QQAD 2006 Revisited

[vineet.nitd]

This thread is for discussing problems that appeared in last year's Quant Question A Day .
The need for this thread was much felt by the eager junta , so here we have it opened
The idea is to assimilate the concepts of last year also in addition to the currently running QQAD.

I would lay some guidelines for this thread here ..

1>We will keep the problems numbered exactly the way they appeared last year.

2>Please dont look into the solutions rightaway .Otherwise no point of having this thread at all.

3>Three problems will be posted each weekday (not weekends)after 12 noon on this thread at one go and every one needs to solve them. Remember that there is no race for solving them fast here , but its always good if you can .

Solve at ur own ease .Everyone should then put forth their solutions.

4>After the discussions are over , we will have look at the official solutions to the problems .


5>Please dont spam the thread with irrelevant posts not meant to serve the purpose of this thread .

[vineet.nitd]

Repeat Post

[amit_vce]

Originally Posted by vineet.nitd

Repeat Post

Great initiative Vineet. Looking forward to participate in this thread. Kudos to you for your continuing efforts to help pagals in there prep. Keep on .

[Aarav]

As the rule says after 12 only, I put the 3 problems for the day.

-----------------------------------------------------------
Quant Question # 1
------------------------------------------------------------
Let n be positive integer and g(n) denotes the gcd of n^2 + 11 and (n+1)^2 + 11,
then max(g(n)) is

(a) 15 (b) 45 (c) 75 (d) 105

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Quant Question # 2
------------------------------------------------------------
The points P, Q, R lie on a line in that order with PQ=9, QR=21. Let O be a point not
on PR such that PO=RO and the distances PO and QO are integral.
Then sum of all possible perimeters of triangle PRO is

(a) 320 (b) 350 (c) 380 (d) 410

------------------------------------------------------------
Quant Question # 3
------------------------------------------------------------
Which of the following statements about the functions

A(x, y, z) = x² + 3y² - 4xy - 2yz + 2zx and B(x,y,z) = |x-|x-y|| + |y-|y-z|| + |z-|z-x|| is necessarily true?

(a) B(x,y,z) <= |x-y| + |y-z| + |z-x| (b) (B(x,y,z))^2 > A(x,y,z) for x > y > z > 0

(c) A(x,y,z) > -6 (d) None of the above

[Destiny's_Child]

Actually i had tried the first few questions of QQAD 06 thread questions long before when i was going thorugh the thread,
So i kinda remeber the approach to them....

1)
GCD [n^2 + 11, (n+1)^2 + 11] = GCD (2n+1, n^2 + 11)
Now we find the GCD by division method till we get a term independent of n , which would denote the maximum GCD..
The independent term comes out tobe 45.

In fact an important conclusion can be drawn that for two numbers of the form [n^2 + a, (n+1)^2 + a ], the max GCD would be (4a +1)

2)
Here PO=RO , so the triangle is an isosceles traingle
Drop a perpendicular from O to PR which would also divide PR into two equal parts at say T.
Join OQ
Now OP^2 - PT^2 = OQ^2 - QT^2
=> OP^2 - 15^2 = OQ^2 - 6^2
=> OP^2 - OQ^2 = 189
=> (OP-OQ)(OP+OQ) =1*189 = 3*63 = 7*27 = 9*21
On Solving 2OP = 190, 66, 34, 30
But OP = 15 violates the condition that the some of two sides of traingle should be greater than the third side
So perimeter of possible traingles = 190 + 30 + 66 + 30 + 34 + 30 = 380


I am stuck with the third question.....
Dont know how to do that......

[vineet.nitd]

Originally Posted by Destiny's_Child

I am stuck with the third question.....
Dont know how to do that......

plug values ...
for e.g
put , (x,y,z)->(-1,0,5) option C is eliminated
put , (x,y,z)->(1,1,1) option A is eliminated

[Destiny's_Child]

Originally Posted by vineet.nitd

plug values ...
for e.g
put , (x,y,z)->(-1,0,5) option C is eliminated
put , (x,y,z)->(1,1,1) option A is eliminated

@Vineet!!
In this question, is there any approach to the selection of the values we substitute to test the options....
For instance, how did you decide to come up with (-1,0,5)
I also tried some values, but unfortunately my choice of values dint help me narrow down the options....

How can we be sure about option (b)
It might just happen that it turned tobe true for the values we selected

Also we have an option None of the above....., which makes things all the more difficult

[AC_here]

Quant Question #1

Answer is (b) 45

The method is similar to Destiny's_Child. I would like to add one point which I had noted from Aarav's solution last year. The tip is HCF ( x, y ) is same as HCF ( x, y-x ) given y > x.

This reduces the effort for finding HCF by division method because using the division method for finding HCF [ (n+1)^2 + 11, n^2 + 11 ] also gives answer 45

Quant Question #2

Answer is (C) 380

Here possible values of OP=OR= 95,33,17.

Summation of the perimeters is 380.

Quant Question # 3

Answer is (b)

Without loss of generality we can assume that x >= y>= z

A(x, y, z) = 2(x-y)^2 + (y-z)^2 -(x-z)^2
= (x-3y-2z)(x-y)

Put y =0 , x=1
A(x, y,z) = 1-2z < - 6 for z > 3

Hence (c) is False

B(x, y, z) = |y| + |z| + | 2z-x|

Checking option (a) and assume x > 2z
We have LHS : B = y + z + x - 2z = x + y - z ------------(1)
And RHS : |x-y| + |y-z| + |z-x| = 2x - 2z ---------(2)

LHS - RHS = y + z - x

Given in option (a) that y + z - x <=0
but for x= 4, y =3, z= 2 ; y + z - x > 0

So (a) is False !

Checking option (b)

B^2 = ( x + y - z)^2
A = (x-y)(x- 3y+ 2z)

B^2 - A = (x + y -z)^2 - (x - y)(x - 3y + 2z)
= -2y^2 + z^2 + 2xy + 4yz
= 2y(x-y) + z(z + 4y)
> 0

Hence (b) is True

[vineet.nitd]

Originally Posted by Destiny's_Child

@Vineet!!
In this question, is there any approach to the selection of the values we substitute to test the options....
For instance, how did you decide to come up with (-1,0,5)
I also tried some values, but unfortunately my choice of values dint help me narrow down the options....

How can we be sure about option (b)
It might just happen that it turned tobe true for the values we selected

Also we have an option None of the above....., which makes things all the more difficult

B = y+x-z or y+3z-x =>B^2-A =-2y^2+z^2+6xy-4xz or -2y^2+9z^2+8yz+2xy-8xz
Put , z=a ,y=pa and x=qa , where q>p>1
Case1 : -2y^2+z^2+6xy-4xz =a^2(1+4q(p-1)+2p(q-p)) >0
Case2: -2y^2+9z^2+8yz+2xy-8xz =a^2(9+8(p-q)+2p(q-p)) >0{since q<2 as 2z>x and also p>1}
Hence , in both cases B^2 >A


Also , A can achieve any value on the number line . I chose , (-1,0,5) as y=0 will remove 3 expressions and for rest two we take -ve value for x and very large alue of z to make the overall exp. more -ve.

[doomsayer]

Originally Posted by Aarav

As the rule says after 12 only, I put the 3 problems for the day.

-----------------------------------------------------------
------------------------------------------------------------
Quant Question # 3
------------------------------------------------------------
Which of the following statements about the functions

A(x, y, z) = x² + 3y² - 4xy - 2yz + 2zx and B(x,y,z) = |x-|x-y|| + |y-|y-z|| + |z-|z-x|| is necessarily true?

(a) B(x,y,z) <= |x-y| + |y-z| + |z-x| (b) (B(x,y,z))^2 > A(x,y,z) for x > y > z > 0

(c) A(x,y,z) > -6 (d) None of the above

ans :b

when x>y>z >0 then b reduces into x+y+z
and (x+y+z)^2 > x² + 3y² - 4xy - 2yz + 2zx
so b is the ans