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RMO - 2015 Paper discussion (Rajasthan Region)

Time - 3hrs.

Instructions :

  • Calculators (in any form) and protector are not allowed.

  • Rulers and compasses are allowed

  • All questions carry equal marks. Maximum marks : 102.

  1. Let ABC be a triangle. Let B' and C' denote respectively the reflection of B and C in the internal angle bisected of \(\angle A \). Show that the ABC and AB'C' have the same incentre.

  2. Let \(P(x)=x^{2}+ax+b \) be a quadratic polynomial with real coefficients. Suppose there are real numbers \(s \neq t\) such that \(P(s) =t\) and \(P(t) =s\). Prove that \(b-st\) is a root of the equation \(x^{2}+ax+b-st=0\)

  3. Find all integers \(a, b, c\) such that \[a^{2}=bc+1 \ \\\, \\\ b^{2}=ca +1\]

  4. Suppose 32 objects placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of three chosen objects are adjacent nor diametrically opposite ?

  5. Two circles \(T\) and \(\sum\) in the plane intersect at two distinct points A and B, and the center of \(\sum\) lies on \(T\). Let points C and D on \(T\) and \(\sum\) respectively such that C, B and D are co-linear. Let point E be on \(\sum\) such that DE is parallel to AC. Show that AE=AB.

  6. Find all real numbers \(a\) such that \(4<a<5\) and \(a(a-3\{a\}) \) is an integer.
    {Here \(\{a \} \) represents fractional part of \(a\)}.

Note by Shubhendra Singh
12 months ago

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Answer to question 3.\[a^{2}-b^{2}=c(b-a)\] This implies \[a+b+c=0 ...or... a=b\] Case 1\[a+b+c=0\] Solutions are \[(a,b,c)=(1,0,-1),(-1,0,1),(0,1,-1),(0,-1,1),(1,1,0),(1,-1,0),(-1,1,0),(-1,-1,0).\] Case 2\[a=b \] Therefore \[(a,b,c)=(1,1,0),(-1,-1,0).\]. Shivam Jadhav · 12 months ago

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@Shivam Jadhav Why are solutions like \((2,-1,-1)\) not included? Anik Mandal · 11 months, 2 weeks ago

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@Anik Mandal Maybe it's not but how do we know that these are the only solutions? Anik Mandal · 11 months, 1 week ago

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@Anik Mandal By subtracting the given equations we get \(a^{2}-b^{2}=-c(a-b)\) \[\Rightarrow (a+b)(a-b) =-c(a-b) \]

This gives that the equation will have a solution when \(a-b=0 \) or \(a+b=-c\). That's how we get to know about the solutions we are gonna get. Shubhendra Singh · 11 months, 1 week ago

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@Shubhendra Singh It means all the solutions of \(a+b+c = 0\) are solutions? Anik Mandal · 11 months, 1 week ago

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@Anik Mandal No, this is just a condition between a, b, c that will help us in getting the solutions. Shubhendra Singh · 11 months, 1 week ago

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@Anik Mandal I think it's not a solution, please check it again. Shubhendra Singh · 11 months, 2 weeks ago

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@Shubhendra Singh What do you think the cut off would be.of GMO 2015 paper? Adarsh Kumar · 11 months, 2 weeks ago

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@Adarsh Kumar I think it's gonna be around 50-55, the results could be declared any time as the result of rmo have been declared. How much are you accepting?? Shubhendra Singh · 11 months, 2 weeks ago

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@Shubhendra Singh Did you get selected? Kushagra Sahni · 11 months, 1 week ago

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@Kushagra Sahni Yes 8-) Shubhendra Singh · 11 months, 1 week ago

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@Shubhendra Singh What About Dev Sharma? Kushagra Sahni · 11 months, 1 week ago

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@Kushagra Sahni nope and u? Dev Sharma · 11 months, 1 week ago

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@Dev Sharma Dev you are in which class ?? Chirayu Bhardwaj · 8 months, 1 week ago

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@Dev Sharma I also didn't get selected. I did 4 but my writing was horrible. Kushagra Sahni · 11 months, 1 week ago

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@Kushagra Sahni Would it be sahni at place of juneja!! Dev Sharma · 11 months, 1 week ago

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@Dev Sharma I wish there was, Next year there will be Sahni and no Juneja. Interestingly my mother's surname was Juneja before marriage. Kushagra Sahni · 11 months, 1 week ago

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@Shubhendra Singh How many marks did you got? And Congrats! Harsh Shrivastava · 11 months, 1 week ago

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@Harsh Shrivastava I don't know how much I have scored, how did you get to know about your marks. Shubhendra Singh · 11 months, 1 week ago

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@Shubhendra Singh Its written in the result list of my state. Harsh Shrivastava · 11 months, 1 week ago

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@Harsh Shrivastava There's nothing like this in rajasthan's list, even the cut-off is not mentioned. Shubhendra Singh · 11 months, 1 week ago

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@Harsh Shrivastava Are you selected for INMO @Harsh Shrivastava Chinmay Sangawadekar · 11 months, 1 week ago

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@Chinmay Sangawadekar Luckily yes. Harsh Shrivastava · 11 months, 1 week ago

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@Harsh Shrivastava Why luckily ? you are intelligent bro . Unfortunately , our principal doesn't allow us for RMO.@Harsh Shrivastava Chinmay Sangawadekar · 11 months, 1 week ago

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@Chinmay Sangawadekar Why? Harsh Shrivastava · 11 months, 1 week ago

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@Harsh Shrivastava Dont know . H e thinks we should focus on our school math and so , Chinmay Sangawadekar · 11 months, 1 week ago

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@Harsh Shrivastava Congrats Harsh, 52 is a good score. All the best for INMO Mayank Chaturvedi · 11 months, 1 week ago

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@Mayank Chaturvedi Thanks! Harsh Shrivastava · 11 months, 1 week ago

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@Harsh Shrivastava Meet you at INMO training camp on 14th-15th. Mayank Chaturvedi · 11 months, 1 week ago

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@Mayank Chaturvedi What's this?? Shubhendra Singh · 11 months, 1 week ago

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@Shubhendra Singh Camp for RMO qualified from Chhattisgarh Mayank Chaturvedi · 11 months, 1 week ago

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@Mayank Chaturvedi Oh you have also qualified? Congo!

Btw where's training camp? Harsh Shrivastava · 11 months, 1 week ago

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@Harsh Shrivastava B.C.S.Govt. P.G. College, Dhamtari C.G on January 14-15, 2016. All the details are written on top of the result list Mayank Chaturvedi · 11 months, 1 week ago

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@Shivam Jadhav How do u know a=b from the equation a+b+c=o Ashish Sacheti · 12 months ago

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@Ashish Sacheti Well here is how he arrived at the conclusion.

\((a-b)(a+b)=-c (a-b) \)

Thus we have two cases:

Case 1:

\((a-b)=0\) or \(a=b\)

Case 2:

If \(a\not=b\) , we cancel out \((a-b)\) from botb sides to obtain \(a+b=-c\) or \(a+b+c=0\) Sualeh Asif · 12 months ago

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@Ashish Sacheti I am also not sure at how you arrive at that conclusion Mardokay Mosazghi · 12 months ago

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@Shivam Jadhav That's the best way. Shubhendra Singh · 12 months ago

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@Shivam Jadhav Did the same !!! Nice and standard solution btw... Chinmay Sangawadekar · 12 months ago

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6)\[\{a\}=a-4\] \[a(a-3a+12)=-2a^2+12a\] \[-2a^2+12a=n\] \[(a-3)^2=\dfrac{-n+18}{2}\] \[(4-3)^2<(a-3)^2<(5-3)^2\] \[1<\dfrac{-n+18}{2}<4\] \[10<n<16\] put values n=11,12,13,14,15 to find \[a=\dfrac{6+\sqrt{36-2n}}{2}=3+\dfrac{\sqrt{14}}{2},3+\dfrac{\sqrt{12}}{2},3+\dfrac{\sqrt{10}}{2},3+\dfrac{\sqrt{8}}{2},3+\dfrac{\sqrt{6}}{2}\] Aareyan Manzoor · 12 months ago

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@Aareyan Manzoor Sorry to say that, substitute a = 4 + 1 / 2 + sqrt 6 Easha Manideep D · 12 months ago

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@Easha Manideep D Do you mean \( 4 + \frac{1}{\sqrt{6} + 2} \)

\( 4 + \frac{1}{\sqrt{6} + 2} = 4 + \frac{\sqrt{6} - 2}{2} = 3 + \frac{\sqrt{6}}{2} \) Siddhartha Srivastava · 12 months ago

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@Siddhartha Srivastava ya Easha Manideep D · 12 months ago

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@Easha Manideep D i am sorry. couldnt understand. that is >5 Aareyan Manzoor · 12 months ago

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@Aareyan Manzoor Hey ! 4 + 1/ something is never greater than 5 and that something is > 1 Easha Manideep D · 12 months ago

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@Easha Manideep D

\[\] i am unable to understand what you are saying. Aareyan Manzoor · 12 months ago

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@Aareyan Manzoor How did you get the RHS in the second line ? Vishal Yadav · 12 months ago

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@Vishal Yadav second line: we deduced \(\{a\}=a-4\). we substitute this value into the given expression to find \[a(a-3\{a\})=a(a-3(a-4))=a(a-3a+12)=a(-2a+12)=-2a^2+12\]. hope this helps. Aareyan Manzoor · 12 months ago

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@Aareyan Manzoor What answer were you getting for Q.4 Shubhendra Singh · 12 months ago

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@Shubhendra Singh sorry man, my combinatorics is weak(very) Aareyan Manzoor · 12 months ago

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@Aareyan Manzoor Perfect, I too got the same answer. Shubhendra Singh · 12 months ago

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Answer to question 2\[P(s)=s^{2}+as+b=t,P(t)=t^{2}+at+b=s\] \[P(s)-P(t)=s^{2}-t^{2}+a(s-t)=t-s\] \[P(s)-P(t)=s+t+a+1=0\] \[P(s)P(t)=(st)^{2}+ast^{2}+bt^{2}+bs^{2}+bas+b^{2}+as^{2}t+a^{2}st+abt=st\] \[(st)^{2}+b^{2}-st=-(ast^{2}+as^{2}t+a^{2}st+bt^{2}+abt+bs^{2}+bas)\] \[(st)^{2}+b^{2}-st=-(ast(a+s+t)+bt(a+t)+bs(a+s))\] \[(st)^{2}+b^{2}-st=(ast(1)+bt(1+s)+bs(1+t))\] \[(st)^{2}+b^{2}-st=(ast+b(s+t)\] \[(st)^{2}+b^{2}-st=(ast-b(1+a)\] \[(st)^{2}+b^{2}-st-ast+b+ab=0\] But \[P(b-st)-st=(st)^{2}+b^{2}-st-ast+b+ab\] Therefore \[P(b-st)-st=0\] Hence proved. Shivam Jadhav · 12 months ago

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@Shivam Jadhav Similar Approach ... Nice and elegant solution. Chinmay Sangawadekar · 12 months ago

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Q2 Consider the polynomial \(Q(x)=x^2+ax+b-st\) When \(Q(x)=0\), we realise that if \(b-st\) is a root then the other root is 1 by Vieta's formulas. So the sum of the roots is \((b-st+1)\). We can factor it to \(Q(x)=(x-(b-st))(x-1).\) Now realise that this means \[a=-(b-st+1) \implies st=a+b+1\]. Thus if we can prove this we are done. Now move to the other pat of the information provided. Realise that \[P(s)-t=P(t)-s, \\ s^2 -t^2 =a(t-s) +(t-s) \\ a=-1-s-t\] We can cancel \(t-s\) out since \(t\not=s\). Substitute \( a=-1-s-t\) back into \[P(s)-t=0 \implies b=st+s+t \\ \therefore a+b+1=st\] And we are done! Sualeh Asif · 12 months ago

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@Sualeh Asif In this question instead of proving b-st as the root it is sufficient to prove that 1 is the root of the eqn. Isn't it?? Mohit Gupta · 12 months ago

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@Mohit Gupta Well using vietas formulas you see that the multipleof the two roots is \(b-st\). Hence the roots should be 1 and \(b-st\). Thus if you prove 1 is a root of the equation \(b-st\) is automatically a root. Sualeh Asif · 12 months ago

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As per the level of paper how much one need to score for clear this level Aakash Khandelwal · 12 months ago

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@Aakash Khandelwal How much u r getting?? Devansh Shah · 12 months ago

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Hey guys I didn't get selected : ( Easha Manideep D · 11 months, 1 week ago

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@Easha Manideep D Don't get sad, work on your weaknesses and try again. Shubhendra Singh · 11 months, 1 week ago

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Who got selected for rmo and how much scores....? Shubham Gupta · 11 months, 2 weeks ago

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@Shubham Gupta There are 33 selections from rajasthan, and luckily I'm one of them. Shubhendra Singh · 11 months, 2 weeks ago

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everybody....the answer to the combinatorics question is here:

go here Vaibhav Prasad · 12 months ago

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@Vaibhav Prasad Where did you get the answers from? I think that your answer is way too small. The answer for 28 object has been posted in a website(resonance) and that answer is bigger than the one you posted. Svatejas Shivakumar · 12 months ago

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@Harsh Shrivastava Wow. Did you get it right in the exam as well? Svatejas Shivakumar · 12 months ago

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@Svatejas Shivakumar Nah the solutions are written by my FIITJEE sir,not me. Harsh Shrivastava · 12 months ago

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@Harsh Shrivastava How many did you solve in the exam? Svatejas Shivakumar · 12 months ago

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@Svatejas Shivakumar I screwed up the paper.( only 2 solutions are perfectly correct rest contain some flaws :( ) Harsh Shrivastava · 12 months ago

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@Svatejas Shivakumar I got that right check out what I have done in my solution below. Shubhendra Singh · 11 months, 4 weeks ago

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I attmpted 5 my logic is correct but I commited calculation mistake in 2 problems Satyam Mani · 12 months ago

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I have done 4 questions correctly Ashutosh Kaul · 12 months ago

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The same questions were asked in Jharkand . What should be the expected cut off ? Wasif Jawad Hussain · 12 months ago

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@Wasif Jawad Hussain 45-55 Aditya Kr · 11 months, 3 weeks ago

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This is the solution of 5th question

This is the solution of 5th question

Hemant Kumae · 12 months ago

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this the solution of 5th question

this the solution of 5th question

Hemant Kumae · 12 months ago

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@Hemant Kumae Yup did the same way but rather with a more tedious approach Mohit Gupta · 12 months ago

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@Hemant Kumae I too did the same way Shubhendra Singh · 12 months ago

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Answer to Q4

Arrange all things in a straight line \(a_{1},a_{2}...….a_{32}\)

See the following arrangement

\(P \\ \boxed{a} \\ Q \\ \boxed{b} \\ R \\ \boxed{c} \\ S \)

Here a, b, c are any three things P, Q, R, S are no of things left in their middle and aside.

Here \(P+Q+R+S =29 \) where \(P, S \geq 0 \& Q, R \geq 1\)

So the no. of solutions of this equation are \(^{27+4-1}C_{3}=4060\). Now here we had taken them in a line.

When arranged in a circle \(a_{1}\) and \(a_{32}\) can't be together so we will remove the cases in which both of them are together, there are 28 cases to be removed.

So now we get 4060-28 = 4032 cases

Now we have to remove the cases of diametrically opposite things.

Here two things \(a_{p}; a_{q} \) p>q are diametrically opposite if \(p-q=16\) we will get 16 pairs with 26 arrangements corresponding to every pair so total cases are 26×16=416

Finally the answer comes out to be 4032-416=3616 Shubhendra Singh · 12 months ago

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Answer to question 6. Let \[a=m+\frac{b}{c}\] where \(m\) is any integer and \[0<b<c\] . Then \[a(a-3{a})=( m+\frac{b}{c})( m-2\frac{b}{c}) \] \[m^{2}-\frac{bm}{c}+\frac{2b^{2}}{c^{2}}\]. \[m^{2}-\frac{2b^{2}-bcm}{c^{2}}\] Now, \(m\) is an integer . Let's consider \[ \frac{2b^{2}-bcm}{c^{2}}=k\] where \(k\) is an integer . After solving we get \[\frac{b}{c}=\frac{m+_{-}\sqrt{m^{2}+8k}}{4}\].....(I) But \[\frac{b}{c}<1\]....(II) Now putting value of \(\frac{b}{c}\) from (I) to (II). We get \[m+k<2\] Therefore there are infinitely many integers \(m,k\) such that \[m+k<2\]. Hence proved. Shivam Jadhav · 12 months ago

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I have a doubt in 5th question. Chinmay Sangawadekar · 12 months ago

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Comment deleted 12 months ago

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@Sualeh Asif There are five solutions: \(3+\sqrt{x}\) where x={1.5,2,2.5,3,3.5} Samarth Agarwal · 12 months ago

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[@Nihar Mahajan , [@Swapnil Das , did you appear for RMO ? Chinmay Sangawadekar · 12 months ago

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how much are you getting? Dev Sharma · 12 months ago

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@Dev Sharma Did you appear for RMO Dev ? Chinmay Sangawadekar · 12 months ago

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@Chinmay Sangawadekar Yeah Dev Sharma · 12 months ago

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@Dev Sharma How much will you score ? And how many questions did you attempt? All? Chinmay Sangawadekar · 12 months ago

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@Dev Sharma Well I was able to do all of them, let's see how the marking is done Shubhendra Singh · 12 months ago

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@Shubhendra Singh Can you please tell what was your answer for the 4th question? A very similar problem had appeared in my region. Svatejas Shivakumar · 12 months ago

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@Svatejas Shivakumar I got 3616 Shubhendra Singh · 12 months ago

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@Shubhendra Singh i was also getting something like this, btw in which class you are? Dev Sharma · 12 months ago

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@Dev Sharma 11th Shubhendra Singh · 12 months ago

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@Shubhendra Singh In 11 and 15 yrs old ? Did you put wrong age while signing up? Chinmay Sangawadekar · 12 months ago

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@Dev Sharma Happy Birthday Dev! Kushagra Sahni · 12 months ago

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@Kushagra Sahni Thanks Dev Sharma · 12 months ago

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@Shubhendra Singh I got 2015!! what was your approach? Samarth Agarwal · 12 months ago

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@Samarth Agarwal First tell me yours Shubhendra Singh · 12 months ago

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@Shubhendra Singh What was your approach Samarth Agarwal · 12 months ago

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@Shubhendra Singh A weird one: There are 30+29+28+...+1 ways to choose the restricted objects and total ways I think should be (32C3)/2 as I studied in a book that for circur permutation we need to divide by two... so 2015 :( Samarth Agarwal · 12 months ago

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@Samarth Agarwal Well I had a long one but here are results that I got. No of ways of selecting 3 such that no 2 are adjacent \(^{30}C_{3}- 28=4032\) and the most of arrangements containing diametrically opposite were \(26 ×16\). Finally getting 4032-416=3616. Shubhendra Singh · 12 months ago

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@Shubhendra Singh How did you solve for the condition that none of the two are adjacent or diametrically opposite ? Vishal Yadav · 12 months ago

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@Dev Sharma Pheww ^_^ got all except the geometry problems... and 4) Chinmay Sangawadekar · 12 months ago

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I got 5 questions correct expect that 4 th one. How much u all have done Shubham Gupta · 12 months ago

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What is the expected cutoff. I did 5 correct questions and am in class 10. Can I know the max marks they give if the solution is completely correct Shubham Gupta · 12 months ago

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@Shubham Gupta That's obvious : \(17\times5\) = 85(lol) Vishal Yadav · 12 months ago

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@Vishal Yadav Yeah that's very funny. 85 of course. Kushagra Sahni · 12 months ago

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@Kushagra Sahni I mean that have u appeared for RMO last year and know that how much lenient checking they do. I have heard that they give max 14 marks in full correct solutions. Shubham Gupta · 12 months ago

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@Shubham Gupta Are you serious? Which region? Svatejas Shivakumar · 12 months ago

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@Svatejas Shivakumar Rajasthan Shubham Gupta · 12 months ago

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