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

[scribbles]

THE SOLUTION THAT APPEARED IN MY INBOX FOR QQAD#13

Let us take the two problems of finding A and B one by one.
Put y = xk in x^3 + y^3 = 4x^2 => x^2(x + k^3x) = 4x^2 => x(1+k^3) = 4. We need to maximize x + xk = x(1+k) = 4(1+k)/(1+k^3) = 4/[(k-1/2)^2 + 3/4)] = 16/3 = A
For the second part, please see that (a-1/3)^2 + (b-1/3)^2 + (c-1/3)^2 = 16 => c can be maximized when a = b = 1/3 and c = 13/3.



My dear fellow puys,
WITH REference TO THE SOLUTION OF QQAD#13... i am stuck at two places
FIRST; HOW DO YOU GET 4/[(k-1/2)^2 + 3/4)] from 4(1+k)/(1+k^3)

4(1+k)/(1+k^3)= 4/(1-k+k^2) ... and then what is the mental algorithm(why and how) for converting it into 4/[(k-1/2)^2 + 3/4)]

SECOND: for the second part how do you get this (a-1/3)^2 + (b-1/3)^2 + (c-1/3)^2 = 16

I understood the alternative part but didnt get these two....
Please Help
Scribbles
PS. This was a really nice question.
PPS. sorry for interrupting the discussions for todays question

[pavanpadekal]

Originally Posted by scribbles

THE SOLUTION THAT APPEARED IN MY INBOX FOR QQAD#13

Let us take the two problems of finding A and B one by one.
Put y = xk in x^3 + y^3 = 4x^2 => x^2(x + k^3x) = 4x^2 => x(1+k^3) = 4. We need to maximize x + xk = x(1+k) = 4(1+k)/(1+k^3) = 4/[(k-1/2)^2 + 3/4)] = 16/3 = A
For the second part, please see that (a-1/3)^2 + (b-1/3)^2 + (c-1/3)^2 = 16 => c can be maximized when a = b = 1/3 and c = 13/3.



My dear fellow puys,
WITH REference TO THE SOLUTION OF QQAD#13... i am stuck at two places
FIRST; HOW DO YOU GET 4/[(k-1/2)^2 + 3/4)] from 4(1+k)/(1+k^3)
4(1+k)/(1+k^3)= 4/(1-k+k^2) ... and then what is the mental algorithm(why and how) for converting it into 4/[(k-1/2)^2 + 3/4)]

4(1+k)/(1+k^3)=4/[(1+k^3)/(1+k)]
But
[(1+k^3]/(1+k)]=[k^2-k +1]
=[k^2-k+1/4+3/4]
=[(k^2 -2*k*1/2+(1/2)^2)+3/4]
=[(k-1/2)^2+3/4]
Hence
4(1+k)/(1+k^3)=4/[(1+k^3]/(1+k)]= 4/[(k-1/2)^2+3/4]

[Aarav]

Originally Posted by scribbles

THE SOLUTION THAT APPEARED IN MY INBOX FOR QQAD#13

My dear fellow puys,
WITH REference TO THE SOLUTION OF QQAD#13... i am stuck at two places
FIRST; HOW DO YOU GET 4/[(k-1/2)^2 + 3/4)] from 4(1+k)/(1+k^3)

4(1+k)/(1+k^3)= 4/(1-k+k^2) ... and then what is the mental algorithm(why and how) for converting it into 4/[(k-1/2)^2 + 3/4)]

SECOND: for the second part how do you get this (a-1/3)^2 + (b-1/3)^2 + (c-1/3)^2 = 16

I understood the alternative part but didnt get these two....
Please Help
Scribbles
PS. This was a really nice question.
PPS. sorry for interrupting the discussions for todays question

Hello Scribbles, f(x) = ax^2 + bx + c when a > 0 assumes min value at x = -b/2a as that can be seen by calculus (first derivative vanishes as x = -b/2a, second derivative is a > 0) or by completing squares f(x) = 1/a[{(ax) + b/2}^2 -{b^2/4 +ac}]. Thus, red is min when ax + b is 0.

For (a-1/3)^2 + (b-1/3)^2 + (c-1/3)^2 = 16, very difficult to pin-point how you can see this unless you are comfortable with symmetry. Please follow alternative approach for sure shot answer.

[rajeev_hts]

Originally Posted by Aarav

Two weeks into QQAD, let me get a quick feedback from you. Please answer these questions.

(1) Is the level of problems in QQAD too much for you to handle?
(2) Are you learning through QQAD, or is it some questions and some answers type stuff?
(3) Is the solution presented comprehensive?
(4) Do you mind if some good old QQAD problems (around 15% in QQAD 2008 season) are reframed and asked again?

Hi Arav,

You are doing great job. Please see my feedback below:-
1) We are handling only one problems that shakes and tests many of our fundas. I think everybody enjoys it.
2) Yes.
3) Yes, and the discussion after the question completes the funda side of question.
4) Yes, If we do not rewise how will we rember old fundas.

Kudos Arav, It feels great to be among PUYs like you guys.

[Aarav]

Originally Posted by rajeev_hts

Hi Arav,

You are doing great job. Please see my feedback below:-
1) We are handling only one problems that shakes and tests many of our fundas. I think everybody enjoys it.
2) Yes.
3) Yes, and the discussion after the question completes the funda side of question.
4) Yes, If we do not rewise how will we rember old fundas.

Kudos Arav, It feels great to be among PUYs like you guys.

Thanks Rajeev. At any point if you have any issues, most welcome to approach me

[implex]

Originally Posted by Aarav

Two weeks into QQAD, let me get a quick feedback from you. Please answer these questions.

(1) Is the level of problems in QQAD too much for you to handle?
(2) Are you learning through QQAD, or is it some questions and some answers type stuff?
(3) Is the solution presented comprehensive?
(4) Do you mind if some good old QQAD problems (around 15% in QQAD 2008 season) are reframed and asked again?

1) level is good, some questions are tricky but i still believe everyone here should get them correctly as there is no exam condition, under exam conditions, we tend to bit a little less sharp

2) They are quite a few things to learn, first being to always be on your toes, keep working hard and yeah, there are quite a few competitors around.

3) Solutions are masterpieces I guess, one of the best if not the best

4) I won't mind that much, but I would prefer new problems

[Aarav]

Originally Posted by implex

4) I won't mind that much, but I would prefer new problems

There are certain old problems in QQAD that I feel everyone should attempt in real time as these are master-piece. Secondly, this exercise gives me better perspective to compare earlier set of students with freshers and set problems accordingly.

[implex]

Originally Posted by Aarav

There are certain old problems in QQAD that I feel everyone should attempt in real time as these are master-piece. Secondly, this exercise gives me better perspective to compare earlier set of students with freshers and set problems accordingly.

the thing is quite a few guys are taking qqad for nth time where n>1 so they are at an advantage or may be some disadvantage if the problems are repeated, or are similar, the sense of achievement is not there. While the newbies may feel that they could have solved it too, so it hurts both the groups. But this is too much of thinking, if you feel you can still test everyone with the same problem, you should.

Batsmen have played yorkers many times, but they still get out!!

[selebratinglife]

Originally Posted by implex

the thing is quite a few guys are taking qqad for nth time where n>1 so they are at an advantage or may be some disadvantage if the problems are repeated, or are similar, the sense of achievement is not there. While the newbies may feel that they could have solved it too, so it hurts both the groups. But this is too much of thinking, if you feel you can still test everyone with the same problem, you should.

Batsmen have played yorkers many times, but they still get out!!

Valid point implex, but Aarav surely has much broader vision.. It doesn't harm if you solve a problem twice rather than missing out on it.May be he can just mail those and we can PM the solutions to him to conform.

[Superstar]

Hi Aarav,

After folowing all the discussions the entire day, the solution to the problem is still not clear to many of us.

After we get the number 196 in base 10 or 524 in base 6, we need to find out the value of 'x' which is the number of bases which would represent this number in two digits.

Now, what is wrong in dealing in decimals and try to find the number of bases which can show 196 in two digits. We get 182 as the answer or 350 in base 7.

I am not able to understand the solution. I am sure others would also like to finally finish this off today itself.

Please provide the detailed solution if possible by also answering my concern.

Thanks,

Your Fan and Admirer
Superstar