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# Which of the following equations is NOT equivalent to 10y^2=

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Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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11 Sep 2012, 04:37
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Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) $$30y^2=3x^2-12$$

(B) $$20y^2=(2x-4)(x+2)$$

(C) $$10y^2+4=x^2$$

(D) $$5y^2=x^2-2$$

(E) $$y^2=\frac{(x^2-4)}{10}$$

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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11 Sep 2012, 04:38
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SOLUTION

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

When $$x=2$$, then $$10y^2=(2+2)(2-2)=0$$ --> $$y=0$$.

Now, plug $$x=2$$ into the answer choices and look for the option which does not give $$y=0$$. Only option D gives the value of $$y$$ different from zero.

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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11 Sep 2012, 09:07
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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

A) dividing by 3 on either side gives the same equation as given in question stem ==> 3* 10y^2=3(x^2-4) ==> 3* 10^y = 3(x + 2)(x - 2) ==> Q
B) diving by 2 on both side ==> 2* 10y^2=2 * (x-2)(x+2) ==> Q
C) taking 4 to right hand side and solving ==> 10y^2=x^2 - 4 ==> 10y^2 = (x+2)(x-2) ==> Q
D) multiplying both side by 2 ==> 10y^2=2x^2-4 ==> left hand side can't be factored to (x-2)(x+2). Hence, this is NOT equivalent to equation given in Q stem.
We can stop here and mark option D.
E) multiply both sides by 10 ==> 10 y^2=10 *(x^2-4)/10 ==> 10 y^2=(x^2-4) ==> 10 y^2=(x+2)(x-2)

** basic formula which should be known prior to solving this question =====> (A^2 - B^2) = (A+B)(A-B)

Other method to solve this Q is by substituting the values. The best value to take for "x" is 2 as it will make right hand side of the equation ZERO.
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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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11 Sep 2012, 09:19
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Apparently (D), since after dividing by 2, we should get 5y^2=(x^2-4)/2
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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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12 Sep 2012, 12:49
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Which of the following equations is NOT equivalent to 10y^2=(x+2)(x-2) ?
(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

The original equation reduces to 10y^2 = x^2 -4
Option 1 - 3 times of equation
Option 2 - 2 times of equation
Option 3 - Same as original of equation
Option 4 - Answer (Skip this option & jump to option 5, which in turn is right. Thus using POE this will be the answer. To cross check one can solve it)
Option 5 - 1/10 times of equation

Hope it helps
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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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10 Dec 2012, 01:14
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$$10y^2=(x+2)(x-2)$$ expands as $$10y^2=x^2-4$$

(A) $$30y^2=3x^2-12$$
Divide by 3 yields $$10y^2=x^2-4$$
Eliminate!

(B) $$20y^2=(2x-4)(x+2)$$ expands as $$20y2=2x^2-8$$
Divide by 2 yields $$10y^2 = x^2-4$$
Eliminate!

(C) $$10y^2+4=x^2$$Exactly the orginal.
Eliminate!

(D) $$5y^2=x^2-2$$
Multiply by 2 expands as $$10y^2=2x^2-4$$
It's totally different!

(E) $$y^2=\frac{{x^2-4}}{{10}}$$
Multiply by 10 expands as $$10y^2=x^2-4$$
Eliminate!

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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12 Oct 2013, 09:09
Bunuel wrote:
SOLUTION

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

When $$x=2$$, then $$10y^2=(2+2)(2-2)=0$$ --> $$y=0$$.

Now, plug $$x=2$$ into the answer choices and look for the option which does not give $$y=0$$. Only option D gives the value of $$y$$ different from zero.

Bunuel could we also use -2??

Thanks,
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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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12 Oct 2013, 09:33
runningguy wrote:
Bunuel wrote:
SOLUTION

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

When $$x=2$$, then $$10y^2=(2+2)(2-2)=0$$ --> $$y=0$$.

Now, plug $$x=2$$ into the answer choices and look for the option which does not give $$y=0$$. Only option D gives the value of $$y$$ different from zero.

Bunuel could we also use -2??

Thanks,
C

Yes we can. We can use the same exact logic.
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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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27 Oct 2014, 00:48
One good thing of this question is that $$y^2$$ term is on LHS of the question as well as all 5 options given.

$$10y^2 = (x+2)(x-2)$$

For y = 0, x = 2 (Ignore -ve sign aspect)

Placing value of y = 0 in the OA

A: $$3x^2 - 12 = 0; x = 2$$

B: (2x-4)(x+2) = 0; x = 2

C: $$x^2 = 4; x = 2$$

D:$$x^2 - 2 = 0; x = \sqrt{2}$$ >> Does not stand >> This is the answer

E: Same as option C

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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07 Sep 2015, 16:57
Bunuel wrote:
SOLUTION

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

When $$x=2$$, then $$10y^2=(2+2)(2-2)=0$$ --> $$y=0$$.

Now, plug $$x=2$$ into the answer choices and look for the option which does not give $$y=0$$. Only option D gives the value of $$y$$ different from zero.

How do you actually solve using smart numbers? When I did it, all of my answers were y=0, x=2.

Let's pick x=2, when this is plugged into the equation we get y=0. So we're going to plug in 2 for all x's that we see, when y = 0 that is our answer.

a. 30y^2=3x^2-12
3(2)^2-12 = 0, therefore y=0 when x=2

b. 20y^2=(2x-4)(x+2)
(2(2)-4)(2+2) = 0, therefore y=0 when x=2

What am I doing wrong?

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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07 Sep 2015, 17:16
aces021 wrote:
Bunuel wrote:
SOLUTION

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

When $$x=2$$, then $$10y^2=(2+2)(2-2)=0$$ --> $$y=0$$.

Now, plug $$x=2$$ into the answer choices and look for the option which does not give $$y=0$$. Only option D gives the value of $$y$$ different from zero.

How do you actually solve using smart numbers? When I did it, all of my answers were y=0, x=2.

Let's pick x=2, when this is plugged into the equation we get y=0. So we're going to plug in 2 for all x's that we see, when y = 0 that is our answer.

a. 30y^2=3x^2-12
3(2)^2-12 = 0, therefore y=0 when x=2

b. 20y^2=(2x-4)(x+2)
(2(2)-4)(2+2) = 0, therefore y=0 when x=2

What am I doing wrong?

I think you are missing out on what the question is asking. It is asking to find the expression that will NOT give you the same value as $$10y^2=(x+2)(x-2)$$

When you do use, x=2, you get the following values of 'y'

A) 0
B) 0
C) 0
D) 2/5
E) 0

From the original expression you see that when you get y=0 for x=2 for options A-C and E.

Thus, with your 'smart numbers', D is the only expression that DOES NOT give you the same values on the left hand side as those on the right hand side. D is thus the correct answer.

FYI, using smart numbers is not the best strategy for this question. You need to realize that $$(A+B)(A-B) = A^2-B^2$$ and thus the original expression becomes $$10y^2 = x^2-4$$

D will not give you this expression and is thus the correct answer.
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Last edited by Engr2012 on 07 Sep 2015, 19:09, edited 2 times in total.
Edited the typos, alternate reasoning

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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07 Sep 2015, 19:05
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Engr2012 wrote:
aces021 wrote:
Bunuel wrote:
SOLUTION

Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

When $$x=2$$, then $$10y^2=(2+2)(2-2)=0$$ --> $$y=0$$.

Now, plug $$x=2$$ into the answer choices and look for the option which does not give $$y=0$$. Only option D gives the value of $$y$$ different from zero.

How do you actually solve using smart numbers? When I did it, all of my answers were y=0, x=2.

Let's pick x=2, when this is plugged into the equation we get y=0. So we're going to plug in 2 for all x's that we see, when y = 0 that is our answer.

a. 30y^2=3x^2-12
3(2)^2-12 = 0, therefore y=0 when x=2

b. 20y^2=(2x-4)(x+2)
(2(2)-4)(2+2) = 0, therefore y=0 when x=2

What am I doing wrong?

I think you are missing out on what the question is asking. It is asking to find the expression that will NOT give you the same value as $$10y^2=(x+2)(x-2)$$

When you do use, y = 0 and x=2, you get the following

A) 15 = 15
B) 10 = 10
C) 9=9
D) 2.5 $$\neq$$ 7
E) 0.5 = 0.5

From the original expression you see that when you use y=0 and x=2, you get 0=0 (LHS=RHS).

Thus, with your 'smart numbers', D is the only expression that DOES NOT give you the same values on the left hand side as those on the right hand side. D is thus the correct answer.

FYI, using smart numbers is not the best strategy for this question. You need to realize that $$(A+B)(A-B) = A^2-B^2$$ and thus the original expression becomes $$10y^2 = x^2-4$$

D will not give you this expression and is thus the correct answer.

Thanks for the response, how are you actually getting for a. 15=15 using x=2, y=0?

30y^2=3x^2-12
30(0)^2 = 3(2)^2 - 12

What is the order of operations? Does the above equal to
0 = 0

Or does it equal to
30*1 = 36-12
30 = 24 ?

Either way I'm not getting 15=15. Thanks!

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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07 Sep 2015, 19:12
aces021 wrote:

Thanks for the response, how are you actually getting for a. 15=15 using x=2, y=0?

30y^2=3x^2-12
30(0)^2 = 3(2)^2 - 12

What is the order of operations? Does the above equal to
0 = 0

Or does it equal to
30*1 = 36-12
30 = 24 ?

Either way I'm not getting 15=15. Thanks!

Sorry. I had solved this using some other set of values and wrote those values. I have updated the solution. Yes, you are correct that option A will give you y=0 for x=2. As a matter of fact, options A-C and E give y=0 when you use x=2. Option D does not and is thus the correct answer.
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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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10 May 2016, 18:11
Attached is a visual that should help.
Attachments

Screen Shot 2016-05-10 at 5.50.11 PM.png [ 76.13 KiB | Viewed 6817 times ]

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Re: Which of the following equations is NOT equivalent to 10y^2= [#permalink]

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25 May 2016, 09:39
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Bunuel wrote:
Which of the following equations is NOT equivalent to $$10y^2=(x+2)(x-2)$$ ?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

To solve this question, we start by FOILing the right hand side of the given equation.

10y^2 = (x+2)(x-2)

10y^2 = x^2 – 4

We will manipulate each of the answer choices to see if it equals 10y^2 = x^2 – 4. Let’s start with A.

A) 30y^2 =3x^2 – 12

If we divide this entire equation by 3 we are left with:

10y^2 = x^2 – 4

Answer choice A is not correct.

B) 20y^2 = (2x-4)(x+2)

FOILing (2x-4)(x+2) we get:

20y^2 = 2x^2 – 8

If we divide this entire equation by 2 we obtain:

10y^2 = x^2 – 4

Answer choice B is not correct.

C) 10y^2 + 4 = x^2

If we subtract 4 from the both sides of the equation, we obtain:

10y^2 = x^2 – 4

Answer choice C is not correct.

D) 5y^2 = x^2 – 2

We should notice that no matter how we try to manipulate 5y^2 = x^2 – 2, it will never be equal to 10y^2 = x^2 – 4. For example, if we multiply both sides of 5y^2 = x^2 – 2 by 2, we will have 10y^2 = 2x^2 – 4. However, that is not the same as 10y^2 = x^2 – 4.

To be certain, we should also test answer E.

E) y^2 = (x^2 – 4)/10

If we multiply the entire equation by 10 we obtain:

10y^2 = x^2 – 4

Answer choice E is not correct.

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Re: Which of the following equations is NOT equivalent to 10y^2=   [#permalink] 25 May 2016, 09:39
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# What is the outer product of a tensor product of vectors?

• I
If one has two single-particle Hilbert spaces ##\mathcal{H}_{1}## and ##\mathcal{H}_{2}##, such that their tensor product ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## yields a two-particle Hilbert space in which the state vectors are defined as $$\lvert\psi ,\phi\rangle =\lvert\psi\rangle\otimes\lvert\phi\rangle\in\mathcal{H}_{1}\otimes\mathcal{H}_{2}$$ where ##\lvert\psi\rangle\in\mathcal{H}_{1}## and ##\lvert\phi\rangle\in\mathcal{H}_{2}##.

Now, the inner product for ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## is defined such that $$\langle\phi ,\psi\vert\psi ,\phi\rangle =\left(\langle\phi\rvert\otimes\langle\psi\lvert\right)\left(\lvert\psi\rangle\otimes\lvert\phi\rangle\right) =\langle\psi\lvert\psi\rangle_{1}\langle\phi\lvert\phi\rangle_{2}$$ where ##\langle\cdot\lvert\cdot\rangle_{1}## is the inner product defined on ##\mathcal{H}_{1}## and ##\langle\cdot\lvert\cdot\rangle_{2}## the inner product defined on ##\mathcal{H}_{2}##.

How though is the outer product defined? Is it simply $$\lvert\psi ,\phi\rangle\langle\phi ,\psi\rvert =\lvert\psi\rangle\otimes\lvert\phi\rangle\langle\phi\rvert\otimes\langle\psi\rvert =\lvert\psi\rangle\langle\psi\rvert_{1} \lvert\phi\rangle\langle\phi\rvert_{2}$$ where ##\lvert\psi\rangle\langle\psi\rvert_{1}## is the outer product in ##\mathcal{H}_{1}## and ##\lvert\phi\rangle\langle\phi\rvert_{2}## is the outer product in ##\mathcal{H}_{2}##.

andrewkirk
Homework Helper
Gold Member
It might be easier to think of more general outer products, rather than the outer product of an element with itself, as you have written.
That is, consider the outer product ##|a,b\rangle\langle c,d|=|a\rangle\otimes |b\rangle\langle c|\otimes\langle d|##.

This is an operator on the product space ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}##. To see what it does, we apply it to an element of that space ##|e,f\rangle=|e\rangle\otimes |f\rangle##.

The result is ##|a,b\rangle \langle c|e\rangle \langle d|f\rangle##

This is an operator on the product space H1⊗H2\mathcal{H}_{1}\otimes\mathcal{H}_{2}. To see what it does, we apply it to an element of that space |e,f⟩=|e⟩⊗|f⟩|e,f\rangle=|e\rangle\otimes |f\rangle.

The result is |a,b⟩⟨c|e⟩⟨d|f⟩

So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then? Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$

I ask in particular as I'm trying to understand the notion of a partial trace.

Last edited:
andrewkirk
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Gold Member
So one can't define it as ##\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\lvert b\rangle\langle d\rvert## then?
That symbol string on the RHS is not intrinsically meaningful. It juxtaposes the two operators ##\lvert a\rangle\langle c\rvert## and ##\lvert b\rangle\langle d\rvert## but:
- since they are not scalars, the juxtaposition cannot be interpreted as scalar multiplication
- since they are operators on different spaces (##\mathcal H_1## and ##\mathcal H_2##), the juxtaposition cannot be interpreted as composition of operation.

Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$
That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above ##(
|a\rangle\otimes |b\rangle)(\langle c|\otimes\langle d|)## (I have added parentheses here that were only implied above, to make it clear what operations are being performed).

That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above (|a⟩⊗|b⟩)(⟨c|⊗⟨d|)

How does one show that the two expressions are equivalent? Would it be something like this:
$$\left(\lvert a\rangle\otimes\lvert b\rangle\langle c\rvert\otimes\langle d\rvert\right)\lvert e\rangle\otimes\lvert f\rangle =\lvert a\rangle\otimes\lvert b\rangle\left(\langle c\vert e\rangle\langle d\vert f\rangle\right)\\ =\left(\lvert a\rangle\langle c\vert e\rangle\right)\otimes \left(\lvert b\rangle\langle d\vert f\rangle\right) =\left(\lvert a\rangle\langle c\lvert\otimes\lvert b\rangle\langle d\lvert\right)\lvert e\rangle\otimes\lvert f\rangle$$

andrewkirk
$$(|a,b\rangle\langle c,d|)|e,f\rangle\equiv (|a\rangle\otimes|b\rangle)(\langle c|\otimes\langle d|)(|e\rangle\otimes|f\rangle)=(\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
$$(|a\rangle\langle c|\otimes|b\rangle\langle d|)|e,f\rangle\equiv (|a\rangle\langle c|\otimes|b\rangle\langle d|)(|e\rangle\otimes |f\rangle)\ =(|a\rangle\langle c|)|e\rangle\otimes (|b\rangle\langle d|)|f\rangle \equiv (\langle c|e\rangle|a\rangle)\otimes (\langle d|f\rangle|b\rangle)$$
$$(|a,b\rangle\langle c,d|)=(|a\rangle\langle c|\otimes|b\rangle\langle d|)$$

format: plain


text:
# Problem with NonLinearModelFit when changing data sample area

It's a physics promblem, I want to confirm an integration without analytic solution to have the form of f[x]=k/x^2. The data set come from a NIntegration:

{xmin, xmax, xstep} = {2, 5000000, 100000};
ydata = Table[NIntegrate[((1 - x*Cos[a])*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(3/2)) - (x - Cos[a])/((x^2 - 2*x*Cos[a] + 1)^(3/2)) + (2*(x^2)*(Sin[a])^2*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(5/2)), {a, 0, 2*Pi}, WorkingPrecision -> 100], {x, xmin, xmax, xstep}];
xdata = Table[x, {x, xmin, xmax, xstep}];
data = Partition[Riffle[xdata, ydata], 2];
fit = NonlinearModelFit[data, k/(x^2), k, x]

data sample look like this:

[![enter image description here][1]][1]

when xmin=2, no matter how large xmax is, the best estimate of coefficient k is the same -0.28, but when change the starting point to say 3 million, every time I change xmin or xmax it returns a different result, and usually it's apparently a bad regression like this:

[![enter image description here][2]][2]

Can anyone help me figure out what's happening? lots of thanks.

Update

@JimBaldwin As suggested I remove the NIntegration and use the new code for generating data:

xdata = Table[Exp[Log[2] + (Log[5000000] - Log[2]) (i - 1)/100], {i, 101}];
integrand = ((1 - x*Cos[a])*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(3/
2)) - (x - Cos[a])/((x^2 - 2*x*Cos[a] + 1)^(3/2)) + (2*(x^2)*(Sin[a])^2*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(5/2));
y = FullSimplify[Integrate[integrand, {a, 0, \[Pi]}, Assumptions -> x > 2] + Integrate[integrand, {a, \[Pi], 2 \[Pi]}, Assumptions -> x > 2]];
ydata = SetPrecision[Table[y, {x, xdata}], 100];
data = Transpose[{xdata, -ydata}];
fit = NonlinearModelFit[Log[data], k - 2 logx, k, logx]
Show[ListLogLogPlot[data],
LogLogPlot[Exp[fit[Log[x]]], {x, 2, 5000000}, PlotStyle -> Red]]

But it still returns the same(need 10 reputation to post the image, it is the same with the first image in the answer)

don't know whether I take the suggested method wrong or there is nothing improved by breaking the integration apart.

• Are you sure the fit function is correct, I can't get it to fit. – Feyre Oct 18 '16 at 19:13
• yes it works, but it doesn't show the result directly due to the high working precision of the previous NIntegration – user43887 Oct 19 '16 at 3:00

In such cases it is usually better to fit the linearized function and have equally spaced predictor values in the log scale than the original scale. The linearized function is

$$\log(-y)=k-2\log(x)$$

Here is some code to provide a fit:

xdata = Table[Exp[Log[2] + (Log[5000000] - Log[2]) (i - 1)/100], {i, 101}];
ydata = Table[
NIntegrate[((1 - xdata[[i]]*Cos[a])*(xdata[[i]] -
Cos[a]))/((xdata[[i]]^2 - 2*xdata[[i]]*Cos[a] + 1)^(3/
2)) - (xdata[[i]] -
Cos[a])/((xdata[[i]]^2 - 2*xdata[[i]]*Cos[a] + 1)^(3/
2)) + (2*(xdata[[i]]^2)*(Sin[a])^2*(xdata[[i]] -
Cos[a]))/((xdata[[i]]^2 - 2*xdata[[i]]*Cos[a] + 1)^(5/
2)), {a, 0, 2*Pi}, WorkingPrecision -> 100], {i,
Length[xdata]}];
data = Transpose[{xdata, -ydata}];
fit = NonlinearModelFit[Log[data], k - 2 logx, k, logx]
Show[ListLogLogPlot[data],
LogLogPlot[Exp[fit[Log[x]]], {x, 2, 5000000}, PlotStyle -> Red]]

So we see that likely the "2" in $-k/x^2$ should probably be another parameter to be estimated. Here's the modified code and fit:

xdata = Table[Exp[Log[2] + (Log[5000000] - Log[2]) (i - 1)/100], {i, 101}];
ydata = Table[
NIntegrate[((1 - xdata[[i]]*Cos[a])*(xdata[[i]] -
Cos[a]))/((xdata[[i]]^2 - 2*xdata[[i]]*Cos[a] + 1)^(3/
2)) - (xdata[[i]] -
Cos[a])/((xdata[[i]]^2 - 2*xdata[[i]]*Cos[a] + 1)^(3/
2)) + (2*(xdata[[i]]^2)*(Sin[a])^2*(xdata[[i]] -
Cos[a]))/((xdata[[i]]^2 - 2*xdata[[i]]*Cos[a] + 1)^(5/
2)), {a, 0, 2*Pi}, WorkingPrecision -> 100], {i,
Length[xdata]}];
data = Transpose[{xdata, -ydata}];
fit = NonlinearModelFit[Log[data], k - b logx, {k, b}, logx]
fit["BestFitParameters"]
(* {k -> -0.188838695582118849038332746255,
b -> 4.0049222758293443525902547347118} *)
Show[ListLogLogPlot[data],
LogLogPlot[Exp[fit[Log[x]]], {x, 2, 5000000}, PlotStyle -> Red]]

So I suspect you really want $y=-k/x^4$ rather than $y=-k/x^2$.

Update

While I think my answer is what you really need, I didn't answer your original question about why the fit changed so much by changing the minimum value. Here's a handwaving explanation:

Because of the wrong model being fit ($-k/x^2$) and because the larger the predictor value the predicted values became very close to zero, the point with the smallest predictor value was fit nearly perfectly as deviations from it would otherwise result in huge increases in the sum of squares making it the most influential data point.

When $-k/x^4$ is fit no single data point has such a dominating influence as before with $-k/x^2$.

2nd update

Hmmm. Maybe there's hope for $-k/x^2$ if getting a better fit from $-k/x^4$ is only because of numerical issues with NIntegrate. There is a solution for the integration when you break the integration into parts:

integrand = ((1 - x*Cos[a])*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(3/2)) -
(x - Cos[a])/((x^2 - 2*x*Cos[a] + 1)^(3/2)) + (2*(x^2)*(Sin[a])^2*(x - Cos[a]))/
((x^2 - 2*x*Cos[a] + 1)^(5/2))
y = FullSimplify[
Integrate[integrand, {a, 0, π}, Assumptions -> x > 2] +
Integrate[integrand, {a, π, 2 π}, Assumptions -> x > 2]]

I don't have time tonight to complete the testing of this tonight but having the above results removes some of the potential numerical integration issues.

3rd Update

It appears that as @Feyre suggested, you have the wrong function (or in my definitely non-physicist's mind you've discovered something new about gravity).

Let

f[x_] = Integrate[((1 - x*Cos[a])*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(3/2)) -
(x - Cos[a])/((x^2 - 2*x*Cos[a] + 1)^(3/2)) +
(2*(x^2)*(Sin[a])^2*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(5/2)),
{a, 0, π}, Assumptions -> x > 1] +
Integrate[((1 - x*Cos[a])*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(3/2)) -
(x - Cos[a])/((x^2 - 2*x*Cos[a] + 1)^(3/2)) +
(2*(x^2)*(Sin[a])^2*(x - Cos[a]))/((x^2 - 2*x*Cos[a] + 1)^(5/2)),
{a, π, 2 π}, Assumptions -> x > 1];

If we take the limit of f[x]/(-k/x^4) as x -> ∞, then we have

Limit[f[x]/(-k/x^4), x -> ∞]
(* π/(4 k) *)

This says that k = π/4 which means that we don't really need NonlinearModelFit and that a LogLogPlot of -(π/4)/x^4 divided by f[x] is all we need to check on the fit.

• Thanks a lot, your method works and explanation is convincing, but I'm still confused because according to the physics model behind the expression, it's supposed to be x^2 not x^4(it's about gravity and x is distant), and i observed the integration before using Mathematica, I'm quite sure that the x^3 and above factors don't exist, and that's why i use the x^2 model without hesitation. I'll examine the equation again. btw, is there any possibility to change the algorithm of nonlinearmodelfit to get a less accurate but more stable and consistent regression of -k/x^2 – user43887 Oct 19 '16 at 3:28
• Thank you very much! I tried your approach but it makes no difference, see above the updated question if interested~ – user43887 Oct 19 '16 at 12:45


text:
# scipy.stats.mstats.hdquantiles¶

scipy.stats.mstats.hdquantiles(data, prob=[0.25, 0.5, 0.75], axis=None, var=False)[source]

Computes quantile estimates with the Harrell-Davis method.

The quantile estimates are calculated as a weighted linear combination of order statistics.

Parameters
dataarray_like

Data array.

probsequence, optional

Sequence of quantiles to compute.

axisint or None, optional

Axis along which to compute the quantiles. If None, use a flattened array.

varbool, optional

Whether to return the variance of the estimate.

Returns
A (p,) array of quantiles (if var is False), or a (2,p) array of quantiles and variances (if var is True), where p is the number of quantiles.


text:
RRB Group D Mock Test - 15

# RRB Group D Mock Test - 15

Test Description

## 100 Questions MCQ Test RRB (Group D) - Mock Tests & Previous Year Papers | RRB Group D Mock Test - 15

RRB Group D Mock Test - 15 for Railways 2022 is part of RRB (Group D) - Mock Tests & Previous Year Papers preparation. The RRB Group D Mock Test - 15 questions and answers have been prepared according to the Railways exam syllabus.The RRB Group D Mock Test - 15 MCQs are made for Railways 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB Group D Mock Test - 15 below.
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RRB Group D Mock Test - 15 - Question 1

### There is a circular pond and a foot-path runs along its boundary. A man walks around it exactly once, keeping close to the edge. His each step is 55 cm long and he takes exactly 440 steps to around once then. The diameter of the pond is

Detailed Solution for RRB Group D Mock Test - 15 - Question 1

Circumference of the outer edge = 55 x440 = 24200 cm
We know that c = 2πr
∴ 24200 = 2 x 22/7 x r
⇒ r = 24200 x 7/2 x 22 = 3850 cm
radius of the outer edge or pond = 3850 cm = 3850/100 = 38.5 m
diameter = 38.5 x 2 = 77 m

RRB Group D Mock Test - 15 - Question 2

### Tanya's grandfather was 8 times older to her 16 years ago. He would be 3 times of her age 8 years from now. Eight years ago, what was the ratio of Tanya's age to that of her grandfather?

Detailed Solution for RRB Group D Mock Test - 15 - Question 2

Let, Tanya age 16 years ago = x
Grandfather's age 16 years ago = 8x
8 years from now, 3(x + 16 + 8) = (8x + 16 + 8)
⇒ x = 48/5
8 years ago ratio was:
⇒ z + 8/8z + 8
⇒ 88/424 = 11/53

RRB Group D Mock Test - 15 - Question 3

### Read the following information relating to a family. There are six persons A, B, C, D, E and F. C is the sister of F. B is the brother of E's husband. D is the father of A, and grandfather of F. There are two fathers, three brother and a mother in the group. How is F related to E ?

Detailed Solution for RRB Group D Mock Test - 15 - Question 3 D is father of A and grandfather or F
So, A is father of F
Thus, D and A are the two fathers. C is the sister of F
So, C is the daughter of A.
Since there is only one mother, it is evident that E is the wife of A nd hence the mother of C and F
So, B is brother of A. There are three brothers
So, F is the brother of C
So, F is the son of E
RRB Group D Mock Test - 15 - Question 4
The average monthly salary of 30 employees is Rs 650. If the manager's salary is added, then the average monthly salary increases by Rs 50. The salary of the manager is
RRB Group D Mock Test - 15 - Question 5

Four of the following five are alike in a certain way and hence from a group. Which one does not belong to the group ?

RRB Group D Mock Test - 15 - Question 6
In a certain code GARIMA is written as 725432 and TINA as 6482. How is MARTINA written in that code ?
RRB Group D Mock Test - 15 - Question 7

A boat goes 6 km an hour in still water, but it takes thrice as much time in going the same distance against the current. The speed of the current (in km/hr) is

Detailed Solution for RRB Group D Mock Test - 15 - Question 7

Let the speed of current be x
and the speed of boat = 6 kmph
⇒ Time = D/S
⇒ 3D/6 = D/6 - x
⇒ 18 - 3x = 6
⇒ x = 4 kmph
Hence, speed of current is 4 kmph

RRB Group D Mock Test - 15 - Question 8
The book 'Cricket My Style' is written by:
RRB Group D Mock Test - 15 - Question 9
Consider the following statem-ents (As per provisional popul-ation data of Census 2011)?
A. India's Literacy rate is 74.04%.
B. India's Males literacy rate is 82.14%.
C. India's Female literacy rate is 65.46%.
D. Kerala state has the highest Literacy Rate with 93.91%.
E. Orissa state has the Lowest Literacy Rate with 63.82%.
Which of the statements given above is/are incorrect?
RRB Group D Mock Test - 15 - Question 10
Choose the analogous pair.
Agreement : Dissent
RRB Group D Mock Test - 15 - Question 11
Complete the analogous pair.
Men : Messers :: ? : Women
RRB Group D Mock Test - 15 - Question 12

What sum of money will become Rs 1352 in 2 yrs at 4% per annum Compound Interest?

Detailed Solution for RRB Group D Mock Test - 15 - Question 12

CI = Rs 1352
T = 2 years
R = 4% per annum
CI = P[1 + R/100 ]T
1352 = P[ 104/100 ]2 = P[ 10816/10000 ]
⇒ P = 1352 x 10000 /10816
⇒ P = Rs 1250

RRB Group D Mock Test - 15 - Question 13
Find the number of triangles in the given figure.
RRB Group D Mock Test - 15 - Question 14
The vast territory known as Louisiana was purchased from :
RRB Group D Mock Test - 15 - Question 15
The sheet of paper shown in the figure (X) in each problem, is folded to form a box. Choose from amongst the alternatives (I), (II), (III) and (IV), the boxes that are similar to the box that will be formed.
Detailed Solution for RRB Group D Mock Test - 15 - Question 15 The figure (X) is similar to fig. V.
So, the cube shown in fig. (II) which has both the shaded faces adjacent to the face bearing the dot, can't be formed.
Therefore, the cubes shown in fig. (I), (II) and (IV) can be formed.
RRB Group D Mock Test - 15 - Question 16
Which country uses the currency Dirham?
RRB Group D Mock Test - 15 - Question 17
'Nation is State plus Nationality'.who said this?
RRB Group D Mock Test - 15 - Question 18
The authority empowered to make laws in respect to the matters not included in any of the three lists is
RRB Group D Mock Test - 15 - Question 19
Which suburban station became the first railway station in India to be operated by an all-woman staff in July 2017?
RRB Group D Mock Test - 15 - Question 20
Rohit walked 25 metres towards South. Then he turned to his left and walked 20 metres. He then turned to his left and walked 25 metres. He again turned to his right and walked 15 metres. At what distance is he from the starting point and in which direction ?
Detailed Solution for RRB Group D Mock Test - 15 - Question 20
He is (20 + 15)35m away in the EAST direction from starting point
RRB Group D Mock Test - 15 - Question 21
Indian Railways has finalised a plan to equip how many railway stations across India (including those in rural and remote areas) with Wi-Fi facilities at an estimated cost of Rs 700 crore?
RRB Group D Mock Test - 15 - Question 22
Which music director had the distinction of being the first Asian to compose a symphony for Royal Philharmonic Orchestra of Landon?
RRB Group D Mock Test - 15 - Question 23
Which was the first Indian Satellite Launched from Soviet Cosmodrome?
RRB Group D Mock Test - 15 - Question 24
Which of the following is NOT required for high level language program before it is executed ?
RRB Group D Mock Test - 15 - Question 25
Which is the most labour intensive amongst the following?
RRB Group D Mock Test - 15 - Question 26
The period of 11th plan is—
RRB Group D Mock Test - 15 - Question 27
The speed of wind is measured by :
RRB Group D Mock Test - 15 - Question 28
Ghatshila is famous for
RRB Group D Mock Test - 15 - Question 29
Who among the following called himself as the 'Second Alexander'?
RRB Group D Mock Test - 15 - Question 30
Who among the following was known as a 'Nirguna' reformer?
RRB Group D Mock Test - 15 - Question 31
W.T.O. came into existence in
RRB Group D Mock Test - 15 - Question 32
Which one of the following situations makes a firm most efficient?
RRB Group D Mock Test - 15 - Question 33
The average salinity of the ocean water is
RRB Group D Mock Test - 15 - Question 34
Which of the following is used for writing on glass?
RRB Group D Mock Test - 15 - Question 35
The colour of a star is an indication of its
RRB Group D Mock Test - 15 - Question 36
When lightwaves pass from air to glass, the variable attected are
RRB Group D Mock Test - 15 - Question 37
Even though an animal is fed with carbohydrate-rich diet, its blood sugar concentration trends to remain constant. This is on account of the fact that in the case of an animal
RRB Group D Mock Test - 15 - Question 38
Haemoglobin is dissolved in the plasma of
RRB Group D Mock Test - 15 - Question 39
Which one of the following features of DNA makes it uniquely suited to store and transmit genetic information from generation to generation?
RRB Group D Mock Test - 15 - Question 40
Consider the following statements:
Which of the following are characteristics shown by a breastfed baby when compared to a bottle-fed baby?
1- It is less obese.
2-It shows greater capaciity to resist diseases.
3- It gets more vitamins and proteins.
4- Its growth in height is abnormal.
Of these statements which option is correct
RRB Group D Mock Test - 15 - Question 41
It is possible to produce seedless tomato fruits by
RRB Group D Mock Test - 15 - Question 42
Which country has suspended more than USD 1.15 billion security assistance to Pakistan, accusing Islamabad of harbouring terror groups like the Afghan Taliban and the Haqqani Network within its border?
RRB Group D Mock Test - 15 - Question 43

∠BOC = 7xo + 20o ; ∠COA = 3xo and AB is a straight line. Find (∠BOC - ∠COA) .

Detailed Solution for RRB Group D Mock Test - 15 - Question 43

∠BOC = 7xo + 20o
∠COA = 3xo
we know that
∠BOC + ∠COA = 180o (Linear pair)
7xo + 20o + 3xo = 180o
⇒ 10xo + 20o = 180o
⇒ 10xo = 180o - 20o = 160o
⇒ xo = 160o /10o = 16o
∠BOC = (7 x 16) + 20 = 112 + 20 = 132o
∠COA = 3 x 16 = 48o
∠BOC - ∠COA = 132o - 48o = 84o

RRB Group D Mock Test - 15 - Question 44
Find the missing character.
RRB Group D Mock Test - 15 - Question 45
The Government of India jointly with the Government of ______has unveiled the ‘Special logo’ to mark the Golden Jubilee of the establishment of formal diplomatic relations between the two nations in the year 2018
RRB Group D Mock Test - 15 - Question 46
Mark the wrong combination
RRB Group D Mock Test - 15 - Question 47
Mark the correct option for missing places to complete the series.
a_bb_abba_ _ baa_ b
RRB Group D Mock Test - 15 - Question 48

If the letters in the word DOLPHIN are rearranged as they appear in English alphabetical order, which of the following letters, will be the fifth from left?

RRB Group D Mock Test - 15 - Question 49
Arrange the words given below in a meaningful sequence.
1. Foetus 2. Child 3. Baby 4. Adult 5. Youth
RRB Group D Mock Test - 15 - Question 50

Study the following diagram :

The value of 'x' is

Detailed Solution for RRB Group D Mock Test - 15 - Question 50

3 × 4 = 12
12 × 5 = 60
3 × 5 = 15 = x

RRB Group D Mock Test - 15 - Question 51

If Δ denotes =, + denotes > , - denotes >, □ denotes ≠, × denotes > and ÷ denotes < then a + b -c denotes :-

RRB Group D Mock Test - 15 - Question 52
By mixing two qualities of pulses in the ratio 2:3 and selling the mixture at the rate of Rs 22 per kg, a shopkeeper makes a profit of 10%. If the cost of the smaller quantity be Rs 14 per kg, the cost per kg of the larger quantity is
RRB Group D Mock Test - 15 - Question 53
Under Pradhan Mantri Swasthya Suraksha Yojana (PMSSY), the central Government has decided to set up an All India Institutes of Medical Sciences (AIIMS) in which state?
RRB Group D Mock Test - 15 - Question 54
Which is the first State in India to launch High Risk Pregnancy (HRP) Portal which helps in early identification of high-risk pregnant cases, ensures timely referral of such cases to the civil hospitals for further management and delivery by specialists?
RRB Group D Mock Test - 15 - Question 55
Complete the analogous pair.
125 : 17 18 21 :: 458 : ?
RRB Group D Mock Test - 15 - Question 56
If p is a prime number such that (p + 2)is also a prime number,then p (p + 2) + 1 is
Detailed Solution for RRB Group D Mock Test - 15 - Question 56 Let p be 3
then (p + 2) will be 5
⇒ p(p + 2) + 1 = 3(5) + 1 = 16
Which is a perfect square
Consider another case,
Let p = 11
the p + 2 = 13
⇒ p(p + 2) + 1 = 143 + 1 = 144
Which is also a perfect square
RRB Group D Mock Test - 15 - Question 57
Radha started a business, investing Rs 75,000. After 3 months, Sunidhi joined her with an amount of Rs 1,25,000 and after 6 months Neha joined them with an amount of Rs 1,50,000. Profit earned at the end of 3 yrs from when Radha started the business should be distributed in what ratio among Radha, Sunidhi and Neha respectively?
RRB Group D Mock Test - 15 - Question 58
45% of 1500 + 35% of 1700 = ?% of 3175
RRB Group D Mock Test - 15 - Question 59
World Book Fair 2018, with focus on environment was hosted by which Indian state?
RRB Group D Mock Test - 15 - Question 60
What is the theme of the 26th New Delhi World Book Fair being organised at Pragati Maidan?
RRB Group D Mock Test - 15 - Question 61
A train covers a distance between Station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. What is the distance between the stations A and B?
RRB Group D Mock Test - 15 - Question 62

A cistern has two taps (which fill it in 12 mins and 15 mins respectively) and an exhaust tap. When all the three taps are opened together, it takes 20 min to fill an emty cistern. How long will the exhaust tap take to empty it?

Detailed Solution for RRB Group D Mock Test - 15 - Question 62

Let Tap A can fill the cistern in 12 mins.
Let Tap B can fill the cistern in 15 mins.
Let Tap C can empty the cistern in x mins.
∴ 1/12 + 1/15 - 1/x = 1/20
⇒ x = 10

RRB Group D Mock Test - 15 - Question 63
The sale price of 2 oranges is equal to the cost price of 3 oranges. The gain % is
RRB Group D Mock Test - 15 - Question 64
Ram ranked ninth from the top and thirty–eighth from the bottom in a class. How many students are there in the class?
RRB Group D Mock Test - 15 - Question 65
A's money is to B's money is 4:5 and B's money is to C's money is 2:3. If A has Rs 800,then C has
RRB Group D Mock Test - 15 - Question 66
Rig Veda, the oldest of Vedas, chiefly contains?
RRB Group D Mock Test - 15 - Question 67
Indian Coast Guard Ship C-161 (ICGS C -161) was commissioned at which city of Gujarat?
RRB Group D Mock Test - 15 - Question 68
Primary gold has a purity of ...... carats
RRB Group D Mock Test - 15 - Question 69
Red blood corpuscles are produced in :
RRB Group D Mock Test - 15 - Question 70
Mica is used in an electric iron because it is a
RRB Group D Mock Test - 15 - Question 71
Which of the following solvents is used for dry cleaning of clothes ?
RRB Group D Mock Test - 15 - Question 72
Which of the following are true regarding the compact fluorescent tubes now available in market for home use? (1) They use less power (about 20%) compared to filament type bulbs for same amount of light (2) They operate at higher voltage (3) They are narrower and shorter than common fluorescent tubes
RRB Group D Mock Test - 15 - Question 73
The relatively greater weight of an object at the poles is attributable to :
RRB Group D Mock Test - 15 - Question 74
The effect of circular movements on a person in a satellite around the earth is that :
RRB Group D Mock Test - 15 - Question 75

The difference between the simple interest on a certain sum of money at 6% per annum for 10 years and at 5% per annum for 2 years is Rs. 100. Find the sum

Detailed Solution for RRB Group D Mock Test - 15 - Question 75

( P x 6 x 10 /100 ) - ( P x 5 x 2/100 ) = 100
⇒ 60P - 10P = 10000
⇒ 50P = 10000
⇒ P = 200
⇒ Principle = Rs 200

RRB Group D Mock Test - 15 - Question 76

(14)2 + 289 ÷ 17 = ?

Detailed Solution for RRB Group D Mock Test - 15 - Question 76

142 + 289 ÷ 17

= 142 + 172 /17

= 196 + 17 = 213

RRB Group D Mock Test - 15 - Question 77

10- ? = 100 - 86

RRB Group D Mock Test - 15 - Question 78
Baron Pierre de Coubertin, father of the modern Olympic Games, belongs to :
RRB Group D Mock Test - 15 - Question 79
The Sports Minister launched the three-stroke Khelo India logo recently in New Delhi at the Nehru Stadium. The logo has been designed by ________
RRB Group D Mock Test - 15 - Question 80
Who won the WTA Auckland Open Tennis title, in Auckland, New Zealand?
RRB Group D Mock Test - 15 - Question 81
How long will a boy take to run round a square field of side 35 metres, if he runs at the rate of 9 km/hr?
RRB Group D Mock Test - 15 - Question 82
Below are the statements followed by three conclusions numbered I,II and III. You have to consider the statements and the following conclusions and decide which of the conclusion(s) follows the statement(s).
Statements :
a. All books are cars.
b. All cars are tigers.
Conclusions :
I. All books are cars.
II. All tigers are books.
III. All cars are books.
Detailed Solution for RRB Group D Mock Test - 15 - Question 82
Conclusions
I) True
II) False
III) False
Only conclusion (I) follows
RRB Group D Mock Test - 15 - Question 83
If A, B, C, D are the angles of a cyclic quadrilateral, then $\mathrm{cos}A+\mathrm{cos}B+\mathrm{cos}C+\mathrm{cos}D$ =
RRB Group D Mock Test - 15 - Question 84

A mother and a doughter working together can complete a work in 4 days. But if the mother works alone, she can complete the work in 6 days. Both of them worked for one day and then the mother had to leave. How long will the daughter take to complete the remaining work?

Detailed Solution for RRB Group D Mock Test - 15 - Question 84

(M + D)'s one day's work = 1/4
Remaining work = 3/4
M's one day's work = 1/6
D's one day's work = 1/4 - 1/6 = 1/12
∴ 1/12 th of the work is done by the daughter in 1 day
3/4 th of the work will be done by daughter in
12 x 3/4 = 9 days

RRB Group D Mock Test - 15 - Question 85
A cylindrical tennis ball container can contain maximum three balls stacked on one another. The top and bottom balls also touch the lid and the base of the base of the container respectively. If the volume of a tennis ball is 240 cm3, then what is the volume of the container?
RRB Group D Mock Test - 15 - Question 86
How many meaningful words can be made from the letters 'LMEA' using each letter only once in each word?
Detailed Solution for RRB Group D Mock Test - 15 - Question 86 The words formed are:
(1) MALE
(2) MEAL
(3) LAME
RRB Group D Mock Test - 15 - Question 87
How many major organs does theU.N have?
RRB Group D Mock Test - 15 - Question 88
Complete the analogous pair.
125 : 17 18 21 :: 458 : ?
RRB Group D Mock Test - 15 - Question 89
Rig Veda, the oldest of Vedas, chiefly contains?
RRB Group D Mock Test - 15 - Question 90

In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer.

I. x2 + 5x + 6 = 0
II. y2 + 3y + 2 = 0

Detailed Solution for RRB Group D Mock Test - 15 - Question 90

I. x2 + 5x + 6 = 0
⇒ x2 + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2(x + 3) = 0
⇒ x = -2, -3
Also,

II. y2 + 3y + 2 = 0
⇒ y2 + 2y + y + 2 = 0
⇒ y(y + 2) + 1(y + 2) = 0
⇒ y = -1, -2
Hence, x ≤ y

RRB Group D Mock Test - 15 - Question 91

Each of the following questions consists of two sets of figures. Figures 1,2,3 and 4 constitute the Problem Set while figures A,B,C,D and E constitute the Answer Set. There is a definite relationship between figures 1 and 2. Establish a similar relationship between figures 3 and 4 by selecting a suitable figure from the Answer Set .

RRB Group D Mock Test - 15 - Question 92

In the following problem, out of the five figures marked 1, 2, 3, 4 and 5, four are similar in a certain manner. However, one figure is not like the other four. Choose the figure which is different from the rest.

RRB Group D Mock Test - 15 - Question 93

In the following questions, select a figure from amongst the four alternatives, which when placed in the blank space of fig. (X) would complete the pattern.

RRB Group D Mock Test - 15 - Question 94

Eight friends A, B, C, D, E, F, G and H are sitting around a circle facing the centre. H is to the immediate left of E who is third to the right of B. C is second to the right of D and is not a neighbour of B. F is second to the right of G and is not a neighbour of C.

Q.  Who is second to the right of E?

Detailed Solution for RRB Group D Mock Test - 15 - Question 94

RRB Group D Mock Test - 15 - Question 95

Find the number in the series which is wrong.

6, 12, 36, 144, 722, 4320, 30240

Detailed Solution for RRB Group D Mock Test - 15 - Question 95

6, 12, 36, 144, 722, 4320, 30240
The pattern should be
6, 6 × 2 = 12, 12 × 3 = 36, 36 × 4 = 144,
144 × 5 = 720, 720 × 6 = 4320, 4320 × 7 = 30240
Hence, the wrong entry is 722

RRB Group D Mock Test - 15 - Question 96

Find out from amongst the four alternatives as to how the pattern would appear when the transparent sheet is folded at the dotted line.

RRB Group D Mock Test - 15 - Question 97

In the following questions, the symbols \$, @, ©, % and # are used with the following meaning as illustrated below:
'P @ Q' means 'P is not smaller than Q'.
'P # Q' means 'P is not greater than Q'.
'P % Q' means 'P is neither greater than nor smaller than Q'.
'P © Q' means 'P is neither smaller than nor equal to Q'.
'P \$ Q' means 'P is neither greater than nor equal to Q'.
Now in each of the following questions assuming the given statements to be true, find which of the two conclusions I and II given below them is/are definitely true?
Give answer (1) if only Conclusions I is true.
Give answer (2) if only Conclusion II is true.
Give answer (3) if either Conclusion I or II is true.
Give answer (4) if neither Conclusion I nor II is true.
Give answer (5) if both conclusions I and II are true.

Q.

Statements : V \$ W, W @ T, T # H
Conclusions : I. V © T
II. H % W

RRB Group D Mock Test - 15 - Question 98

Q. What was the grand average marks of the five subjects in 1996 ?

Detailed Solution for RRB Group D Mock Test - 15 - Question 98

Average marks of the five subjects in 1996

RRB Group D Mock Test - 15 - Question 99

Q. The difference in the average marks in History between 1944 and 1995 was exactly equal to the difference in the highest marks in Hindi between which of the following pairs of years ?

Detailed Solution for RRB Group D Mock Test - 15 - Question 99

Difference in avg marks in History between 1994 and 1995 was 58 - 49 = 9
Difference in highest marks in Hindi between pair of years = 84 - 75 = 9
i.e. 1995 and 1992

RRB Group D Mock Test - 15 - Question 100

Q. What was the approximate percentage increase in average marks in History from 1992 and 1993 ?

Detailed Solution for RRB Group D Mock Test - 15 - Question 100

Difference of Averge marks in History from 1992 and 1993 is = 55 - 46 = 9.
Approximate percentage increase = 9/46 × 100 = 19.56 ≅ 20

## RRB (Group D) - Mock Tests & Previous Year Papers

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text:
# Making a new year

I know it may seem as if I’m ready to throw in the towel on blogging—only 78 posts last year, and 15 of those were in January—but I’m not. Last year was very busy in both my professional and personal life, and my internet life had to take a back seat because of that. Also, there is the siren song of Twitter, which makes it so easy to say nearly what you want, especially if you’re willing to cheat and let your thoughts spill over into a second or third tweet.1 My goal for the year is to focus my tweeting on stupid jokes and replies to stupid jokes. Other stuff should go here, even when it’s relatively short.

Like this post would have been if I hadn’t done so much throat-clearing in the first paragraph.

To start the blogging year, I needed to make this year’s folder and monthly subfolders for the Markdown source files. This isn’t hard to do in the Finder, but it’s much more fun to do it in the Terminal. I go to the source directory and issue these commands

mkdir 2017
cd 2017
jot -w %02d 12 | xargs mkdir

The “2017” in the second line doesn’t have to be typed out: Escape-period (⎋.) is the shortcut that enters the final item of the last line into the current line, much like typing ↑ enters entire previous command. And like ↑, typing ⎋. repeatedly works backward through your command history, entering the final item of each previous line in turn.

The interesting line, and the one that’s the most fun, is the third line, which populates the 2017 folder with subfolders for each month, like this:

The jot command prints out the numbers from 1 to 12, with the -w %02d switch formatting the single-digit numbers with a leading zero (the -w switch uses the printf formatting directives). Then the xargs command takes that output and turns it into arguments for mkdir, which creates all the subdirectories. Boom.

For many years, I thought xargs was this really complicated command that only true Unix wizards could control. And while it can be used in complex ways, it is at heart a pretty simple utility. It’s sort of like a pipe, but instead of turning the standard output of one command into the standard input of another, it turns that standard output into a list of arguments for the second command. Once I understood that concept, it didn’t scare me anymore.

1. Strings of tweets longer than that are an abomination, worse than a textshot of your own words. When I see someone tweeting “Thread,” I stay away.


text:
# Polynomial division with reference to CRC32

The polynomials we're using will always have coefficients of either 1 or 0.

We have a divisor polynomial which will always remain the same, we'll call this polynomial $D$. We have a dividend polynomial, which is variable, and will be called polynomial $X$.

We'll divide $X$ by $D$ and take the remainder, $R_1$ $$X \equiv R \pmod D$$

We'll then modify $X$ in the following way:

We'll add some terms to it, for example three terms. Before we do this, we'll promote all the existing terms by the number of terms we wish to add. So in this case we'll multiply $X$ by $x^3$. Then we'll add, for example, $x^2 + x + 1$.

Finally we'll divide this modified $X$ (which we'll call $X_1$) by $D$ again, giving us remainder $R_2$.

For example:

$$X \equiv R_1 \pmod D$$ $$x^{32} + x^{14} + x^{12} + x^3 \equiv R_1 \pmod D$$

Terms to add: $x^4 + x^2 + 1$

$$X_1 \equiv R_2 \pmod D$$ $$x^{37} + x^{19} + x^{17} + x^8 + x^4 + x^2 + 1 \equiv R_2 \pmod D$$

What we would like to derive is a formula whereby we can predict $R_2$ given only $R_1, D$and any terms we wish to add to $X$.

Thanks for your time. If I haven't been clear but you still wish to help, please don't hesitate to ask for clarification.

• Thanks for the edit dextero4, makes everything much clearer. – user116446 Dec 18 '13 at 18:46

Since the $\pmod D\;$ operation is linear, you get from $X_1 = X + Y,\;$ where $Y$ are the additional terms with $R_Y \equiv Y \pmod D$, the relation $$R_2 \equiv R_1 + R_Y \pmod D$$


text:
# Tetrahedron splitting/subdivision

Given a regular Tetrahedron A (i.e. each edge of A has same length), is it possible to split A into several smaller regular tetrahedra of equal size? I.e. smaller tetrahedra should completely fill volume of A, and they should not overlap.

This can be done in 2D with a triangle and square, and it can be done in 3D with cube (i.e. you can split cube into several smaller cubes of equal size). But I see no way to do same thing in 3D with tetrahedron.

If this can be done, how (how smaller tetrahedra should be positioned)?
If this cannot be done, is there a proof that this is impossible?

P.S. I'm not a mathematician, and this is not a homework, but I'd like to know how/if this can be done.

-
See also here: mathoverflow.net/questions/131176/… –  Benjamin Dickman May 22 '13 at 0:07

The answer is: No. There is a somehwat rambling discussion here. Let $B$ be a smaller tetrahedron that is jammed into the apex of $A$. It fills the solid angle there completely. Let $e$ be a base edge of $B$. Then one cannot fill the neighborhood of $e$ by gluing in further regular tetrahedra along it. One way to see this is that the dihedral angle of the tetrahedron is $\delta = \cos^{-1}(1/3) \approx 70.5^\circ$, and the dihedral angle along $e$ to be filled is $\pi - \delta \approx 109.5^\circ$, which cannot be formed from copies of $\delta$.

-
Did not see Anton's answer while I was writing mine. Mine is a version of his Answer 1. –  Joseph O'Rourke Jun 18 '10 at 12:14
Accepting this one because of the link to discussion. –  SigTerm Jun 18 '10 at 12:43

Answer 1: Look what happens on a face of the big tetrahedron where some edges of small ones come together: you have to make angle 180° out of some dihedral angles of the tetrahedra (which is about 70°) --- that is impossible.

Answer 2: There is the so-called Dehn invariant. If a polyhedron $X$ is split into a number of polyhedra $Y_1,Y_2,\dots,Y_n$ then the Dehn invariant of $X$ is equal to the sum of the Dehn invariants of $Y_i$.

For the regular tetrahedron, the Dehn inveriant is nonzero and proportional to the length of a side. Suppose you could split a regular tetrahedron with side $a$ into a number of tetrahedra with sides $a_1, a_2,\dots, a_n$. Then from the volume you have $$a_1^3+a_2^3+\dots+a_n^3=a^3$$ and from the Dehn invariant you have $$a_1+a_2+\dots+a_n=a.$$ It follows that there is no nontrivial splitting.

-
Anton, why does the first answer not work for cubes? Secondly, I am just curious whether subdivision of "1/3" of cube (halfs of three faces with one joint vertex) into equal congruent parts is possible. –  Wadim Zudilin Jun 18 '10 at 12:03
The Dehn invariant of cube is $0$. –  Anton Petrunin Jun 18 '10 at 12:04
And for the first answer, the dihedral angle of a cube is 90 degrees, which evenly divides 180 degrees. –  Gerald Edgar Jun 18 '10 at 13:32
The volume should involve cubes of $a_i$s. –  Victor Protsak Jun 19 '10 at 2:30

It is however possible to split a regular tetrahedron into four smaller regular tetrahedra and one regular octahedron, and there are other possibilities based on the tetrahedral-octahedral honeycomb.

-

This question has already been pretty handily answered by Anton, but I figured I'd add another (somewhat ridiculous) proof to his list:

Answer 3: Suppose that we had an infinite subdivision of regular tetrahedra. Then we would also have a crystallographic lattice with tetrahedral symmetry, since this structure would have to tile across 3D space. However, the classification of crystallographic lattices rules out this possibility!

http://en.wikipedia.org/wiki/Crystallographic_group

-
Classification of crystallographic lattices relies on some angles dividing $\pi$, so this is a convoluted way of phrasing Anton's Answer 1. –  Victor Protsak Jun 19 '10 at 2:28


text:
# Thread: quick fraction simplification help

1. ## quick fraction simplification help

Hey, just need some help with a fraction simplification problem.

how can i make [k(k+1)(k+2)]/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1)

and then take that step and have it equal [(k+1)(k+2)(k+3)]/3

i slightly understand the second to third step but not how to go from the first to the second (factor out a k but how?) thanks.

2. Originally Posted by Xender1
Hey, just need some help with a fraction simplification problem.

how can i make [k(k+1)(k+2)]/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1)

and then take that step and have it equal [(k+1)(k+2)(k+3)]/3

i slightly understand the second to third step but not how to go from the first to the second (factor out a k but how?) thanks.

$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

$=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$

$=\frac{(k+1)(k+2)(k+3)}{3}$

$=(k+1)(k+2)(\frac{k+3}{3})$

$
=(k+1)(k+2)(k/3 + 1)
$


text:
## Photon States

It is now obvious that the integer is the number of photons in the volume with wave number and polarization . It is called the occupation number for the state designated by wave number and polarization . We can represent the state of the entire volume by giving the number of photons of each type (and some phases). The state vector for the volume is given by the direct product of the states for each type of photon.

The ground state for a particular oscillator cannot be lowered. The state in which all the oscillators are in the ground state is called the vacuum stateand can be written simply as . We can generate any state we want by applying raising operators to the vacuum state.

The factorial on the bottom cancels all the we get from the raising operators.

Any multi-photon state we construct is automatically symmetric under the interchange of pairs of photons. For example if we want to raise two photons out of the vacuum, we apply two raising operators. Since , interchanging the photons gives the same state.

So the fact that the creation operators commute dictates that photon states are symmetric under interchange.

Jim Branson 2013-04-22


text:
# .

In differential geometry, Liouville's equation, named after Joseph Liouville, is the nonlinear partial differential equation satisfied by the conformal factor f of a metric $$f^2(\mathrm{d}x^2 + \mathrm{d}y^2)$$ on a surface of constant Gaussian curvature K:

$$\Delta_0\log f = -K f^2,$$

where $$\Delta_0 is the flat Laplace operator. \( \Delta_0 = \frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \bar z}$$

Liouville's equation appears in the study of isothermal coordinates in differential geometry: the independent variables x,y are the coordinates, while f can be described as the conformal factor with respect to the flat metric. Occasionally it is the square $$f^2$$ that is referred to as the conformal factor, instead of f itself.

Liouville's equation was also taken as an example by David Hilbert in the formulation of his nineteenth problem.[1]

Other common forms of Liouville's equation

By using the change of variables " $$\log \,f = u$$ ", another commonly found form of Liouville's equation is obtained:

$$\Delta_0 u = - K e^{2u}.$$

Other two forms of the equation, commonly found in the literature,[2] are obtained by using the slight variant " $$2\log \,f = u$$ " of the previous change of variables and Wirtinger calculus:[3]

$$\Delta_0 u = - 2K e^{u}\quad\Longleftrightarrow\quad \frac{\partial^2 u}{{\partial z}{\partial \bar z}} = - \frac{K}{2} e^{u}.$$

Note that it is exactly in the first one of the preceding two forms that Liouville's equation was cited by David Hilbert in the formulation of his nineteenth problem.[1][4]
A formulation using the Laplace-Beltrami operator

In a more invariant fashion, the equation can be written in terms of the intrinsic Laplace-Beltrami operator

$$\Delta_{\mathrm{LB}} = \frac{1}{f^2} \Delta_0$$

as follows:

$$\Delta_{\mathrm{LB}}\log\; f = -K.$$

Properties
Relation to Gauss–Codazzi equations

Liouville's equation is a consequence of the Gauss–Codazzi equations when the metric is written in isothermal coordinates.
General solution of the equation

In a simply connected domain $$\Omega$$ , the general solution of Liouville's equation can be found by using Wirtinger calculus.[5] Its form is given by

$$u(z,\bar z) = \frac{1}{2} \ln \left( 4 \frac{ \left|{\mathrm{d} f(z)}/{\mathrm{d} z}\right|^2 }{ ( 1+K \left|f(z)\right|^2)^2 } \right)$$

where f(z) is any meromorphic function such that

$${\mathrm{d} f(z)}/{\mathrm{d} z}\neq 0 for every z \in \Omega$$ .[5]
f(z) has at most simple poles in $$\Omeg$$ a.[5]

Application

Liouville's equation can be used to prove the following classification results for surfaces:

Theorem.[6] A surface in the Euclidean 3-space with metric $$\mathrm{d}l^2 = g(z,{\bar z}){\mathrm{d}z}{\mathrm{d} \bar z}$$ , and with constant scalar curvature K is locally isometric to:

the sphere if K > 0;
the Euclidean plane if K = 0;
the Lobachevskian plane if K < 0.

Notes

See (Hilbert 1900, p. 288): Hilbert does not cite explicitly Joseph Liouville.
See (Dubrovin, Novikov & Fomenko 1992, p. 118) and (Henrici, p. 294).
See (Henrici, pp. 287–294).
Hilbert assumes K = -1/2, therefore the equation appears as the following semilinear elliptic equation:

$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = e^f$$

See (Henrici, p. 294).

See (Dubrovin, Novikov & Fomenko 1992, pp. 118–120).

References

Mathematics Encyclopedia


text:
# Small-Step Walks

(May 2016)

We provide here a few mathematical tables that present the data computed in our work:

We also provide some of the data obtained for $Q(x,y;t)$ as Maple source files:

Each of these files contains a Maple table, with index of the form [$ss$, $vx$, $vy$, $mode$], where $ss$ is the step set index (between 1 and 19), $vx$ and $vy$ are the respective values of $x$ and $y$ for which creative telescoping was applied, and $mode$ represents the ordre of execution of the two nested integrations (either "xy", for $x$ after $y$, or "yx", for $y$ after $x$).

Finally, those data are also provided as a Maple library.


text:
## Thursday, August 13, 2009

### More on coins

Geeky Mom writes:

Yesterday, I filled a bag with about half of the coins and trudge to the bank on my way to the farmer's market. The bank lets you estimate how much money you have and if you're close, you win a prize. As I poured the money into the machine, I saw mostly pennies. So, I estimated about 7.50. When it was all said in done, I had slightly over \$52 in coins. Mostly dimes, it turned out. I'm really bad at estimating, especially now that I rarely handle cash, much less coins. All my cash is digital, exchanged either via electronic transfer or similarly, using my debit card. I used to keep tips in a mayonnaise jar on my dresser, saving up for the deposit on an apartment in graduate school town. I know about how much was there, in part because I knew how much I made in tips, but also because I dealt with cash all the time.

I spent a little more than half of my new found cash at the farmer's market. Even though I always take cash there (most of the vendors don't take other forms of payment), I felt a little giddy at having such a large amount, created, it seemed, out of thin air.

Money now does seem to come out of thin air, arriving in bank accounts without anyone having to touch anything. I used to work at a bank during the summers. One summer I filed loan applications, the 3 attached parts left after everything was signed off. Another summer I filed the checks people deposited into their accounts, checks that were then sent to other banks to be filed and then placed in an envelope to be sent to the customer with her statement. Even then, the real transaction happened electronically, with a machine reading routing and account numbers, a human inputting the amounts, which were then coded onto the check to be read by another machine. For a brief time each summer, it was my job to count money coming in from the vendors at the annual summer festival. Bags of coins and dollar bills showed up at the bank and I stood behind the tellers, counting it all by hand, recording amounts on deposit slips, amounts that were later entered into computers while the money itself went into the vault, to be redistributed to banks or to customers withdrawing money.

Her post captures how a lot of people have lost touch with their "number sense" about money, because most monetary transactions are about electrons zapping around from place to place.

It's actually very hard to pay your income tax bill in cash. Even if you walk into one of the few remaining IRS taxpayer assistance centers with a wad of twenty dollar bills trying to pay off your tax bill, they are reluctant to accept it. Basically, the IRS is just not set up to take cash. They really want you to use their electronic payments system or mail them a check. And don't even think about trying to pay your tax bill with a wheelbarrel full of pennies!

Ideally, what makes it as simple as possible for the IRS is for your employer to (slightly) overpay your taxes through withholding from your wages throughout the year, and then they'll send you a refund (preferably by direct deposit) after you file your tax return at the end of the year.

Today's Wall Street Journal has an op-ed from Charles Murray saying that this approach is "bad for Democracy."

He proposes that we replace the current system under which most American pay most of their taxes through withholding from their wages with a system under which all American taxpayers would instead just send in quarterly tax payments to the IRS.


text:
# Solution: When comparing acids with a formula HA, where A can be any o...

###### Question

When comparing acids with a formula HA, where A can be any one element, what is the trend in terms of acid strength?

A. Increases as A goes toward the top, left of the periodic table

B. Increases as A goes toward the top, right of the periodic table

C. Increases as A goes toward the bottom, left of the periodic table

D. Increases as A goes toward the bottom, right of the periodic table


text:
1. ## Proof Help!!

I'm not sure on how to prove statements with "if and only if"
Prove these using only field axioms, order axioms and the fact that 1>0 and x0=0
a>0 iff a^-1>0
a>1 iff 0<a^-1<1

2. Hello amm345
Originally Posted by amm345
I'm not sure on how to prove statements with "if and only if"
Prove these using only field axioms, order axioms and the fact that 1>0 and x0=0
a>0 iff a^-1>0
a>1 iff 0<a^-1<1
We need to establish first that $\Big((a>0) \land (b<0)\Big) \Rightarrow (ab < 0)$.

Now one of the axioms of an ordered field is that $\Big((a>0) \land (b>0)\Big) \Rightarrow (ab >0)$; and it is well established that $a(-b)=-(ab)$ (see here for instance.)

So $(a>0) \land (b<0)$

$\Rightarrow (a>0) \land ((-b)>0)$

$\Rightarrow a(-b)>0$

$\Rightarrow -(ab)>0$

$\Rightarrow ab < 0$, since $-(ab) +ab = 0$

We can now prove the first of the required statements by contradiction,as follows:

First consider the case where $a>0$ and assume that $a^{-1} < 0$. Then:

$\Big((a >0) \land (a^{-1}<0)\Big) \Rightarrow (a.a^{-1} <0)$ using the result above.

But $a.a^{-1} = 1>0$. Contradiction. Therefore $(a>0) \Rightarrow (a^{-1} >0)$.

Then prove it the other way around, and consider the case where $a^{-1} >0$, and assume that $a<0$.

$\Big((a^{-1} >0) \land (a<0\Big)) \Rightarrow (a^{-1}.a<0)$, again using the same result.

But $a^{-1}.a=1>0$. Contradiction. Therefore $(a^{-1}>0) \Rightarrow (a>0)$

We have now proved the implication in both directions. Hence $(a^{-1}>0) \iff (a>0)$

To prove the second of your two statements, establish first that $\Big((a>1)\land(b>1)\Big) \Rightarrow (ab >1)$. Hint:
$(b>1) \Rightarrow (b=1+c)$ for some $c>0$. Then use the distributive law on $a.(1+c)$.

Then use a contradiction method as I did for the first part, to show that $(a>1)\Rightarrow (a^{-1}<1)$. Hence, using the result above that $(0.

Finally, prove the result in the other direction to show that $(a>1) \iff (0.


text:
## Stupid Law Suits

This is our lounge area. Feel free to come in and get acquainted!
ed
Posts: 39391
Joined: Tue Jun 08, 2004 11:52 pm
Title: G_D

### Re: Stupid Law Suits

Rob Lister wrote:On going to Lowes website, this is what I find

Home Depot lists it in the specifications
Common: 2 in. x 4 in. x 8 ft.; Actual: 1.5 in. x 3.5 in. x 96 in.
Note that my answer does not have to rely on visual aids, it stands on its own. That is why it is better
Rob Lister
Posts: 23119
Joined: Sun Jul 18, 2004 7:15 pm
Title: Incipient toppler
Location: Swimming in Lake Ed

### Re: Stupid Law Suits

Yours is better. But can you make wood?
gnome
Posts: 24517
Joined: Tue Jun 29, 2004 12:40 am
Location: New Port Richey, FL

### Re: Stupid Law Suits

I can gripe about W all day, but this moment in the 2004 debates was a masterpiece of comedic timing.
"If fighting is sure to result in victory, then you must fight! Sun Tzu said that, and I'd say he knows a little bit more about fighting than you do, pal, because he invented it, and then he perfected it so that no living man could best him in the ring of honor. Then, he used his fight money to buy two of every animal on earth, and then he herded them onto a boat, and then he beat the crap out of every single one. And from that day forward any time a bunch of animals are together in one place it's called a zoo! (Beat) Unless it's a farm!"
--Soldier, TF2
Rob Lister
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Title: Incipient toppler
Location: Swimming in Lake Ed

### Re: Stupid Law Suits

An Alabama man who said he tripped and broke his hip while buying a watermelon at a Walmart store has won a $7.5 million verdict in his lawsuit against the retailer. Henry Walker on Wednesday was awarded the damages after a jury trial in Phenix City, Alabama, just west of the Georgia state line, court records show. Walker had sued Arkansas-based Wal-Mart Stores Inc., saying his foot became trapped in a pallet beneath the watermelons as he reached for one of the fruits at a Phenix City Walmart on June 25, 2015. Al.com reports that when the then-59-year-old turned back toward his shopping cart, he fell. One of his lawyers says that Walker's days of playing basketball three times a week ended with the injury, which now requires him to use a walker. http://www.newser.com/story/251409/he-f ... icher.html I too have had my foot caught in a pallet. Pallets are hazardous like that. I think there is a shared responsibility here. Poor man broke his hip and will have to use a walker until the case is utterly settled, wherein he can toss it out. 7.5 million is a little much. I award him$100K and he can keep the watermelon.
asthmatic camel
Posts: 19706
Joined: Sat Jun 05, 2004 1:53 pm
Title: Forum commie nun.
Location: Stirring the porridge with my spurtle.

### Re: Stupid Law Suits

Judge camel says:

"You, sir, are a dumbass. If you are incapable of carrying out such a simple task as buying a watermelon without seriously injuring yourself, then I recommend that you stay at home and send somebody else who is. Either that or book yourself a room at the Hotel California. Case dismissed, costs awarded to the defendant."
Shit happens. The older you get, the more often shit happens. So you have to try not to give a shit even when you do. Because, if you give too many shits, you've created your own shit creek and there's no way out other than swimming through the shit. Oh, and fuck.
gnome
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Location: New Port Richey, FL

### Re: Stupid Law Suits

7.5m seems scaled not to the plaintiff's need or portion of responsibility, but as punitive damages. This usually means at some point the defendant pulled an ill-advised dick move and the jury got angry when they were informed of it.

It's not 100% reliable as a maxim, but I'd be willing to bet a small amount on it this time.

This is why (I would think) that large businesses usually make a settlement with plaintiffs injured on the premises even if they were not necessarily in the wrong, and carry liability insurance for such occasions. Something broke that script this time.
"If fighting is sure to result in victory, then you must fight! Sun Tzu said that, and I'd say he knows a little bit more about fighting than you do, pal, because he invented it, and then he perfected it so that no living man could best him in the ring of honor. Then, he used his fight money to buy two of every animal on earth, and then he herded them onto a boat, and then he beat the crap out of every single one. And from that day forward any time a bunch of animals are together in one place it's called a zoo! (Beat) Unless it's a farm!"
--Soldier, TF2
Rob Lister
Posts: 23119
Joined: Sun Jul 18, 2004 7:15 pm
Title: Incipient toppler
Location: Swimming in Lake Ed

### Re: Stupid Law Suits

gnome wrote:7.5m seems scaled not to the plaintiff's need or portion of responsibility, but as punitive damages. This usually means at some point the defendant pulled an ill-advised dick move and the jury got angry when they were informed of it.

It's not 100% reliable as a maxim, but I'd be willing to bet a small amount on it this time.

This is why (I would think) that large businesses usually make a settlement with plaintiffs injured on the premises even if they were not necessarily in the wrong, and carry liability insurance for such occasions. Something broke that script this time.
Meh. Fuck you gnome. He gets $100K. That's it. Fuck the jury. Their biased because their all foot-hangers. On a serious note: it is easy to get ones foot caught in a pallet. That's why the 100K. This an't their first foot-hanger. They've been warned before. THAT's why the 7.5 mil. Still ... 100K. Not a penny more. P.S. I am channeling CH. Anaxagoras Posts: 28291 Joined: Wed Mar 19, 2008 5:45 am Location: Yokohama/Tokyo, Japan ### Re: Stupid Law Suits We now have a resolution to that lumber lawsuit I mentioned earlier: Lumber lawsuit against Menards dismissed; judge says retailer didn't lie about its 4x4s A federal judge has slammed the door on the Illinois lumber shoppers who sued Menards claiming it deceived them about the size of its 4x4s. Saying no reasonable consumer would regard Menards’ descriptions of its lumber the way plaintiffs Michael Fuchs and Vladislav Krasilnikov said they did, the judge last week dismissed the would-be class action lawsuit against the Wisconsin-based home-improvement chain. The decision by U.S. District Judge Edmond Chang throws out a case in which Menards was accused of deception because it marketed and labeled its 4x4s without specifying that the boards measure 3½ by 3 ½ inches. So-called dimensional lumber — 2x4s, 4x4s, 2x6s and such — is commonly sold by names that do not specify the measurements of the pieces. The longstanding industry convention is recognized by the U.S. Department of Commerce, which distinguishes between the “nominal” designations for pieces of lumber and their actual size. The department says a 2x4, for example, can measure 1½ inches thick by 3½ inches wide. The distinction between the name and the actual dimensions stems from the fact that lumber, when it is produced, typically is trimmed to smooth it after the initial rough cut, Chang said in his decision. The government endorsement of the industry practice supports Menards’ argument that it didn’t falsely market its lumber, Chang said. Further, he said, the labels on the lumber “are literally true,” because they do not show inch-mark symbols after the customary trade names of the lumber. “Without a literally untrue statement, combined with the government-recognized distinction between nominal sizes and actual sizes, no reasonable consumer would think that the labels showed the exact dimensions of the lumber,” Chang said. Beyond that, the judge said, Fuchs and Krasilnikov — who alleged they discovered they had been deceived when they got home and measured their lumber — easily could have used their tape measures in the store. “It would be one thing if packaging prevented access to the height and width. … It is another thing where the plaintiffs can readily see if there is a mismatch between what they perceive as the size on the label and the height and width of the lumber,” Chang said. In a brief statement, Menards applauded the decision by Chang, who sits in the northern district of Illinois. “We feel that it was a frivolous lawsuit at best and we’re glad that common sense prevailed,” Menards spokesman Jeff Abbott said by email. An attorney with McGuire Law, the Chicago-based class action firm that represented Fuchs and Krasilnikov, did not respond to a request for comment. McGuire Law also represents the plaintiff in a similar lawsuit against Home Depot. That case is pending. Home Depot lawsuit still pending but I don't see why the result should be any different. I think the defendants should be awarded attorney costs too. A fool thinks himself to be wise, but a wise man knows himself to be a fool. William Shakespeare Doctor X Posts: 73004 Joined: Fri Jun 04, 2004 8:09 pm Title: Collective Messiah Location: Your Mom ### Re: Stupid Law Suits They have no comment since they lost their profit. Remind me to bring a tape measure when buying lumber. --J.D. Mob of the Mean: Free beanie, cattle-prod and Charley Fan Club! "Doctor X is just treating you the way he treats everyone--as subhuman crap too dumb to breathe in after you breathe out." – Don DocX: FTW. – sparks "Doctor X wins again." – Pyrrho "Never sorry to make a racist Fucktard cry." – His Humble MagNIfIcence "It was the criticisms of Doc X, actually, that let me see more clearly how far the hypocrisy had gone." – clarsct "I'd leave it up to Doctor X who has been a benevolent tyrant so far." – Grammatron "Indeed you are a river to your people. Shit. That's going to end up in your sig." – Pyrrho "Try a twelve step program and accept Doctor X as your High Power." – asthmatic camel "just like Doc X said." – gnome WS CHAMPIONS X4!!!! NBA CHAMPIONS!! Stanley Cup! SB CHAMPIONS X6!!!!!! sparks Posts: 16935 Joined: Fri Oct 26, 2007 4:13 pm Location: Friar McWallclocks Bar -- Where time stands still while you lean over! ### Re: Stupid Law Suits 2 by 4s haven't been 2 by 4s for a long time... So much so, that when you do come across them, it's usually a big fucking problem. And how 'bout those new-fangled metal studs? "I smell a big fat commie rat Mr. President!" You can lead them to knowledge, but you can't make them think. Doctor X Posts: 73004 Joined: Fri Jun 04, 2004 8:09 pm Title: Collective Messiah Location: Your Mom ### Re: Stupid Law Suits THANKS OBAMA! --J.D. Mob of the Mean: Free beanie, cattle-prod and Charley Fan Club! "Doctor X is just treating you the way he treats everyone--as subhuman crap too dumb to breathe in after you breathe out." – Don DocX: FTW. – sparks "Doctor X wins again." – Pyrrho "Never sorry to make a racist Fucktard cry." – His Humble MagNIfIcence "It was the criticisms of Doc X, actually, that let me see more clearly how far the hypocrisy had gone." – clarsct "I'd leave it up to Doctor X who has been a benevolent tyrant so far." – Grammatron "Indeed you are a river to your people. Shit. That's going to end up in your sig." – Pyrrho "Try a twelve step program and accept Doctor X as your High Power." – asthmatic camel "just like Doc X said." – gnome WS CHAMPIONS X4!!!! NBA CHAMPIONS!! Stanley Cup! SB CHAMPIONS X6!!!!!! ed Posts: 39391 Joined: Tue Jun 08, 2004 11:52 pm Title: G_D ### Re: Stupid Law Suits I thought that the measurement was becasue they were measured green and that after drying they shrank. Perhaps our resident expert on shrinkage, Mr. Lister, would care to put in his 6" (2.5 actual)? About that stereo shemp Posts: 6783 Joined: Thu Jun 10, 2004 12:16 pm Title: inbred shit-for-brains Location: Planet X ### Re: Stupid Law Suits This means my house should be 25% bigger! I'm suing! "It is not I who is mad! It is I who is crazy!" -- Ren Hoek "what dicking deep shit i produce" -- pillory Freedom of choice Is what you got Freedom from choice Is what you want People are shitting themselves to death Crap so much they fail to take a breath But even when their kids are starvin' They thought Trump would throw them Charmin. Rob Lister Posts: 23119 Joined: Sun Jul 18, 2004 7:15 pm Title: Incipient toppler Location: Swimming in Lake Ed ### Re: Stupid Law Suits A federal judge in Chicago has tossed a class-action lawsuit that argued McDonald's was duping consumers because a single "Extra Value Meal" costs more than the sum total of the individual components of the meal bought separately. The Chicago Daily Law Bulletin reported Monday that Kelly Killeen said she bought a sausage burrito breakfast Extra Value Meal in Chicago for$5.08. But she said the "value" label was deceptive because buying the same two sausage burritos, hash browns, and coffee individually would have cost a total of \$4.97. Judge Elaine Bucklo noted that prices for combo meals and individual items are easily visible from the counter, the AP reports. She said just because some consumers don't want to bother to compare prices doesn't mean they can claim they've been fooled.
I don't know if this actually falls under stupid but it's pretty stupid.
Grammatron
Posts: 35355
Joined: Tue Jun 08, 2004 1:21 am
Location: Los Angeles, CA

### Re: Stupid Law Suits

Kelly Killeen also testified that the individual prices were legible and visible.
Rob Lister
Posts: 23119
Joined: Sun Jul 18, 2004 7:15 pm
Title: Incipient toppler
Location: Swimming in Lake Ed

### Re: Stupid Law Suits

So it's a better value for McDonald
gnome
Posts: 24517
Joined: Tue Jun 29, 2004 12:40 am
Location: New Port Richey, FL

### Re: Stupid Law Suits

I mean, fuck McDonalds for being sneaky, but I don't see it as actionable. I'd get my money back by getting customer service and saying I'm dissatisfied with the experience. That would probably do more to change it anyway. Isn't McDonalds one that will send you coupons or something if you complain?
"If fighting is sure to result in victory, then you must fight! Sun Tzu said that, and I'd say he knows a little bit more about fighting than you do, pal, because he invented it, and then he perfected it so that no living man could best him in the ring of honor. Then, he used his fight money to buy two of every animal on earth, and then he herded them onto a boat, and then he beat the crap out of every single one. And from that day forward any time a bunch of animals are together in one place it's called a zoo! (Beat) Unless it's a farm!"
--Soldier, TF2
sparks
Posts: 16935
Joined: Fri Oct 26, 2007 4:13 pm
Location: Friar McWallclocks Bar -- Where time stands still while you lean over!

### Re: Stupid Law Suits

Simple: Don't go there anymore.

Hell, we all know clowns are evil, right?
You can lead them to knowledge, but you can't make them think.
Grammatron
Posts: 35355
Joined: Tue Jun 08, 2004 1:21 am
Location: Los Angeles, CA

### Re: Stupid Law Suits

sparks wrote:Simple: Don't go there anymore.

Hell, we all know clowns are evil, right?
They got rid of the clown
sparks
Posts: 16935
Joined: Fri Oct 26, 2007 4:13 pm
Location: Friar McWallclocks Bar -- Where time stands still while you lean over!

### Re: Stupid Law Suits

Really? How do you know this? I mean really, for sure.

He's still there, pulling strings. You know he is.
You can lead them to knowledge, but you can't make them think.


text:
# Observational Evidence of the Dark matter Existence

### Problem 1

Analyze the problem of a small satellite galaxy moving in the gravitational field of a large galaxy (the problem of non-decreasing behavior of the rotational curves.)

### Problem 2

What radial dependence of the density of a spherically symmetric galactic halo corresponds to the constant velocity of satellite galaxies?

### Problem 3

We have noted above that outside the optically observed range of a galactic disk the rotation curves of the dwarf satellite galaxies do not depend on the galacto-centric distance (the flatness of the rotational curves). External regions of the galaxies are mostly filled by cold neutral hydrogen. Evidently the rotation velocity of the gas gives important information about the mass distribution in the galaxy. How can one measure that velocity?

### Problem 4

Using the virial theorem, express the mass of a galaxy cluster through the observed quantities---the average velocity of galaxies in the cluster and its size. Estimate the mass of the Coma cluster (Berenice's hair) for $R\approx10^{23}m$, $\langle v^2\rangle^{1/2}\approx2\cdot10^6m/s$.

### Problem 5

Show that the age of the matter--dominated Universe contradicts observations.

### Problem 6

A galaxy has the visible mass of $10^{11}M_\odot$ and a horizontal rotation curve up to distance of $30kpc$ at velocity $250km/s$. What is the ration What is the dark to visible mass ratio in the galaxy?


text:
# Producing antiprotons

1. Apr 28, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
What is the minimum proton energy needed in an accelerator to produce antiprotons $\bar{p}$ by the reaction

$$p+p \to p+p + (p+\bar{p})$$

The mass of a proton and antiproton is m_p.

2. Relevant equations

3. The attempt at a solution
I said 2 m_p c^2 because that is the sum of the rest energies of a proton and an antiproton. Apparently that is wrong.

2. Apr 28, 2008

### kamerling

The velocities of the reaction products can't all be 0. Use conservation of momentum.

3. Apr 28, 2008

### ehrenfest

Hmm. So, the momentum four-vector of the system has to conserved, right? Is it true that the total energy of the system has to be conserved, right? I was always confused about that.

The total energy is relativistic kinetic energy + rest energy =$$\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}mc^2$$.

OK. I see now. The four-momentum contains the total energy as one of its components. And if the four-momentum is conserved then all of its components are conserved. But please confirm that four-momentum is conserved here.

I am not really sure how to implement that though. I will probably need

$$E^2 = p^2 c^2 +m^2 c^4$$

What is the first thing I should do?

Last edited: Apr 28, 2008
4. Apr 28, 2008

### tnho

I think you may try to use four-momentum vector approach.
This will never wrong~

and i think the answer is something like $$7m_p c^2$$

Last edited: Apr 28, 2008
5. Apr 28, 2008

### Hootenanny

Staff Emeritus
Four momentum is always conserved. You can make the calculation much easier by noting that the square of the energy-momentum four-vector is Lorentz invariant (i.e. doesn't change with a Lorentz boost).

6. Apr 28, 2008

### Staff: Mentor

Does one assume that the two initial protons collide head on? That would produce the minimal energy. If one proton is at rest then there is recoil to consider.

Also, when a particles kinetic energy is equal to rest mass, then one must consider relativistic effects. Total energy and momentum are conserved.

7. Apr 28, 2008

### Hootenanny

Staff Emeritus
That would be the case if one of the protons were stationary, but that isn't the most efficient method of creating a particle-anti-particle pair

Note also that your value contains the rest mass of the proton.

Edit: It seems Astro beat me to it

8. Apr 28, 2008

### ehrenfest

OK good so four-momentum is conserved. I am still trying to figure out what the first thing I should do is! So, if we label the six particles in the order from left to right that they appear in the reaction we get

$$\sum_{i=1}^2 \mathbf{p}_i - \sum_{i=3}^6 \mathbf{p}_i = 0$$

by conservation of four-momentum. What should I do with that? That is 4 equations containing 24 different quantities. How do I deal with that?

Last edited: Apr 28, 2008
9. Apr 28, 2008

### nrqed

There is no way to answer the question before you say in what frame you want the energy to be calculated!! If the reaction is in the center of mass frame, then the energy of each colliding proton must be 2 mc^2 but you said that was wrong?

10. Apr 28, 2008

### ehrenfest

The back of my book says the answer is
.

Honestly, I cannot tell you what frame I want the energy calculated. I posted the question in its entirety! Can you tell me what frame to calculate the energy now that I gave you the answer?

11. Apr 28, 2008

### reilly

1.The center of mass coordinate system will give the easiest approach to the problem--why?

2.Simply apply the appropriate conservation rules -- how many?-- to relate before and after -- everything will work out just fine. (Think about vectors, four vectors, and/or components thereof, and which might be simplest descriptive tool available to you.

Good luck,
Reilly Atkinson

(This is a very standard problem; an important one as well.)

12. Apr 28, 2008

### nrqed

What they are doing is considering a frame in which one proton is smashing against a second proton initially at rest.

Call P_1 the four-momentum of the moving proton, p_2, the four-momentum of the proton at rest.

Call P_3,P_4,P_5, P_6 the four-momenta of the produced particles. The minimum energy corresponds to the case when they have all the same four-momentum.

we have $$P_1 = (E_1, \vec{p_1}), P_2 = (m_p c^2, \vec{0}), P_3=P_4=P_5=P_6 = (E', \vec{p'})$$

Conservation of four-momentum gives

$$P_1 + P_2 = 4 P'$$

squaring, you get

$$P_1^2 +P_2^2 + 2 P_1 \cdot P_2 = 16 P'^2$$

all the squares give P^2 = m_p^2 c^4.

Finish the calculation and isolate E_1.

Patrick

13. Apr 28, 2008

### ehrenfest

OK. So we go to the CM coordinate system. In that system the 3-momentum is 0 before the collision, so it has to be zero after the collision. The energy in that system before the collision is given by

$$E^2 = p_1^2+p_2^2 + 2m_p c^2$$

where p_1 and p_2 are the relativistic momentum of protons 1 and 2.

E^2 must also be conserved.

Is that right?

EDIT: I didn't read the post above when I wrote this.

14. Apr 28, 2008

### ehrenfest

Do you know that because you have done the problem before or because that is obvious or what? To prove that statement, should I first assume that the four-momenta are different and then use Lagrange multipliers or calculus or something?

What do you mean by this? Should that be P'?

15. Apr 28, 2008

### nrqed

I know that because in the center of mass frame, it's obvious that the minimum energy corresponds to all the particles in th efinal state being at rest. If you boost to the lab frame, they all have the same four-momentum
I meant that for *any* of the P you get that. So that P could be P' or P_1 or P_2 , etc.

16. Apr 28, 2008

### pam

That problem is generally given for the target proton at rest, which agrees with the answer in the book. Use the invariance of E^2-p^2, so (4m)^2=(E_lab + m)^2-(p_lab)^2.
Solve for E_lab.

17. Apr 28, 2008

### tnho

See if i have done anything wrong or not~

First do it in the frame where the target proton (marked 2 below) is stationary.
$$P_1 = (E_1/c,p_1,0,0) \mbox{ ; } P_2 = (mc,0,0,0)$$

By conservation of four-momentum,
$$P_1 + P_2 = P_3 + P_4 +P_5 +P_6$$

$$(P_1 + P_2)^2 = (P_3 + P_4 +P_5 +P_6)^2$$

$$(P_1 + P_2)^2 = P_1^2 + 2P_1 \cdot P_2 + P_2^2 = -2m^2c^2 -2mE_1$$

$$(P_3 + P_4 +P_5 +P_6)^2 = -4m^2c^2$$ (evaluate in CM frame for minimum energy)

Then,
$$-2m^2 c^2 -2mE_1 = -16 m^2c^2$$

$$E_1 = 7mc^2$$

=)

Last edited: Apr 28, 2008
18. Apr 29, 2008

### ehrenfest

OK. I understand everything except how you proved that the 4-momenta of the four products of the reaction should be the same to obtain the minimum energy E_1.

So nrqed and tnho say they go to the center-of-mass frame to do this. What do we know about the center-of-mass frame in special relativity? In the nonrelativistic case, we know that the sum of the linear momenta of the particles is 0 in the center-of-mass frame. In special relativity, is it true that the sum of the relativistic momenta is 0 in the center-of-mass frame?

My book talks about the "center-of-momentum" frame although I am not really sure what that means. Is that the same as the center-of-mass frame here?

19. Apr 29, 2008

### tnho

Yes. I think so. The sum of linear momentum of the particles is zero in the centre of mass frame in SR.

20. Apr 30, 2008

### reilly

For relatively simple scattering problems, elastic scattering for example, the lab frame - target at rest -- works nicely, and is the standard frame used in most particle physics.

However, for more complex situations involving particle production, the Center of Mass/Momentum frame is preferred, because of the simplicity it affords. (In practice, the Center of Mass and Center of Momentum mean the same thing, here the frame in which the two protons have equal and opposite momentum.) But, to get back to the lab frame requires a Lorentz transformation.

A hint for simplification: let the initial protons move along the z axis - can this be true in both the lab and CM frames?

With this approach, the answer given by nqrd should be very apparent.

Regards,
Reilly Atkinson