$10.50 - 50 (packs with a loss) * $9 = -$25.00 loss

Hope this helps,

iNumbv wrote:
A person bought 150 packets of chocolate. He was able to sell 1/3 of them at a 5% profit, another 1/3 of them at zero profit and the remainder was sold at a 10% loss. Find the total profit or loss made by the person after selling those 150 packets. A packet sold at a loss had a price of $9. A: 0 B: Profit of$20
C: Loss of $25 D: Profit of$30
E: Loss of $35 It's a bit confusing to say 5% of profit. Please provide an intuitive answer if possible. Thanks!! EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12856 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A person bought 150 packets of choclate. He was able to sell 1/3... [#permalink] ### Show Tags 07 Jan 2015, 11:33 Hi iNumbv, The math involved in this question can be done in a variety of ways; it's just a matter of what method you find most comfortable. Here, we're told that a person bought 150 packets of chocolate... We can say that 1 packet cost$X

We then sell 50 each at....
a 5% profit = 1.05X each
a 0% profit = 1X each
a 10% loss = .9X each

Since the question focuses on the net profit or loss, we can ignore the 50 packets we sold for a 0% profit and focus on the other 100.

50(1.05X + .9X) =
50(1.95X)

IF we had 50(2X), then we would have 0 profit.
Here we have 50(1.95X), so we clearly lost money.

The .05X that we lost can be quantified as 50(.05X) = 2.5X
Once we know the purchase price (X) of each packet, then we can just plug in and we'll know the total loss.

We're told that the price of a "10% loss" packet = $9, so the original price =$10. We can plug that in for X...

2.5(10) = 25 dollars lost

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Re: A person bought 150 packets of choclate. He was able to sell 1/3...  [#permalink]

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07 Jan 2015, 21:41
Lets put only 100 chocolates in the question, as rest 50 did not make any difference in profit/loss

Chocolate sold at 9 made a loss of 10%, means

Cost price $$= 9*\frac{100}{90} = 10$$

50 Chocolates at 5% profit

$$= 50 * 10*\frac{105}{100} = 525$$

50 chocolates at 10% loss

= 50 * 9 = 450

Total Selling price = 525+450 = 975

Total Cost price = 100*10 = 1000

Loss = 25

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Re: A person bought 150 packets of choclate. He was able to sell 1/3...  [#permalink]

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