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.

3 years, 2 months ago

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when will the results of gmo will be declared

- 3 years, 1 month ago

Is the answer to question 4 2268?

- 3 years, 2 months ago

Yes it is

- 3 years, 2 months ago

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

- 3 years, 2 months ago

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

- 3 years, 2 months ago

Can anyone tell the answer of 5th question

- 3 years, 2 months ago

it is 4/3

- 3 years, 1 month ago

I m getting 1/3

- 3 years, 1 month ago

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

- 3 years, 1 month ago

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.

- 3 years, 2 months ago

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

- 3 years, 1 month ago

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

- 3 years, 2 months ago

@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.

- 3 years, 2 months ago

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

- 3 years, 2 months ago

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

- 3 years, 2 months ago

Answer to question number 2. $P_{1}(m)-P_{2}(n)=a_{1}-a_{2}=\frac{2(b_{2}-b_{1})}{m-n}$ This implies that $$a_{1}-a_{2}=even$$. $P_{1}(m)-P_{2}(n)=a_{1}+a_{2}=-2(m+n)$ This implies that $$a_{1}+a_{2}=even$$. Hence proved.

- 3 years, 2 months ago

Shivam dud you gave GMO or RMO?

- 3 years, 2 months ago

RMO

- 3 years, 2 months ago

How many did you solve?

- 3 years, 2 months ago

5

- 3 years, 2 months ago

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

- 3 years, 2 months ago

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

- 1 year, 4 months ago

Can anybody please post a proper solution for question $$4$$ with explanation?

- 3 years, 1 month ago

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

- 3 years, 1 month ago