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1. ## define the domian

I'm having problems definning the value of the function with absoute value on top:

|x^2-1|/x^2-1

for the interval -1<x<1

because the square root of 1 is +1 or -1 , how do you define which value to
use in the above interval?

Because for x>=+1,-1 and x<+1,-1

2. ## Re: define the domian

I'm not sure what you are asking exactly. I'll try to help though.

The domain only depends on the denominator of that fraction, since the absolute value doesn't have any restriction as to what values of x you can put in it. But for the denominator:

$x^2-1\neq 0$
$x\neq \pm1$

So the domain of the function is $D=\left \{ x\in \mathbb{R} ~|~x\neq \pm 1 \right \}$

I'm also not sure what you mean when you say you can't find the value of the function for the interval (-1,1). Just put in a value, say x=0. And you get y=-1. That is the value of the function in the interval.


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3 UFD case; added 2 characters in body

It is also true for local rings. Let $A$ be a matrix with coefficients in $K = \mathrm{Frac}(R)$, such that $A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular, $A (I_n + E_{i,j})A^{-1} \in M_n(R)$ and $A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$ for all $i \neq j$, so that $a_{i,j} b_{k,l} \in R$ for all $i,j,k,l$, where $a$, $b$ denote the coefficients of $A$, $B:=A^{-1}$.

Since $\sum_j a_{1,j}b_{j,1}=1$, there is a $j_0$ such that $a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let $A'=a_{1,j_0}^{-1} A$. Then the coefficients of $A'^{-1} = a_{1,j_0} B$ are in $R$, and those of $A'$ also because $\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$.

For a general $R$, this just shows that $A \in K^{\times} \mathrm{GL}_n (R_m)$ for any maximal ideal $m$ of $R$.

Edit: This implies that the normalizer is indeed $K^{\times} \mathrm{GL}_n(R)$ in the case $R$ is a UFD. Indeed we have that $I^n=(\det (A))$ where $I=\sum_{i,j} a_{i,j} R$ by localization ($i/\det(A) \in \cap_m R_m = R$ for any $i$ product of $n$ coefficients of $A$). Looking at decompositions, we see that $\mathrm{gcd}(a_{i,j})^n = \det(A)$, and so $A/\mathrm{gcd}(a_{i,j}) \in \mathrm{GL}_n(R)$.

2 added 3 characters in body

It is also true for local rings. Let $A$ be a matrix with coefficients in $K = \mathrm{Frac}(R)$, such that $A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular, $A (I_n + E_{i,j})A^{-1} \in M_n(R)$ and $A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$ for all $i \neq j$, so that $a_{i,j} b_{k,l} \in R$ for all $i,j,k,l$, where $a$, $b$ denote the coefficients of $A$, $A^{-1}$.B:=A^{-1}$. Since$\sum_j a_{1,j}b_{j,1}=1$, there is a$j_0$such that$a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let$A'=a_{1,j_0}^{-1} A$. Then the coefficients of$A'^{-1} = a_{1,j_0} B$are in$R$, and those of$A'$also because$\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$. For a general$R$, this just shows that$A \in K^{\times} \mathrm{GL}_n (R_m) $for any maximal ideal$m$of$R$. 1 It is also true for local rings. Let$A$be a matrix with coefficients in$K = \mathrm{Frac}(R)$, such that$A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular,$A (I_n + E_{i,j})A^{-1} \in M_n(R)$and$A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$for all$i \neq j$, so that$a_{i,j} b_{k,l} \in R$for all$i,j,k,l$, where$a$,$b$denote the coefficients of$A$,$A^{-1}$. Since$\sum_j a_{1,j}b_{j,1}=1$, there is a$j_0$such that$a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let$A'=a_{1,j_0}^{-1} A$. Then the coefficients of$A'^{-1} = a_{1,j_0} B$are in$R$, and those of$A'$also because$\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$. For a general$R$, this just shows that$A \in K^{\times} \mathrm{GL}_n (R_m) $for any maximal ideal$m$of$R\$.

format: plain


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# finite automata – Constructing a NFA that accept complement of language L of another NFA

Is there a general way to do it? The answer is yes: one way to do it is to find a DFA that accepts $$L$$ (for example with the powerset construction), make it complete (by adding a sink state), and swap final states and non-final states. The automaton is deterministic, but it is a special case of non deterministic.

Is there a polynomial time way to do it? I don’t know, since the construction above can be exponential time in the number of states (because of the powerset construction).


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Home / Transformer

# Transformer

## Three Phase Transformer Connections

In this section, we’ll review the operating characteristics of basic delta-delta, wye-wye, delta-wye and wye-delta connections of the transformer. The construction of a three-phase transformer can be represented as shown in Figure 1. The shell type core has three sets of primary and secondary windings. How these windings are connected …

## Ideal Transformer | Theory | Equations | Example Problems

In all transformers, single phase or three phase, the primary winding gets energy from the mains and the secondary winding is connected to the load(s). If the secondary circuit is turned off by a switch, there is no load on the secondary winding. Ideal Transformer Theory Nevertheless, the primary winding is …

## Single-Phase Transformer | Construction | Theory

In the same way that single-phase AC and three-phase AC are different, although both are alternating current electricity, single-phase and three-phase transformers cannot be interchanged. Nevertheless, most of what can be said for single-phase transformers is later on extended to three-phase transformers because they work based on the same principle. …

## Power Transformer Differential Protection

A power transformer’s main task is to transform an electrical power from one voltage level to the other voltage level. A power transformer is the most important part of the electrical system as well as well the most expensive part. The function of all other electrical equipment (e.g. circuit breakers, …

## Exciting Current in Transformer | Inrush Current in Transformer

A practical transformer is not ideal and will require current into the primary winding to establish the flux in the core. The current that establishes the flux is called the exciting current. The exciting current magnitude is usually about 1%-5% of the rated current of the primary. By Faraday’s law, …

## Buck-Boost Transformer Working Principle

A buck-boost transformer is a type of transformer which is primarily used to adjust the voltage level applied to various electric equipment. Buck-boost transformers are utilized in in several applications such as uninterruptible power supplies (UPS) units for computers. When an existing AC electrical circuit suffers from excessive voltage drop …

## Three Phase Transformer Construction

A transformer consists of two or more electrical windings that are linked together by a magnetic ﬁeld. Except for special-purpose transformers, the coupling is enhanced with a ferromagnetic core. Transformers are often drawn as shown in Figure 1, with the primary coil on one leg and the secondary coil on …

Autotransformer Definition The autotransformer has a single winding on an iron core. One of the coil terminals is common to both input and output, and the other output terminal is movable so that it can make contact with any turn on the winding. Step-Up/ Step-Down Autotransformer An autotransformer can be …

## Three Phase Transformer Connections Phasor Diagrams

Three Phase Transformer Connections A three-phase transformer is built for a specific connection and voltage transformation and the unit will have a nameplate with the internal connections shown. When a single unit or bank of three is used, there are four types of connections. The four basic connections are: Y-Y, Y-∆, ∆-Y, and ∆-∆. …

## Transformer Nameplate Details Explained

Transformer nameplates contain several standard items of information and other optional information. Transformer nameplate must specify the following parameters: Volt-Ampere (VA) or kilovolt-amperes (kVA) rating The voltage rating of both the primary and secondary circuits The impedance rating of the transformer (normally restricted to 25 kVA or larger) The required …

## Transformer Polarity Test

Transformer windings are marked to identify terminals with the same polarity. Polarity marks may either be dots or + marks. Alternatively, polarity marks may be shown by assigning the same subscripts to corresponding primary and secondary labels, H and X, respectively. Thus, when the primary terminals are named H1 and …

## Transformer Losses and Efficiency Calculation

Transformer Efficiency Definition The efficiency of the transformer is defined as the ratio of the power output Poutput and power input Pinput, both expressed in watts. Thus, $\begin{matrix} \eta =\frac{{{P}_{out}}}{{{P}_{in}}} & {} & \left( 1 \right) \\\end{matrix}$ Or since the input is the output plus losses, \[\begin{matrix}   \begin{align}  & \eta …

## Determination of Transformer Equivalent Circuit Parameters

In order for a model to be useful, there must be a way to determine the values of the model parameters. Two simple tests are used to determine the values for the parameters of the transformer equivalent circuit of Fig. 1 (a) and (b). The two tests are the short-circuit …


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# The Derivative Of Lnx And Examples

Derivative Of Lnx: Part of calculus is mastering the basic derivative rules like the production rule, the power rule, or the chain rule. One of the rules you will see come up often is the rule for the outgrowth of lnx. In the accompanying lesson, we will look at some illustrations of how to apply this rule to finding extraordinary types of by-products. We will also see how using the laws of logarithms can help make taking these kinds of by-products even easier.

The aggressive function has an inverse gathering, which is called the natural logarithm, and is stood for ln(x). Our goal on this page is to verify that the derivative of the natural logarithm is a sensible function. Precisely, we show that the following is true.

ln(x) = 1/x

## What Is The Derivative Of Lnx

Our next task is to resolve what is the derivative of the genuine logarithm.  We begin with the contrary definition.  If

y  =  ln x

then

ey  =  x

Now essentially take the derivative of both sides with relation to x reminding to multiply by dy/dx on the left hand side from the time of it is given in terms of y not x.

ey dy/dx  =  1

From the inverse definition, we can substitute x in for ey to get

x dy/dx  =  1

Finally, divide by x to get

dy/dx  =  1/x

We have proven the following theorem

If f(x)  =  ln x, then

f ‘(x)  =  1/x

## Derivative Of Lnx^2

ln(x) = lim(d->0) [ ln(x+d) – ln(x) ] / d = lim ln((x+d)/x) / d
= lim (1/d) ln(1 + d/x) = lim [ ln (1 + d/x)^(1/d) ].

Set u=d/x and substitute:

lim(u->0) [ ln (1 + u)^(1/(ux)) ] = 1/x ln [ lim(u->0) (1 + u)^(1/u) ]
= 1/x ln (e) (Definition of e)
= 1/x.

## Anti Derivative Of Lnx

Remember that when taking the outgrowth, you can break the by-product up over addition/subduction, and you can take out guideline. This allows us to find the following.

### Example

(3ln(x))=3(1x)=3x(3ln(x))′=3(1x)=3x

### Example

(ln(x)5)=15(ln(x))=(15)(1x)=15x(ln(x)5)′=15(ln(x))′=(15)(1x)=15x

### Example

(2x2ln(x))=4x1x(2×2–ln(x))′=4x–1x

These show you the more straightforward types of derivatives you can find using this rule. But, if we combine this with the laws of logarithms we can do even more.

## Derivative Of 1/Lnx Example

Example

find the derivative of

f(x)  =  ln[(1 + x)(1 + x2)2(1 + x3)]

Solution

The last thing that we want to do is to use the product rule and chain rule multiple times.  Instead, we first simplify with properties of the natural logarithm.  We have

ln[(1 + x)(1 + x2)2(1 + x3)]  =  ln(1 + x) + ln(1 + x2)2 + ln(1 + x3)

=  ln(1 + x) + 2 ln(1 + x2) + 3 ln(1 + x3)

Now the derivative is not so daunting.  We have use the chain rule to get

1                 4x                   9x2
f ‘(x)  =                   +                     +
1 + x            1 + x2             1 + x3

## Derivative Of Lnx

Find the derivative of the function:
y=ln(x2)y=ln(x2)

### Solution

Before applying any calculus rules, first expand the expression using the laws of logarithms. Here, we can use rule (1). This step is all algebra; no calculus is done until after we expand the expression.

y=ln(x2)=2ln(x)y=ln(x2)=2ln(x)

Now, take the derivative. This is the calculus step.

y=(2ln(x))=2(ln(x))=2(1x)=2xy′=(2ln(x))′=2(ln(x))′=2(1x)=2x

In the example above, only one rule was needed to fully expand the expression. The next example shows you how to apply more than one rule.

### Example

Find the derivative of the function.
y=ln(5x4)y=ln(5×4)

Before taking the derivative, we will expand this expression. Since the exponent is only on the x, we will need to first break this up as a product, using rule (2) above. Then, we can apply rule (1).

y=ln(5x4)=ln(5)+ln(x4)=ln(5)+4ln(x)y=ln(5×4)=ln(5)+ln(x4)=ln(5)+4ln(x)

Now take the derivative of the expanded form of the function, and then simplify.

y=(ln(5)+4ln(x))=(ln(5))+4(ln(x))=4(1x)=4xy′=(ln(5)+4ln(x))′=(ln(5))′+4(ln(x))′=4(1x)=4x

You may be wondering what happened to ln(5)ln(5). Remember – this is a constant. If you check your calculator, you will find that ln(5)1.61ln(5)≈1.61. The derivative of a constant is zero.

None of these examples have used rule (3), so let’s look at one more example to see how that might be applied.

### Example

Find the derivative of the function.
y=ln(6x2)y=ln(6×2)

Here we have a fraction, which we can expand with rule (3), and then a power, which we can expand with rule (1). Remember that this is just algebra – no calculus is involved just yet.

y=ln(6x2)=ln(6)ln(x2)=ln(6)2ln(x)y=ln(6×2)=ln(6)–ln(x2)=ln(6)–2ln(x)

Now that we have ln(x)ln(x) by itself, we can apply the derivative rule for the natural log.

y=(ln(6)2ln(x))=(ln(6))2(ln(x))=2(1x)=2x

## What is derivative of natural log?

DERIVATIVE OF THE NATURAL LOGARITHM. y = ln(x) The exponential function has an inverse function, which is called the natural logarithm, and is denoted ln(x).

## What is the derivative of Lnx 2?

Thus, the derivative of ln x2 is 2/x.

## How do you differentiate LN?

Solution
1. We use the formula. ln x. f(x) = ln 4. so that. f ‘(x) = x ln 4.
2. We again use the formula. ln(3x + 4) f(x) = ln 10. now use the chain rule to get. f ‘(x) = (3x + 4) ln 10.
3. Use the product rule to get. f ‘(x) = log(2x) + x(log(2x))’ Now use the formula to get. ln(2x) log (2x) = ln 10. The chain rule gives. 2 1.


text:
Study guide and 3 practice problems on:

## Right hand rule

The right hand rule helps qualitatively determine the direction of a cross product.
If you align the fingers of your right hand with the vector $\bfa$ and you can curl them directly to align with $\bfb$, then your right thumb points in the direction of $\bfa \times \bfb$.


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# Proper Average Treatment Effect estimators (and standard errors) for Generalized Linear Models with log link?

What is the proper formula for estimating the Average Treatment Effect with a simple main effects generalized linear model?

My first pass at defining the Average Treatment Effect for a GLM with a log link:

$$ATE = \biggl[\frac{\mathbb{E}(Y|A=1,\mathbf{X})}{\mathbb{E}(Y|A=0,\mathbf{X})}\biggr]\\ = \biggl[\frac{e^{\beta_0+\beta_A\cdot 1+\mathbf{\beta}\mathbf{X}}}{e^{\beta_0+\beta_A\cdot0+\mathbf{\beta}\mathbf{X}}}\biggr] \\ =e^{\beta_A}$$

where the 95% confidence interval for the ATE estimate would be $$e^{\beta_A}\pm e^{1.96 \times SE(\beta_{A})}$$.

For a project at work I'm doing a difference-in-differences case-control healthcare analysis to estimate the % change in episode-level medical costs from period=0 to period=1 for treatment=1 relative to treatment=0, using pooled cross-sectional observational data. (i.e., not longitudinal records but independent observations in each period $$\times$$ treatment group). Expanding the above ATE formula:

$$ATE = \biggl[\frac{\mathbb{E}(Y|P=1,A=1,\mathbf{X})/\mathbb{E}(Y|P=0,A=1,\mathbf{X})}{\mathbb{E}(Y|P=1,A=0,\mathbf{X})/\mathbb{E}(Y|P=0,A=0,\mathbf{X})}\biggr]\\ = \biggl[\frac{e^{\beta_0+\beta_P\cdot1+\beta_A\cdot1+\beta_{PA}\cdot1+\mathbf{\beta}\mathbf{X}}/e^{\beta_0+\beta_P\cdot0+\beta_A\cdot1+\beta_{PA}\cdot0+\mathbf{\beta}\mathbf{X}}}{e^{\beta_0+\beta_P\cdot1+\beta_A\cdot0+\beta_{PA}\cdot0+\mathbf{\beta}\mathbf{X}}/e^{\beta_0+\beta_P\cdot0+\beta_A\cdot0+\beta_{PA}\cdot0+\mathbf{\beta}\mathbf{X}}}\biggr] \\ =(e^{\beta_P +\beta_{PA}}/e^{\beta_P})\\ =e^{\beta_{PA}}$$

where the 95% confidence interval for the ATE estimate would be $$e^{\beta_{PA}}\pm e^{1.96 \times SE(\beta_{PA})}$$.

If I want to use inverse propensity score weighting, doubly robust methods, etc. for estimating the ATE, observed records would fall in the "treatment" group for the above difference-in-differences study if P=1 and A=1, with the "control" group containing records where P=0 and A=1, P=1 and A=0, or P=0 and A=0.

Is my reasoning on-track here?

I believe there is a parameteric solution to estimating the Average Treatment Effect for a difference-in-differences study using the $$g$$-computation method:

1) Fit a GLM to the data, including terms for the time period and treatment group plus interactions:

$$\mathbb{E}[Y|P,T,\mathbf{X}]=e^{\beta_0+\beta_P\cdot P+\beta_T\cdot T+\beta_{PT}\cdot P\cdot T+. . .}$$

2) For each record in the data compute the three counterfactual expected values of $$Y$$ (in addition to the existing expected value in the original data) by substituting alternative values for the the time period and treatment variables into the regression formula: $$\mathbb{E}[Y_i|P_i=0, T_i=0, \mathbf{X}_i]$$$$\mathbb{E}[Y_i|P_i=0, T_i=1,\mathbf{X}_i]$$$$\mathbb{E}[Y_i|P_i=1, T_i=0,\mathbf{X}_i]$$$$\mathbb{E}[Y_i|P_i=1, T_i=1,\mathbf{X}_i]$$

3) Compute the Average Treatment Effect by taking the geometric mean of the individual treatment effects:

$$ATE = \biggl[\prod_{i=1}^n\frac{\mathbb{E}(Y_i|P_i=1,A_i=1,\mathbf{X}_i)/\mathbb{E}(Y_i|P_i=0,A_i=1,\mathbf{X}_i)}{\mathbb{E}(Y_i|P_i=1,A_i=0,\mathbf{X}_i)/\mathbb{E}(Y_i|P_i=0,A_i=0,\mathbf{X}_i)}\biggr]^{\frac{1}{n}}$$

The 95% confidence interval for the $$ATE$$ estimate can be estimated with bootstrapping, computer resources permitting.


text:
# Peter Stone's Selected Publications

Classified by TopicClassified by Publication TypeSorted by DateSorted by First Author Last NameClassified by Funding Source

## A Multi-Robot System for Continuous Area Sweeping Tasks

Mazda Ahmadi and Peter Stone. A Multi-Robot System for Continuous Area Sweeping Tasks. In Proceedings of the IEEE International Conference on Robotics and Automation, pp. 1724–1729, May 2006.
Some videos of the robot referenced in the paper.
ICRA 2006

### Abstract

As mobile robots become increasingly autonomous over extended periods of time, opportunities arise for their use on repetitive tasks. We define and implement behaviors for a class of such tasks that we call continuous area sweeping tasks. A continuous area sweeping task is one in which a group of robots must repeatedly visit all points in a fixed area, possibly with non-uniform frequency, as specified by a task-dependent cost function. Examples of problems that need continuous area sweeping are trash removal in a large building and routine surveillance. In our previous work we have introduced a single-robot approach to this problem. In this paper, we extend that approach to multi-robot scenarios. The focus of this paper is adaptive and decentralized task assignment in continuous area sweeping problems, with the aim of ensuring stability in environments with dynamic factors, such as robot malfunctions or the addition of new robots to the team. Our proposed negotiation-based approach is fully implemented and tested both in simulation and on physical robots.

### BibTeX Entry

@InProceedings(ICRA06,
title="A Multi-Robot System for Continuous Area Sweeping Tasks",
booktitle="Proceedings of the {IEEE} International Conference on Robotics and Automation",
pages="1724--1729",
month="May",year="2006",
abstract={As mobile robots become increasingly autonomous over
extended periods of time, opportunities arise for
their use on repetitive tasks. We define and
implement behaviors for a class of such tasks that
we call \emph{continuous area sweeping} tasks. A
continuous area sweeping task is one in which a
group of robots must repeatedly visit all points in
a fixed area, possibly with non-uniform frequency,
as specified by a task-dependent cost
function. Examples of problems that need continuous
area sweeping are trash removal in a large building
and routine surveillance.  In our previous work we
have introduced a single-robot approach to this
problem. In this paper, we extend that approach to
multi-robot scenarios. The focus of this paper is


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06 Jun 2019, 17:57
MathRevolution wrote:
[GMAT math practice question]

(number properties) If two integers have no common factors other than $$1$$, they are called relatively prime. Are $$x$$ and $$z$$ relatively prime?

1) $$x$$ and $$y$$ are relatively prime.

2) $$y$$ and $$z$$ are relatively prime.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have $$3$$ variables ($$x, y$$ and $$z$$) and $$0$$ equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
If $$x = 2, y = 3, z = 5$$, then $$x$$ and $$z$$ are relatively prime, and the answer is ‘yes’.

If $$x = 2, y = 3, z = 2,$$ then $$x$$ and $$z$$ are not relatively prime, and the answer is ‘no’.

Both conditions together are not sufficient, since they don’t yield a unique answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
_________________
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 06 Jun 2019, 17:58 [GMAT math practice question] (number properties) We define ‘a mod n’ to be the remainder when a is divided by n. What is the value of ‘a mod 12’? 1) a mod 3=1 2) a mod 4=a mod 2 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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07 Jun 2019, 02:03
MathRevolution wrote:
[GMAT math practice question]

(absolute value) Is $$x<y<z$$?

$$1) |x+2|<y<z+2$$

$$2) |x-2|<y<z-2$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have $$3$$ variables ($$x, y$$ and $$z$$) and $$0$$ equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Now, by condition 1),$$x < x + 2 ≤ | x + 2 | < y < z + 2,$$ and $$x < y$$.
By condition 2), $$|x-2|<y<z-2 < z,$$ and so $$y < z$$.
Thus, both conditions together are sufficient since they yield $$x < y < z$$, and the unique answer is ‘yes’.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If $$x = 1, y = 4$$, and $$z = 7$$, then the answer is ‘yes’.
If $$x = 1, y = 4,$$ and $$z = 3,$$ then the answer is ‘no’.
Condition 1) is not sufficient since it doesn’t yield a unique answer.

Condition 2)
If $$x = 1, y = 4,$$ and $$z = 7$$, then the answer is ‘yes’.
If $$x = 5, y = 4,$$ and$$z = 7,$$ then the answer is ‘no’.
Condition 2) is not sufficient since it doesn’t yield a unique answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 07 Jun 2019, 02:03 [GMAT math practice question] (function) If the symbol $$#$$ represents one of addition, subtraction, multiplication, or division, what is the value of $$(2 # 1)$$? 1) 2 # 2 = 1 2) (–1/2) # (-1) = $$\frac{1}{2}$$ _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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09 Jun 2019, 18:20
MathRevolution wrote:
[GMAT math practice question]

(number properties) We define ‘a mod n’ to be the remainder when a is divided by n. What is the value of ‘a mod 12’?

1) a mod 3=1
2) a mod 4=a mod 2

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have $$1$$ variable ($$n$$) and $$0$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
The numbers satisfying condition 1) are $$1, 4, 7, 10, 13, 16, 19, 22, …$$ .

Their remainders, when they are divided by $$12$$, are $$1, 4, 7$$ and $$10.$$

Condition 1) is not sufficient since it doesn’t yield a unique answer.

Condition 2)
The numbers satisfying condition 2) are $$0, 1, 4, 5, 8, 9, 12, 13, 16, 17, …$$ .

Their remainders, when they are divided by $$12$$, are $$0, 1, 4, 5, 8$$ and $$9$$.

Condition 2) is not sufficient, since it doesn’t yield a unique answer.

Conditions 1) & 2)
The numbers satisfying conditions 1) & 2) are $$1, 4, 13, 16, … .$$

Their remainders, when they are divided by $$12$$, are $$0$$ and $$1.$$

Both conditions together are not sufficient, since they don’t yield a unique answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 09 Jun 2019, 18:22 MathRevolution wrote: [GMAT math practice question] (function) If the symbol $$#$$ represents one of addition, subtraction, multiplication, or division, what is the value of $$(2 # 1)$$? 1) 2 # 2 = 1 2) (–1/2) # (-1) = $$\frac{1}{2}$$ => Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Since we have $$1$$ variable ($$#$$) and $$0$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first. Condition 1) $$#$$ is the division operation since $$\frac{2}{2} = 1.$$ So, $$2#1 = \frac{2}{1} = 2.$$ Condition 1) is sufficient since it yields a unique answer. Condition 2) $$#$$ could be the multiplication, subtraction or the division operation since $$(-\frac{1}{2}) * (-1) = \frac{1}{2}, (-\frac{1}{2})-(-1) = (\frac{1}{2})$$ and (-$$\frac{1}{2})/(-1) = \frac{1}{2}.$$ Since $$2 * 1 = 2$$ and $$\frac{2}{1} = 2$$, but $$2-1 = 1$$, condition 2) does not yield a unique value for $$2#1$$. Condition 2) is not sufficient since it doesn’t yield a unique answer. Therefore, A is the answer. Answer: A If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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10 Jun 2019, 00:16
[GMAT math practice question]

(inequality) $$0 < a < b < c$$. Is $$a < 3$$?

$$1) \frac{1}{c} > \frac{1}{3}$$

$$2) \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$
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12 Jun 2019, 01:26
MathRevolution wrote:
[GMAT math practice question]

(inequality) $$0 < a < b < c$$. Is $$a < 3$$?

$$1) \frac{1}{c} > \frac{1}{3}$$

$$2) \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

We have $$\frac{1}{a} > \frac{1}{b} > \frac{1}{c}$$, since we are given that $$0 < a < b < c.$$

Condition 1) implies that $$c < 3$$. Therefore $$a < 3$$, and Condition 1) is sufficient.

Condition 2)
Since $$\frac{1}{a} > \frac{1}{b} > \frac{1}{c} > 0$$, and $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$, we must have $$\frac{1}{a} + \frac{1}{a} + \frac{1}{a} = \frac{3}{a} > 1$$. Therefore, $$a < 3$$, and the answer is ‘yes’.
Condition 2) is sufficient since it yields a unique answer.

This question is a CMT4(B) question: condition 1) is easy to work with and condition 2) is difficult to work with. For CMT4(B) questions, D is most likely to be the answer.
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12 Jun 2019, 19:23
MathRevolution wrote:
[GMAT math practice question]

(inequality) Is $$x^3-x^2+x-1 > 0?$$

$$1) x^5 > x^2$$

$$2) x^3 + x > x^2 + 1$$

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient.

$$x^3-x^2+x-1 > 0$$

$$=> x^2(x-1) + (x-1) > 0$$

$$=> (x^2+1)(x-1) > 0$$

$$=> (x-1) > 0$$, since $$x^2+1 > 0$$

$$=> x>1$$

The question asks if $$x > 1.$$

Since we have $$1$$ variable ($$x$$) and $$0$$ equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
$$x^5 > x^2$$

$$=> x^5 - x^2 > 0$$

$$=> x^2(x^3 – 1) > 0$$

$$=> x^2(x-1)(x^2+x+1) > 0$$

$$=> x – 1 > 0$$ since $$x^2+x+1 >0,$$ and $$x^2 > 0$$ if $$x ≠ 0$$
$$x > 1$$
This condition is equivalent to the question. Therefore, condition 1) is sufficient.

Condition 2)
$$x^3 + x > x^2 + 1$$

$$=> x^3 - x^2 + x – 1 > 0$$

$$=> x^2(x-1)+ (x-1) > 0$$

$$=> (x^2+1)(x-1) > 0$$

$$=> (x-1) > 0$$, since $$x^2+1 > 0$$

$$=> x > 1$$

This condition is equivalent to the question. Therefore, condition 2) is sufficient.

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 12 Jun 2019, 22:32 [GMAT math practice question] (number properties) If p and q are prime numbers, what is the number of the different factors of p^2q^3? 1) pq=143 2) p and q are different _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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13 Jun 2019, 23:53
MathRevolution wrote:
[GMAT math practice question]

(number properties) If $$x$$ and $$y$$ are positive integers, are $$x$$ and $$y$$ consecutive?

$$1) x+y=3$$

$$2) xy=2$$

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since x and y are positive integers, condition 1) tells us that either x = 1 and y = 2, or x = 2 and y= 1. Thus, x and y are consecutive integers.

Since x and y are positive integers, condition 2 tells us that either x = 1 and y = 2, or x = 2 and y= 1. Thus, x and y are consecutive integers.

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 14 Jun 2019, 00:05 [GMAT math practice question] (function) At how many points does the parabola y=f(x)=ax^2+bx+c (a≠0) intersect the x-axis? 1) x=f(y) intersects the y-axis at two points. 2) y=f(x-1) intersects the x-axis at two points. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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16 Jun 2019, 22:55
MathRevolution wrote:
[GMAT math practice question]

(number properties) If p and q are prime numbers, what is the number of the different factors of p^2q^3?

1) pq=143
2) p and q are different

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since p and q are prime numbers, condition 1) tells us that p = 11 and q = 13, or p = 13 and q = 11. Therefore, since p and q are different prime numbers, the number of different factors of p^2q^3 is (2+1)(3+1) = 12. Condition 1) is sufficient since it yields a unique solution.

Condition 2)
Since condition 2) tells us that p and q are different prime numbers, the number of factors of p^2q^3 is (2+1)(3+1) = 12.
Condition 2) is sufficient since it yields a unique solution.

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 16 Jun 2019, 23:04 MathRevolution wrote: [GMAT math practice question] (function) At how many points does the parabola y=f(x)=ax^2+bx+c (a≠0) intersect the x-axis? 1) x=f(y) intersects the y-axis at two points. 2) y=f(x-1) intersects the x-axis at two points. Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. Condition 1) Condition 1) is equivalent to the statement that y=f(x) intersects the x-axis at two points, simply by switching the roles of x and y. This means that y=f(x) intersects the x-axis at two points. Condition 1) is sufficient since it yields a unique solution. Condition 2) The graph of y = f(x-1) is the same as the graph of f(x), moved to the right by one unit. This horizontal movement does not change the number of roots of y. Therefore, y = f(x) also has two roots. Condition 2) is sufficient since it yields a unique solution. Therefore, D is the answer. Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2). _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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16 Jun 2019, 23:13
[GMAT math practice question]

(inequality) Is x<0?

1) x^3+1<0
2) x^3+2x^2+x+2=0
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 13 Apr 2018 Posts: 23 Location: Zambia Concentration: Finance, Real Estate GPA: 2.98 WE: Analyst (Investment Banking) Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 17 Jun 2019, 11:10 MathRevolution wrote: Ask me Anything about GMAT Data Sufficiency Hello, I am Max Lee, Founder and Lead Math Tutor at Math Revolution. I have over 6,000 posts and almost 5,000 Kudos. This topic is a new feature on the DS Forum and a way for you to directly interact with me and ask anything about the DS, e.g. if you want a certain concept explained or have a particular you question you want me addressed, this is the place to post a link to it or your question. I intend to have this thread be as a "Everything You Need to Know about DS" type of thread. I will keep updating this post with links and resources that are helpful for the DS. Meanwhile, you can ask me anything My other discussions you may be interested in: Thank you all - good luck on the GMAT and look forward to seeing you in the DS forum! - Max Lee. Hi, Do you have a matrix with variables and equations? i.e. if there are 2 variables and 2 equations you can solve the equation etc Best Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7612 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS [#permalink] Show Tags 18 Jun 2019, 02:43 [GMAT math practice question] (absolute value) If |2x|>|3y|, is x >y? 1) x>0 2) y>0 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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19 Jun 2019, 00:18
MathRevolution wrote:
[GMAT math practice question]

(inequality) Is x<0?

1) x^3+1<0
2) x^3+2x^2+x+2=0

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
x^3+1<0
⇔ (x+1)(x^2-x+1) < 0
⇔ x+1<0 since x^2-x+1 > 0
⇔ x < -1
In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
Thus, since the solution set of the question, ‘x<0’, includes that of condition 1), ‘x<-1’, condition 1) is sufficient.

Condition 2)
x^3+2x^2+x+2=0
⇔ x^2(x+2)+(x+2)=0
⇔ (x^2+1)(x+2)=0
⇔ x+2=0 since x^2+1 > 0
⇔ x = -2
Thus, x < 0 and condition 2) is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
_________________
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Re: Math Revolution DS Expert - Ask Me Anything about GMAT DS   [#permalink] 19 Jun 2019, 00:18

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text:
# Set up sidewaysfigure to have the bottom of the figure toward the spine

I am typesetting a book. There are a couple figures that are wider than they are tall, and appear too small if placed on a page in the usual fashion. I have found that using the sidewaysfigure environment from the rotating package, I can put these figures in landscape orientation taking up entire pages. This is satisfactory except for one issue: in the compiled pdf, the figures are oriented with the bottom away from the spine; I want the bottom toward the spine. If I use the package option figuresright or figuresleft, I can make the figure orientation independent of the page number, but this is not what I want. The clockwise and counterclockwise options do not appear to change anything (which makes sense, since the documentation states that these options are for one-sided documents, but I had hoped).

In using the sidewaysfigure environment from the rotating package in a two-sided document, how do I orient the figures with the bottom toward the spine?

-

Strange that there seems to be no such package option ... so I took the definition of \end@rotfloat and partially swapped the code for left and right hand pages. (And \let \endsidewaysfigure and \endsidewaystable to the new definition of \end@rotfloat.)

\documentclass{book}

\usepackage{rotating}

\makeatletter
\def\end@rotfloat{%
\end{minipage}\end{lrbox}%
\stepcounter{r@tfl@t}%
\rot@label{RF\ther@tfl@t}%
\rot@pageref{RF\ther@tfl@t}{\R@@page}%
\edef\@tempa{Adding sideways \@captype\space on page \R@@page\space}
\rot@mess@toks\expandafter{\@tempa}
\wd\rot@float@box\z@
\ht\rot@float@box\z@
\dp\rot@float@box\z@
\vbox to \textheight{%
\setkeys{Grot}{units=360}%
\if@rot@twoside
\else
\let\R@@page\rot@LR
\fi
\ifthenelse{\isodd{\R@@page}}{%
\if@rot@twoside
\rot@mess@toks\expandafter{\the\rot@mess@toks (right hand page)}%
\fi
% BEGIN OF ORIGINAL LEFT HAND PAGE
\@@line{%
\hskip\rotFPbot
\rotatebox{-90}{\box\rot@float@box}%
\hskip\rotFPtop
}%
\vfill
% END OF ORIGINAL LEFT HAND PAGE
}{%
\if@rot@twoside
\rot@mess@toks\expandafter{\the\rot@mess@toks (left hand page)}%
\fi
% BEGIN OF ORIGINAL RIGHT HAND PAGE
\vfill
\@@line{%
\hskip\rotFPtop
\rotatebox{90}{\box\rot@float@box}%
\hskip\rotFPbot
}%
% END OF ORIGINAL RIGHT HAND PAGE
}%
\rot@message{\the\rot@mess@toks}
}%
\end@float
}
\let\endsidewaysfigure\end@rotfloat
\let\endsidewaystable\end@rotfloat
\makeatother

\begin{document}

\begin{sidewaysfigure}
\centering
\rule{15cm}{1cm}
\caption{First figure}
\end{sidewaysfigure}

\begin{sidewaysfigure}
\centering
\rule{15cm}{1cm}
\caption{Second figure}
\end{sidewaysfigure}

\end{document}

-
There is not a package option because rotated tables should have the top towards the spine. –  egreg Jul 20 '11 at 21:01
@egreg: Yet there are the package options figuresleft and figuresright. –  lockstep Jul 20 '11 at 21:04
What happens in "one-sided" documents has no bearing with typography and this explains the counterclockwise option. :) I suppose that figuresleft and figuresright are thought for documents that have many rotated figures or tables, in particular if they are on a spread. Anyway, nice answer! –  egreg Jul 20 '11 at 21:15
Thanks!________ –  Charles Staats Jul 21 '11 at 1:18
One reference is my A Few Notes on Book Design : try texdoc memdesign to view it. The relevant section is 4.2.2 Mixed portrait and landscape pages. In double sided documents on verso (even) pages the top should be towards the fore edge and on recto (odd) ages the top should be towards the spine. The basic idea is that the reader should always have to turn the book in a consistent direction to read any landscape pages --- to the right is the commonly accepted direction. –  Peter Wilson Jul 26 '12 at 18:38


text:
Delbert’s Arena

We are privileged here at RigWorld to have Delbert Hall amongst our membership. I’ve created this page for Delbert as he has graciously agreed to share some of his knowledge here with us.

Delbert has recently published another new book with Brian Sickels

With nearly 400 pages of text and over 550 illustrations, The Theatre Riggers’ Handbook is the most comprehensive book on theatre rigging available. Written by experienced riggers Hall and Sickels, authors of The Arena Riggers’ Handbook, this book clearly describes all aspects of theatre rigging, including hardware, rigging math and techniques, installations, fire curtains, concert shells, hemp and manual counterweight rigging, automated systems, aerial rigging, rigging safety and inspections, and more. Whether you are a beginning rigger or a seasoned professional, The Theatre Riggers’ Handbook is a “must have” for your professional library.

Delbert Hall, Ph.D., is head of the design and technical theatre program in the Division of Theatre and Dance at East Tennessee State University, where he has taught since 1986. Dr. Hall received his Ph.D. from the University of Florida, his MFA from the University of North Carolina at Greensboro, and his B.S.Ed. from Western Carolina University. He is the U.S. Institute for Theatre Technology -Southeast Region’s Outstanding Educator in Theatrical Design and Technology for 2000-2001, and received an Outstanding Achievement Award for special effects rigging from the Kennedy Center for the Arts in 2000. Delbert is the president of D2 Flying Effects, LLC, and has been rigging performer flying effects and aerial dance apparatus for over 30 years. He is an ETCP certified rigger (Theatre) and an ETCP Recognized Trainer. Delbert regularly teaches workshops on rigging and rigging math and is the originator of RigCalc. Delbert also does destructive testing of hardware used in the entertainment industry and fabrics used in aerial dance in order to better understand how these items break and at what loads.

This month we are lucky enough to have a preview lesson from the second edition of Rigging Math Made Simple. I had the pleasure of being one of the Technical editors of this book and it is a valuable tool to have in your library, it will be available in mid January and can be purchased through SpringKnollPress.com or Amazon.com. If you have any questions, please post in the comments section at the bottom of the page and Delbert will be happy to respond.

Delbert has released several great Rigging Tools with instructions in the new edition of the book. Make sure to download Delbert’s Excel workbooks here: Rigging Math Downloads

Lesson 19:

Two-Point Bridles in an Arena

In Lesson 5 we discussed how to calculate the length of a bridle based on knowing the Horizontal and Vertical distance the Bridle Point would be from the Hanging points. In other words, the head rigger needs to determine the exact position of the Bridle Point and then calculate the lengths of the bridle legs that would create this position. While this will work, it is not always the way it is done.

For example, the head rigger will need the Bridle Point at a specific location in plan view, but instead of needing to put the Bridle Point at an exact height, the head rigger may want it at a minimum height above the deck (high enough to get the truss at the need height). There is also a maximum height above the deck for the Bridle Point, which is determined by the point where the bridle angle reaches 120 degrees (90 degrees is preferred). As long as the Bridle Point is between the minimum and maximum heights, all is well because the chain hoist will control the height of the truss. To understand why the head rigger does not just choose an arbitrary point between the minimum and maximum heights, we need to understand the resources for making bridle leg.

As discussed in Lesson 18, common lengths of steel slings that rigging companies have include: 2’, 2.5’, 5’, 10’, 20,’ 30’, and 50’ lengths. Baskets and bridle legs will be made from a combination of these standard length slings. When a bridle leg is an “odd length,” one that cannot just be made from the standard lengths listed above, links of Deck Chain are added to make the desired length bridle leg.

Note: Actual deck chain is very seldom used anymore – it has been replaced by Special Theatrical Alloy Chain (sometimes called Special Theatrical Adjusting Chain) or STAC chain. Many, if not most, riggers use the term “Deck Chain” when referring to “STAC chain.” Throughout the rest of this book, I will use the term “Deck Chain” when referring to “STAC chain.”

STAC chain, which can be purchased in 3-foot and 5-foot lengths, is VERY expensive, especially when compared to the price of cable slings, so rigging companies typically carry only a minimum amount of this chain for rigging. The fewer components in the rig, the fewer chances for mistakes. Because of this, it is usually necessary that at least one bridle leg be able to be made by using ONLY the standard length steel slings. Not using any Deck Chain in a bridle is even better.

In the drawing below, we have a bridle where the Bridle Point (also called the “apex”) needs to be 10 feet from one I-beam and 15 feet from the second I-beam. This establishes a vertical line on which the Bridle Point must be placed, but where on that line? We see that the I-beams are 80 feet above the deck, and have determined that the minimum height of the bridle point needs to be 40 feet above the deck, and the maximum height of the bridle point is 65 feet above the deck. So, the Bridle Point can be anywhere along the 25-foot long space between these points.

As you see, I have indicated that the first bridle leg is 25 feet long. You might be asking how did I come to choose this length? First, it is a bridle length that I could make with standard length slings (a 20 footer and a 5 footer); and second, it is a distance that I knew from experience would reach from the I-beam to “somewhere” on the line, between the minimum and maximum height points. Of course, if you are not sure if a certain length will work, make a quick, scaled drawing and measure. I will give you more hints on estimating this length shortly, but first let’s use this length.

It is important to note now that this leg will be longer than 25 feet. Why? Because I have not added-in the Effective Length of Hitch (ELOH) for the basket, and because I have not added-in the additional length of the two shackles in the leg (remember, each piece of steel will require a shackle). Assuming we use a basket make with a 5-foot sling that is going around a 12” x 9” I-Beam (the example in Lesson 18), we get:

Leg 1 = 25 + 1.6093 + 0.3966

Leg 1 = 27.01

Remember, this bridle leg, as calculated, is made without any Deck Chain.

Not only do we know the length of the leg, but also we now have two of the three lengths that make up the bridle triangle. The bridle leg (the hypotenuse) is 27.01 feet long and one leg (the Horizontal distance) is 10 feet long. With this information we can calculate the third leg of the triangle (the Vertical distance). Why this distance is important will become clear soon.

So, using the following variation of the Pythagorean Theorem to calculate this distance using the equation:

V =$\sqrt(Leg x Leg) – (H x H)$

So, plugging in the numbers, we get…

V =$\sqrt(27.1 x 27.1) – (10 x 10)$

V =$\sqrt729.5401 – 100$

V =$\sqrt629.5401$

V = 25.09 feet

Before we go on the next step, as promised, here are a couple of suggestions that might help you in estimating the length of the first bridle leg, if you choose not to make a drawing and measure.

Option 1: After you estimate the bridle length and calculate the height of the Bridle Point from the beams, subtract that number from the height of the beam and compare it to the minimum and maximum Bridle Point heights. Using the problem above:

Bridle Point Height = 80 – 22.9

Bridle Point Height = 57.1 feet

Since this measure is greater that the minimum Bridle Point height (40 feet) and less than the maximum Bridle Point Height (65 feet), it is “Good.”

Option 2: Take the beam height and subtract the minimum Bridle Point height (80 – 40 = 40 feet).  This is going to the somewhere near the maximum length of the bridle. We also know that the bridle length must be greater than the distance from the beam to the maximum Bridle Point height (80-65 = 15 feet). A good compromise length is often somewhere in between, so 40 – 15 = 25 feet.

As we now are beginning to see, several standard bridle lengths would have put the Bridle Point between the minimum and maximum Bridle Point heights.

Option 3: We can see the span of lengths that would work by calculate the bridle lengths if the Bridle Point was at both the minimum and maximum Bridle points heights that falls between these numbers.

Leg Length (Min) =$\sqrt(15 x 15) + (10 x 10)$

Leg Length (Min) =$\sqrt(225) + (100)$

Leg Length (Min) =$\sqrt325$

Leg Length (Min) = 18 feet

Leg Length (Max) =$\sqrt(40 x 40) + (10 x 10)$

Leg Length (Max) =$\sqrt(1600) + (100)$

Leg Length (Max) =$\sqrt1700$

Leg Length (Max) = 41.23 feet

So, the following bridle lengths would all have worked: 20’, 25’, 30’, 35’, and 40 feet. Whichever one you choose, you will need to calculate the distance from the beam to the bridle point.

So, why is the 25.09 foot distance so important? It is because we need to know this distance in order to find the length of the other bridle leg. We now know the Horizontal distance (15 feet) and the Vertical distance (25.09 feet), so we can plug them into the Pythagorean Theorem and find the length of the leg. (This assumes both beams are at the same height above the deck).

Leg =$\sqrt(15 x 15) + (25.09 x25.09)$

Leg =$\sqrt(225) + (629.54)$

Leg =$\sqrt854.54$

Leg = 29.23 feet

Finally, we need to figure out how to make a bridle leg that is 29.23 feet long.

Figuring out how to create this bridle will take a bit of trial and error. However, before you begin you should know that there is no one correct answer, and how you do it will be based on what lengths of Steel and Deck Chain are available to you.  But, let’s look at you might approach this problem.

We can get most of the way to 29.23 feet by using one 20-foot and one 5-foot pieces of Steel (and the lengths of two shackles). So,

L2 = 20 + 5 + 0.1983 + 0.1983

L2 = 25.396 feet

We still need 3.8334 feet of leg.

We know we need a basket, so let’s add 1.6093 feet to our leg for a 5’ basket.

L2= 25.3966 + 1.6093

L2 = 27.0059 feet

That still leaves us 2.2241 feet short. The big question now is “how do we create this length?”

The answer depends on what we have to use. Deck (STAC) chain is a common material, but if we have a 2-foot long piece of Steel, that could be used as well. No matter what we use, we need to remember that each new piece of hardware will require an additional shackle (and we can also use shackles to add length). Let’s look at two ways two ways to create that additional length of the leg.

Since we know that we will need a shackle, subtract the length of the shackle (0.1983 feet) from desired length (2.2241 feet). We get 2.0258 feet. If you added a 2-foot Steel here, the remainder would be a mere 0.0258 feet, or 0.3096 inches.  That is REALLY close.

If we do not have a 2-foot Steel, we will need to make this length entirely from Deck (STAC) chain. Again, we subtract the length of the shackle (0.1983 feet) from desired length (2.2241 feet to get 2.0258 feet. Now, to determine the number of needed links of chain, we divide this length by the internal measurement of one link of Deck Chain. One common size of Deck Chain had 3.174 inches (or 0.2645 feet) links, so we will use this measurement.

Since you cannot add a fraction of a link, you have to choose either 7 or 8 links. Seven links would be a little short of the desired length and eight would be a little longer than the desired. Of these two, eight is the closer, but it is still more than an inch off.

Another option is to add seven links plus an additional shackle. This would create 2.0498 feet of addition length. This is only 0.024 feet (0.288 inches) longer than the desired addition length. Again, REALLY close to our desired length.

If a finer adjustment was needed, you could have added the shackle to the basket. This would have added only half of the length of the shackle to the length of the leg.

As stated earlier, there is more than one possible solution to this problem. Had we selected to use a longer basket, the ELOH would have been greater, which would have resulted in different lengths of Steel and Deck Chain needed. It would be extremely time consuming to manually calculate every single configuration for making up a bridle legs. In Lesson 20, we will look at tools that can help you look at lots of options very quickly.

4 thoughts on “Delbert’s Arena”

1. i like the calculation you guys did, but will like you guys to show more on how to calculate the size of shackle to use for each basket.

2. I’ll forward this to Delbert for his opinion.
For the most part, in arena rigging, you will be using 5/8″ shackles with 3/8″ wire rope for all hoists that are 1 ton or less and 3/4″ shackles with 1/2″ wire rope for two ton hoists.

• The WLL is not the issue when choosing a shackle for arena rigging – shackles are extremely strong (much stronger than the wire rope slings we use). The real consideration is the size of the eye (thimble) on the wire rope sling. 5/8″ shackles work best on 3/8″ wire rope slings and 3/4″ shackles work best with 1/2″ wire rope slings. Now, the wire rope diameter of the sling is based on the max load you should be lifting (basically, the capacity of the chain hoist). Almost universally, folks use 3/8″ wire rope slings with 1-Ton hoist (or smaller), and 1/2″ wire rope slings with 2-Ton hoist. I hope that helps.

3. Nice post. I was checking constantly this blog and I am impressed!
Extremely helpful information particularly the last part I
care for such info much. I was seeking this certain info for a very long time.
Thank you and best of luck.


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# If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,

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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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27 Nov 2013, 23:26
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If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

[Reveal] Spoiler:
my approach towards statement 2:
x=11 and u=6,
thus 11=6^2-v^2
25=v^2
v=5 or v=-5
Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55
Please correct me if something is wrong!/spoiler]
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Nov 2013, 05:40, edited 2 times in total.
Renamed the topic and edited the question.

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Re: If x=u^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink]

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28 Nov 2013, 01:33
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Since y=60 in st (1) then uv=30, and the values of u*v is 6,5 or -6,-5 if x=11.I.E 6^2- 5^2=11 and (-6)^2-(-5)^2=11. Therfore the value of z=36+25=61

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Re: If x=u^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink]

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28 Nov 2013, 02:06
vogelleblanc wrote:
If x=u^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is the value of z?

(1) y=60
(2) u=6

my approach towards statement 2:
[Reveal] Spoiler:
x=11 and u=6,
thus 11=6^2-v^2
25=v^2
v=5 or v=-5
Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55
Please correct me if something is wrong!

Answer has to be "B" ... its not explicitly mentioned that U and V are integer so. A is not option....

B because X and U are given now... so V can be calculated and so the Z

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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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28 Nov 2013, 06:46
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@sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

@sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question

For (1) we have:
$$11 = u^2 - v^2$$;

$$60= 2uv$$;

and we need to find the value of $$u^2 + v^2$$.

If you solve $$11 = u^2 - v^2$$ and $$60= 2uv$$, you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case $$u^2 + v^2=36+25$$.

Thus the answer is D.

Hope it's clear.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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01 Dec 2013, 02:16
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... ah. So, the fact that x=11, limits the possible values of u and v to (6,5) or (-6,-5), right?

bunuel, thanks a lot so far!

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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

Another approach, for F.S 1, in which you don't have to calculate for anything

We know that $$x = u^2 - v^2$$ = 11,Squaring on both sides, we have $$u^4 + v^4 - 2*v^2*u^2$$ = 121

Thus, adding$$4*v^2*u^2$$ on both sides, we have $$(u^2 + v^2)^2 = 121+4*v^2*u^2 = z^2$$

Hence, this statement is sufficient to calculate the value of z.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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03 Dec 2013, 01:26
vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

[Reveal] Spoiler:
my approach towards statement 2:
x=11 and u=6,
thus 11=6^2-v^2
25=v^2
v=5 or v=-5
Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55
Please correct me if something is wrong!/spoiler]

We have that x= u^2-v^2 and x =11
z= u^2+ v^2
and y = 2 uv

from St 1 we have 2uv= 60 -----> uv =30-------> u=30/v

Putting the above in equation of x we have

11= (30/v)^2 - v^2---------> Solving, we get an equation in degree 4

v^4- 11v^2-900=0, let v^2= a....so the equation becomes

a^2+ 11a-900=0

Solving for quadratic equation, we get roots as

= (-11 +/- \sqrt{11^2 +4*1*900}) / 2*1

or (-11 +/- 61)/ 2

so possible values of a = -72/2 or 50/2....since a = v^2 and therefore neglecting negative value as square of a number is greater than or equal to zero

therefore v^ 2= 25 or v = +5/-5
So whether v= 5 or v= -5 it does not change the value of z as

z = u^2 +v^2 or (30/+ or -5)^2 + (+/- 5)^2 --------> z= 61

So St1 is sufficient

Clearly St 2 is also sufficient as u = 6 then 11= 36 - v^2 or v^2 =25 or +/5

Ans D
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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29 Mar 2014, 07:26
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mau5 wrote:
vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

Another approach, for F.S 1, in which you don't have to calculate for anything

We know that $$x = u^2 - v^2$$ = 11,Squaring on both sides, we have $$u^4 + v^4 - 2*v^2*u^2$$ = 121

Thus, adding$$4*v^2*u^2$$ on both sides, we have $$(u^2 + v^2)^2 = 121+4*v^2*u^2 = z^2$$

Hence, this statement is sufficient to calculate the value of z.

As Z can be negative as well as +ve, Z value cannot be uniquely determined by using the above method.

It is better to solve equations and determine.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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17 Apr 2014, 13:13
Could someone please elaborate a little bit more on this problem?

How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one?

Thanks
Cheers!
J

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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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jlgdr wrote:
Could someone please elaborate a little bit more on this problem?

How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one?

Thanks
Cheers!
J

Given in the stem:

$$11 = u^2 - v^2 = (u + v)(u - v)$$
$$y = 2uv$$
$$z = u^2 + v^2$$

$$z + y = u^2 + v^2 + 2uv = (u + v)^2$$
$$z - y = u^2 + v^2 - 2uv = (u - v)^2$$

$$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$

121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0)
So z can be 61 or 11 depending on whether y is 60 or 0.

Statement 1 gives y = 60. So z must be 61. Sufficient

Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 16894 [12], given: 230 Current Student Joined: 06 Sep 2013 Posts: 1988 Kudos [?]: 705 [0], given: 355 Concentration: Finance Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 02 May 2014, 16:58 VeritasPrepKarishma wrote: jlgdr wrote: Could someone please elaborate a little bit more on this problem? How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one? Thanks Cheers! J Given in the stem: $$11 = u^2 - v^2 = (u + v)(u - v)$$ $$y = 2uv$$ $$z = u^2 + v^2$$ $$z + y = u^2 + v^2 + 2uv = (u + v)^2$$ $$z - y = u^2 + v^2 - 2uv = (u - v)^2$$ $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ 121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11 So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0) So z can be 61 or 11 depending on whether y is 60 or 0. $$Statement 1 gives y = 60. So z must be 61. Sufficient$$ Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient Answer (D) Hi Karishma, Assume you meant $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ Please confirm Thanks for this approach, its the best i've seen so far on this master problem Cheers! J Kudos [?]: 705 [0], given: 355 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7608 Kudos [?]: 16894 [0], given: 230 Location: Pune, India Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 02 May 2014, 19:05 jlgdr wrote: VeritasPrepKarishma wrote: jlgdr wrote: Could someone please elaborate a little bit more on this problem? How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one? Thanks Cheers! J Given in the stem: $$11 = u^2 - v^2 = (u + v)(u - v)$$ $$y = 2uv$$ $$z = u^2 + v^2$$ $$z + y = u^2 + v^2 + 2uv = (u + v)^2$$ $$z - y = u^2 + v^2 - 2uv = (u - v)^2$$ $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ 121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11 So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0) So z can be 61 or 11 depending on whether y is 60 or 0. $$Statement 1 gives y = 60. So z must be 61. Sufficient$$ Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient Answer (D) Hi Karishma, Assume you meant $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ Please confirm Thanks for this approach, its the best i've seen so far on this master problem Cheers! J I am not sure what the problem is. I have written exactly the same thing: $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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28 May 2014, 11:40
I am not sure what the problem is.
I have written exactly the same thing: $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$[/quote]

Not so, you wrote.
$$(z + y)(z - y) = (u + v)^2*(u + v)^2 = 11^2 = 121$$

Anyways, just wanted to be clear as intended meaning. Do not worry, this tiny error will not change my high esteem for your work in quant

Thanks again Karishma

Cheers
J

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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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29 Jun 2014, 05:37
VeritasPrepKarishma wrote:
jlgdr wrote:
Could someone please elaborate a little bit more on this problem?
..........
121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
.........

Why do you assume that (u+v)^2 and (u-v)^2 are integers ?
If we knew that u,v are integers, then the simplest way to find the figures in my opinion is:
uv = 30 = 1*30 = 2*15 = 3*10 = 5*6 = 6*5 = ...
(u-v)(u+v) = 11

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.

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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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29 Jun 2014, 12:56
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BOOKMARKED
VeritasPrepKarishma wrote:
jlgdr wrote:
Could someone please elaborate a little bit more on this problem?
..........
121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
.........

Why do you assume that (u+v)^2 and (u-v)^2 are integers ?
If we knew that u,v are integers, then the simplest way to find the figures in my opinion is:
uv = 30 = 1*30 = 2*15 = 3*10 = 5*6 = 6*5 = ...
(u-v)(u+v) = 11

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.

If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

Given: $$u^2 - v^2=11$$ and $$y = 2uv$$.
Question: $$u^2 + v^2=?$$

(1) y = 60 --> $$2uv=60$$ --> $$4u^2v^2=3,600$$.

Square $$u^2 - v^2=11$$: $$u^4-2u^2v^2+v^2=121$$;

Add $$4u^2v^2$$ to both sides: $$u^4+2u^2v^2+v^4=121+3,600$$;

Apply $$a^2+2ab+b^2=(a+b)^2$$: $$(u^2+v^2)^2=121+3,600$$.

$$u^2+v^2=\sqrt{121+3,600}=61$$.

Sufficient.

(2) u = 6 --> $$36 - v^2=11$$ --> $$v^2=25$$ --> $$u^2+v^2=36+25=61$$. Sufficient.

Hope it helps.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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29 Jun 2014, 20:58
1
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Expert's post
Why do you assume that (u+v)^2 and (u-v)^2 are integers ?
If we knew that u,v are integers, then the simplest way to find the figures in my opinion is:
uv = 30 = 1*30 = 2*15 = 3*10 = 5*6 = 6*5 = ...
(u-v)(u+v) = 11

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.

Also note that you are given that x is an integer (11). Also statement 1 tells you that y is an integer (60) and statement 2 tells you that u is an integer. So using any one of the statements gives you enough hint that we are dealing with integers only.
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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15 Sep 2014, 12:53
Bunuel wrote:
vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question

For (1) we have:
$$11 = u^2 - v^2$$;

$$60= 2uv$$;

and we need to find the value of $$u^2 + v^2$$.

If you solve $$11 = u^2 - v^2$$ and $$60= 2uv$$, you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case $$u^2 + v^2=36+25$$.

Thus the answer is D.

Hope it's clear.

Given that $$x =11 = u^2 - v^2 = (u+v)(u-v)$$
Thus there we can be 4 possibilities of (u,v) = (6,5)OR(6,-5)OR(-6,5)OR(-6,-5)
From this only we have sufficient info to answer " what is the value of z, given $$z = u^2 + v^2$$ "
Because in each case, answer will be 61.
So we do not even need any statement!
Please explain if i am missing anything here..

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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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15 Sep 2014, 14:07
I don't know why, but I started to solve this DS-question like a PS-question without considering the information given in (1) and (2).

$$x = u^2 - v^2$$ ----> $$x = (u+v)*(u-v)$$

We're given $$x = 11$$ , since 11 is a prime number the only factors of $$x$$ are 1 and 11, this tells us $$u$$ and $$v$$ can take any combination of +-5 and+-6

Hence, $$z = u^2 + v^2$$ will always take the value 61

Therefore we don't need any data to solve the problem!

edit: saw tushain came to the same conclustion
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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15 Sep 2014, 21:05
tushain wrote:
Bunuel wrote:
vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question

For (1) we have:
$$11 = u^2 - v^2$$;

$$60= 2uv$$;

and we need to find the value of $$u^2 + v^2$$.

If you solve $$11 = u^2 - v^2$$ and $$60= 2uv$$, you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case $$u^2 + v^2=36+25$$.

Thus the answer is D.

Hope it's clear.

Given that $$x =11 = u^2 - v^2 = (u+v)(u-v)$$
Thus there we can be 4 possibilities of (u,v) = (6,5)OR(6,-5)OR(-6,5)OR(-6,-5)
From this only we have sufficient info to answer " what is the value of z, given $$z = u^2 + v^2$$ "
Because in each case, answer will be 61.
So we do not even need any statement!
Please explain if i am missing anything here..

TehMoUsE wrote:
I don't know why, but I started to solve this DS-question like a PS-question without considering the information given in (1) and (2).

$$x = u^2 - v^2$$ ----> $$x = (u+v)*(u-v)$$

We're given $$x = 11$$ , since 11 is a prime number the only factors of $$x$$ are 1 and 11, this tells us $$u$$ and $$v$$ can take any combination of +-5 and+-6

Hence, $$z = u^2 + v^2$$ will always take the value 61

Therefore we don't need any data to solve the problem!

edit: saw tushain came to the same conclustion

Both of you incorrectly assume that the variables are integers only: 11 = u^2 - v^2 has infinitely many non-integer solutions for u and v.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,   [#permalink] 15 Sep 2014, 21:05

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# Work and Energy : Practice Questions

1 Mark Questions

1. Define SI unit of work.

2. A student is writing a three hours science paper. How much work is done by the student? Give reasons to your answer.

3. List two essential conditions for work to be done.

4. Give one example when work done by a body negative.

5. A 2 m high person is holding a 25 kg trunk on his head and is standing at a roadways bus-terminus. How much work is done by the person?

6. Calculate the work done when a force of 15 N moves a body by 5 m in its direction.

7. Moon is experiencing a gravitational force due to earth and is revolving around the earth in a circular orbit. How much work is done by moon?

8. State law of conservation of energy.

9. Name the term used for the sum of kinetic energy and potential energy of a body.

10. If the heart works 60 joules in one minute, what is its power?Â

11. Define 1 KWh.

12. Define 1 W of power.

13. Name the type of energy possessed by a raised hammer.

14. In an oscillating pendulum, at what position the potential and kinetic energy are maximum?

15. A coolie is walking on a railway platform with a load of 30 kg on his head. How much work is done by coolie?

16. Identify the kind of energy possessed by a running athlete.

17. At what speed a body of mass 1 kg will have a kinetic energy of 1 J.?

18. If the speed of the body is halved, what is the change in its kinetic energy?

19. A horse of mass 210 kg and a dog of mass 25 kg are running at the same speed. Which of the two possesses more kinetic energy? How?

20. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

21. What will cause greater change in Kinetic energy of a body. Changing its mass or changing its velocity?

22. What is power? What is its SI unit ?

23. To what height should a body of mass 5 kg be raised so that its potential energy is 490 J? [g = 9.8 m/${s^2}$]

24. On pushing a mighty stone, a child is unable to move it. If work done is zero, where is the energy, he, lost?

25. State the SI unit of

(a) work,
(b) power.

26. Relate watt to joule.

27. What is one unit of electrical energy consumed equal to?

28. Relate one horsepower to the SI unit of power?

29. A ball of mass 5 kg is dropped from a tower d, height 2 m. What is its K.E. half way down?

30. Define one joule.

31. Define one Watt.

32. In what situation(s) is work done equal to zero?

33. When a body falls freely, its potential energy

(A) .........and simultaneously,
(B) its kinetic energy...........

34. A machine of 500 W works for 2 hours. Find the number of units of electricity consumed.

35. How is energy conversion occurring in a Hydel power station?

36. If a force of 50 N moves a body with constant speed of 10 m/s, how much power is spent?

37. State the CGS unit of energy.

38. Relate the CGS unit of energy to its SI unit.

39. Derive a relation between kinetic energy of a body and its momentum.

Solutions

1. SI unit of work is Joule. 1 Joule of work is done when a force of 1 N displaces a body by 1 m.Â

2. Work done = Force Ã— displacement as displacement is zero. No work is done by the student.Â

3. (i) Force should be applied on body.

(ii) Body should move in the direction of force.Â

4. When a body is raised vertical upward, displacement and gravitation force are is in opposite direction. The work done by gravity is negative.Â

5. Zero. As displacement is zero.Â

6. Work done = 15 Ã— 5 = 75 J.Â

7. No work is done by moon as the gravity force and displacement are perpendicular to each other.Â

8. Law of conservation of energy states that energy can only be converted from one form to another, it can neither be created nor destroyed.Â

9. Mechanical Energy.Â

10. Power = ${{60J} \over {60s}}$ = 1 W. [1 min = 60 s]

11. 1 KWh is the energy consumed when power of 1000 w is consumed for 1 hour.

12. When one joule of work is done in one second, the power is said to be 1 watt.

13. Potential Energy.

14. At the highest point of the bob, potential energy is maximum, while at the lowest point kinetic energy is maximum.

15. Since force due to load is downward, and the coolie is moving in horizontal direction, no work is done by coolie.

16. Kinetic energy.

17. K.E = ${1 \over 2}m{v^2}$

$V\, = \,\sqrt {{{2KE} \over m}} \, = \sqrt {{{2 \times 1} \over 1}\,} = \,\sqrt {2\,} \,m\,/s$

18. Since $K.E\,\alpha {v^2}$

If v is halved; KE = 1/4 of the original.

19. $K.E\,\alpha \,Mass$.

Since mass of horse is greater than that of dog, the horse will posses greater kinetic energy.

20. A freely falling body loses potential energy , but it gains equal amount of kinetic energy such that the sum total of energy remains constant. Hence , this does not violate the law of conservation of energy.

21. Changing velocity of a body will cause greater change in the kinetic energy of the body.

22. Power is the rate of doing work. Its unit is watt.

1 Watt = 1 Joule/s.

23. P.E = mg.h

$\,h\, = \,{{P.E} \over {mg\,}}\, = \,{{490} \over {5 \times \,9.8}}\, = \,10\,m.$

24. The energy is lost in the form of heat energy.

25. (a) SI unit of work is Joule.

(b) SI unit of power is Watt

26. 1 watt = 1 Joule / 1 sec

27. 1 unit of electrical energy = $3.6\, \times \,{10^6}\,$ Joule

28. 1 horse power = 746 Watts

29. A ball is dropped from 2 m. The halfway point is when it has moved through 1m.

By III equation of motion
${v^2}\, - \,{\mu ^{2\,}}\, = \,2 gh$
$\,{v^2}\, = \,2\, \times \,10\, \times \,1$
K.E = ${1 \over 2}m{v^2}\, = \,{1 \over 2}\, \times \,2\, \times 10\, \times \,1\, = \,1\,J.$

30. 1 Joule of work is said to be done when a body is displaced by 1m due to a force of 1N

31. 1 watt is the power of an object which does work at the rate of 1 Joule per second

32. Work is equal to zero when-

(i) Force is zero
(ii) Displacement is zero
(iii) Displacement is perpendicular to the direction of force.

33. (A) decreases, (B) increases

34. Work done 500 Ã— 2 = 1000 Wh

= 1 KWh.
unit of electricity consumed = 1.

35. Potential Energy of water in dam $\to$ Kinetic energy of water passing through gates $\to$ Kinetic energy of turbines $\to$ Electrical Energy.

36. Power = ${{Work\,done} \over {Time}} = {{F \times d} \over t}$ = F Ã— V.

= 50 Ã— 10 = 500 W

37. CGS unit of energy is ergs

38. 1 Joule = ${10^7}$ ergs

39. Kinetic energy = ${1 \over 2}m{v^2}$

= ${1 \over 2}{{{m^2}{v^2}} \over m}\, = {{{p^2}} \over {2m}}$ , where p is the momentum.

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