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 A\angle A . Show that the ABC and AB'C' have the same incentre.

  2. Let P(x)=x2+ax+bP(x)=x^{2}+ax+b be a quadratic polynomial with real coefficients. Suppose there are real numbers sts \neq t such that P(s)=tP(s) =t and P(t)=sP(t) =s. Prove that bstb-st is a root of the equation x2+ax+bst=0x^{2}+ax+b-st=0

  3. Find all integers a,b,ca, b, c such that a2=bc+1  b2=ca+1a^{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 TT and \sum in the plane intersect at two distinct points A and B, and the center of \sum lies on TT. Let points C and D on TT 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 aa such that 4<a<54<a<5 and a(a3{a})a(a-3\{a\}) is an integer.
    {Here {a}\{a \} represents fractional part of aa}.

Note by Shubhendra Singh
4 years ago

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1 vote

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Answer to question 3.a2b2=c(ba)a^{2}-b^{2}=c(b-a) This implies a+b+c=0...or...a=ba+b+c=0 ...or... a=b Case 1a+b+c=0a+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).(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 2a=ba=b Therefore (a,b,c)=(1,1,0),(1,1,0).(a,b,c)=(1,1,0),(-1,-1,0)..

Shivam Jadhav - 4 years ago

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Did the same !!! Nice and standard solution btw...

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That's the best way.

Shubhendra Singh - 4 years ago

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How do u know a=b from the equation a+b+c=o

Ashish Sacheti - 4 years ago

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I am also not sure at how you arrive at that conclusion

Mardokay Mosazghi - 4 years ago

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Well here is how he arrived at the conclusion.

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

Thus we have two cases:

Case 1:

(ab)=0(a-b)=0 or a=ba=b

Case 2:

If aba\not=b , we cancel out (ab)(a-b) from botb sides to obtain a+b=ca+b=-c or a+b+c=0a+b+c=0

Sualeh Asif - 4 years ago

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Why are solutions like (2,1,1)(2,-1,-1) not included?

Anik Mandal - 3 years, 11 months ago

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I think it's not a solution, please check it again.

Shubhendra Singh - 3 years, 11 months ago

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@Shubhendra Singh What do you think the cut off would be.of GMO 2015 paper?

Adarsh Kumar - 3 years, 11 months 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 - 3 years, 11 months ago

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@Shubhendra Singh Did you get selected?

Kushagra Sahni - 3 years, 11 months ago

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@Kushagra Sahni Yes 8-)

Shubhendra Singh - 3 years, 11 months ago

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@Shubhendra Singh How many marks did you got? And Congrats!

Harsh Shrivastava - 3 years, 11 months ago

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@Harsh Shrivastava Congrats Harsh, 52 is a good score. All the best for INMO

Mayank Chaturvedi - 3 years, 11 months ago

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@Mayank Chaturvedi Thanks!

Harsh Shrivastava - 3 years, 11 months ago

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@Harsh Shrivastava Meet you at INMO training camp on 14th-15th.

Mayank Chaturvedi - 3 years, 11 months ago

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@Mayank Chaturvedi Oh you have also qualified? Congo!

Btw where's training camp?

Harsh Shrivastava - 3 years, 11 months 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 - 3 years, 11 months ago

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@Mayank Chaturvedi What's this??

Shubhendra Singh - 3 years, 11 months ago

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@Shubhendra Singh Camp for RMO qualified from Chhattisgarh

Mayank Chaturvedi - 3 years, 11 months ago

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@Shubhendra Singh What About Dev Sharma?

Kushagra Sahni - 3 years, 11 months ago

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@Kushagra Sahni nope and u?

Dev Sharma - 3 years, 11 months ago

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@Dev Sharma I also didn't get selected. I did 4 but my writing was horrible.

Kushagra Sahni - 3 years, 11 months ago

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@Kushagra Sahni Would it be sahni at place of juneja!!

Dev Sharma - 3 years, 11 months 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 - 3 years, 11 months ago

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@Dev Sharma Dev you are in which class ??

Chirayu Bhardwaj - 3 years, 8 months ago

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@Kushagra Sahni I don't know how much I have scored, how did you get to know about your marks.

Shubhendra Singh - 3 years, 11 months ago

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@Shubhendra Singh Its written in the result list of my state.

Harsh Shrivastava - 3 years, 11 months ago

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@Harsh Shrivastava Are you selected for INMO @Harsh Shrivastava

A Former Brilliant Member - 3 years, 11 months ago

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@A Former Brilliant Member Luckily yes.

Harsh Shrivastava - 3 years, 11 months ago

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@Harsh Shrivastava Why luckily ? you are intelligent bro . Unfortunately , our principal doesn't allow us for RMO.@Harsh Shrivastava

A Former Brilliant Member - 3 years, 11 months ago

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@A Former Brilliant Member Why?

Harsh Shrivastava - 3 years, 11 months ago

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@Harsh Shrivastava Dont know . H e thinks we should focus on our school math and so ,

A Former Brilliant Member - 3 years, 11 months 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 - 3 years, 11 months ago

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Maybe it's not but how do we know that these are the only solutions?

Anik Mandal - 3 years, 11 months ago

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@Anik Mandal By subtracting the given equations we get a2b2=c(ab)a^{2}-b^{2}=-c(a-b) (a+b)(ab)=c(ab)\Rightarrow (a+b)(a-b) =-c(a-b)

This gives that the equation will have a solution when ab=0a-b=0 or a+b=ca+b=-c. That's how we get to know about the solutions we are gonna get.

Shubhendra Singh - 3 years, 11 months ago

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@Shubhendra Singh It means all the solutions of a+b+c=0a+b+c = 0 are solutions?

Anik Mandal - 3 years, 11 months 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 - 3 years, 11 months ago

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6){a}=a4\{a\}=a-4 a(a3a+12)=2a2+12aa(a-3a+12)=-2a^2+12a 2a2+12a=n-2a^2+12a=n (a3)2=n+182(a-3)^2=\dfrac{-n+18}{2} (43)2<(a3)2<(53)2(4-3)^2<(a-3)^2<(5-3)^2 1<n+182<41<\dfrac{-n+18}{2}<4 10<n<1610<n<16 put values n=11,12,13,14,15 to find a=6+362n2=3+142,3+122,3+102,3+82,3+62a=\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 - 4 years ago

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Perfect, I too got the same answer.

Shubhendra Singh - 4 years ago

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What answer were you getting for Q.4

Shubhendra Singh - 4 years ago

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sorry man, my combinatorics is weak(very)

Aareyan Manzoor - 4 years ago

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How did you get the RHS in the second line ?

Vishal Yadav - 4 years ago

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second line: we deduced {a}=a4\{a\}=a-4. we substitute this value into the given expression to find a(a3{a})=a(a3(a4))=a(a3a+12)=a(2a+12)=2a2+12a(a-3\{a\})=a(a-3(a-4))=a(a-3a+12)=a(-2a+12)=-2a^2+12. hope this helps.

Aareyan Manzoor - 4 years ago

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Sorry to say that, substitute a = 4 + 1 / 2 + sqrt 6

easha manideep d - 4 years ago

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i am sorry. couldnt understand. that is >5

Aareyan Manzoor - 4 years ago

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@Aareyan Manzoor Hey ! 4 + 1/ something is never greater than 5 and that something is > 1

easha manideep d - 4 years ago

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@Easha Manideep D i am unable to understand what you are saying.

Aareyan Manzoor - 4 years ago

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Do you mean 4+16+2 4 + \frac{1}{\sqrt{6} + 2}

4+16+2=4+622=3+62 4 + \frac{1}{\sqrt{6} + 2} = 4 + \frac{\sqrt{6} - 2}{2} = 3 + \frac{\sqrt{6}}{2}

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From where did (a-3) come.....I think the way I be done is quite different.....And I think the answer is 6-√2,6-√3

Anubhav Mahapatra - 2 years, 2 months ago

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Sorry.... I got it.....

Anubhav Mahapatra - 2 years, 2 months ago

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Answer to question 2P(s)=s2+as+b=t,P(t)=t2+at+b=sP(s)=s^{2}+as+b=t,P(t)=t^{2}+at+b=s P(s)P(t)=s2t2+a(st)=tsP(s)-P(t)=s^{2}-t^{2}+a(s-t)=t-s P(s)P(t)=s+t+a+1=0P(s)-P(t)=s+t+a+1=0 P(s)P(t)=(st)2+ast2+bt2+bs2+bas+b2+as2t+a2st+abt=stP(s)P(t)=(st)^{2}+ast^{2}+bt^{2}+bs^{2}+bas+b^{2}+as^{2}t+a^{2}st+abt=st (st)2+b2st=(ast2+as2t+a2st+bt2+abt+bs2+bas)(st)^{2}+b^{2}-st=-(ast^{2}+as^{2}t+a^{2}st+bt^{2}+abt+bs^{2}+bas) (st)2+b2st=(ast(a+s+t)+bt(a+t)+bs(a+s))(st)^{2}+b^{2}-st=-(ast(a+s+t)+bt(a+t)+bs(a+s)) (st)2+b2st=(ast(1)+bt(1+s)+bs(1+t))(st)^{2}+b^{2}-st=(ast(1)+bt(1+s)+bs(1+t)) (st)2+b2st=(ast+b(s+t)(st)^{2}+b^{2}-st=(ast+b(s+t) (st)2+b2st=(astb(1+a)(st)^{2}+b^{2}-st=(ast-b(1+a) (st)2+b2stast+b+ab=0(st)^{2}+b^{2}-st-ast+b+ab=0 But P(bst)st=(st)2+b2stast+b+abP(b-st)-st=(st)^{2}+b^{2}-st-ast+b+ab Therefore P(bst)st=0P(b-st)-st=0 Hence proved.

Shivam Jadhav - 4 years ago

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Similar Approach ... Nice and elegant solution.

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Q2 Consider the polynomial Q(x)=x2+ax+bstQ(x)=x^2+ax+b-st When Q(x)=0Q(x)=0, we realise that if bstb-st is a root then the other root is 1 by Vieta's formulas. So the sum of the roots is (bst+1)(b-st+1). We can factor it to Q(x)=(x(bst))(x1).Q(x)=(x-(b-st))(x-1). Now realise that this means a=(bst+1)    st=a+b+1a=-(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,s2t2=a(ts)+(ts)a=1stP(s)-t=P(t)-s, \\ s^2 -t^2 =a(t-s) +(t-s) \\ a=-1-s-t We can cancel tst-s out since tst\not=s. Substitute a=1st a=-1-s-t back into P(s)t=0    b=st+s+ta+b+1=stP(s)-t=0 \implies b=st+s+t \\ \therefore a+b+1=st And we are done!

Sualeh Asif - 4 years ago

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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 - 4 years ago

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Well using vietas formulas you see that the multipleof the two roots is bstb-st. Hence the roots should be 1 and bstb-st. Thus if you prove 1 is a root of the equation bstb-st is automatically a root.

Sualeh Asif - 4 years ago

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As per the level of paper how much one need to score for clear this level

Aakash Khandelwal - 4 years ago

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How much u r getting??

Devansh Shah - 4 years ago

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how much are you getting?

Dev Sharma - 4 years ago

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Well I was able to do all of them, let's see how the marking is done

Shubhendra Singh - 4 years ago

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Can you please tell what was your answer for the 4th question? A very similar problem had appeared in my region.

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@A Former Brilliant Member I got 3616

Shubhendra Singh - 4 years ago

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@Shubhendra Singh i was also getting something like this, btw in which class you are?

Dev Sharma - 4 years ago

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@Dev Sharma 11th

Shubhendra Singh - 4 years ago

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@Shubhendra Singh In 11 and 15 yrs old ? Did you put wrong age while signing up?

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@Dev Sharma Happy Birthday Dev!

Kushagra Sahni - 4 years ago

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@Kushagra Sahni Thanks

Dev Sharma - 4 years ago

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@Shubhendra Singh I got 2015!! what was your approach?

Samarth Agarwal - 4 years ago

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@Samarth Agarwal First tell me yours

Shubhendra Singh - 4 years 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 - 4 years 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 30C328=4032^{30}C_{3}- 28=4032 and the most of arrangements containing diametrically opposite were 26×1626 ×16. Finally getting 4032-416=3616.

Shubhendra Singh - 4 years 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 - 4 years ago

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@Shubhendra Singh What was your approach

Samarth Agarwal - 4 years ago

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Did you appear for RMO Dev ?

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Yeah

Dev Sharma - 4 years ago

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@Dev Sharma How much will you score ? And how many questions did you attempt? All?

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Pheww ^_^ got all except the geometry problems... and 4)

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[@Nihar Mahajan , [@Swapnil Das , did you appear for RMO ?

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I have a doubt in 5th question.

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Answer to question 6. Let a=m+bca=m+\frac{b}{c} where mm is any integer and 0<b<c0<b<c . Then a(a3a)=(m+bc)(m2bc)a(a-3{a})=( m+\frac{b}{c})( m-2\frac{b}{c}) m2bmc+2b2c2m^{2}-\frac{bm}{c}+\frac{2b^{2}}{c^{2}}. m22b2bcmc2m^{2}-\frac{2b^{2}-bcm}{c^{2}} Now, mm is an integer . Let's consider 2b2bcmc2=k \frac{2b^{2}-bcm}{c^{2}}=k where kk is an integer . After solving we get bc=m+m2+8k4\frac{b}{c}=\frac{m+_{-}\sqrt{m^{2}+8k}}{4}.....(I) But bc<1\frac{b}{c}<1....(II) Now putting value of bc\frac{b}{c} from (I) to (II). We get m+k<2m+k<2 Therefore there are infinitely many integers m,km,k such that m+k<2m+k<2. Hence proved.

Shivam Jadhav - 4 years ago

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Answer to Q4

Arrange all things in a straight line a1,a2....a32a_{1},a_{2}...….a_{32}

See the following arrangement

PaQbRcSP \\ \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=29P+Q+R+S =29 where P,S0&Q,R1P, S \geq 0 \& Q, R \geq 1

So the no. of solutions of this equation are 27+41C3=4060^{27+4-1}C_{3}=4060. Now here we had taken them in a line.

When arranged in a circle a1a_{1} and a32a_{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 ap;aqa_{p}; a_{q} p>q are diametrically opposite if pq=16p-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 - 4 years ago

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this the solution of 5th question this the solution of 5th question

Hemant Kumae - 4 years ago

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I too did the same way

Shubhendra Singh - 4 years ago

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Yup did the same way but rather with a more tedious approach

Mohit Gupta - 4 years ago

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This is the solution of 5th question This is the solution of 5th question

Hemant Kumae - 4 years ago

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The same questions were asked in Jharkand . What should be the expected cut off ?

Wasif Jawad Hussain - 4 years ago

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45-55

Aditya Kr - 3 years, 12 months ago

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I have done 4 questions correctly

Ashutosh Kaul - 4 years ago

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I attmpted 5 my logic is correct but I commited calculation mistake in 2 problems

satyam mani - 4 years ago

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

go here

Vaibhav Prasad - 4 years ago

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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.

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See this

Harsh Shrivastava - 4 years ago

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@Harsh Shrivastava Wow. Did you get it right in the exam as well?

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@A Former Brilliant Member Nah the solutions are written by my FIITJEE sir,not me.

Harsh Shrivastava - 4 years ago

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@Harsh Shrivastava How many did you solve in the exam?

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@A Former Brilliant Member I screwed up the paper.( only 2 solutions are perfectly correct rest contain some flaws :( )

Harsh Shrivastava - 4 years ago

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I got that right check out what I have done in my solution below.

Shubhendra Singh - 3 years, 12 months ago

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Who got selected for rmo and how much scores....?

Shubham Gupta - 3 years, 11 months ago

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There are 33 selections from rajasthan, and luckily I'm one of them.

Shubhendra Singh - 3 years, 11 months ago

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Hey guys I didn't get selected : (

easha manideep d - 3 years, 11 months ago

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Don't get sad, work on your weaknesses and try again.

Shubhendra Singh - 3 years, 11 months ago

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I got 5 questions correct expect that 4 th one. How much u all have done

Shubham Gupta - 4 years 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 - 4 years ago

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That's obvious : 17×517\times5 = 85(lol)

Vishal Yadav - 4 years ago

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Yeah that's very funny. 85 of course.

Kushagra Sahni - 4 years 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 - 4 years ago

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@Shubham Gupta Are you serious? Which region?

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@A Former Brilliant Member Rajasthan

Shubham Gupta - 4 years ago

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