Aarav Says

Till Blue, it's ok ... but you made major mistakes after that. Arnab, the problem is still 60% to be done from where the blue ends.

Q) Let p, q, r, s be real numbers such that p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21. Prove that ab-cd >= 2 for some permutation (a, b, c, d) of (p, q, r, s).

My approach this problem is:


(p + q + r + s)^2 = (p^2 + q^2 + r^2 + s^2) + 2(pq + pr + ps + qr + qs + rs)
81 = 21 + 2(pq + pr + ps + qr + qs + rs).
(pq + pr + ps + qr + qs + rs) = (81 - 21)/2 = 30.
(pq + rs) + (p + q)(r + s) = 30.
(pq + rs) + (r + s) = 30.
(pq + rs) + 9(r + s) = 30.
(pq + rs) + 9(r + s) = 30.
(pq rs) + 9(r + s) = 30.
(pq rs) = 30 + 9(r + s).

In order to find the maximum & minimum value of either of (p, q, r, s) satisfying both the given conditions, other three must be equal (Say y)

So, the modified conditions are: x + 3y = 9 ---(I) & x^2 + 3y^2 = 21---(II)

Solving we get (x = 3, y = 2) & (x = 1.5, y = 2.5).

So, Value of (r + s) will always vary between 4 & 5.

Considering the cases of extremities:

For (r = 1.5 & s = 2.5)
=> (pq-rs) = 30 + (2.25 + 6.25) 9(1.5 + 2.5) = 30 + 8.5 36 = 2.5
For (r = 2 & s = 2)
=> (pq-rs) = 30 + (4 + 4) 9(2 + 2) = 30 + 8 36 = 2
For (r = 3 & s = 2)
=> (pq-rs) = 30 + (9 + 4) 9(3 + 2) = 30 + 13 45 = (-2)

Here we can see that for values of (1.5<= r <= 2), (pq-rs)>=2 & increases as r approaches downward extremity in the range.
Again, for values of (2< r <= 3), (pq-rs)<2 & decreases as r approaches towards upward extremity in the range.


So, for some permutations (a, b, c, d) of (p, q, r, s)=> =====Concluded.



Arav/Vineet, Check & please lemme know whether my explanation is okay. Or I've to work harder on this?
Cheers,
Arnab

Is the answer (c) ?

The answer is there in front of everyone, wonder what is stopping people from posting the right option. Isn't there anyone who has got this?

arnabsadhu Says

Neo, the nth term is and not equal to as U've considered here.

Ohh..Indeed it is..that was a err i agree..

Will give a fresh try..Seems like my pencil and paper are unavoidable.

Neo

See the one in Red.. You are wrong as (a+b)(a-b)= a2-b2. you can't factorise it this way.
When you multiply thoe two brackets you get n4-n2+1 & not n4+n2+1.

Thakurji..
In a hurry i have erred in the nth term but not in the factorization as u pointed out.

(n^2+n+1)(n^2-n+1) = (n^4+n^2+1)

I have no intention to calculate and derive it here. plz check ur calculations:)

Neo

So far.. all the answers are wrong. Somebody has factored n^4 + n^2 + 1 which was a previous problem in this forum in place of n^4 + n^2 + n which is required here.

Finally..the easiest one i got to c after ages..believe me did it without using my pencil n paper..not a big deal though.:)

Sum = SIGMA[n/(n^4+n^2+1)]

The main catch is to factorise the denominator
n^4+n^2+1 = (n^2+n+1)(n^2-n+1)

=>Sum = SIGMA[ (n^2+n+1)-(n^2-n+1) / 2(n^2+n+1)(n^2-n+1) ]

= 1/2 * SIGMA [ 1/ (n^2-n+1) - 1/(n^2+n+1) ]

= 1/2 * 1 = 1/2


Neo

Great Method Yaar

------------------------------------------------------------
Quantitative Question # 82
------------------------------------------------------------
Question:
The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...upto infinity is
(a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these

I think the option is (a). Is this correct. because the first term -> 1/3 so the whole will tends to just a li'll greater than this i.e 1/2

Finally..the easiest one i got to c after ages..believe me did it without using my pencil n paper..not a big deal though.:)

Sum = SIGMA[n/(n^4+n^2+1)]

The main catch is to factorise the denominator
n^4+n^2+1 = (n^2+n+1)(n^2-n+1)

=>Sum = SIGMA[ (n^2+n+1)-(n^2-n+1) / 2(n^2+n+1)(n^2-n+1) ]

= 1/2 * SIGMA [ 1/ (n^2-n+1) - 1/(n^2+n+1) ]

= 1/2 * 1 = 1/2


Neo

Ok ... sure on this?

Finally..the easiest one i got to c after ages..believe me did it without using my pencil n paper..not a big deal though.:)

Sum = SIGMA[n/(n^4+n^2+1)]

The main catch is to factorise the denominator
n^4+n^2+1 = (n^2+n+1)(n^2-n+1)

=>Sum = SIGMA[ (n^2+n+1)-(n^2-n+1) / 2(n^2+n+1)(n^2-n+1) ]

= 1/2 * SIGMA [ 1/ (n^2-n+1) - 1/(n^2+n+1) ]

= 1/2 * 1 = 1/2


Neo

Neo, the nth term is [n/(n+n^2+n^4)]and not equal to [n/(n^4+n^2+1)] as U've considered here.

Finally..the easiest one i got to c after ages..believe me did it without using my pencil n paper..not a big deal though.:)

Sum = SIGMA[n/(n^4+n^2+1)]

The main catch is to factorise the denominator
n^4+n^2+1 = (n^2+n+1)(n^2-n+1)

=>Sum = SIGMA[ (n^2+n+1)-(n^2-n+1) / 2(n^2+n+1)(n^2-n+1) ]

= 1/2 * SIGMA [ 1/ (n^2-n+1) - 1/(n^2+n+1) ]

= 1/2 * 1 = 1/2


Neo

See the one in Red.. You are wrong as (a+b)(a-b)= a2-b2. you can't factorise it this way.
When you multiply thoe two brackets you get n4-n2+1 & not n4+n2+1.

Finally..the easiest one i got to c after ages..believe me did it without using my pencil n paper..not a big deal though.:)

Sum = SIGMA[n/(n^4+n^2+1)]

The main catch is to factorise the denominator
n^4+n^2+1 = (n^2+n+1)(n^2-n+1)

=>Sum = SIGMA[ (n^2+n+1)-(n^2-n+1) / 2(n^2+n+1)(n^2-n+1) ]

= 1/2 * SIGMA [ 1/ (n^2-n+1) - 1/(n^2+n+1) ]

= 1/2 * 1 = 1/2


Neo

sunit_chotu Says

It can never be 1/2 yaar as the first term itself is 1 and u r adding positive quantities to it. it will always be greater than 1.

U r mistaken Sunit.....the first term is 1/3

wrong post.

value of nth term in the series tends to zero as n-> infinity. add the first few terms. answer is 1/2

------------------------------------------------------------
Quantitative Question # 82
------------------------------------------------------------
Question:
The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...upto infinity is
(a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these

ANSWER IS a)1/2

------------------------------------------------------------
Quantitative Question # 82
------------------------------------------------------------

Question:


The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...upto infinity is
(a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these

hey thambi even let me knw ya.. even ill like 2 knw dem.. thanks ya..

Hi Aarav,

Please look into my post n tell me where I am going wrong.................(I am sure I might have made a lot of blunders )


Thanks

Yes Arun, you have. Please look at the hint and then try.

Ok ... Hint for the problem

If a >= b >= c>= d then prove the asked thing with a+b >=5 and show that a+b < 5 is never possible. We will be done

Hi Aarav,

Please look into my post n tell me where I am going wrong.................(I am sure I might have made a lot of blunders )


Thanks

For a rhombus, d1*d2/2 = Area of the rhombus=A.

Let QS= d2 and let PR=d1.

In triangle PRS:- (d2/2-25)^2 + (d1/2)^2 = (25)^2

In triangle PQS:- (d1/2-25/2)^2 + (d2/2)^2 = (25/2)^2

From these two equations we get :- d1=2*d2.

Substituting the value of d1 in the first equation we get:-

5*d2^2 - 100*d2=0
=> 5*d2(d2-20)=0
=>d2=20.
Thus d1=2*20=40

Thus A= 0.5 * 20 *40
=400

Thus correct answer option is (d)....

I could not participate (except for the first day) in the contest due to my site visit. Seen the problems. Very interesting problems those are. Thanks Arav & Vineet for such commendable effort.
My approach this problem is:
(p + q + r + s)^2 = (p^2 + q^2 + r^2 + s^2) + 2(pq + pr + ps + qr + qs + rs)
81 = 21 + 2(pq + pr + ps + qr + qs + rs).
(pq + pr + ps + qr + qs + rs) = (81 - 21)/2 = 30.
(pq + rs) + (p + q)(r + s) = 30.
(pq + rs) + [ 9 (r + s)] (r + s) = 30.
(pq + rs) + 9(r + s) [(r + s)^2] = 30.
(pq + rs) + 9(r + s) [(r^2 + s^2) + 2rs] = 30.
(pq rs) + 9(r + s) [(r^2 + s^2)] = 30.
(pq rs) = 30 + [(r^2 + s^2)] 9(r + s).
Now for any real value of r & s, (r^2 + s^2) >= 0.
(pq rs) >= 30 9(r + s).
When (r + s) <= (28/9). (pq rs) >= 2.
Is this alright???? Whats Say Arav/Vineet/Sunit??
Cheers,
Arnab

Till Blue, it's ok ... but you made major mistakes after that. Arnab, the problem is still 60% to be done from where the blue ends.

Hi For the problem
We have been given the conditions :
p+q+r+s=9;
=> p+q = 9 - (r+s)

Squaring both sides we get :

p^2+q^2 + 2pq = 81 + r^2 +s^2 + 2rs - 18(r+s)

=> 21-r^2-s^2 + 2 pq = 81 +r^2 +s^2 + 2rs - 18(r+s)
=>2(pq-rs) = 61 + 2 (r^2 + s^2) - 18 (r +s)
=> 2(pq-rs) = K
where
K =61 + 2 (r^2 + s^2) - 18 (r +s);

we have to prove K >= 4 {since we want to prove for some arrangement of( p,q,r,s) pq-rs >=2}
now in the above equation in order to find the minimum value of L.H.S we have to find the minimum value of 2 (r^2 + s^2) - 18 (r +s) :
since p^2 + q^2 + r^2 + s^2 =21 the range in which the values of four variables can lie is (-sq.root 21 , +sq.root 21 ) , r + s can go upto maximum of +sq.root 21 but in that case r^2 + s^2 becomes 21 which gives me a value of K greater than 4 .
Alternately if we assume r^2 + s^2 =0 than in that case r+s =0 hence again K > 4;similarly for any other values it can be observed that K>4
hence proved

Perfectly ok

Since, today's problem is finished, we come back to the unsolved problem from the contest.

Let p, q, r, s be real numbers such that p+q+r+s = 9 and p^2 + q^2 + r^2 + s^2 = 21. Prove that ab-cd >= 2 for some permutation (a, b, c, d) of (p, q, r, s).

I could not participate (except for the first day) in the contest due to my site visit. Seen the problems. Very interesting problems those are. Thanks Arav & Vineet for such commendable effort.

My approach this problem is:

(p + q + r + s)^2 = (p^2 + q^2 + r^2 + s^2) + 2(pq + pr + ps + qr + qs + rs)
81 = 21 + 2(pq + pr + ps + qr + qs + rs).
(pq + pr + ps + qr + qs + rs) = (81 - 21)/2 = 30.
(pq + rs) + (p + q)(r + s) = 30.
(pq + rs) + [ 9 (r + s)] (r + s) = 30.
(pq + rs) + 9(r + s) [(r + s)^2] = 30.
(pq + rs) + 9(r + s) [(r^2 + s^2) + 2rs] = 30.
(pq rs) + 9(r + s) [(r^2 + s^2)] = 30.
(pq rs) = 30 + [(r^2 + s^2)] 9(r + s).
Now for any real value of r & s, (r^2 + s^2) >= 0.
(pq rs) >= 30 9(r + s).
When (r + s) <= (28/9). (pq rs) >= 2.

Is this alright???? Whats Say Arav/Vineet/Sunit??

Cheers,
Arnab

Is there an archive of all the prev questions of QQAD 07 or a link where I can get it.
Thanks in advance
- Thambi

attaching the diagram for today's problem see the soln. above by me.

answer is d)400

my approach:
area of triangle having same base and the third vertices on the same line which is parallel to the base is equal.
let the side of rhombus be a.

hence area of tiangle PQS = PRS
=> (PR * a^2)/(4*25/2) = (QS*a^2)/ (4*25)
=> PR = QS/2
let the diagonal meet at O.
=> now by pythagoras theorem in triangle ROS.
PR=v(4/5)*a
also area of triangle PRS*2=1/2*(PR*QS)
which gives PR=a^2/25
equating both we get, a=10v5.
put this in above equation PR=20 and QS=40,
hence A = 400.

@Aarav maine zyaada lumba method to use nahin kiya na.

Perfectly ok

Since, today's problem is finished, we come back to the unsolved problem from the contest.

Let p, q, r, s be real numbers such that p+q+r+s = 9 and p^2 + q^2 + r^2 + s^2 = 21. Prove that ab-cd >= 2 for some permutation (a, b, c, d) of (p, q, r, s).

Hi,can anybody put the above solution with the help of a figure????

Hi All ,
This is my first post to the forum

The solution for Quant Question no.81 goes like this (The answer is 400:))

In A rhombus out of four triangle formed inside it each has equal area ,So

if the diagonals are d1 and d2 and the side of the rhombus is a than
A/2 = a*a*d1/4*(25/2) = a*a*d2/(4*25) (As area of a triangle is abc/4R)

this gives the ratio of the diagonals as 1:2
Now we have the area as A=1/2*d1*d2=d1^2 ( since d2=2 d1)
now if you draw the geometrical diagram you can find that

(25/2)^2 = (d1/2)^2+ (d1-25/2)^2

which gives d1=20 and hence Area A=20^2=400.:grab:

area of a triangle = abc/4R

here let the diagonals be d1 and d2 and side of rhombus = a

so we have a.a.d1/4R1 = A/2 = a.a.d2/4R2

so d1/d2 =2

A= 1/2d1.d2 = d2^2

only 400 fits in

hi;regarding question 81

and = 400
well R= a^2*x/4*area of half rhombus = 25
similarly a^2*y/4*area of half rhombus = 25/2
we get x=2y

putting it back in any equation we get x and y = 20 and 40 respectively
hence are comes out to be 400

Good people:).Most of you cracked it even before I reached office.:)
What better opener than this.We are looking for variety of approaches to this problem today:)

Carry on..!!

Here is the complete solution

Given that Circumradius of PRS is 25 , So area of PRS is PS*SR*PR /4 R which is PS*SR*PR /100 ....................................................... (1)

Given that Circumradius of PSQ is 25/2 , So area of PSQ is PS*PQ*QS /4 R which is PS*PQ*QS /50 ....................................................... (2)

Now these two areas should be the same so equating (1) and (2) we get PR= 2 SQ ( SR=PQ)

Now teh area of a rhombus is 1/2*PR*SQ which gives us SQ^2 ..........(3)

Only 400 fits in . So the answer is 400

answer is d)400

my approach:
area of triangle having same base and the third vertices on the same line which is parallel to the base is equal.
let the side of rhombus be a.

hence area of tiangle PQS = PRS
=> (PR * a^2)/(4*25/2) = (QS*a^2)/ (4*25)
=> PR = QS/2
let the diagonal meet at O.
=> now by pythagoras theorem in triangle ROS.
PR=v(4/5)*a
also area of triangle PRS*2=1/2*(PR*QS)
which gives PR=a^2/25
equating both we get, a=10v5.
put this in above equation PR=20 and QS=40,
hence A = 400.

@Aarav maine zyaada lumba method to use nahin kiya na.

Answer should be d ) 400.

Ratio of two diagonals = 2 / 1

Find the relation between the side and the diagonals.

d1 = 4*a / 5^1/2

d2 = 2*a / 5^1/2

Use cosine formula and get the angle between the sides and use pythagoras theorem to get the relation between the diagonal and side.

Area = a*a*sine of angle between the sides of two sides.

and hence answer is 400

You dont even need to use the cosine formula , Once you get the ratio of diagonals as 2:1 then chip in the formula for area , we get one diagonal as c and other other as 2c. So the area is 1/2 * c* 2c = c^2 . Only Option (d) 400 satisfies

Answer should be d ) 400.

Ratio of two diagonals = 2 / 1

Find the relation between the side and the diagonals.

d1 = 4*a / 5^1/2

d2 = 2*a / 5^1/2

Use cosine formula and get the angle between the sides and use pythagoras theorem to get the relation between the diagonal and side.

Area = a*a*sine of angle between the sides of two sides.

and hence answer is 400

Quantitative Question # 81
------------------------------------------------------------ Question:
Consider two triangles PRS and PQS.Let two circles circumscribe these two triangles and their radii be 25 and 25/2.Also,PQRS is a rhombus whose area is equal to A. Then the value of A is
(a) 200 (b) 125 (c) 375 (d) 400 (e) 325

Is the answer Option(d) 400 .......Will post the method if it is correct

And of course.. there is that thrill that you sometimes get when you cheat..!

There are no shortcuts in life.. I am pretty sure about that.. but I think one can use shortcuts when it comes to MBA questions..!

Lol.. I could try the 'proper' method.. though it's just that I am kinda bummed right now.. haven't slept the whole night.. Another trouble is that I am pretty bad at Euclidean geometry. Could still try though.. but if my guess is correct.. I would probably spare myself the trouble of going 'proper'..!

This is a semi-intelligent guess.. let's see if I am right

Come on sloth, you are much better than taking semi-XXX guesses. Go by proper method, I'm sure you would crack this in no time.

Quantitative Question # 81
------------------------------------------------------------ Question:

Consider two triangles PRS and PQS.Let two circles circumscribe these two triangles and their radii be 25 and 25/2.Also,PQRS is a rhombus whose area is equal to A. Then the value of A is

(a) 200 (b) 125 (c) 375 (d) 400 (e) 325


Answer : (e) 325

Ok.. well I suck at geometry. And I am not preparing for CAT. It appeared to me that if we denote the longer and shorter diagonals of the rhombus as l and s respectively, then the diameter of one of the circles, in my case the PQS circle will fall short of the longer diagonal by some amount x. The diameter of the other circle, in my case the PRS circle will exceed the shorter diagonal by the same amount x.

We have,

25 = l - x

50 = s + x

adding..
l + s = 75
Area of the rhombus is ls/2.
I tried substituting the given values of the area in the above formula.
I got 5 quadratic equations of the form
l(75 - l) = area * 2
For the first 4 areas.. the roots of this equation i.e the length of the longer diagonal is not a whole number.
But for the last area i.e. 325, we get
l(75 - l) = 650
This is a sweet equation..and solves to give l = 65 or 10!

i.e. the length of the longer diagonal can be either 65 or 10. Hmm..seems interesting. Since I am calling it the longer diagonal, I am forced to consider its length as 65 in which case the length of the shorter diagonal becomes 10. I am guessing that since it's an MBA question.. the diagonals are likely to be whole numbers and not rationals.



P.S. -- (This is a semi-intelligent guess.. let's see if I am right..!)

atshubha Says

@Aarav - Are the QQAD not mailed anymore ?

QQAD was on hold for last 5 days owing to NL software problem.Today's Nl has already hit 7500 mail boxes and 2 people have already clicked and opened it.Hopefuly it should reach you in sometime:)

@Aarav - Are the QQAD not mailed anymore ?

------------------------------------------------------------
Quantitative Question # 81
------------------------------------------------------------
Question:

Consider two triangles PRS and PQS.Let two circles circumscribe these two triangles and their radii be 25 and 25/2.Also,PQRS is a rhombus whose area is equal to A. Then the value of A is

(a) 200 (b) 125 (c) 375 (d) 400 (e) 325

Sorry for posting it in this thread...
calling u all puys.... i've an imp announcement. so, LISTEN UP !!!
am quite sure most of u'll be giving AIMCATS, SIMCATS, CL and other mocks. it'll be really really useful for all of us if we analyse a paper for a week. so, here goes the new game cum learning experience :
u'll be posting a set of 3 to 5 questions (negotiable) which u found tough/interesting/tricky. the other members will try solving it. those who get da correct answer will b awarded 10 points for each correct answer. no negative marking.
the solutions to these sets will be posted by the person who asked these questions, after some considerable time ( say next day, or tht day itself if many have already submitted their answers)
we've nothing 2 lose here, n dont worry if u dont get much points.. it'll be an awesome learning experience, coz no single person can enroll 2 all the mock cats of all insti.
so, u all, first task : POST THE NAMES OF THE INSTITUTES WHERE U ARE GIVING MOCK CATS. NOW !!!!!!!!!!!THE WORK LOAD WILL BE DISTRIBUTED BY ME.
and please reply if u'll cooperate in this attempt. without ur cooperation it wont succeed.

we'll start this in a new thread.. not this one. i dunno how 2 start a new thread so posting here. one of u please take this initiative of starting this new thread, n post that new thread link here. we'll all subscribe. n i'll do the work distribution n monitoring there.
what say aarav, vineet, sunit, arnav, apple, fermions n all?

Please pardon me for breaking the flow of quiz..........
I was busy since lst month and couldnot solve the question but now facing a problem in one of the questions .........
Four persons go to a birthday party. They leave their top-coats and hats in the lounge and pick them while returning back. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q.Please elaborate how to get q???

I saw the solution 2*6+2*6-2*2 bit could not find the logic:neutral::neutral::neutral:
it was written ...out of three hats 2 ways for selecting wrong hat and 6 ways for selecting wrong Coat How come 6 ways..........??????

vineet,
pardon me my laziness.
I couldnt follow the posts on this thread 4 abt a week or so.
could u plz tell me in brief as to what these contest qs. r all about ??
went thru some of the earlier posts but not many, hence couldn't figure out exactly.

The contest was just to keep everyone in flow and in the process put forth a bunch of conceptual questions to make the problem solving more of a pleasure and also so to fill the void of 5 days.You may have a look at those problems by browsing last few pages and give them a try.You shall start getting QQAD problems Monday onwards.:)

Post deleted!

vineet,
pardon me my laziness.
I couldnt follow the posts on this thread 4 abt a week or so.
could u plz tell me in brief as to what these contest qs. r all about ??
went thru some of the earlier posts but not many, hence couldn't figure out exactly.

exams were going on.. so could visit PG.. they've ended now..
yeh sab contest questions kya hai? what contest? n wht bout da question-a-day ? am not gettin them in my emails anymore...

NL software was malfunctioning and so we decided to procrastinate the QQAD delivery by 5 days.NL for QQAD shall resume from Monday.
In the meantime to utilise the weekdays we started a problem solving contest.Few questions are still pending unsolved.You can go for them.:)

exams were going on.. so could visit PG.. they've ended now..
yeh sab contest questions kya hai? what contest? n wht bout da question-a-day ? am not gettin them in my emails anymore...

answer to 6)
join MQ and PN Let the orthocentre be point O. let FQ and AD meet at point G and BE and FP meet at H.
now rt angled triangle GQA and GMF are similar
in the same way rt angled triangle HNF and HPB are similar
also they are all similar to each other as FGOH is a parallelogram and hence opposite angles FGM and FHN are equal.
angle QGM = angle NOM = angle PHN (alternate interior angles)
triangle AGF is similar to FHB
and so triangle PHN is similar to MGQ this is only possible when all the four points are collinear.

Thanks a lot aarav and vineet for such a wonderful set of questions :)
These questions will definitely help us think in different directions for a problem in the exam ... awesome i must say :satisfie::)
At the end of the day : winning is not my priority : solving few questions among them and learning a lot from others is wot is imp :)

That's the attitude Vikram. In my books, the winner is the one who got the maximum takeaway from this initiative

Aarav Says

I think it will be fair if we give 15 full points to Amit for his effort on the problem. He seemed to have solved it thoroughly on his own. Vineet, is 15 points to him ok?

Yeah definitely.Infact he deserves bonus points for both the attempts.

LATEST STANDINGS
1>Amit=62.5
2>Sunit Kumar=49.5
3>Vikram=20
4>Nipun Jain=20
5>Superstar=15
6>Pratyush Sinha=15

Thanks a lot aarav and vineet for such a wonderful set of questions :)
These questions will definitely help us think in different directions for a problem in the exam ... awesome i must say :satisfie::)
At the end of the day : winning is not my priority : solving few questions among them and learning a lot from others is wot is imp

I think it will be fair if we give 15 full points to Amit for his effort on the problem. He seemed to have solved it thoroughly on his own. Vineet, is 15 points to him ok?

a^2 + b^2 -4ab + b = 0
a = (4b +/- 2root(3b^2 - b))/2
For the above equality to hold for integer positive a, b(3*b-1) must be a perfect square.
This is possible when,
(i)Both b and 3b-1 are individually perfect squares
or
(ii)gcd(b,3b-1)>1

Clearly 3b-1 can never be a perfect square and
gcd(b,3b-1)=gcd{b,3b-(3b-1)}=gcd(b,1)=1

Hence b and 3b-1 are coprimes and so b(3*b-1) can never be a perfect square.
Therefore,there do not exist such positive integers a, b

Well, precisely what we were looking for in the day. I'm not sure if you know that we are running a mini-contest here, but these answers fetch you some points. Very well done indeed