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
2 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 \[ 1,\frac{43}{31},\frac{5}{3},\frac{37}{27},\frac{13}{9},\frac{19}{13},\frac{61}{43},\frac{55}{39}\]

Shivam Jadhav - 2 years, 10 months ago

Log in to reply

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

Gaurav Manwani - 1 year ago

Log in to reply

Can anybody please post a proper solution for question \(4\) with explanation?

Saurabh Mallik - 2 years, 9 months ago

Log in to reply

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 - 2 years, 9 months ago

Log in to reply

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.

Shivam Jadhav - 2 years, 10 months ago

Log in to reply

Shivam dud you gave GMO or RMO?

Aakash Khandelwal - 2 years, 10 months ago

Log in to reply

RMO

Shivam Jadhav - 2 years, 10 months ago

Log in to reply

@Shivam Jadhav How many did you solve?

Brilliant Member - 2 years, 10 months ago

Log in to reply

@Brilliant Member 5

Shivam Jadhav - 2 years, 10 months ago

Log in to reply

when will the results of gmo will be declared

Nikhil Shah - 2 years, 9 months ago

Log in to reply

Is the answer to question 4 2268?

Divyansh Choudhary - 2 years, 10 months ago

Log in to reply

Yes it is

Adarsh Kumar - 2 years, 10 months ago

Log in to reply

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

Divyansh Choudhary - 2 years, 10 months ago

Log in to reply

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

Divyansh Choudhary - 2 years, 10 months ago

Log in to reply

Can anyone tell the answer of 5th question

Devansh Shah - 2 years, 10 months ago

Log in to reply

it is 4/3

Aryan Goyat - 2 years, 9 months ago

Log in to reply

I m getting 1/3

Devansh Shah - 2 years, 9 months ago

Log in to reply

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

Aryan Goyat - 2 years, 9 months ago

Log in to reply

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 - 2 years, 10 months ago

Log in to reply

@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 - 2 years, 10 months ago

Log in to reply

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

Devansh Shah - 2 years, 10 months ago

Log in to reply

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

Shivam Jadhav - 2 years, 10 months ago

Log in to reply

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

Aryan Goyat - 2 years, 9 months ago

Log in to reply

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

Yatharth Chowdhury - 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...