CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[pavanpadekal]

p+q-3=x1+x2
x2=p+q-3-x1
x1=p+q-3-x2
x1.x2=p
(p+q-9-x1)x1=p
x1^2-q+3=0
x2^2-p+3=0
x3^2-q+3=0
x4^2-p+3=0
x1^2-x3^2=0
(x1-x3)(x1+x3)=0
As x1+x3>0
x1=x3
Similarly
x2=x4
Roots of equation 1 (x1,x2)
Roots of equation 2 (x3,x4)=(x1,x2)
Now since both equations have common roots the coefficients must be same
Hence p=q

Not sure if this is also wrong!
And this is a little roundabout,Nevertheless tried!
Beautifully crafted problem ! I am sure we will get to learn lots from whatever is the right solution to this problem
Thanks to Aarav

[Aarav]

Originally Posted by nbangalorekar

Aarav id solved a similar problem (not this one) but the logic i applied here... n it worked..
the problem id solved was similar in the sense that it too had 4 variables interconnected in the same way... thats y i could solve it..
but i guess more solutions could be possible..
ciao

Fair enough. What was the problem you had solved?

[Aarav]

Originally Posted by pavanpadekal

p+q-3=x1+x2
x2=p+q-3-x1
x1=p+q-3-x2
x1.x2=p
(p+q-9-x1)x1=p
x1^2-q+3=0
x2^2-p+3=0
x3^2-q+3=0
x4^2-p+3=0
x1^2-x3^2=0
(x1-x3)(x1+x3)=0
As x1+x3>0
x1=x3
Similarly
x2=x4
Roots of equation 1 (x1,x2)
Roots of equation 2 (x3,x4)=(x1,x2)
Now since both equations have common roots the coefficients must be same
Hence p=q

Not sure if this is also wrong!
And this is a little roundabout,Nevertheless tried!
Beautifully crafted problem ! I am sure we will get to learn lots from whatever is the right solution of this problem
Thanks to Aarav

p = 7, q = 8 is also a solution.

[mejogi]

Either x=0 or p=q
x=0 not possible. hence p=q
T.f. x^2-px+2P-3=0,which gives more than 6 ordered pairs, where p>0,x>0 and{p-(D)^(1/2)}>0

Hence option (5) None of these

[nbangalorekar]

Originally Posted by Aarav

Fair enough. What was the problem you had solved?

.. please see my edited post..
henceforth il avoid posting if iv solved similar questions...
but the solution is too good n i give all the credit to u that it stuck has with me all along..
simply superb...

[sabsebadapaagal]

Originally Posted by implex

x^2 -px + p + q - 3 = 0
P^2>=4(p+q-3)
or p+q-3>=0
p+q>=3

but roots are positive so p+q>3

now from AM and GM
PQ<=9/4
but P and Q are integers
gives PQ=1 or 2

if p=q
we get
p<=2 or p>=6
one solution is p=q=2
now for p=q=6
.
..
...
so total solution is 2
option a)

I guess total no of solution possible is 1 as PQ<=9/4 , so P=Q=6 gets discarded(does not satisfies pq<=9/4).... so the total no of ordered pair in this case should be 1 ,hence the correct option should be

5) None of these

[Aarav]

Originally Posted by nbangalorekar

.. please see my edited post..
henceforth il avoid posting if iv solved similar questions...
but the solution is too good n i give all the credit to u that it stuck has with me all along..
simply superb...

Not an issue actually the idea is that if people have attempted the problem before then I shouldn't be asking these again. Since I was not very sure on this, took a chance for today's problem. You can post the solution even if you have seen similar problem to any QQAD problem, there is absolutely nothing wrong with it.

[implex]

Originally Posted by sabsebadapaagal

I guess total no of solution possible is 1 as PQ<=9/4 , so P=Q=6 gets discarded(does not satisfies pq<=9/4).... so the total no of ordered pair in this case should be 1 ,hence the correct option should be

5) None of these

this approach is flawed!! I know it because its mine!

[nbangalorekar]

Aarav please dont give me any points for that... I seriously dont deserve...
Il also try to think of another method... but i guess itl be impossible in the wake of this..
n my apologies to u for posting t so early... i was very excited at being able to solve this

[implex]

Originally Posted by Aarav

Not an issue actually the idea is that if people have attempted the problem before then I shouldn't be asking these again. Since I was not very sure on this, took a chance for today's problem. You can post the solution even if you have seen similar problem to any QQAD problem, there is absolutely nothing wrong with it.

I have known this trick but I always forget to use it

for example if we have xy+yz+zx=xyz where x,y,z are positive integers
this trick will do it for us..
but I was too stupid to use it in the problem in hand!!