Hii everyone. This note is a type of quiz game where you can solve questions and post them too. The advantage of this note is you can practice both solving and posting good questions. Your solutions and questions will be appreciated too by community members. So, enjoy the beauty of solving the questions. Go ahead :

**Rules And Regulations :**

In this game the questions will be posted and by the community members itself.

The one who solves the latest question first will be able to post a new question. You should provide a clear solution to the question you have solved.

**Only the one who first solves the latest question correctly will have a chance to post the new question.**If any member has a different solution to a question which has already been solved he can also post the solution but he cannot post a new question.

There is also a leader board (below) in which the top 10 highest solvers will be shown. So try to solve as many questions as possible.

The one who posted the question should not be the first one to solve that question. However he can post his solution if he wants only after anybody solves his question.

**Summary**

**Total Questions Solved** : 6

**Latest Question Posted** : Problem 6

**Leader Board (Top 10 highest Solvers)** :

**NAME - PROBLEMS SOLVED**

$1$.Vilakshan Gupta - 2

$2$. Ram Mohith - 2

$3$. Vaibhav Priyadarshi and R Mathe and Michael Mendrin and Tom Clancy - 1

$4$. Matin Naseri - 1

If any improvements are required in this game kindly inform them to me.

Thank You and **All the Best**

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$</code> ... <code>$</code>...<code>."> Easy Math Editor

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestThis is a nice discussion. Really enjoyed

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Problem 1(solved by Vaibhav Priyadarshi)Find the value of $\log_2 10 - \log_8 125$

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$log_{2}10-log_{8}125$

=$log_{2}10-log_{2^3}125$

=$log_{2}10-(log_{2}125)/3$

=$(3log_{2}10-log_{2}125)/3$

=$(log_{2}10^3-log_{2}125)/3$

=$(log_{2}(1000/125))/3$

=$(log_{2}8)/3$

=3/3 =1.

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Now can I post a new question?

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Please use \log in your solution.

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Another solution for Problem 1 :$\log_8 125 = \log_{2^3} 125 = \dfrac{1}{3}\log_2 125 = \log_2 \sqrt[3]{125} = \log_2 5$

$\log_2 10 - \log_8 125 = \log_2 10 - \log_2 5$

$\implies \log_2 \dfrac{10}{5} = \log_2 2 = 1$

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$\textbf{Problem 2 :}$

Find the value of $(5+2\sqrt{13})^{1/3}+(5-2\sqrt{13})^{1/3}$

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$\frac{1}{4}(1+3\sqrt{13}+i(\sqrt{39}-\sqrt{3}))$?

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Sir, if this is correct you can post a new question but whether it is correct or not Vibhav should tell it.

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Some things look complicated but just do the math and you will get the answer.

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hint, note that $a^3+b^3=10$. Rest is upto youLog in to reply

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$(a^3+b^3)=(a+b)(a^2-ab+b^2)=10$, but I don't see where the "1" comes up. Both factors are complex conjugates.

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If you let the expression to be equal to $x$ and then cube it, then you will get the equation $10-9x=x^3$ and from here, by

observation, we can conclude that the answer has to be $1$.The other two roots have to be complex conjugates, but the answer in real numbers is $1$.

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Can you give explanation for it? Why isn't this a real number? Principal Cube root of real number should be real.

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The answer is 1.

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Yes it is correct!

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How is it "1"? This expression is a root of a 4th degree polynomial with no real roots.

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$(5+2\sqrt{13})^{1/3}$ + $(5-2\sqrt{13})^{1/3}$ = $x$

LetCubing both sides, we get

$10+3(5^2-(2√13)^2)^{1/3}x$ = $x^3$

$10-9x=x^3$

It have 1 as a solution.

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$a^3+b^3+3ab(a+b)$

By usingHere a+b=x.

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$(5^2-(2\sqrt{13})^2))^{\frac{1}{3}}$ is complex. You slipped up on sign. Or at least about fractional powers of $-1$.

Let me try to see if I can save this....

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$(-27)^{1/3}$ not real?

IsLog in to reply

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$(-27)^{\frac13}$ is an imaginary number , but we are concerned about the real valued root , and that is , $-3$.

I agree that the principal root ofCheck this

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Vaibhav shall I update the note as Vilakshan Gupta solved your question. What do you think?

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Consider an arbitrary polynomial $X^{3}+aX+b\in\mathbb{R}[X]$. Then, provided $\Delta:=(a/3)^{3} + (b/2)^{2} > 0$, letting

$C:=-b/2+\sqrt{\Delta},\quad D:=-b/2-\sqrt{\Delta}$

one has (taking real cubed roots) that $C^{1/3}+D^{1/3}$ is

the unique real rootof $X^{3}+aX+b$.Now choosing $b:=-10$ and $a:=9$, one has $5+2\sqrt{13}=5+\sqrt{52}=C$ and $5-2\sqrt{13}=5-\sqrt{52}=D$ (one may obtain the values of $a,b$ by simply solving $-b/2=5$ and $\Delta=52$. Thus the expression in the problem, $\breve{x}:=(5+2\sqrt{13})^{1/3}+(5-2\sqrt{13})^{1/3}$, is equal to the unique real root of

$X^{3}+9X-10$

Now, note that $1$ is a root of this polynomial. Thus it is the unique real root. Hence $\boxed{\breve{x}=1}$.

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Good. Now you too solved this question but you cannot post the new question because Vilakshan Gupta already solved it. But this solution comes into your account.

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$(5+2\sqrt{13})^{\frac{1}{3}}-(-5+2\sqrt{13})^{\frac{1}{3}}$, then, yes, the value becomes $1$. But what is done is to assume that $(-a)^{\frac{1}{n}}=-(a)^{\frac{1}{n}}$, which is not true for all $n>1$. That's not a safe step to take when doing computations, because what that does is to "convert a complex quantity into a real".

If we rewrote the expression toThis is a controversal subject, involving fractional powers of complex numbers, which are multi-valued, and this sort of thing crops up on Brilliant from time to time.

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@Ram Mohith

But this solution comes into your account.I have no idea what this sentence means. Regardless. I solved it for the sake of solving it. I don’t pay attention to who does what first nor do I look at other people’s solutions until I’ve done things myself.@Michael Mendrin I worked completely in $\mathbb{R}$. For odd $n\in\mathbb{N}$ the unique real $n$th root of $-a$

doescoincide with $-(a^{1/n})$. I do appreciate that if we work in $\mathbb{C}$ it is all a matter of convention: there is no compelling reason to choose one argument over another (iethe choice of $k\in\mathbb{Z}$ in $\frac{\theta+2\pi k}{n}$).Log in to reply

$(-2)(-2)=4$ doesn't mean it's safe to assume that $\sqrt{4}=-2$. Likewise fractional powers of negative numbers. Whenever I see something like that arise, it's never safe to "convert a complex into a real".

Well, right, what do we do if we have multiple values for this sort of thing? Just like just because we know thatWhat's the value of $(-1)^{\frac{1}{3}}$?

I meant to explore this more in detail this morning, but this Question Game moves on.

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$(-1)^{\frac{1}{3}}$ should be $(-1)^{\frac{1}{3}}$, which denotes a multiple-valued complex quantity. It's kind of analogous to some quantities in quantum physics, huh? "It can have a range of values, depending on how you look at it".

The value ofLog in to reply

$n$th root function $f:z\in\mathbb{C}\mapsto \sqrt[n]{|z|}\exp(\imath\frac{\arg(z)+2\pi k}{n})$ is holomorphic except along a semi-infinite line.

It is not a multivalued object (except maybe in set theory). In algebra, it’s just ill-defined. And in complex analysis one acknowledges the existence of many inverses, and the consequences of this being any choice of theLog in to reply

Why is my value $\frac{1}{4}(1+3\sqrt{13}+i(\sqrt{39}-\sqrt{3}))$ any less valid than "1"?

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$f:\mathbb{C}\longrightarrow\mathbb{C}$. Being multi-valued means a function of the form $f:\mathbb{C}\longrightarrow\wp(\mathbb{C})$. Both of these are technically true. But in algebra, it is not sensible to work with the latter.

Being ill-defined here means there be no single-valued (inverse) functionYour result is fine. The whole point of this discussion, was you disputing mine. I defended my result by stating that I’m working in the framework $\mathbb{R}$. You’re working in the framework $\mathbb{C}$. Which framework one chooses is another issue. As soon as these are fixed, the answers are what they are—within the framework $\mathbb{R}$ the only valid result is $1$. In your framework, which is perfectly legitimate your result is one of several perfectly valid answers.

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This is not the first time I've run to this sort of thing in Brilliant.org, where powers of complex numbers are involved. Several problems have already been posted to explore this topic.

An annoyance that I have is that for the sake of rigor, a function of the form $y=f(x)$ is, by convention, defined to be single valued. This is the reason why $y=\sqrt{x}$ by convention only yields positive $y$, i.e. only the top half of the parabola. But lots of implicit functions of the form $f(x,y)=0$ are multi-valued in both the $x$ and $y)$ directions. This is the reason why whenever complex quantities are involved, especially powers of them, it raises red flags for me and I move with caution, and ask myself, "okay, what's the convention with this, and in what context is this? " This is just from habit now.

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issensible preference however, when one demands restricting a solution to say a particular field. Nonetheless, as I said, one always needs to fix a framework and work in it. If things happen to have more solutions in a larger framework, then so be it. For example asking, whether $X^{2}+X+1$ be irreducible without specifying an algebraic structure, is simply begging for problems, as the answers is yes or no, depending on the context (if the field is $\mathbb{R}$, $\mathbb{Q}$, $\mathbb{C}$, $\mathbb{F}_{2}$, … or if one works in a ring, etc.).Log in to reply

odd$n\in\mathbb{N}$. Foroddpowers there is only one real root.For $n\in\mathbb{N}$ odd, the map $f:x\in\mathbb{R}\mapsto x^{n}\in\mathbb{R}$ is continuous and unbound above and below, thus the image (as a continuous image of a connected set) is all of $\mathbb{R}$, moreover the map is injective. Hence there is a unique Inverse in $\mathbb{R}$. Thus for all $a\in\mathbb{R}$ there is exactly one value $b\in\mathbb{R}$ such that $b^{n}=a$. This is

the unique real $n$th root. Eg, the uniquereal $3$rd rootof $-1$ is $-1$.I disagree with this. What is sensible is

it is never safe to not fix a framework. And fix a framework I did. There is no compulsion to work in one particular field. Sinceeg$\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}\subseteq\mathrm{GL}(\mathbb{C},3)\subseteq\ldots$ are field extensions, one could by your logic also argue, that it’s never safe to work in $\mathbb{C}$ and that one should always work in some $\mathrm{GL}(\mathbb{C},3)$.Log in to reply

$(-1)^{\frac{1}{3}}$"? There is no one unique answer,

Let me try to explain what I think the problem is. Suppose instead of something as complicated as the original expression, the problem asks, "what's the value ofunlessthe problem asks for the unique real root. If one argues that "1" is a valid value of the original complicated expression, I can equally argue that my complex value is equally valid.What I am saying, in ordinary mathematical computations, one must be careful about assuming only the unique real roots of such fractional powers of negative quantities. Otherwise, I could "show" that $-1 = \frac{1}{2} +\frac{\sqrt{3}}{2}i$, for example (and therefore -1= 3!)

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It's creative. I want to join into your discussion.

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You can. In fact I am encouraging everyone to join this. Try to solve any question the first and you can post a new question.

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$\log_2 10 - \log_8 125$

Thus $\log_2 5 ={\log_8 125}$

$\log_2 10 -$$\log_2 5 =$$\log_2{\frac{10}{5}}=$$\log_2 2 = 1$

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Also I have fixed my code,My last line had Errors.

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$\text{Problem 3:}$

In the $\triangle ABC$ ,$O$ is any arbitrary point in the interior of the triangle. Lines $OA$ , $OB$ , $OC$ are drawn such that the angles $OAB$,$OBC$ and $OCA$ are each equal to $\omega$ , then prove that $\cot A+\cot B+\cot C=\cot \omega$ and $\csc^2A+\csc^2B+\csc^2C=\csc^2\omega$

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Ok. It is tough but no problem. But from next time onwards try to post problems of Easy and Medium difficulty.

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Let $a,b,c$ be the angles of the triangle and $x$ be the angle common to all 3 vertices. Then using Law of Sines, we have

$\dfrac{Sin(x)}{OB}=\dfrac{Sin(b-x)}{OA}$

$\dfrac{Sin(x)}{OC}=\dfrac{Sin(c-x)}{OB}$

$\dfrac{Sin(x)}{OA}=\dfrac{Sin(a-x)}{OC}$

Hence

$4{ \left( Sin(x) \right) }^{ 3 }=4Sin(a-x)Sin(b-x)Sin(c-x)$

And from there on, using product-to-sum and sum-to-product trig identities, we have

$3Sin(x)-Sin(3x)=Sin(2a+x)+Sin(2b+x)-Sin(2(a+b)-x)-Sin(3x)$

then after rearrangement and adding the term $-Sin(2a-x)$ on both sides

$-Sin(2a-x)+Sin(x)+Sin(2(a+b)-x)-Sin(2b+x)=-Sin(2a-x)-2Sin(x)+Sin(2a+x)$

then

$-4Cos(x)Sin(a)Sin(b)Sin(a+b)+4Cos(a)Sin(b)Sin(a+b)Sin(x)=$ $-4Cos(b)Sin(a)Sin(a+b)Sin(x)+4Cos(a+b)Sin(a)Sin(b)Sin(x)$

and finally

$Sin(a)Sin(b)Sin(c)Sin(x)(Cot(a)+Cot(b)+Cot(c)-Cot(x))=0$

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Good solution sir. If it is correct you are the one who can post the next question. But let's see what Vilaskhan Gupta says.

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Can you explain your first step, from where did the first equation come?

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Two unit disks are randomly thrown on the plane. If they overlap, what's the expected distance between their centers?

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$\text{Problem 4:}$

Two unit disks are randomly thrown on the plane. If they overlap, what's the expected distance between their centers?

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1

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Ah, that's what's wrong with this problem, if I say your answer is wrong, then somebody else will probably figure out the correct answer.

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I solved it like this: If they overlap, maximum distance between their centres would be 2(if they just touches each other). Minimum distance between their centres would be 0, when they are coincident. So expected distance can be average of maximum and minimum which is 1.

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The answer is $\boxed{0}$.

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$\boxed{0.5}$.

Okay, thus I think the final answer isWhat about $\boxed{0.5}$?

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It's not 0.5 and it's not 1/2

Pick a fraction between 0 and 2.

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I know two methods of doing it, but I know none!

One method is by using an excel sheet and other is by integration (both of which I don't know)

Just another guess....0.67

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Yes, the correct answer is 2/3

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Sir, I will mark this as solved by Vilakshan Gupta. I am giving chance for Mohammad Farhan to post the latest problem.

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I think this game lacks number of solvers

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Yes, probably.

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76 members have seen this note but only 6 people are participating.

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If many people start to participate then this game would be really creative.

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Yes,Maybe. but How to we should solve the issue?

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Do you have any idea ?

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Where is the 5th problem

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It is not posted till because nobody solved question 4.

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Problem 5: (Solved By Ram Mohith)

There were 5 friends namely Beatricia, Jing Kai, Emmanuel, Freddy and Benjamin.

Beatricia : 9 sweets

Jing Kai: 7 sweets

Benjamin : ? sweets

Emmanuel : 8 sweets

Freddy : 6 sweets

What is ?.

(LaTeX is hard)

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7 sweets or 5 sweets. I had both explanations. I you tell the correct answer I will post the relevant one.

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WRONG! YOU LOSE A LIFE! Sorry TWO LIVES! 1 MORE TRY

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The correct answer is $8$ sweets.

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Explain

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Solution for Problem 5The number of letters in each person's name is the count of sweets they have.

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That's Right! That's what the Gifted Education Programme in Singapore has

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You are now tied With Vilakshan Gupta

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But Vilakshan Gupta answer two questions which are not posted by himself. But I answered one question which I itself posted and now your question. So, first priority is given to him.

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Ram, You forgot to follow (Combinatorics) and (Electricity and Magnetism)

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Because those chapters are not started yet in our academics.

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Oh

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But do you know permutations?

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$\text{Problem - 6}$

A ball falls on surface from $10~m$ height and rebounds to $2.5~m$. If the duration of contact with floor is $0.01$ seconds. Find the average acceleration during the contact ?

$(A) 2100~m/s^2$

$(B) 1400~m/s^2$

$(C) 700~m/s^2$

$(D) 400~m/s^2$

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Wow, you are even kinder. You gave options

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How many times can we pick the options?

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It not based on how many times you attempt. The first one who gives the correct solution with explanation is said to be as the first solver.

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(D) because 10/2.5 =4

according to acceleration = m/s^2

so 4/0.01=400

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No. wrong explanation.

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I think the correct answer is $2100{\frac{m}{s^2}}$ but I'm not sure about it.

Note: you have a typo in your first line. Find no fid

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Yes you are correct. But you should provide your explanation.

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$2.5×10=25;{\frac{25}{0.01}}=2500$

$10÷2.5= 4; {\frac{4}{0.01}}=400$

$2500 - 400 = 2100$

Hence the answer is $2100{\frac{m}{s^2}}$

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@Ram Mohith, When will @Tom Clancy post a new question

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Don't know. As he is a new user he came to brilliant only 2 or 3 days and then he stopped.

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@Ram Mohith, can you please post another question?

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