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# Pricing Under Rough Volatility

@article{Bayer2015PricingUR,
title={Pricing Under Rough Volatility},
author={Christian Bayer and Peter K. Friz and Jim Gatheral},
journal={ERN: Other Econometric Modeling: Derivatives (Topic)},
year={2015}
}
• Published 28 January 2015
• Economics
• ERN: Other Econometric Modeling: Derivatives (Topic)
From an analysis of the time series of volatility using recent high frequency data, Gatheral, Jaisson and Rosenbaum previously showed that log-volatility behaves essentially as a fractional Brownian motion with Hurst exponent H of order 0.1, at any reasonable time scale. The resulting Rough Fractional Stochastic Volatility (RFSV) model is remarkably consistent with financial time series data. We now show how the RFSV model can be used to price claims on both the underlying and integrated…
171 Citations
Volatility is rough
• Economics
• 2018
Estimating volatility from recent high frequency data, we revisit the question of the smoothness of the volatility process. Our main result is that log-volatility behaves essentially as a fractional
Short-time at-the-money skew and rough fractional volatility
The Black–Scholes implied volatility skew at the money of SPX options is known to obey a power law with respect to the time to maturity. We construct a model of the underlying asset price process
The characteristic function of rough Heston models
• Mathematics
• 2016
It has been recently shown that rough volatility models, where the volatility is driven by a fractional Brownian motion with small Hurst parameter, provide very relevant dynamics in order to
Perfect hedging in rough Heston models
• Economics
The Annals of Applied Probability
• 2018
Rough volatility models are known to reproduce the behavior of historical volatility data while at the same time fitting the volatility surface remarkably well, with very few parameters. However,
Weak error rates for option pricing under linear rough volatility
• Mathematics
• 2020
In quantitative finance, modeling the volatility structure of underlying assets is vital to pricing options. Rough stochastic volatility models, such as the rough Bergomi model [Bayer, Friz,
Multiscaling and rough volatility: an empirical investigation
• Economics
• 2022
Pricing derivatives goes back to the acclaimed Black and Scholes model. However, such a modeling approach is known not to be able to reproduce some of the financial stylized facts, including the
Volatility Is Rough
• Mathematics, Economics
• 2014
Estimating volatility from recent high frequency data, we revisit the question of the smoothness of the volatility process. Our main result is that log-volatility behaves essentially as a fractional
Option pricing under fast‐varying long‐memory stochastic volatility
• Mathematics
Mathematical Finance
• 2018
Recent empirical studies suggest that the volatility of an underlying price process may have correlations that decay slowly under certain market conditions. In this paper, the volatility is modeled
Persistent and Rough Volatility
• Economics
SSRN Electronic Journal
• 2020
This paper contributes to an ongoing debate on volatility dynamics. We introduce a discrete-time fractional stochastic volatility (FSV) model based on the fractional Gaussian noise. The new model has
Multifactor Approximation of Rough Volatility Models
• Mathematics
SIAM J. Financial Math.
• 2019
This paper designs tractable multi-factor stochastic volatility models approximating rough volatility models and enjoying a Markovian structure and applies this procedure to the specific case of the rough Heston model.

## References

SHOWING 1-10 OF 22 REFERENCES
Volatility Is Rough
• Mathematics, Economics
• 2014
Estimating volatility from recent high frequency data, we revisit the question of the smoothness of the volatility process. Our main result is that log-volatility behaves essentially as a fractional
Affine fractional stochastic volatility models
• Mathematics
• 2012
By fractional integration of a square root volatility process, we propose in this paper a long memory extension of the Heston (Rev Financ Stud 6:327–343, 1993) option pricing model. Long memory in
Asymptotic analysis for stochastic volatility: martingale expansion
A general class of stochastic volatility models with jumps is considered and an asymptotic expansion for European option prices around the Black–Scholes prices is validated in the light of Yoshida’s
On the short-time behavior of the implied volatility for jump-diffusion models with stochastic volatility
• Mathematics, Economics
Finance Stochastics
• 2007
Abstract In this paper we use Malliavin calculus techniques to obtain an expression for the short-time behavior of the at-the-money implied volatility skew for a generalization of the Bates model,
The Volatility Surface: A Practitioner's Guide
List of Figures. List of Tables. Foreword. Preface. Acknowledgments. Chapter 1: Stochastic Volatility and Local Volatility. Stochastic Volatility. Derivation of the Valuation Equation, Local
Realized Volatility: A Review
• Mathematics
• 2006
This article reviews the exciting and rapidly expanding literature on realized volatility. After presenting a general univariate framework for estimating realized volatilities, a simple discrete time
Rough fractional diffusions as scaling limits of nearly unstable heavy tailed Hawkes processes
• Mathematics
• 2015
We investigate the asymptotic behavior as time goes to infinity of Hawkes processes whose regression kernel has $L^1$ norm close to one and power law tail of the form $x^{-(1+\alpha)}$, with
Smile Dynamics IV
This paper introduces a new quantity, which is called the Skew Stickiness Ratio, and shows how, at order one in the volatility of volatility, it is linked to the rate at which the at-the-money-forward skew decays with maturity.
Multivariate High-Frequency-Based Volatility (HEAVY) Models
• Economics
• 2011
This paper introduces a new class of multivariate volatility models that utilizes high-frequency data. We discuss the models dynamics and highlight their di¤erences from multivariate GARCH models. We


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# Estimating $\pi$ very accurately [duplicate]

Possible Duplicate:
Do We Need the Digits of $\pi$?

I often hear about people who compute a lot of digits of $\pi$.Does estimating $\pi$ to a large degree of precision have any importance (or potential use) in mathematics ?

Thank you

-
No.${}{}{}{}{}{}$ –  Graphth Dec 15 '12 at 23:14
Do you have any idea why its done ? –  Amr Dec 15 '12 at 23:14
Because we're nerds. –  vermiculus Dec 15 '12 at 23:16
It is of great importance, but the reasoning involved is circular. –  copper.hat Dec 15 '12 at 23:18
Knowing many digits of pi has no practical significance, but developing the methods needed to compute many digits of pi in a reasonable amount of time is of use, as those methods have applications in other places. Also, while attempting to compute digits of pi, we have created some beautiful mathematics, for example the Bailey–Borwein–Plouffe formula. –  Potato Dec 15 '12 at 23:25

## marked as duplicate by Henning Makholm, sdcvvc, Argon, Martin Argerami, Asaf KaragilaDec 16 '12 at 0:03

In my opinion, the value in itself is of as much importance to mathematics as the value of any number, say $\sqrt{2}$.

On the other hand, estimating the value of $\pi$ is of great importance due to the vast amount of techniques it generates.

Think of all the cute power series and inverse trig relations of $\pi$.

1) Machin's formula:

$$\frac{\pi}{4} = 4 \arctan\frac{1}{5} - \arctan\frac{1}{239}$$

2) Leibniz formula for $\pi$:

$$\frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \arctan{1} = \frac{\pi}{4}$$

3) Euler Formula for $\pi$ $$\frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \cdots$$

4) Bailey-Borwein-Plouffe formula

$$\pi = \sum_{i = 0}^{\infty}\left[ \frac{1}{16^i} \left( \frac{4}{8i + 1} - \frac{2}{8i + 4} - \frac{1}{8i + 5} - \frac{1}{8i + 6} \right) \right]$$

Apparently, the above formula can be used to extract the digits of $\pi$ from an arbitrary location!! I am reading it now (see spigot algorithms) due to a suggestion from Potato and I am curious about its behavior.

Consider the memorable sharp integral bounds you generate:

$$\frac{1}{1260} = \int_0^1\frac{x^4 (1-x)^4}{2}\,dx < \int_0^1\frac{x^4 (1-x)^4}{1+x^2}\,dx = \frac{22}{7} - \pi < \int_0^1\frac{x^4 (1-x)^4}{1}\,dx = {1 \over 630}$$

See Lucas for interesting extensions. Especially the error margin of $\frac{355}{113}$ approximation.

Think of all the sneaky ways $\pi$ can crop up surprisingly, like in the Buffon's needle problem. We can use this experiment to empirically estimate the value of $\pi$.

Euler apparently proved that if you pick two integers at random, the probability that they are co-prime is $\frac{6}{\pi^2}$. The first thing I asked myself the first time I saw it was 'how did $\pi$ appear?'

With normal distribution containing $\pi$ in it's p.d.f and due to central limit theorem, I wont be surprised if there are so many other ways of estimating $\pi$ empirically.

Estimating $\pi$ seems to be an interesting hobby that has given rise to some beautiful methods ranging from the Gauss-Legendre algorithm, continued fractions, empirical probabilistic techniques, complex numbers, geometry, integrals and infinite series.

So I think the spirit of estimating a number is very important to mathematics, while the value of the number in itself may not have much importance.

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Great answer. One small suggestion: Add the Bailey-Borwein-Plouffe formula! –  Potato Dec 15 '12 at 23:48
Nice post, are there any other examples similar to Buffon's needle problem where $\pi$ shows up unexpectedly? –  Michael Boratko Dec 15 '12 at 23:48
@Potato: Woah!! The idea of Bailey-Borwein-Plouffe formula is wonderful.Let me add it. –  Isomorphism Dec 15 '12 at 23:55
@Michael, loosely speaking, the probability that two randomly-chosen integers are relatively prime is $\frac{6}{\pi^2}$. –  Dan Brumleve Dec 16 '12 at 0:03

format: plain


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# Solving a Linear ODE

Solve the following linear ODE $$3t^{2}y'+t^{3}y=cos(t)$$

What i tried

Since this is a linear equation, i used the integrating factor method. First i didide both the LHS and RHS of the equation by $3t^{2}$ so that $y'$ have a coefficient of $1$. Then i find the integrating factor $u(t)=t^{1/3}$. I the multiply the integrating factor of $u(t)$ to both the LHS and RHS of the equation and then integreating both sides of the equation, getting $$yt^{1/3}=\int(1/3)t^{-5/3}cos(t)$$ Im stuck from here onwards though,I tried integration by parts but it seems to lead me to nowhere. Could anyone explain. Thanks

• The integrating factor is exp $( \int \alpha (t) dt)$, where $\alpha$ is the coefficient of $y$. So I think your integrating factor is wrong. Jan 24 '15 at 6:32
• My integrating factor is $e^{1/3ln(t)}$ which simplifies to $t^{1/3}$ Jan 24 '15 at 6:35

You have \begin{align} 3t^2\frac{dy}{dt}+t^3y=\cos\left(t\right).\tag{1} \end{align} Now put the equation in standard form as follows: \begin{align} 3t^2\frac{dy}{dt}+t^3y=\cos\left(t\right)\Leftrightarrow \frac{dy}{dt}+\left(\frac{t}{3}\right)y=\frac{\cos\left(t\right)}{3t^2}\tag{2}. \end{align} Notice that the LHS looks a lot like the product rule of differentiation. Therefore, we must find an integrating factor $\mu\left(t\right)$: \begin{align} &\mu\left(t\right) = \exp\left(\int \frac{t}{3}\:dt\right)=e^{t^2/6},\tag{3} \\ \therefore e^{t^2/6}\frac{dy}{dx}+\left(\frac{t}{3}\right)&e^{t^2/6}y=\frac{d}{dt}\left[\left(e^{t^2/6}\right)\left(y\right)\right]=\frac{e^{t^2/6}\cos\left(t\right)}{3t^2} \tag{4} \\ &\implies \int\frac{d}{dt}\left[\left(e^{t^2/6}\right)\left(y\right)\right]\:dt=\int\frac{e^{t^2/6}\cos\left(t\right)}{3t^2}\:dt\tag{5}\\ &\implies y\left(t\right)=e^{-t^2/6}\int\frac{e^{t^2/6}\cos\left(t\right)}{3t^2}\:dt.\tag{6} \end{align} Now solve for the integral (it's not too bad).


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# Find identity and inverses of the group $(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$.

Let $$(\Bbb{Z}_6,+)$$ and $$(\Bbb{Z}_5,\cdot)$$ be two groups.

Define an operation for $$(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$$ such that $$(\Bbb{Z}_6,+)\times(\Bbb{Z}_5,\cdot)$$ is a group. Find the identity and inverses of the product.

I know the following:

Property If $$(G_1,*_1)$$ and $$(G_2,*_2)$$ are two GROUPS, then the set $$G_1\times G_2$$ with the operation $$(a,b)*(c,d)=(a*_1c,b*_2d)$$ is also a GROUP.

But here, $$(\Bbb{Z}_5,\cdot)$$ is NOT a group since the inverse element of $$0$$ does not exist. So what should we do here?

**Should we consider another operation like $$(a,b)*(c,d)=(a+_6c,b+_5d)?$$

But the hypothesis is not true because $$(\Bbb{Z}_5,\cdot)$$ is not a group, so I do not know if I can define such operation of the product of "two" groups.

Any help?

• Perhaps the author of the question meant $\cdot$ to mean addition also, and since $+$ is taken, they chose $\cdot$. Besides, it even says "Let $(\Bbb Z_6,+)$ and $(\Bbb Z_5,\cdot )$ be two groups". – Shaun Nov 21 '19 at 0:29
• Ah, just realized your issue is the operation of $G$ is addition in the first component but multiplication in the second. You're probably fine once you realize this – oshill Nov 21 '19 at 0:29
• Yes $(\Bbb Z_5, \cdot)$ is not a group so we cannot define the product ...However $(\Bbb Z_5 \setminus \{0\}, \odot)$ is a group – Chinnapparaj R Nov 21 '19 at 0:29
• @Shaun I don't think so; the author defines (\cdot) as a multiplication (saw in other exercises; this is one of them). – manooooh Nov 21 '19 at 0:34
• @oshill yeah! The problem is in the second component! Because all the elements of the form $(x,0)$ (where $x\in\Bbb{Z}_6$) does not have inverse! – manooooh Nov 21 '19 at 0:35

Assuming $$(\Bbb Z_5,\cdot)=U(5)$$ is the group of units modulo five, the identity of $$G:=(\Bbb Z_6, +)\times U(5)$$ is $$([0]_6, [1]_5)$$ and the inverse of $$([a]_6, [b]_5)$$ in $$G$$ is $$([-a]_6, [c]_5)$$, where $$c$$ is such that $$5x+bc=1$$ for some $$x\in \Bbb Z$$ and $$[d]_n:=\{k\in \Bbb Z\mid n \text{ divides } (k-d)\}.$$

Assuming $$(\Bbb Z_6,+)=(\Bbb Z_6, +_6)$$ and $$(\Bbb Z_5,\cdot)=(\Bbb Z_5, +_5)$$, then the identity of $$H:=(\Bbb Z_6, +_6)\times(\Bbb Z_5, +_5)$$ is $$([0]_6, [0]_5)$$ and the inverse of $$([a]_6, [b]_5)$$ in $$H$$ is $$([-a]_6, [-b]_5)$$.

NB: Since five is prime, there is only one group of order five up to isomorphism. The question specifies that $$(\Bbb Z_5,\cdot)$$ is a group.

• Yes $\Bbb{Z}_{5}=\{0,1,2,3,4\}$. What does $5x+bc=1$ have to do with $[d]_n$? Thank you! – manooooh Nov 21 '19 at 0:44
• Well, $5x=1-bc$, so $5\mid 1-bc$. Hence $[1]_5=[bc]_5=[b]_5\cdot [c]_5$, so $[b]_5$ has $[c]_5$ as an inverse. – Shaun Nov 21 '19 at 0:48
• But where do we put $[d]_n$? – manooooh Nov 21 '19 at 0:49
• No worries, @manooooh. Please don't forget to upvote & accept this answer if it works for you. – Shaun Nov 21 '19 at 0:54
• But with your definition of inverse I cannot figure what should be the inverse of $(0,0)$. It should be $(0,c)$, where $c$ is such that $5x+0c=5x=1$ for some $x\in\Bbb{Z}$. But $x=1/5\notin\Bbb{Z}$. So $(0,0)$ has no inverse. What am I doing wrong? – manooooh Nov 21 '19 at 0:54


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# Curve fitting project

Exercises that approximate the cosine curve from $–\pi/2$ to $\pi/2$ with best-fitting parabolas satisfying varioius constraints. Uses the Hilbert space of square integrable functions with its natural inner product (integral of product of functions). Computes each approximation along with its squared error. Solutions available to instructors from author on request.

Identifier:
http://www.math.wayne.edu/~drucker/MAALinAlgResources/curve%20fitting%20project.pdf
Rating:
Creator(s):
Daniel Drucker
Cataloger:
Douglas Meade
Publisher:
Wayne State University website
Rights:
Daniel Drucker

## Comments

Nice write up.

### Nice exposure to use of linear algebra in optimization.

Nice exposure to use of linear algebra in optimization.


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# Other Techniques

This chapter is a grab-bag of various techniques that have a latent variable interpretation to the models. Only brief descriptions are provided at present, though more may be added in the future. In addition, you can see some more techniques in the associated notes that were used to give a workshop on factor analysis and related techniques, though the bulk of it is covered in this document

## Recommender Systems

Practically everyone has been exposed to recommender systems such as collaborative filtering and related models. That’s how Netflix, Amazon and others make their recommendations to you given the information you’ve provided about likes and dislikes, what other similar people have provided, and how similar the object of interest is to others.

The following image, taken from Wikipedia (click the image to go there), conceptually shows how a user-based collaborative filtering method would work, where a recommendation is given based on what other similar users have given.

Let’s go with movies as an example. You might only rate a handful, and indeed most people will not rate most movies. But at some point most movies will have been rated. So how can one provide a recommendation for some movie you haven’t seen? If we group similar movies into genres, and similar people into demographic categories and based on taste, one can recommend something from a similar genre of movies that you like, that people in the same demographic category seem to like as well.

If you think of genres as latent variables for movies, you can employ the factor analytic techniques we’ve talked about. Similarly, we can find clusters of people using cluster analytic techniques. In short, collaborative filtering/recommender systems are using latent variable techniques to a specific type of data, e.g. ratings. More modern approaches will incorporate user and item characteristics, recommendations from other systems, and additional information. The following provides some code for you to play with, using a straightforward singular value decomposition on movie ratings, which is the same technique used in Base R’s default prcomp function for PCA. You might compare it with method = 'POPULAR'.

library(recommenderlab)
data("MovieLense")
MovieLense
barplot(table(getRatings(MovieLense)), col = '#ff5500', border='#00aaff') # rating frequencies
recommenderRegistry$get_entries(dataType = "realRatingMatrix") # the methods available recom_svd <- Recommender(MovieLense, method = "SVD") recom_svd After running it try some predictions. # predicted ratings for two users recom <- predict(recom_svd, MovieLense[2:3], type="ratings") recom as(recom, "matrix")[,1:10] # comparison model recom_popular <- Recommender(MovieLense, method = "POPULAR") getModel(recom_popular)$topN
recom <- predict(recom_popular, MovieLense[2:3], type="topNList")
recom
as(recom, "list")

All in all, thinking about your data in terms of a recommendation system might not be too far-fetched, especially if you’re already considering factor analysis of some sort.

## Hidden Markov Models

Hidden Markov models can be used to model latent discrete states that result in a sequence of observations over time. In terms of a graphical model, we can depict it as follows:

In this sense, we have latent variable $$z$$ that represents the hidden state of a system, while the outcome is what we actually observe. There are three latent states above, and the relative width of the edge reflects the transition probability of moving from one state to the other. The icons are the categories of observations we can potentially see, and there is some probability, given a latent state of seeing a particular category. Such a situation might lead to the following observed sequence of observations:

We start in state 1 where the heart and yellow smiley are most probable. Let’s say we observe ♥. The second state is most likely so we get there, and because it has some probability of staying in that state, we observe § and §. We finally get to state 3 and see , where we go back to state 1, where we see ☺, jump to latent state 3 etc. And such a process continues for the length of sequence we see.

This can be seen as a mixture model/latent class situation that we’ll talk about more later. The outcomes could also be continuous, such that the latent state determines the likelihood of the observation in a manner more directly akin to the latent linear model for standard factor analysis.

## “Cluster analysis”

Aside from mixture models, when people use the term ‘cluster analysis’ they are typically referring to distance-based methods. Given a distance matrix that informs how dissimilar observations are from one another, the methods try to create clusters of observations that are similar to one another, and clusters that are more distinct from other clusters.

### K-means

K-means cluster analysis is probably the most commonly used clustering method out there. Conceptually it’s fairly straightforward- find $$k$$ clusters that minimize the variance of its members from the mean of its members. As such it’s easy to implement in standard data settings.

K-means can actually be seen as a special case of the Gaussian mixture model described in a previous chapter, and it also has connections to PCA and ICA. The general issue is trying to determine just how many clusters one should retain. The following plot shows both a two and three cluster solution using the kmeans function in base R, e.g. kmeans(faithful, 2).

### Hierarchical

Other methods can be thought of as a clustering the data in a hierarchical fashion. These can start at the bottom of the hierarchy (agglomerative), allowing every observation into its own cluster, and successively combining them. For example, first choose a measure of dissimilarity, and combine the two observations that are most alike, then add one to those or if another pair are closer, make a new cluster. Conversely, one can start with every observation in one cluster (divisive), and take the most dissimilar and split it off, continuing on until every observation is in its own cluster.

Practically at every turn you’re faced with multiple options for settings to choose (distance, linkage method, cluster determination, general approach), and most decisions will be arbitrary . While these are actually still commonly used, you always have better alternatives. They are fine to use in a quick visualization to sort the data more meaningfully though, as above.

## ICA

The latent linear model versions of PCA and factor analysis assume the observed variables are normally distributed (even standard PCA won’t work nearly as well if the data aren’t. This is not required, and independent components analysis (ICA) does not. This visualization duplicates that seen in Murphy (2012) where we have two (uniform) independent sources. We can see that the ICA correctly recovers those components.

If you believe that truly independent sources of signal underlie your data, ICA would be an option. It is commonly applied to deal with images or sound.


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Compute the spectrum for a operator

Find the spectrum of the operator $$\begin{split} A & \colon C[0,1] \rightarrow C[0,1] \\ & f \mapsto (Af)(x) := f(x) + \int_0^x f(t)dt \end{split}$$

P.S.: I know the spectrum of $A_1\colon f \mapsto \int_0^x f(t)dt$

-

The spectrum comes from analyzing $A-\lambda I$. Since $A = A_1+I$, you can see that $A-\lambda I = A_1-(\lambda-1)I$. It follows that $\sigma(A) = \sigma(A_1) +\{1\}$.

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$\sigma(A) = \lbrace 0 \rbrace + \lbrace 1 \rbrace = \lbrace 1\rbrace?$ –  daisy Dec 18 '12 at 18:01
Yes. A fairly standard notation for sets $A,B \subset \mathbb{R}$ is $A\square B = \{a \square b| a\in A, b \in B \}$, where $\square$ is some binary operator such as $+$. –  copper.hat Dec 18 '12 at 18:10


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## Algebraic & Geometric Topology

### Surface links which are coverings over the standard torus

Inasa Nakamura

#### Abstract

We introduce a new construction of a surface link in 4–space. We construct a surface link as a branched covering over the standard torus, which we call a torus-covering link. We show that a certain torus-covering $T2$–link is equivalent to the split union of spun $T2$–links and turned spun $T2$–links. We show that a certain torus-covering $T2$–link has a nonclassical link group. We give a certain class of ribbon torus-covering $T2$–links. We present the quandle cocycle invariant of a certain torus-covering $T2$–link obtained from a classical braid, by using the quandle cocycle invariants of the closure of the braid.

#### Article information

Source
Algebr. Geom. Topol., Volume 11, Number 3 (2011), 1497-1540.

Dates
Revised: 1 March 2011
Accepted: 2 March 2011
First available in Project Euclid: 19 December 2017

https://projecteuclid.org/euclid.agt/1513715237

Digital Object Identifier
doi:10.2140/agt.2011.11.1497

Mathematical Reviews number (MathSciNet)
MR2821433

Zentralblatt MATH identifier
1230.57022

#### Citation

Nakamura, Inasa. Surface links which are coverings over the standard torus. Algebr. Geom. Topol. 11 (2011), no. 3, 1497--1540. doi:10.2140/agt.2011.11.1497. https://projecteuclid.org/euclid.agt/1513715237

#### References

• E Artin, Theory of braids, Ann. of Math. $(2)$ 48 (1947) 101–126
• S Asami, S Satoh, An infinite family of non-invertible surfaces in $4$–space, Bull. London Math. Soc. 37 (2005) 285–296
• I Berstein, A L Edmonds, On the construction of branched coverings of low-dimensional manifolds, Trans. Amer. Math. Soc. 247 (1979) 87–124
• I Berstein, A L Edmonds, On the classification of generic branched coverings of surfaces, Illinois J. Math. 28 (1984) 64–82
• J S Birman, Braids, links, and mapping class groups, Annals of Math. Studies 82, Princeton Univ. Press (1974)
• O Bogopolski, Introduction to group theory, EMS Textbooks in Math., European Math. Soc., Zürich (2008) Translated, revised and expanded from the 2002 Russian original
• J Boyle, Classifying $1$–handles attached to knotted surfaces, Trans. Amer. Math. Soc. 306 (1988) 475–487
• J Boyle, The turned torus knot in $S\sp 4$, J. Knot Theory Ramifications 2 (1993) 239–249
• G Burde, K Murasugi, Links and Seifert fiber spaces, Duke Math. J. 37 (1970) 89–93
• G Burde, H Zieschang, Eine Kennzeichnung der Torusknoten, Math. Ann. 167 (1966) 169–176
• J S Carter, D Jelsovsky, S Kamada, L Langford, M Saito, Quandle cohomology and state-sum invariants of knotted curves and surfaces, Trans. Amer. Math. Soc. 355 (2003) 3947–3989
• J S Carter, D Jelsovsky, S Kamada, M Saito, Computations of quandle cocycle invariants of knotted curves and surfaces, Adv. Math. 157 (2001) 36–94
• J S Carter, S Kamada, M Saito, Geometric interpretations of quandle homology, J. Knot Theory Ramifications 10 (2001) 345–386
• J S Carter, M Saito, Knotted surfaces and their diagrams, Math. Surveys and Monogr. 55, Amer. Math. Soc. (1998)
• E Fadell, L Neuwirth, Configuration spaces, Math. Scand. 10 (1962) 111–118
• H Gluck, The embedding of two-spheres in the four-sphere, Trans. Amer. Math. Soc. 104 (1962) 308–333
• J Hillman, $2$–knots and their groups, Australian Math. Soc. Lecture Ser. 5, Cambridge Univ. Press (1989)
• M W Hirsch, Differential topology, Graduate Texts in Math. 33, Springer, New York (1994) Corrected reprint of the 1976 original
• D Joyce, A classifying invariant of knots, the knot quandle, J. Pure Appl. Algebra 23 (1982) 37–65
• S Kamada, Surfaces in ${\bf R}\sp 4$ of braid index three are ribbon, J. Knot Theory Ramifications 1 (1992) 137–160
• S Kamada, $2$–dimensional braids and chart descriptions, from: “Topics in knot theory (Erzurum, 1992)”, (M E Bozhuyuk, editor), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 399, Kluwer, Dordrecht (1993) 277–287
• S Kamada, A characterization of groups of closed orientable surfaces in $4$–space, Topology 33 (1994) 113–122
• S Kamada, An observation of surface braids via chart description, J. Knot Theory Ramifications 5 (1996) 517–529
• S Kamada, Braid and knot theory in dimension four, Math. Surveys and Monogr. 95, Amer. Math. Soc. (2002)
• R A Litherland, Symmetries of twist-spun knots, from: “Knot theory and manifolds (Vancouver, B.C., 1983)”, (D Rolfsen, editor), Lecture Notes in Math. 1144, Springer, Berlin (1985) 97–107
• C Livingston, Stably irreducible surfaces in $S\sp 4$, Pacific J. Math. 116 (1985) 77–84
• T Mochizuki, Some calculations of cohomology groups of finite Alexander quandles, J. Pure Appl. Algebra 179 (2003) 287–330
• J M Montesinos, On twins in the four-sphere. I, Quart. J. Math. Oxford Ser. $(2)$ 34 (1983) 171–199
• H Neumann, Generalized free products with amalgamated subgroups. II, Amer. J. Math. 71 (1949) 491–540
• L Rudolph, Braided surfaces and Seifert ribbons for closed braids, Comment. Math. Helv. 58 (1983) 1–37
• M Teragaito, Symmetry-spun tori in the four-sphere, from: “Knots 90 (Osaka, 1990)”, (A Kawauchi, editor), de Gruyter, Berlin (1992) 163–171
• T Yanagawa, On ribbon $2$–knots: The $3$-manifold bounded by the $2$–knots, Osaka J. Math. 6 (1969) 447–464


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# Moving last digit to first

Is it possible to find all positive integers $n$ such that if we move its last digit to the first digit, we get $2n$? I.e $2(a_m\cdot 10^m+\ldots +a_0)=a_0\cdot 10^m+a_m\cdot 10^{m-1}\cdots+a_1$

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Some simple playing around shows that this is equivalent to $2(a \cdot 10 + b) = b \cdot 10^k + a$, where $10^{k-1} \le a < 10^k$, and hence, $19 a = (10^k-2) b$. Therefore, a neccessary condition is that $b$ is a multiple of 19, and $a$ is even. Since $1 \le b \le 9$, this should answer the question. –  Johannes Kloos Apr 30 '12 at 12:01
@Johannes, why must $b$ be a multiple of 19? Why can't $10^k-2$ be a multiple of 19? –  Gerry Myerson Apr 30 '12 at 12:43
Um, you're right. It can. –  Johannes Kloos Apr 30 '12 at 12:52
OEIS A146088 gives the first ten of an infinite sequence –  Henry Apr 30 '12 at 13:03

There are at least two ways to solve this. One is the straightforward algebraic way given by Johannes Kloos and Gerry Myerson. The other is trickier but easier.

Let's guess—just guess—that $2n$ might begin with, say, 6, and see what that would imply. If we divide $2n$ by 2, using the usual short division algorithm, we see at once that $n$ must then begin with 3:

      3 ...      6
-------------
2 ) 6??? ...

But the digits of the dividend, $2n$, are exactly the same as the digits of the quotient, $n$, except that the quotient has the 6 on the end instead of on the beginning. So the second digit of the dividend must be 3:

      3 ...      6
-------------
2 ) 63?? ...

So continuing with short division, the second digit of the quotient (and the third digit of the dividend) is 1:

      31 ...     6
-------------
2 ) 631? ...

This time the short division algorithm says that since there was a remainder of 1 when we divided by 2, the next digit of the quotient is 11/2 = 5:

      315 ...    6
-------------
2 ) 6315 ...

Continuing in this way quickly yields:

      315789473684210526
-------------------
2 ) 631578947368421052

When we get to a 6 in the quotient and no remainder from the previous division, we can stop, and we have an answer: 315789473684210526 × 2 = 631578947368421052.

All the solutions can be found this way, by selecting some initial digit other than 6. Had we selected 2, we would have gotten the smallest solution, $2n$ = 210526315789473684, and $n=$105263157894736842. (Starting with 1 yields the degenerate solution $2n=$105263157894736842, $n=$052631578947368421; starting with 0 yields the even more degenerate solution $2n=n=0$.)

Had we decided not to stop the first time the initial digit repeated, but to continue dividing until the next stopping point, we would have gotten a solution twice as long; there are arbitrarily long solutions.

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Incidentally, solutions in radices other than base 10 are sometimes much simpler. For example, in base 8, 25×2=52, and in base 3, 1012×2=2101. The algebraic technique will show you where the simple solutions are. –  MJD Apr 30 '12 at 13:10

Rewrite the equation as $$19(10^{m-1}a_m+10^{m-2}a_{m-1}+\cdots+a_1)=(10^m-2)a_0$$ and you get $19$ divides $10^m-2$. The smallest solution is $m=17$. So you get $$10^{m-1}a_m+10^{m-2}a_{m-1}+\cdots+a_1=5263157894736842a_0$$. Taking $a_0=2$ you get the solution $$105263157894736842$$

You can get others solutions by taking other values of $a_0$ and/or taking $m=17+18t$, $t=1,2,3,\dots$. Many references will be found by typing 105263157894736842 into Google.

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"The smallest solution is $m=17$" is something of a rabbit from a hat, since finding such $m$ is a discrete logarithm problem, and discrete logarithms are believed to be hard to find. I do not know any elementary method for finding this one other than trial multiplication mod 19. –  MJD Apr 30 '12 at 17:16
@Mark: Finding discrete logarithms modulo 19 is not hard (by exhaustive search even). Finding discrete logarithmg modulo a 100-digit prime OTOH... If you want something quicker for the present case: $$10^{18}\equiv1\pmod{19}$$ by little Fermat. Combine that with $2\cdot10=20\equiv1\pmod{19}$ and get $$10^{18}\equiv 2\cdot 10\pmod{19}.$$ Hmm... –  Jyrki Lahtonen Apr 30 '12 at 21:21
@Jykri Yes, I did not mean to say that this particular question was hard, only that it should have been accompanied by some explanation. –  MJD May 1 '12 at 14:26


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Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

6

In case of NYSE equities rule 7.6 reads: "The minimum price variation ("MPV") for quoting and entry of orders in securities traded on the NYSE Arca Marketplace is USD 0.01, with the exception of securities that are priced less than USD 1.00 for which the MPV for quoting and entry of orders is USD 0.0001." For NASDAQ equities rule 4701 (k) reads: "The term "...

4

You said:"I understand that the generated ticks will be generated using interpolation (so they won't be exacts)". You are very optimistic, they will not only be far away from being exact, they (the tick data) will be completely removed from reality, the only parameters known for the tick data will be boundary conditions, such as open high low close. You ...

2

What Tick data you have in mind? NASDAQ ITCH is tick data but you have to construct the limit order book yourself to keep track of the best bid and ask price for each stocks. Not a trivial task. If you get TAQ data, you will get the best bid and ask (NBBO) but TAQ data has some issues like no odd-lot trades and trades are not mark buyer or seller initiated.

2

http://tsp.finra.org/finra_org/ticksizepilot/TSPilotSecurities.txt The relevant field being "Tick_Size_Pilot_Program_Group", Gx or C categorization. There is no formula to determine this as this rule is new, on a trial basis and the participating listings aren't chosen completely objectively. It's designed to help market making firms at the expense of ...

2

The way I managed to solve this was by scrapping the broker website and compile a list containing Symbol, Primary Exchange, Tick Increment, ISIN, CONID, ASSETID of 6589 Equities from NASDAQ, NYSE and AMEX. Here's a gist of the list:

2

We've done this before (*), but it is trivial: one tick for open, one tick for high, one tick for low, one tick for close. For tick timestamps, open is 08:01:00, close is 08:01:59.999, high is 08:01:20 and low is 08:01:40. (Or if that gives strange results, open is 08:01:00.001 and close is 08:02:00.000; it depends how your system makes bars.) If you want ...

1

You probably can't find a definitive resource. You can however past this together using sources on line. https://www.finra.org/industry/tick-size-pilot-program https://www.sec.gov/oiea/investor-alerts-bulletins/ia_ticksize.html I know that securities above \$1 have \$0.01 tick while those below \$1 have \$0.0001 tick since Jan 29, 2001 but how about ...

1

(1) The tick size puts a lower bound on the bid-ask spread (and often is equal to the bid-ask spread) Tick size too big: the bid-ask spread (which the market maker earns but the customers pay) is too big. Favors the market-makers at the expense of the customers. Tick size too small: vice versa. Therefore the tick size influences the transaction cost. (2) ...

1

From a feed perspective, you can take a look at the feed spec, it should be mentioned among descriptions of the data types e.g. fixed point integer with 4 implied decimal places and be applied across all symbols. This should be universal so you could hardcode it. Thereafter, you can look up the MPV in the rule books - see this answer: https://quant....

1

I do not know of anything that is already made and works out of the box, but I would recommend writing a small script in Python using csv.reader, this would probably be as fast as trying to plug something on your csv, especially if your format is a bit exotic ?

1

Supply and demand... If you want an event that produce a change in the value of a currency, just look at the ruble. As Russia, gets more and more isolated and inflation spins out of control the ruble lose its value against other currencies.

1

That exchange spec implies that there isn't a legal price of 1001, 1002, ..., 1009. The next available price after 1000 is just 1010. If your pricing engine determines a fair value of, say, 1005, then your application will need to apply some rounding logic. For example, you can always round away from the inside price, or you can simply round to the nearest ...

Only top voted, non community-wiki answers of a minimum length are eligible


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Lets consider it case by case:

Case 1: Charge particle is at rest. It has electric field around it. No problem. That is its property.

Case 2: Charge particle started moving (its accelerating). We were told that it starts radiating EM radiation? Why? What happened to it? What made it do this?

Follow up question: Suppose a charge particle is placed in uniform electric field, it accelerates because of electric force it experiences. Then work done by the electric field should not be equal to change in its kinetic energy right? It should be equal to change in K.E + energy it has radiated in the form of EM Waves. right? But then, why don't we take energy radiated into consideration when solving problem (I'm tutoring grade 12 students. I never encountered a problem in which energy radiated is considered.)

How does moving charges produce magnetic field?

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A question that comes to my mind right now is, given an acceleration of magnitude $a$, what is the resulting frequency of the EM wave? I am almost a physicist now, I did not continue because the amazing world of programming has captured me so I am working right now! But someday I will go back and continue studying physics, and it also was mentioned to us but I haven't seen any mathematical derivation. – iharob Nov 11 '15 at 0:08

A diagram may help:

Here, the charged particle was initially stationary, uniformly accelerated for a short period of time, and then stopped accelerating.

The electric field outside the imaginary outer ring is still in the configuration of the stationary charge.

The electric field inside the imaginary inner ring is in the configuration of the uniformly moving charge.

Within the inner and outer ring, the electric field lines, which cannot break, must transition from the inner configuration to the outer configuration.

This transition region propagates outward at the speed of light and, as you can see from the diagram, the electric field lines in the transition region are (more or less) transverse to the direction of propagation.

Also, see this Wolfram demonstration: Radiation Pulse from an Accelerated Point Charge

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Could you kindly comment on my "Follow up question" – claws May 21 '13 at 12:55
@claws, my only comment for now (I'll save more for later when I have time) is to refer you to this on the question of "Does a uniformly accelerated charge radiate?": mathpages.com/home/kmath528/kmath528.htm – Alfred Centauri May 21 '13 at 13:13
@Alfred Centauri: Purcell also followed the same procedure. Check this. – MAFIA36790 Jan 11 at 3:14

The second problem is quite tough. J. D. Jackson comments, in the introductory remarks of his chapter on 'Radiation Damping, Classical Models of Charged Particles', that we know how to solve classical electrodynamics problems in two ideal conditions - a) given charge and current densities, how to compute the fields and b) given the fields, how to find the motion of charged particles in their presence. When charged particles accelerate, they do produce radiation which in turn affects the motion of all other charged particles. However, that problem is, Jackson says, still unsolved.

Coming to the first problem, if you calculate $\vec{E}$ and $\vec{B}$ for a moving charged particle, you will see that they depend on the acceleration $\vec{a}$ of the charged particle. Now calculate the Poynting vector $\vec{S}$. You will observe that $\vec{S}$, depends on acceleration but not velocity. Integrating it to get power radiated gives the famous Larmor formula. You may want to refer to Griffiths' chapter on 'Electromagnetic Radiation'.

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For Jackson, the radiation also affects the particle itself. For Griffiths, the Poynting vector does have terms that contain velocity, they just fall off faster than 1/r and so don't contribute a finite energy to an infinitely distant surface. But they are carrying power away, just not to super far away. So back to Jackson, that energy loss also affects the motion of the charge. – Timaeus Nov 12 '15 at 1:40

## protected by Qmechanic♦Jan 27 '14 at 19:23

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.


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# Placing text under an entire math expression

I want to place text under an entire multivariate expression, exactly the way one is able with \underbrace, but without the brace. I want it to specifically describe the first variable in the expression, but for the text to span the entire expression. I have scoured the webs and it's all about placing text under single symbols or commands. For example, \underset, \smashoperator, and \limits can't seem to function with more than one command or variable, even when I use brackets around the command line.

I got what I wanted using this:

    \begin{align*}

\end{document}

Since you are using the command \lcurvearrownw I assumed that you were loading the MnSymbol package; this package changes many symbols.

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The usual caveat that MnSymbol changes many symbols should appear. –  egreg Sep 3 '12 at 10:23
@egreg remark added. Thanks! –  Gonzalo Medina Sep 3 '12 at 13:16
Thank you @ Gonzalo, that is much simpler than my original solution and I've noted MnSymbol's potential problems. –  Brazos Wolf Sep 8 '12 at 20:04


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Thread: Tex to LRF? View Single Post
04-06-2008, 08:39 PM   #3
AprilHare
Wizard

Posts: 2,954
Karma: 11862367
Join Date: Apr 2008
Device: Sony Reader PRS-T2
Quote:
 Originally Posted by HarryT The Reader's screen size is 120x90mm.
According to Sony and their guide to generating PDFs, the screen size is 90.6mm by 122.4 mm (3.57" by 4.82").

I tried using an installation of TeXmacs to generate such PDFs (for later conversion to LRF) however it had this tendency to produce A4 size documents anyway.
Anyone know of a way to do this conversion that actually works?
I'm using the Pride and Prejudice TeX document from Project Gutenberg for test purposes (download the zipped file - the unzipped one is actually a PDF as the uploader clearly stuffed up..)

The command \special{papersize=xx,yy} didn't seem to do the trick. I have access to Ubuntu Linux and Windows, if it helps.


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$a_1,a_2,...,a_{2014}$ is a permutation of $1,2,3,...,2014$. What is the greatest number of perfect squares can have a set ${ a_1^2+a_2,a_2... 1.$a_1,a_2,...,a_{2014}$is a permutation of$1,2,3,...,2014$. What is the greatest number of perfect squares can have a set${ a_1^2+a_2,a_2^2+a_3,a_3^2+a_4,...,a_{2013}^2+a_{2014},a_{2014}^2+a_1 }?$2. Do there exist positive integers$a$and$b$such that$a^n+n^b$and$b^n+n^a$are relatively prime for all natural$n$? 3. The triangle$ABC$is inscribed in a circle$w_1$. Inscribed in a triangle circle touchs the sides$BC$in a point$N$.$w_2$— the circle inscribed in a segment$BAC$circle of$w_1$, and passing through a point$N$. Let points$O$and$J$— the centers of circles$w_2$and an extra inscribed circle (touching side$BC$) respectively. Prove, that lines$AO$and$JN$are parallel. 4. Given a scalene triangle$ABC$. Incircle of$\triangle{ABC{}}$touches the sides$AB$and$BC$at points$C_1$and$A_1$respectively, and excircle of$\triangle{ABC}$(on side$AC$) touches$AB$and$BC$at points$ C_2$and$A_2$respectively.$BN$is bisector of$\angle{ABC}$($N$lies on$BC$). Lines$A_1C_1$and$A_2C_2$intersects the line$AC$at points$K_1$and$K_2$respectively. Let circumcircles of$\triangle{BK_1N}$and$\triangle{BK_2N}$intersect circumcircle of a$\triangle{ABC}$at points$P_1$and$P_2$respectively. Prove that$AP_1$=$CP_2$5.$\mathbb{Q}$is set of all rational numbers. Find all functions$f:\mathbb{Q}\times\mathbb{Q}\rightarrow\mathbb{Q}$such that for all$x$,$y$,$z\in\mathbb{Q}$satisfy $$f(x,y)+f(y,z)+f(z,x)=f(0,x+y+z)$$ 6. Prove that, for all$n\in\mathbb{N}$, on$ [n-4\sqrt{n}, n+4\sqrt{n}]$exists natural number$k=x^3+y^3$where$x$,$y\$ are nonnegative integers.

Name


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# Finding the matrix for a linear transformation on a vector space when the basis changes

Let B={$u_1,u_2,u_3$} as basis of Vector Space V, and Let T: V→V be the linear operator defined by,
$$[T]_B=\begin{bmatrix} -3 & 4 & 7 \\ 1 & 0 & -2 \\ 0 & 1 & 0 \\ \end{bmatrix}$$ Find $[T]_{B'}$. B'={$v_1,v_2,v_3$} is basis of V defined by $v_1 = u_1, v_2 = u_1 + u_2, v_3 = u_1 + u_2 + u_3$.

I have solved questions by my effort, but I'm not sure for this answer. Please check and give me a hint for this question. Thank you.


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# How To Calculate The Number Of Molecules In Any Piece of DNA

Lab math – it’s a dreaded topic, isn’t it? Calculating molecules in DNA samples, calculating the concentration of a solution, or working out serial dilutions were tasks that always filled me with dread as an undergrad. And to this day I am still not a huge fan! Fear not – there is light at the end of the tunnel!

In this article, we will go through counting molecules in DNA samples – for instance, how many templates/copies of a 50-base pair (bp) fragment do you have in a 500 ng sample?

You will be a whiz at these calculations in no time!

## Calculating Molecules in DNA – Why Would I Need To?

This information may come in useful when doing the following types of experiments:

1. DNA ligations.
2. Biochemical assays.
3. Real-time PCR (absolute quantification).
4. Measuring viral load/numbers of viral genomes.

Whatever you need this information for, you can go about it in the same step-by-step fashion.

## Here Comes the Science…

• The average weight of a single DNA bp is 650 daltons. This can also be written as 650 g/mol (= molar mass).
• This is the same as saying that one mole of a bp weighs 650 g.
• The molecular weight or molar mass of any double-stranded DNA fragment can therefore be calculated by multiplying its length (in bp) by 650 and the answer will be expressed as daltons or g/mol.

As long as you know the sequence length (i.e. the length/size of the genome), this calculation will work for the genomic DNA of any species. To get the length of your sequence in bp, count the nucleotides in the DNA sequence – either by hand or using character count in a word processing program. If you need some help with converting your calculations between different units, check out our guide for solving your lab math problems.

Question: What is the molecular weight of a 5.2-kb plasmid? Remember 5.2 kb = 5200 bp

Calculation: 5200 bp x 650 daltons = 3,380,000 daltons or g/mol

Now we know how to calculate the molecular weight of a DNA template. Using Avogadro’s number, which is 6.022 x 1023 molecules/mole, the number of molecules of the template per gram can be calculated using the following formula:

$\textrm{Number of copies} = (\textup{\textrm{}ng} \times [6.022 \times 10^{23}])/(\textrm{length }\times [1 \times 10^{9}] \times 650)$

• ng is the amount of DNA (plasmid, primer, etc.) you have in nanograms.
• 6.022 x 1023 = Avogadro’s number.
• Length is the length of your DNA fragment in base pairs. Just multiply by 1000 if you are working in kb.
• We multiply by 1 x 109 to convert our answer to nanograms.

Note: the Avogadro’s number or constant (denoted as NA) is defined as the number of constituent particles (usually atoms or molecules) per mole of a given substance.

## Calculating Molecules in DNA: Useful Examples

### Example 1: Plasmid

Question: You have 300 ng of a 5.2-kb plasmid.

How many copies of your plasmid do you have?

Calculation: Number of copies =

$(300 \times [6.022 \times 10^{23}])/(5200 \times [1 \times 10^{9}] \times 650)$

### Example 2: Primer

Question: You have 50 ng of a 20-bp primer.

How many copies of your primer do you have?

Calculation: Number of copies =

$(50 \times [6.022 \times 10^{23}])/(20 \times [1 \times 10^{9}] \times 650)$

### Example 3: PCR product

Question: You have 150 ng of a 670-bp purified PCR product.

How many copies of your PCR product do you have?

Calculation: Number of copies =

$(150 \times [6.022 \times 10^{23}])/(670 \times [1 \times 10^{9}] \times 650)$

Note:  It’s important to remember that this formula is based on the assumption that you are working with a single DNA species, so when applying to PCR products or plasmids, make sure you only have 1 band or that you have a clean plasmid miniprep.

See, that wasn’t so bad after all. Once you know the length of your DNA fragment and the concentration in nanograms, you can easily apply the formula to get the number of copies in any type of DNA sample.

Happy calculating molecules in DNA. If you have any comments, queries, or suggestions, get in touch using the comment box below.

Originally published September 17, 2014. Reviewed and updated April 27, 2021.

1. Anu on October 11, 2019 at 8:02 pm

The molecular weight (Da) of the zwitterionic form of the peptide GAGAGAGA?

2. fan xi wen on June 2, 2019 at 5:01 am


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# Classes of an equivalence relation

Let $$R$$ be the equivalence relation on the real numbers given by $$R = \{(x, y) \in \Bbb R^2: (x−y)(x+y) = 0 \}$$ What are the equivalence classes of $$R$$?

So I wrote that, for every $$x \in \Bbb R$$, the corresponding equivalence class is $$[x]_R = \{x, -x\}$$.

Is my answer correct and complete or there is something I omitted? Thanks!

• To clarify - by just $R$, you mean the relation, and by $R^2$ or R you're referring to $\mathbb{R}^2$ and $\Bbb R$ respectively, correct? – Eevee Trainer Feb 20 at 1:03
• @EeveeTrainer exactly – JBuck Feb 20 at 1:08
• I believe you are correct but I'd like to see a proof or an argument about why you believe this to be true. (A simple line that $ab = 0 \iff$ one of $a$ or $b$ is zero so....) Also it'd be good if you pointed out that the one exception is $[0]_R = \{0\}$ (although technically $\{0,-0\} = \{0\}$ so it is arguably correct. – fleablood Feb 20 at 1:09
• I edited you question to try and clarify the ambiguities around $R$ and $\Bbb R$; I think I got it right. Did I? Cheers! – Robert Lewis Feb 20 at 1:10
• @fleablood the justification is: (x+y)(x-y) = 0 ⟺ x = y or x = -y. Then the class of any $x ∈ R$ is defined: $[x]_R$ = {$y ∈ R: (x − y)(x + y) = 0$} = {x, -x}. – JBuck Feb 20 at 1:20

1) Make sure that your relation is indeed an equivalence relation, i.e. it is reflexive, symmetric and transitive. Indeed, it is (I just checked, but as an exercise, you should, too.)

2) Now find the ordered pairs , i.e. elements of R. For example, when x = y, or x = -y, R does indeed contain the elements (x, y), (x, -y). Does it contain (y, x), (-y, x)?

• Equivalence is given in the exercise, also proving it wasn't a problem, so I did not include it in the question. Also, the ordered pairs you mention are included in R, by symmetry. – JBuck Feb 20 at 1:51

Yes, I believe you're correct.

We know

$$(x-y)(x+y) = x^2 - y^2$$

as it is the difference of two squares. From that,

$$(x-y)(x+y) = 0 \iff x^2 = y^2 \iff |x| = |y| \iff x = \pm y$$

Thus,

$$(x,y) \in R \iff x = \pm y$$

Thus, any two real numbers $$x,y$$ are related if and only if one is the other, or its negative. Accordingly, for each real number, you have the equivalence class given by it and its negative.

As noted by fleablood in the comments it might not hurt to specifically set $$0$$ aside and thus let the equivalence classes be

$$[x]_R = \left\{\begin{matrix} \{ x,-x \} & x\neq 0\\ \{0\} & x = 0 \end{matrix}\right.$$

just to remove any potential ambiguity, but it's not like it's really wrong presented as-is since $$0=-1 \cdot 0=-0$$. Just depends on how paranoid you want to be about points on the assignment. :p

You can of course justify this by considering $$x=y=0$$ to show $$(x,y)\in R$$ and showing no other $$x,y$$ when one is $$0$$ can be in the same equivalence class. I maintain that, personally, it's fine as-is, but up to you.

Alternative Justification: (thanks to JMoravitz for pointing this out in the comments)

One recalls, when dealing with real numbers, this situation -- if $$ab=0$$, then what happens? At least one of $$a,b$$ are zero, i.e. $$a=0,b=0$$ or $$a=b=0$$.

This, as a bit of a tangent, comes as a consequence of the real numbers, when equipped with traditional addition and multiplication, forming what is known as a field. Fields are a special case of a certain, slightly more general algebraic structure known as an integral domain, itself a special case of another structure known as rings. I won't go into detail on these - the elaboration isn't necessary to understand for what follows, but it does provide a means of rigorously justifying this property. So some further reading if you're curious and have time to burn:

Anyhow, we essentially define an integral domain thusly: a commutative ring, in which there are no zero divisors. Equivalently, an integral domain $$(R,\oplus,\otimes,0',1')$$ is a commutative ring in which $$a\otimes b=0'$$ if and only if $$a=0',b=0',$$ or $$a=b=0'$$. Here, $$0'$$ denotes the "additive identity" for the ring $$R$$, the prime being used to distinguish it from the "actual" number zero. The notion of $$0'$$ being the additive identity simply means that $$a\oplus 0' = 0' \oplus a = a$$ for all $$a \in R$$.

In the case of the real numbers, $$\Bbb R$$, $$0'$$ is in fact the number zero, $$1'$$ the number one, $$\oplus$$ is the tradition addition operation $$+$$, and $$\otimes$$ is the traditional multiplication operation ($$\cdot, \times$$, however you choose to denote it).

Now we tie this back into where it becomes relevant. Or if you didn't care about the tangent into abstract algebra, I get back to the point. You're probably not going to be expected to justify all of this in your class since we just know from experience and such that the property holds for real numbers, but it's useful to know where it comes from, you know?

Anyhow.

So, since $$\Bbb R$$ is a field, we know by definition it has the property that

$$ab = 0 \iff a=0, b=0, \; \text{or} \; a=b=0$$

Thus, since $$xRy$$ if and only if $$(x-y)(x+y)=0$$, we can immediately conclude by this property

$$x-y = 0 \;\;\; \text{and/or} \;\;\; x+y=0$$

If only the first holds, we know $$xRy$$ if and only if $$x=y$$.

If only the second holds, we know $$xRy$$ if and only if $$x=-y$$.

If both hold, we know $$xRy$$ if and only if $$x=y=-y$$, which only holds if $$x=y=0$$.

From there, it's a pretty clear exercise on how to construct the equivalence classes and so on, it more or less parallels the original approach.

• I would skip the steps of expanding the product of $(x-y)(x+y)$ and simply cite or use the fact that $\Bbb R$ is an integral domain and so $(x-y)(x+y)=0$ directly implies (with no inbetween steps needed) that $x=y$ or $x=-y$. In doing so the proof could easily be extended to other integral domains like $\Bbb C$ where the implication $x^2=y^2\iff |x|=|y|$ might have been false. – JMoravitz Feb 20 at 1:25
• @JMoravitz I haven't encountered integral domains yet, so I wouldn't know how to use them in my proof, but thanks anyway. – JBuck Feb 20 at 1:56
• @JBuck Integral domains are a type of algebraic structure - a set equipped with a pair of operations in this case. You might study them later on. One property of integral domains is that if $a \cdot b = 0$, then one of $a,b$ are zero. (The $\cdot$ here does not necessarily mean "multiplication" as you know it, could be some other operation.) That's just glazing over the details. But basically the point JMoravitz is getting at is the fact that you know if $ab=0$ then $a=0, b=0,$ or $a=b=0$. [cont.] – Eevee Trainer Feb 20 at 2:16
• (This is just an obvious fact when dealing with real numbers, rigorously justifiable if you want to; you don't need to appeal to integral domains since you're dealing with familiar territory, though.) In that sense, either $x-y=0$ or $x+y=0$ or both. The first yields $x=y$, the second $x=-y$, and the third yields $x=y=0$. That could be a very simple way to deal with this as opposed to expanding the product, so kudos to @JMoravitz for pointing it out. I'll edit it into my post in a sec. – Eevee Trainer Feb 20 at 2:17

The condition which defines $$R$$,

$$(x - y)(x + y) = 0, \tag 1$$

is clearly equivalent to

$$x^2 - y^2 = 0 \equiv y^2 = x^2 \equiv y = \pm x. \tag 2$$

Is $$R$$ and equivalence relation on $$\Bbb R$$? Well, writing in the usual way

$$xRy \; \text{for} \; (x, y) \in R, \tag 3$$

we check the usual conditions: reflexivity, symmetry, transitivity. Clearly

$$x^2 = x^2 \equiv x = \pm x \equiv xRx; \tag 4$$

$$xRy \equiv x^2 = y^2 \Longleftrightarrow y^2 = x^2 \equiv yRx; \tag{5}$$

so these two give us that $$R$$ is both reflexive and symmetric; that $$R$$ is transitive is just as easily seen:

$$[xRy] \wedge [yRz] \equiv [x^2 = y^2] \wedge [y^2 = z^2] \Longrightarrow x^2 = z^2 \equiv xRz; \tag{6}$$

(4)-(6) give us the reflexive, symmetric and transitive properties we need to accept that $$R$$ is in fact an equivalence relation on $$\Bbb R$$.

So, what are the equivalence classes determined by $$R$$? I think that's pretty easy to deduce at this point; since

$$xRy \equiv x^2 = y^2 \equiv y = \pm x, \tag 7$$

we have

$$[x]_R = \{ x, -x \}, \; \forall x \in \Bbb R; \tag 8$$

of course, this binds even in the event $$x = 0$$, there is nothing in the definition of equivalence classes which requires then all to be of the same cardinality.


text:
## AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.5 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.5

Question 1.
Verify the identity (a + b)2 ≡ a2 + 2ab + b2 geometrically by taking
(i) a = 2 units, b = 4 units
(ii) a = 3 units, b = 1 unit
(iii) a = 5 units, b = 2 unit
Solution:
(i)

= 4 × 4 + 4 × 2 + 2 × 2 + 4 × 2
= 16+ 8 + 4 + 8 = 36 sq.units
[∵ (2 + 4)2 = 62 = 36]

(ii)

Area of a square AEGI
= area of square ABCD + area of rectangle CDEF + area of square CFGH + area of rectangle BIHC.
= 3 × 3 + 3 × 1 + 1 × 1+3 × 1
= 9 + 3 + 1 + 3
= 16 sq. units
[∵ (3 + 1)2 = 42 = 16]

(iii)

= 5 × 5 + 2 × 5 + 2 × 2 + 5 × 2
= 25 + 10 + 4 + 10
= 49 sq.units
[∵ (5 + 2)2 = 72 = 49]

Question 2.
Verify the identity (a – b)2 ≡ a2 – 2ab+ b2 geometrically by taking
(i) a = 3 units, b= 1 unit
(ii) a = 5 units, b = 2 units
Solution:
(i)

Area of AIFE + Area of FGCH = (a – b)2 = a2 – 2ab + b2 [area of AIFE – area of IBGF – area of EFHD + area of FGCH]
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
= 9 – 3 – 3 + 1 = 4
∴ (a – b)2 = 4 sq. units
[∵ (3 – 1 )2 = 22 = 4]

ii)

∴ (a – b)2 = a2 – 2ab + b2
Area of ABCD + Area of CYZS = a2 – 2ab + b2
area of ABCD – area of BXYC – area of DCST + area of CYZS
=5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)2 = (3)2 = 9]

Question 3.
Verify the identity(a + b)(a – b) ≡ a2 – b2 geometrically by taking
(i) a = 3 units, b = 2 units
(ii) a = 2 units, b = 1 unit
Solution:
i)

a2 – b2 = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a – b) (a + b)
= 3 × 3 – 2 × 2
a2 – b2 = 9 – 4= 5sq . units
[ ∵ 32 – 22 = 9 – 4 = 5]

ii)

a2 – b2 = Area of Fig I. + Area of Fig II
= a(a – b) + b(a – b)
= (a + b) (a – b)
=(2 + 1)(2 – 1)
= 3 × 1 = 3
a2 – b2 = 3 sq. units
[∵ (22 – 12) = 4 – 1 = 3]

## AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.4 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.4

Question 1.
Select a suitable identity and find the following products
(i) (3k + 4l)(3k + 4l)
(ii) (ax2 + by2)(ax2 + by2)
(iii) (7d – 9e)(7d – 9e)
(iv) (m2 – n2)(m2 + n2)
(v) (3t + 9s) (3t – 9s)
(vi) (kl – mn) (kl + mn)
(vii) (6x + 5)(6x + 6)
(viii) (2b – a)(2b +c)
Solution:
(3k + 4l) (3k 4l) = (3k + 4l)2 is in the form of (a + b)2.
=(3k)2 + 2 × 3k × 4l+ (4l)2 [ (a+ b)2 = a2 + 2ab + b2
= 3k × 3k + 24kl + 4l × 4l
= 9k2 + 24kl + 16l2

ii) (ax2 + by2) (ax2 + by2) = (ax2 + by2)2 is in the form of (a + b)2.
= (ax2)2 + 2 × ax2 × by2 + (by2)2 [ ∵ (a + b)2 = a2 + 2ab + b2]
= ax2 × ax2 + 2abx2y2 + by2 × by2
= a2x4 + 2ab x2y2 + b2y4

iii) (7d – 9e) (7d – 9e)
= (7d – 9e)2 is in the form of (a – b)2.
= (7d)2 – 2 × 7d × 9e + (9e)2 [ ∵ (a – b)2 = a2 – 2ab + b2]
= 7d × 7d – 126de + 9e × 9e
= 49d2 – 126de + 81e2

iv) (m2 – n2) (m2 + n2) is in the form of (a + b) (a – b).
∴ (a + b) (a – b) = a2 – b2
∴ (m2 + n2) (m2 – n2) = (m2)2 – (n2)2 = m4 – n4

v) (3t + 9s) (3t – 9s) = (3t)2 – (9s)2 [ ∵ (a + b) (a – b) = a2 – b2 ]
= 3t × 3t – 9s × 9s
= 9t2 – 81s2

vi) (kl – mn) (kl + mn) = (kl)2 – (mn)2 [ ∵(a + b) (a – b) = a2 – b2 ]
= kl × kl – mn × mn
= k2l2 – m2n2

vii) (6x + 5) (6x + 6) is in the form of
(ax + b) (ax + c).
(ax + b) (ax + c) = a2x2 + ax(b + c) + bc
(6x + 5) (6x + 6) = (6)2x2 + 6x (5 + 6) + 5 × 6
= 36x2 + 6x × 11 + 30
= 36x2 + 66x + 30

viii) (2b – a) (2b + c) is in the form of (ax – b) (ax + c).
(ax – b) (ax + c) = a2x2 + ax(c – b) – cb
(2b – a) (2b + c) = (2)2(b)2 + 2b (c – a) – ca
= 4b2 + 2bc – 2ab – ca

Question 2.
Evaluate the following by using suitable identities:
(i) 3042
(ii) 5092
(iii) 9922
(iv) 7992
(v) 304 × 296
(vi) 83 × 77
(vii) 109 × 108
(viii) 204 × 206
Solution:
i) 3042 = (300 + 4)2 is in the form of (a + b)2.
∵ (a+b)2 = a2 + 2ab + b2
a = 300, b = 4
(300 + 4)2 = (300)2 + 2 × 300 × 4 + (4)2
= 300 × 300+ 2400 + 4 × 4
= 90,000 + 2400 + 16
= 92,416

ii) 5092 = (500 + 9)2
a  = 500, b = 9
= (500)2 + 2 × 500 × 9 + (9)2
[ ∵ (a + b)2 = a2 + 2ab + b2]
= 500 × 500 + 9000 + 9 × 9
= 2,50,000 + 9000 + 81
= 2,59,081

iii) 9922 = (1000 – 8)2
a = 1000, b = 8
= (1000)2 – 2 × 1000 × 8 + (8)2 [∵ (a-b)2 = a2 – 2ab + b2]
= 1000 × 1000 – 16,000 + 8 × 8
= 10,00,000 – 16000 + 64
= 10,00,064 – 1600
= 9,98,464

iv) 7992 = (800 – 1)2
a = 800, b = 1
= (800)2 – 2 × 800 × 1 + (1)2
= 800 × 800 – 1600 + 1
= 6,40,000 – 1600 + 1
= 6,40,001 – 1600
= 6,38,401

v) 304 × 296 = (300 + 4) (300 – 4) is in the form of (a + b) (a – b).
(a + b) (a – b) = a2 – b2
∴ (300 + 4) (300 – 4) = (300)2 – (4)2
= 300 × 300 – 4 × 4
= 90,000 – 16
= 89,984

vi) 83 × 77 = (80 + 3) (80 – 3)
= (80)2 – (3)2 [ ∵ (a + b) (a – b) = a2 – b2]
= 80 × 80 – 3 × 3
= 6400 – 9
= 6391

vii) 109 × 108 = (100 + 9) (100 + 8)
= (100)2 + (9 + 8)100 + 9 × 8
= 10,000 + 1700 + 72
= 11,772

viii) 204 × 206 = (205 – 1) (205 + 1)
= (205)2 – (1)2 [∵ (a + b)(a-b) = a2 – b2]
= 205 × 205 – 1 × 1
= 42,025 -1
= 42,024

## AP State Syllabus 8th Class Maths Solutions 3rd Lesson Construction of Quadrilaterals InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 3 Construction of Quadrilaterals InText Questions and Answers.

### 8th Class Maths 3rd Lesson Construction of Quadrilaterals InText Questions and Answers

Do this

Question 1.
Take a pair of sticks of equal length, say 8 cm. Take another pair of sticks of equal length, say 6 cm. Arrange them suitably to get a rectangle of length 8 cm and breadth 6 cm. This rectangle is created with the 4 available measurements. Now just push along the breadth of the rectangle. Does it still look alike? You will get a new shape of a rectangle Fig (ii), observe that the rectangle has now become a parallelogram. Have you altered the lengths of the sticks? No! The measurements of sides remain the same.

Give another push to the newly obtained shape in the opposite direction; what do you get? You again get a parallelogram again, which is altogether different Fig (iii). Yet the four measurements remain the same. This shows that 4 measurements of a quadrilateral cannot determine its uniqueness. So, how many measurements determine a unique quadrilateral? Let us go back to the activity!
You have constructed a rectangle with two sticks each of length 8 cm and other two sticks each of length 6 cm. Now introduce another stick of length equal to BD and put it along BD (Fig iv). If you push the breadth now, does the shape change? No!
It cannot, without making the figure open. The introduction of the fifth stick has fixed the rectangle uniquely, i.e., there is no other quadrilateral (with the given lengths of sides) possible now. Thus, we observe that five measurements can determine a quadrilateral uniquely. But will any five measurements (of sides and angles) be sufficient to draw a unique quadrilateral? (Page No. 60)
Yes, any 5 individual measurements are needed to construct a quadrilateral.

Question 2.
Equipment (Page No. 61)
You need: a ruler, a set square, a protractor.
Remember: To check if the lines are parallel.
Slide set square from the first line to the second line as shown in adjacent figures.

Now let us investigate the following using proper instruments. For each quadrilateral,
a) Check to see if opposite sides are parallel.
b) Measure each angle.
c) Measure the length of each side.

Record your observations and complete the table below.

Question 3.
Can you draw the angle of 60°?    (Page No. 63)

Using a scale and compass,
we can construct 60°.

Question 4.
Construct the parallelogram above (Refer text book page no: 75) BELT by using other properties of parallelogram. (Page No. 75)

We can construct a parallelogram using the measurements of a side, a diagonal and an angle.
BE = 5 cm ⇒ LT = 5 cm
∠B = 110° ⇒ ∠E = 180° – 110° = 70°
TE= 7.2 cm

Try these

Question 1.
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm?    (Page No. 70)

In a parallelogram BATS, opposite sides are equal.
BA = ST = 5 cms
AT = BS 6 cms
AS = 6.5 cms

∴ So, we can construct BATS parallelogram. It needs only three measurements.

Question 2.
A student attempted to draw a quadrilateral PLAY given that PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm. But he was not able to draw it why ?
Try to draw the quadrilateral yourself and give reason. (Page No. 70)
PL = 3 cm LA = 4 cm AY = 4.5 cm
PY = 2 cm LY = 6 cm
Here YP + PL < YL [∵ 2 + 3 < 6 ⇒ 5 < 6]

But in a △YPL, the sum of two sides is less than the third side.
∴ We are unable to construct a quadrilateral PLAY [∵ YL > YP]
[∵ The arcs do not intersect which are drawn from L and P, also Y, P, L are collinear points]

Think, discuss and write

Question 1.
Is every rectangle a parallelogram? Is every parallelogram a rectangle?    (Page No. 63)
Yes, every rectangle is a parallelogram. But every parallelogram is not a rectangle.

Question 2.
Uma has made a sweet chikki. She wanted it to be rectangular. In how many different ways can she verify that it is rectangular?    (Page No. 63)
If the sweet chikki is to made into a rectangular shape, she has to verify the following shapes:

2. Trapezium
3. Parallelogram

Question 3.
Can you draw the quadrilateral ABCD with AB = 4.5 cm, BC = 5.2 cm, CD = 4.8 cm and diagonals AC = 5 cm, BD = 5.4 cm by constructing △ABD first and then fourth vertex ‘C’ ? Give reason.       (Page No. 72)
We cannot construct △ABD. So, if we start first from △ABD, it is impossible to construct □ ABCD.
[∵ The length of $$\overline{\mathrm{AD}}$$ is not given]

Question 4.
Construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm. Justify your result.      (Page No. 72)

PQ = 3 cm
RS = 3 cm
PS = 7.5 cm
PR = 8 cm
SQ = 4 cm
With the given measurements △PQS is not possible to construct.
∵ PQ + QS < PS
The arcs which drawn from P and Q are not intersecting.
∴ We can’t obtain vertex ‘S’.
∴ Without vertex ‘S’ we can’t get a quadrilateral PQRS.

Question 5.
Can you construct the quadrilateral PQRS, if we have an angle of 100° at P instead of 75°? Give reason. (Page No. 74)

PQ = 4 cm,
QR = 4.8 cm,
ZP = 100°,
ZQ = 100°,
ZR = 120°
∴ We can construct a quadrilateral with the given measurements.
Since the sum of 4 angles is equal to 360°.

Question 6.
Can you construct the quadrilateral PLAN if PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15° and ∠A = 140°?
(Draw a rough sketch in each case and analyse the figure). State the reasons for your conclusion. (Page No. 74)
PL = 6 cm, ∠A = 9.5 cm, ∠P = 75°, ∠L = 15°, ∠A = 140°

∴ With the given measurements it is not possible to construct a quadrilateral.

Question 7.
Do you construct the given quadrilateral ABCD with AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75° by taking BC as base instead of AB? If so, draw a rough sketch and explain the various steps involved in the construction. (Page No. 77)
AB = 5 cm, BC = 4.5 cm, CD = 6 cm, ∠B = 100°, ∠C = 75°

Construction Steps:

1. Construct a line segment with radius 4.5 cms as $$\overline{\mathrm{BC}}$$
2. With the centres B and C draw two rays with 100°, 75° respectively.
3. With the centres B and C, two arcs are drawn with radius 5 cm and 6 cm respectively. The arcs and the rays are intersected.
4. Let the intersecting points be keep as A, D.
5. Join A, D.
6. ∴ ABCD quadrilateral is formed.

Question 8.
Can you construct the given AC = 4.5 cm and BD = 6 cm quadrilateral (rhombus) taking BD as a base instead of AC? If not give reason. (Page No. 79)

We can construct a rhombus taking BD as base instead of base AC.

Question 9.
Suppose the two diagonals of this rhombus are equal in length, what figure do you obtain? Draw a rough sketch for it. State reasons. (Page No. 79)
In a rhombus if the two diagonals are equal then it becomes a square.
∴ ABCD is a square.
[∵ AB = BC = CD = DA Also AC = BD]

## AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.3

Question 1.
Multiply the binomials:
(i) 2a – 9 and 3a + 4
(ii) x – 2y and 2x – y
(iii) kl + lm and k – l
(iv) m2 – n2 and m + n
Solution:
i) 2a – 9 and 3a + 4
(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)
= 6a2 + 8a – 27a – 36
= 6a2 – 19a – 36

ii) x – 2y and 2x – y
(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)
= 2x2 – xy – 4xy + 2y2
= 2x2 – 5xy + 2y2

iii) kl + lm and k – l
(kl + lm) (k – l) = kl(k – l) + lm(k – l)
= k2l – l2k + klm – l2m

iv) m2 – n2 and m + n
(m2 – n2) (m + n) = m2(m + n) – n2(m + n)
= m3 + m2n – n2m – n3

Question 2.
Find the product:
(i) (x + y)(2x – 5y + 3xy)
(ii) (mn – kl + km) (kl – lm)
(iii) (a – 2b + 3c)(ab2 – a2b)
(iv) (p3 + q3)(p – 5q+6r)
Solution:
i) (x + y) (2x – 5y + 3xy)
= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)
= 2x2 – 5xy + 3x2y + 2xy – 5y2 + 3xy2
= 2x2 – 5y2 – 3xy + 3x2y + 3xy2

ii) (mn – kl + km) (kl – lm)
= kl(mn – kl + km) – lm(mn – kl + km)
= klmn – k2l2 + k2lm – lm2n + kl2m – klm2

iii) (a – 2b + 3c) (ab2 – a2b) = a(ab2 – a2b) – 2b(ab2 – a2b) + 3c(ab2– a2b)
= a2b2 – a3b – 2ab3 + 2a2b2 + 3ab2c – 3a2bc
= 3a2b2 – a3b – 2ab3 + 3ab2c – 3a2bc

iv) (p3 + q3) (p – 5q + 6r) = p3(p – 5q + 6r) + q3(p – 5q + 6r)
= p4 – 5p3q + 6p3r + pq3 – 5q4 + 6rq3
= p4 – 5q4 – 5p3q + 6p3r + pq3 + 6rq3

Question 3.
Simplify the following:
(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)
(ii) (m + n) (m2 – mn + n2)
(iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) – (pr-i-pq) (p-i-q – r)
Solution:
i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)
= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)
= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)
= (y – 3x) (3y – 3x) + (x + y) (x – 3y)
= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)
= 3y2 – 3xy – 9xy + 9x2 + x2 – 3xy + xy – 3y2
= 10x2 – 14xy

ii) (m + n) (m2– mn + n2)
= m(m2 – mn + n2) + n(m2 – mn + n2)
= m3 – m2n + n2m + nm2 – mn2 + n3
= m3 + n3

iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
= a(a – 2b + 5c) – b(a – 2b + 5c) – 2a(a – b – c) – 3c(a – b – c) + 6a(2c – 3a – 5b) + b(2c – 3a – 5b)
= a2 – 2ab + 5ac – ab + 2b2 – 5bc – 2a2 + 2ab + 2ac – 3ac + 3bc + 3c2 + 12ac – 18a2 – 30ab + 2bc – 3ab – 5b2
= – 19a2 – 3b2 – 34ab + 16ac + 3c2

iv) (pq – qr + pr) (pq + qr) – (pr + pq) (p + q – r)
= pq(pq – qr + pr) + qr(pq – qr + pr) – pr(p + q – r) – pq(p + q – r)
= p2q2 – pq2r + p2qr + pq2r – q2r2 + pqr2 – p2r – pqr + pr2 – p2q – pq2 + pqr
= p2q2 – q2r2 + p2qr + pqr2 – p2r + pr2 – p2q – pq 2

Question 4.
If a, b, care positive real numbers such that $$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$$ ,find the value of $$\frac{(a+b)(b+c)(c+a)}{a b c}$$
Solution:
$$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$$ = k then
$$\frac{a+b-c}{c}$$ = k ⇒ a + b – c = kc
⇒ a + b = (ck + c) = c(k + 1) …………… (1)
Similarly b + c = a(k + 1) ……………(2)
c + a = b(k + 1) ………………..(3)

## AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.2 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.2

Question 1.
Complete the table:

Solution:

Question 2.
Simplify: 4y(3y + 4)
Solution:
4y(3y + 4) = 4y × 3y + 4y × 4
= 12y2 + 6y

Question 3.
Simplify x(2x2 – 7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0
Solution:
x(2x2 – 7x + 3)
= x × 2x2 – x × 7x + x × 3
= 2x3 – 7x2 + 3x
= 3 × 3 – 1 × 3 – 3 × 1 + 1 × 1
=9 – 3 – 3 + 1 = 4
∴(a – b)2 = 4sq.units
[∵ (3 – 1)2 = 22 = 4]

(ii)

∴ (a – b)2 = a2 – 2ab + b2
Area of ABCD + Area of CYZS
= a2 – 2ab + b2
area of ABCD – area of BXYC – area of DCST + area of CYZS
= 5 × 5 – 2 × 5 – 2 × 5 + 2 × 2
= 25 – 10 – 10 + 4
= 9 sq.units
[∵ (5 – 2)2 = (3)2 = 9]

Question 4.
Add the product: a(a – b), b(b – c), c(c – a)
Solution:
a(a – b) + b(b – c) + c(c – a)
=a × a – a × b + b × b – b × c + c × c – c × a
=a2 – ab + b2 – bc + c2 – ca
=a2 + b2 + c2 – ab – bc – ca

Question 5.
Add the product: x(x + y – r), y(x – y+r), z(x – y – z)
Solution:
x(x + y – r) +y(x – y + r) + z(x – y – z)
= x2 + xy – xr + xy – y2 + yr + zx – yz – z2
= x2 – y2 – z2 + 2xy – xr + yr + zx – yz

Question 6.
Subtract the product of 2x(5x – y) from product of 3x(x+2y)
Solution:
3x(x + 2y) – 2x(5x – y)
=(3x × x + 3x × 2y)-(2x × 5x – 2x × y)
= 3x2 + 6xy – (10x2 – 2xy)
= 3x2 + 6xy- 10x2 + 2xy
= 8xy – 7x2

Question 7.
Subtract 3k(5k – l + 3rn) from 6k(2k + 3l – 2rn)
Solution:
6k(2k + 3l – 2m) – 3k(5k – l + 3m)
= 12k2+ 18kl – 12km – 15k2 + 3kl – 9km
= -3k2 + 21kl – 21km

Question 8.
Simplify: a2(a – b + c) + b2(a + b – c) – c2(a – b – c)
Solution:
a2(a – b + c) + b2(a + b – c) – c2(a – b – c)
= a3 – a2b + a2c + ab2 + b3 – b2c – ac2 + bc2 + c3
= a3 + b3 + c3 – a2b + a2c + ab2 – b2c – ac2 – bc2

## AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.1 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.1

Question 1.
Find the product of the following pairs:
(i) 6, 7k
(ii) – 31, – 2m
(iii) -5t2 – 3t2
(iv) 6n, 3m
(v) – 5p2, – 2p
Solution:
The product of 6, 7k = 6 × 7k = 42k
ii) The product of – 3l, – 2m = (- 3l) × (- 2m) = 6/m
iii) The product of – 5t2, – 3t2 = (- 5t2) × (- 3t2) = 15t4
iv) The product of 6n, 3m = 6n × 3m = 18mn
v) The product of – 5p2, – 2p = (- 5p2) × (- 2p) = 10p3

Question 2.
Complete the table of the products.

Solution:

Question 3.
Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Solution:

Question 4.
Find the product of the following monomials
(i) xy, x2y , xy, x
(ii) a, b, ab, a3 b, ab3
(iii) kl, lm, km, klm
(iv) pq ,pqr, r
(v) – 3a, 4ab, – 6c, d
Solution:
i) The product of xy, x2y, xy, x = xy × x2y × xy × x
= x5 × y3= x5y3

ii) The product of a, b, ab, a3b, ab3 = a × b × ab × a3b × ab3
= a6 × b6 = a6 b6

iii) The product of kl, lm, km, klm = kl × lm × km × klm
k3 × l3 × m3 =k3l3m3

iv) The product of pq, pqr, r = pq × pqr × r
= p2 × q2 × r2 – p2q2r2

v) The product of – 3a, 4ab, – 6c, d = (- 3a) × 4ab × (- 6c) x d
= + 72a2 × b × c × d
= 72a2bcd

Question 5.
If A = xy,B = yz and C = zx, then find ABC=
Solution:
ABC = xy × yz × zx = x2y2z2

Question 6.
If P = 4x2, T = 5x and R = 5y, then $$\frac{\mathrm{PTR}}{100}$$ =
Solution:
$$\frac{P^{\prime} \Gamma R}{100}=\frac{4 x^{2} \times 5 x \times 5 y}{100}=\frac{100 x^{3} y}{100}$$ = x3 y

Question 7.
Write some monomials of your own and find their products.
Solution:
The product of,some monomials is given below :
i) abc × a2bc = a3b2c2
ii) xy × x2z × yz2 = x3y2z3
iii) p × q × r = p3q3r3

## AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable InText Questions and Answers.

### 8th Class Maths 2nd Lesson Linear Equations in One Variable InText Questions and Answers

Do this

Question 1.
Which of the following are linear equations:        [Page No. 35]
i) 4x + 6 = 8
ii) 4x – 5y = 9
iii) 5x2 + 6xy – 4y2 = 16
iv) xy + yz + zx = 11
v) 3x + 2y – 6 = 0
vi) 3 = 2x + y
vii) 7p + 6q + 13s = 11
(i), (ii), (v), (vi), (vii) are the linear equations.

Question 2.
Which of the following are simple equations?        [Page No. 36]
i) 3x + 5 = 14
ii) 3x – 6 = x + 2
iii) 3 = 2x + y
iv) $$\frac{x}{3}$$ + 5 = 0
v) x2 + 5x + 3 = 0
vi) 5m – 6n = 0
vii) 7p + 6q + 13s = 11
viii) 13t – 26 = 39
(i), (ii), (iv), (viii) are the simple equations.
Since these are all in the form of ax + b = 0.

## AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers InText Questions and Answers.

### 8th Class Maths 1st Lesson Rational Numbers InText Questions and Answers

Do this

Question 1.
Consider the following collection of numbers 1, $$\frac{1}{2}$$, -2, 0.5, 4$$\frac{1}{2}$$, $$\frac{-33}{7}$$, 0, $$\frac{4}{7}$$, $$0 . \overline{3}$$, 22, -5, $$\frac{2}{19}$$, 0.125. Write these numbers under the appropriate category. [A number can be written in more than one group]  (Page No. 2)
i) Natural numbers 1, 22
ii) Whole numbers 0, 1, 22
iii) Integers 0, 1, 22, -5, -2
iv) Rational numbers 1, $$\frac{1}{2}$$, -2, 0.5, 4$$\frac{1}{2}$$, $$\frac{-33}{7}$$, 0, $$\frac{4}{7}$$, $$0 . \overline{3}$$, 22, -5, $$\frac{2}{19}$$, 0.125 etc.
Would you leave out any of the given numbers from rational numbers? No
Is every natural number, whole number and integer is a rational number? Yes

Question 2.
Fill the blanks in the table.     (Page No. 6)

Question 3.
Complete the following table.     (Page No. 9)

Question 4.
Complete the following table.      (Page No. 13)

Question 5.
Complete the following table.      (Page No. 16)

Question 6.
Complete the following table.      (Page No. 17)

Question 7.
Represent – $$\frac{13}{5}$$ on the number line.     (Page No. 22)
Representing – $$\frac{13}{5}$$ on the number line.

Try These

Question 1.
Hamid says $$\frac{5}{3}$$ is a rational number and 5 is only a natural number. Shikha says both are rational numbers. With whom do you agree?       (Page No. 3)
I would not agree with Hamid’s argument. Since $$\frac{5}{3}$$ is a rational number. But ‘5’ is not only
a natural number, it is also a rational number.
Since every natural number is a rational number,
According to Shikha’s opinion $$\frac{5}{3}$$, 5 are rational numbers.
∴ I agree with Shikha’s opinion.

Question 2.
Give an example to satisfy the following statements.        (Page No.3)
i) All natural numbers are whole numbers but all whole numbers need not be natural numbers.
ii) All whole numbers are integers but all integers are not whole numbers.
iii) All integers are rational numbers but all rational numbers need not be integers.
i) ‘0’ is not a natural number.
∴ Every whole number is not a natural number. (∵ N ⊂ W)
ii) -2, -3, -4 are not whole numbers.
∴ All integers are not whole numbers. (∵ W ⊂ Z)
iii) $$\frac{2}{3}$$, $$\frac{7}{4}$$ are not integers.
∴ Every rational number is not an integer. (∵ Z ⊂ Q)

Question 3.
If we exclude zero from the set of integers is it closed under division? Check the same for natural numbers.    (Page No. 6)
If ‘0’ is subtracted from the set of integers then it becomes Z – {0}.
Closure property under division on integers.
Ex: -4 ÷ 2 = -2 is an integer.
3 ÷ 5 = $$\frac{3}{5}$$ is not an integer.
∴ Set of integers doesn’t satisfy closure property under division.
Closure property under division on natural numbers.
Ex: 2 ÷ 4 = $$\frac{1}{2}$$ is not a natural number.
∴ Set of natural numbers doesn’t satisfy closure property under division.

Question 4.
Find using distributivity.     (Page No. 16)
A) $$\left\{\frac{7}{5} \times\left(\frac{-3}{10}\right)\right\}+\left\{\frac{7}{5} \times\left(\frac{9}{10}\right)\right\}$$
B) $$\left\{\frac{9}{16} \times 3\right\}+\left\{\frac{9}{16} \times-19\right\}$$
Distributive law: a × (b + c) = ab + ac
A)

B)

Question 5.
Write the rational number for the points labelled with letters, on the number line.       (Page No. 22)
i)

ii)

i) A = $$\frac{1}{5}$$, B = $$\frac{4}{5}$$, C = $$\frac{5}{5}$$ = 1, D = $$\frac{7}{5}$$, E = $$\frac{8}{5}$$, F = $$\frac{10}{5}$$ = 2.
ii) S = $$\frac{-6}{4}$$, R = $$\frac{-6}{4}$$, Q = $$\frac{-3}{4}$$, P = $$\frac{-1}{4}$$

Think, discuss and write

Question 1.
If a property holds good with respect to addition for rational numbers, whether it holds good for integers? And for whole numbers? Which one holds good and which doesn’t hold good?     (Page No. 15)
Under addition the properties which are followed by set of rational numbers are also followed by integers.

Question 2.
Write the numbers whose multiplicative inverses are the numbers themselves.      (Page No. 15)
The number T is multiplicative inverse of itself.
∵ 1 × $$\frac{1}{1}$$ = 1 ⇒ 1 × 1 = 1
∴ The multiplicative inverse of 1 is 1.

Question 3.
Can you find the reciprocal of ‘0’ (zero)? Is there any rational number such that when it is multiplied by ‘0’ gives ‘1’?
(Page No. 15)
The reciprocal of ‘0’ is $$\frac{1}{0}$$.
But the value of $$\frac{1}{0}$$ is not defined.
∴ There is no number is found when it is multiplied ‘0’ gives 1.
∵ 0 × (Any number) = 0

∴ No, there is no number is found in place of ‘A’.

Question 4.
Express the following in decimal form.     (Page No. 28)
i) $$\frac{7}{5}$$, $$\frac{3}{4}$$, $$\frac{23}{10}$$, $$\frac{5}{3}$$,$$\frac{17}{6}$$,$$\frac{22}{7}$$
ii) Which of the above are terminating and which are non-terminating decimals?
iii) Write the denominators of above rational numbers as the product of primes.
iv) If the denominators of the above simplest rational numbers has no prime divisors other than 2 and 5 what do you observe?
i) $$\frac{7}{5}$$ = 0.4,
$$\frac{3}{4}$$ = 0.75,
$$\frac{23}{10}$$ = 2.3,
$$\frac{5}{3}$$ = 1.66… = $$1 . \overline{6}$$,
$$\frac{17}{6}$$ = 2.833… = $$2.8 \overline{3}$$,
$$\frac{22}{7}$$ = 3.142
ii) From the above decimals $$\frac{7}{5}$$, $$\frac{3}{4}$$, $$\frac{23}{10}$$ are terminating decimals.
While $$\frac{5}{3}$$,$$\frac{17}{6}$$,$$\frac{22}{7}$$ are non-terminating decimals
iii) By writing the denominators of above decimals as a product of primes is

iv) If the denominators of integers doesn’t have factors other than 2 or 5 and both are called terminating decimals.

Question 5.
Convert the decimals $$0 . \overline{9}$$, $$14 . \overline{5}$$ and $$1.2 \overline{4}$$ to rational form. Can you find any easy method other than formal method?     (Page No. 31)
Let x = $$0 . \overline{9}$$
⇒ x = 0.999 ……. (1)
The periodicity of the above equation is ‘1’. So it is to be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 0.999
10x = 9.999 …….. (2)
From (1) & (2)

∴ x = 1 or $$0 . \overline{9}$$ = 1
Second Method:
$$0 . \overline{9}$$ = 0 + $$0 . \overline{9}$$
= 0 + $$\frac{9}{9}$$
= 0 + 1
= 1

Let x = $$14 . \overline{5}$$
⇒ x = 14.55 …….. (1)
The periodicity of the equation (1) is 1.
So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 14.55
10x = 145.55 …….. (2)

Second Method:

Let x = $$1.2 \overline{4}$$
⇒ x= 1.244 …….. (1)
Here periodicity of equation (1) is 1. So it should be multiplied by 10 on both sides.
⇒ 10 × x = 10 × 1.244
10 x = 12.44 …….. (2)

Second Method:

## AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.4

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.4 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.4

Question 1.
Rice costing ₹480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Solution:
Method – 1: Number of men and rice required to them are in inverse proportion.
Number of men ∝ $$\frac{1}{\text { No. of days }}$$

⇒ Compound ratio of 8:12 and 20: 15
= $$\frac{8}{12}=\frac{20}{15}$$ = $$\frac{8}{9}$$ …………….. (2)
From (1), (2)
480 : x = 8 : 9
⇒ $$\frac{480}{x}=\frac{8}{9}$$
⇒ x = $$\frac{480 \times 9}{8}$$ = ₹540
∴ The cost of required rice is ₹ 540

Method – II :
$$\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}$$
M1 = No. of men
D1 = No .of days
W1 = Cost of rice
∴ M1 = 8
D1 = 20
W1 = ₹ 480
M2 = 12
D2 = 15
W2 = ? (x)

⇒ x = 45 x 12 = ₹ 540
The cost of required rice = ₹ 540/-

Question 2.
10 men can lay a road 75 km. long in 5 days. In how many days can 15 men lay a road 45 km. long?
Solution:
$$\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}$$
∴ M1 = 10
D1 = 5
W1 = 75
M2 = 15
D2 = ?
W2 = 45

∴ x = 2
∴ No. of days are required = 2

Question 3.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days can 20 men working at 9 hours per day do the same work?
Solution:
M1D1H1 = M2D2H2
∴ M1 = 24
D1 = 15 days
H1 = 8 hrs
M2 = 20
D2 = ?
H2 = 9 hrs
⇒ 24 × 15 × 8 = 20 × x × 9

∴ No. of days are required = 16
[ ∵ No. of men and working hours are in inverse]

Question 4.
175 men can dig a canal 3150 m long in 36 days. How many men are required to dig a canal 3900 m. long in 24 days?
Solution:
$$\frac{M_{1} D_{1}}{W_{1}}=\frac{M_{2} D_{2}}{W_{2}}$$
M1 = 175
D1 = 36
W1 = 3150
M2 = ?
D2 = 24
W2 = 3900

∴ No. of workers are required = 325

Question 5.
If 14 typists typing 6 hours a day can take 12 days to complete the manuscript of a book, then how many days will 4 typists, working 7 hours a day, can take to do the same job?
Solution:
M1D1H1 = M2D2H2
M1 = 14
D1 = 12 days
H1 = 6
M2 = 4
D2 = ?
H2 = 7
⇒ 14 × 12 × 6 = 4 × x × 7

⇒ x = 36
∴ No. of days are required = 36
[ ∵ No of men and working hours are in inverse proportion]

## AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.3

Question 1.
Siri has enough money to buy 5 kg of potatoes at the price of ₹ 8 per kg. How much can she buy for the same amount if the price is increased to ₹ 10 per kg?
Solution:
Number of kgs of potatoes to their price are in inverse proportion.
∴ x1y1 = x2 y2
⇒ 8 × 5 = 10 × x
⇒ x = $$\frac{8 \times 5}{10}$$ = 4 kgs
∴ 4 kgs of potatoes will be purchased at the rate of ₹ 10 per kg.

Question 2.
A camp has food stock for 500 people for 70 days. ¡f200 more people join the camp, how long will the stock last?
Solution:
Number of persons and their food stock are in inverse proportion.
⇒ x1y1 = x2 y2 (Let y2 = x say)
⇒ 500 × 70 = (500 + 200) × x
⇒  x = $$\frac{500 \times 70}{700}$$ = 5 × 10
∴ x = 50
∴ The food will be stock for (200 + 500) 700 men = 50 days

Question 3.
36 men can do a piece of work in 12 days. ¡n how many days 9 men can do the same work?
Solution:
Number of workers and number of days are in inverse proportion
∴ x1y1 = x2 y2 let y2 = x (say)
= 36 × 12 = 9 × x
x = $$\frac{36 \times 12}{9}$$ = 48
∴ x = 48 days

Question 4.
A cyclist covers a distance of28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Solution:
Time and distance are in direct proportion.
∴ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$ , Let y2 = x (say)
⇒ $$\frac{28}{2}$$ = $$\frac{56}{x}$$
⇒ x = $$\frac{56}{14}$$
∴ x = 4 hours

Question 5.
A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 metres).
Solution:
Speed and distance are in inverse proportion.
⇒ x1y1 = x2 y2 , Let x2 = x (say)
⇒ 16 × 10 = x × 8
⇒ x = $$\frac{16 \times 10}{8}$$= 20
∴ x = 20
∴ The speed to be increased
= 20 – 16 = 4 nautical miles

Question 6.
5 pumps are required to fill a tank in 1$$\frac { 1 }{ 2 }$$ hours. How many pumps of the same type are used to fill the tank in half an hour.
Solution:
Number of pumps and time to fill the tanks are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ 5 × 1$$\frac { 1 }{ 2 }$$ = x x 1$$\frac { 1 }{ 2 }$$
⇒ 5 × $$\frac { 3 }{ 2 }$$ = x x $$\frac { 1 }{ 2 }$$
⇒ x = 5 × 3 = 15
∴ Number of pumps required = 15

Question 7.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Number of workers and time are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ 15 × 48 = x × 30
⇒ x = $$\frac{15 \times 48}{30}$$ = 24
∴ Number of workers required = 24

Question 8.
A School has 8 periods a day each of45 minutes duration. How long would each period become ,if the school has 6 periods a day? ( assuming the number of school hours to be the same)
Solution:
Time and number of periods are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ 45 × 8 = x × 6
⇒ $$\frac{45 \times 8}{6}$$
⇒ 60 minutes

Question 9.
If z varies directly as xand inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.
Solution:
Given that
z varies directly as x and inversely as y So, z ∝ x (1); z ∝ 1/y ……………… (2)
From (1) & (2), z ∝ $$\frac{\mathrm{x}}{\mathrm{y}}$$

Let x1 = 100x, x2 = 112x
(∵ It increases 12%)
y1 = 100y, y2 = 80y
(∵ It decreases 20%)
From (3),

∴ z is increased in 40%

Question 10.
If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.
Solution:
Number of workers and number of days are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ (x + 1) (x + 1) = (x + 2) x k
⇒ k = $$\frac{(x+1)(x+1)}{(x+2)}$$
∴ k = $$\frac{(x+1)^{2}}{(x+2)}$$

Question 11.
Given a rectangle with a fixed perimeter of 24 meters, if we increase the length by 1 m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.

Solution:

## AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R1:R2 = C2:C1?
(ii) Is R3:R4 = C4:C3?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R1:R2 = C2:C1
⇒ 2 : 3 = 16 : 24

(ii) Is R3:R4 = C4:C3

R1:R2 = C2:C1

(iii) R3:R4 = C4:C3
⇒ 4 : 6 = 8 : 12
$$\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}$$ =
∴ R3:R4 = C4:C3

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

## AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) $$\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}$$
Here x1 = 5, y1 = 210
x2 = 2, y2 = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$ , x1 = 16,800
x2 = 36 y2 = ?

= 3 × 4200
y2 = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$ , x1 = 4
y1 = 2,800
x2 = 3 y2 = ?

= 3 × 700 = 2100
y2 = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$ , x1 = 28
y1 = 12
x2 = ? y2 = 9

x2 = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$

i) x1 = 5.6
y1 = 3.2
x2 = 10.5 y2 = ?

∴The length of the shadow = 6 cm

ii) x1 = 5.6 m x2 = ?
y1 = 3.2 m y2 = 5
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$

∴ x2 = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$
x1 = 14 km , x2 = ?
y1 = 25min = $$\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}$$ = y2 = 5hrs
⇒ $$x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}$$
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 $$\frac { 2 }{ 3 }$$ kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$
x1 = 12 km , x2 = ?
y1 = 40 gm
y2 = 16 $$\frac { 2 }{ 3 }$$ gm = $$\frac { 50 }{ 3 }$$ x 1000 gm
= $$\frac { 50000 }{ 3 }$$ gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= $$75 \times \frac{20}{60}=\frac{75}{3}$$ = 25 km

ii) Time taken to travel 250 km
t = $$\frac{d}{s}=\frac{250}{75}$$
t = $$\frac{10}{3}$$ hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$
x1 = 40 km , x2 = 18
y1 = 1
y2 = ?
⇒ $$\frac{40}{1}=\frac{18}{y_{2}}$$
⇒ $$\frac{18}{40}=\frac{9}{20}$$ cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
$$\frac{35 x+50 y}{x+y}$$ = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ $$\frac{x}{y}=\frac{10}{5}$$ (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1