CBSE GMO 2015 Paper discussion

Group Mathematical Olympiad is conducted by CBSE . It is only for CBSE students . Every CBSE school can send 5 students for it. Around 30 students are selected from the country and are eligible to write INMO.

Note by Aditya Chauhan
3 years, 10 months ago

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when will the results of gmo will be declared

Nikhil Shah - 3 years, 9 months ago

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Is the answer to question 4 2268?

Divyansh Choudhary - 3 years, 10 months ago

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Yes it is

Adarsh Kumar - 3 years, 10 months ago

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Can you tell me what is the expected cutoff for gmo?

Divyansh Choudhary - 3 years, 10 months ago

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I was able to do 4.5 questions do i stand a chance to get selected...

Divyansh Choudhary - 3 years, 10 months ago

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Can anyone tell the answer of 5th question

Devansh Shah - 3 years, 10 months ago

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it is 4/3

aryan goyat - 3 years, 9 months ago

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I m getting 1/3

Devansh Shah - 3 years, 9 months ago

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@Devansh Shah see i can't post my solution because it is too long but you can verify it by construction.

aryan goyat - 3 years, 9 months ago

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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 - 3 years, 10 months ago

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you can try the 6 question posted by me they all are of gmo.

aryan goyat - 3 years, 9 months ago

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m cant be 'any' integer as 3<a<4. So a will be 3.something or a = 3 + b/c

Yatharth Chowdhury - 3 years, 10 months ago

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@Shivam Jadhav This ques is a little different from the RMO one , we have to find all values of 'a' between 3 & 4. @Devansh Shah is right , there will be 4 values.

Aditya Chauhan - 3 years, 10 months ago

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Bro ur equation which is quadratic in m is wrong also we will get 4 values of a

Devansh Shah - 3 years, 10 months ago

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A general solution a=(2n+1)+0.5a=(2n+1)+0.5 where nn is a integer.

Shivam Jadhav - 3 years, 10 months ago

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Answer to question number 2. P1(m)P2(n)=a1a2=2(b2b1)mnP_{1}(m)-P_{2}(n)=a_{1}-a_{2}=\frac{2(b_{2}-b_{1})}{m-n} This implies that a1a2=evena_{1}-a_{2}=even. P1(m)P2(n)=a1+a2=2(m+n)P_{1}(m)-P_{2}(n)=a_{1}+a_{2}=-2(m+n) This implies that a1+a2=evena_{1}+a_{2}=even. Hence proved.

Shivam Jadhav - 3 years, 10 months ago

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Shivam dud you gave GMO or RMO?

Aakash Khandelwal - 3 years, 10 months ago

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RMO

Shivam Jadhav - 3 years, 10 months ago

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@Shivam Jadhav How many did you solve?

A Former Brilliant Member - 3 years, 10 months ago

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@A Former Brilliant Member 5

Shivam Jadhav - 3 years, 10 months ago

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Answer to question 3. 6l14l3=7k55k3\frac{6l-1}{4l-3}=\frac{7k-5}{5k-3} After cross multiplication we get 8k+l+lk=68k+l+lk=6 Adding 8 to both sides 8k+8+l+lk=148k+8+l+lk=14 8(k+1)+l(1+k)=148(k+1)+l(1+k)=14 (8+l)(1+k)=14(8+l)(1+k)=14 Since l,kl,k are integers therefore the solutions of (l,k)=(1,1),(6,6),(6,0),(7,13),(15,3),(22,2),(10,8),(9,15)(l,k)=(-1,1),(-6,6),(6,0),(-7,13),(-15,-3),(-22,-2),(-10,-8),(-9,-15) Therefore the required fractions are 1,4331,53,3727,139,1913,6143,5539 1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}

Shivam Jadhav - 3 years, 10 months ago

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1 will be rejected as it isn't a fraction.

Gaurav Manwani - 2 years ago

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Can anybody please post a proper solution for question 44 with explanation?

Saurabh Mallik - 3 years, 9 months ago

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Number of ways of selecting points from points is 28C3=3276

Now, we will eliminate some conditions..

Number of ways of selecting all 3 as adjacent points is 28.

Number of ways of selecting 2 adjacent points and one not adjacent with them is 28×24=672

Number of ways of selecting two points opposite diametrically along with the third point not adjacent to the former points is 14×22=308.

Hence, the desired result will be 3276-28-672-308=2268

Manisha Garg - 3 years, 9 months ago

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