Sending the NL in the evening as well is agreeable to us provided others also feel comfortable with this. Instead of sending 2 problems in the morning, we can send one in the morning at 6 AM, and another at 3 PM.

The question of stopping doesn't arise. Nov. 2 is the last day as told earlier and 2 practice tests are pending to be delivered. I was trying to convey that it dampens our spirits when we see our effort not even attracting the audience, leave alone producing results.

Hey Aarav,Vineet
Never stop QQAD.Ques. in this thread are very beneficial for all of us--practice test also.Explaination given by both of u are awesome.
If possible--plzz post some fundae frm different topics like geometry, algebra---it will also help. Sometimes I feel questions are very difficult (sitters) and when not able to solve in 2-3 attempts, then get demoralize.
Such a less time is remaining for D-day. Dont stop this--plzz.

Sending the NL in the evening as well is agreeable to us provided others also feel comfortable with this. Instead of sending 2 problems in the morning, we can send one in the morning at 6 AM, and another at 3 PM.

The question of stopping doesn't arise. Nov. 2 is the last day as told earlier and 2 practice tests are pending to be delivered. I was trying to convey that it dampens our spirits when we see our effort not even attracting the audience, leave alone producing results.

Aarav,i come 2 the office at 1 pm and then go thru' the qns and discussions.I post only if i feel my soln has something new to offer.And am sure there are hundreds if not thousands of ppl who atleast go through the qns and try to solve them once.Two of them are in my cubicle itself!So plz keep up the pace and frequency of qns,we are all following

Hii aarav n vineet:
firstly I would like to thank u ppl for this wonderful thread.
but plzz dont stop it.
regd the thin response, am just one of those who dont get time in the morning when the qs. r posted b'coz i have to go to college/work as it is with many other ppl.
n when I do log in the evening after coming back home the probs r already solved, in fact it seems like it can be answered in time only by the whole- timers for CAT preps.

Nevertheless i do collect all the qs posted in the whole week each weekend or so(i havent done this for the last 3 weeks :wow:!) n solve them in the weekend.

so the BOTTOM LINE IS:
just b'coz ppl dont respond in the morning when the qs. r posted DOES NOT mean that ppl r not benefited from this most useful thread at PG.
so plz dont stop it.

rather a REQUEST :
why cant we have in the evening when everyone' surely back home n infront of their pc/lappy.

plz deliberate on this n see if it can be done this way.

Sending the NL in the evening as well is agreeable to us provided others also feel comfortable with this. Instead of sending 2 problems in the morning, we can send one in the morning at 6 AM, and another at 3 PM.

The question of stopping doesn't arise. Nov. 2 is the last day as told earlier and 2 practice tests are pending to be delivered. I was trying to convey that it dampens our spirits when we see our effort not even attracting the audience, leave alone producing results.

Hii aarav n vineet:
firstly I would like to thank u ppl for this wonderful thread.
but plzz dont stop it.
regd the thin response, am just one of those who dont get time in the morning when the qs. r posted b'coz i have to go to college/work as it is with many other ppl.
n when I do log in the evening after coming back home the probs r already solved, in fact it seems like it can be answered in time only by the whole- timers for CAT preps.

Nevertheless i do collect all the qs posted in the whole week each weekend or so(i havent done this for the last 3 weeks :wow:!) n solve them in the weekend.

so the BOTTOM LINE IS:
just b'coz ppl dont respond in the morning when the qs. r posted DOES NOT mean that ppl r not benefited from this most useful thread at PG.
so plz dont stop it.

rather a REQUEST :
why cant we have in the evening when everyone' surely back home n infront of their pc/lappy.

plz deliberate on this n see if it can be done this way.

Vineet was telling me if we see any merit in running QQAD any further. Frankly, I have no answer for this. The thin response out here for last few days makes me think that Vineet is right.

are u kidding???.....the thin response may b due to the fact that, every one is very busy with work....myself is busy with work so much that since last 15 or 20 days i havent touched a book or any material or tried QQAD problem sincerely...but ya the thing that i do is to visit this thread whenever i find anytime and try to learn something from u guys...n as my work timings satrts from 2pm, by the time i come ,usually the problem remains well discussed, but still i find so much to learn....

so, keep up the good work!!!

Aarav/Vineet,

I should tell you one thing. Its coz of you and Vineet that has pulled me back here. After spending 10 hrs in office and cooking I make sure I spend atleast an hour here. Your effort is amazing. When ever I feel low and I find my boss screwing me I come here once. I might not be a regular solver of the problems you guys post, but I do spend quite a bit of time trying them. I guess its just a matter of time. We had seen this happen last year also but then we requested for 2 problems at different times every day.

I would just say you guys ROCK.

I remembered you a lot Prade after last year's QQAD, but never had the courage to ask how you are doing after CAT. Nothing will make me more delighted than final converts for you people.

Vineet was telling me if we see any merit in running QQAD any further. Frankly, I have no answer for this. The thin response out here for last few days makes me think that Vineet is right.


Adios for the day!

Aarav/Vineet,

I should tell you one thing. Its coz of you and Vineet that has pulled me back here. After spending 10 hrs in office and cooking I make sure I spend atleast an hour here. Your effort is amazing. When ever I feel low and I find my boss screwing me I come here once. I might not be a regular solver of the problems you guys post, but I do spend quite a bit of time trying them. I guess its just a matter of time. We had seen this happen last year also but then we requested for 2 problems at different times every day.

I would just say you guys ROCK.

part A
option 4
18+3=AD*AM/2
Ar QRN + 20+2=MB*BC/2=MB*AD/2
so 43+QRN=AD*AB/2 ----(1)

similarly
AB*BN/2=Ar PMQ + 20+3
18+2=DC*NC/2=AB*NC/2
So 43+PMQ=AB*BC/2=AB*AD/2----(2)
So we see Ar PMQ=Ar QRN (This is redundant info for me)

Also AB*AD/2=Ar PMQ+18+Ar PQRD----(3)
Using only (2) and (3) Ar PQRD=25

Part B
If D is a mid pt on BC,any line passing thru D will have same perp lengths from B and C. Drawing a fig i can also see that if D is closer to B and AB>AC then X

Vineet was telling me if we see any merit in running QQAD any further. Frankly, I have no answer for this. The thin response out here for last few days makes me think that Vineet is right.

The answer to all the 3 problems put in the day is (1).
Solution to first 2 is already given by members.

This one for the extra DS problem
By Ceva, the lines are concurrent iff (AF/FB)(BD/DC)(CE/EA) = 1 or 1(BD/DC)(BC/AB) = 1, or (BD/DC) = c/a. But BD + DC = a, so BD = ac/(a+c), DC = a^2/(a+c). By Pythagoras, BD^2 = c^2 - AD^2, CD^2 = b^2 - AD^2, so CD^2 - BD^2 = b^2 - c^2. Hence b^2 - c^2 = a^2(a^2 - c^2)/(a + c)^2 = a^2(a - c)/(a + c) is the condition when 3 lines are congruent.
(1) can answer the question as the theird side comes out to be 6.XX if the lines are to be congruent. (2) can not.

Adios for the day!

samsing Says

AD, BE and CF can be concurrent in case of isosceles triangle when angle bisector BE and median CF intersect perpendicular AD at same point. So, both A and B state whether the triangle is isosceles or not. Hence, still the answer is (c) i.e. can be answered by both A or B alone.

The concurrency can occur in scalene triangle as well. That was the whole catch in this question

AD, BE and CF can be concurrent in case of isosceles triangle when angle bisector BE and median CF intersect perpendicular AD at same point. So, both A and B state whether the triangle is isosceles (i.e. AB = AC) or not. Hence, still the answer is (c) i.e. can be answered by both A or B alone.

Ans) c

Since they are concurrent only for an eqilateral traingle, hence could be answer by both.

This one is not so trivial. Just see if anything strikes you in this.

Ans) c

Since they are concurrent only for an eqilateral traingle, hence could be answer by both.

I think this can happen only if triangle is an equilateral triangle. Both A and B are enough to show that triangle is not equilateral, so cannot happen. So, choice is (c) i.e. can be answered by A or B alone.

Please take this extra DS question as well for Practise. This was planned to be put in the NL today but we changed it in the last minute.

In the triangle ABC, AD is an altitude, BE is an angle bisector and CF is a median. Are AD, BE and CF concurrent?
(A) BC = 4 and AC = 5 < AB < 9, where length of AB is a natural number
(B) AC : AB = 5 : 6

@ mehtamanas
Sorry ...was busy working in office..Here's the solution..:)

mehtamanas Says

explain the approach mate

got it!!!thanks for the help

PART B:
ANS-1
let the foot of perpendicular from B,C be M,N respectively.
then triangles BMD and CND are similar.
and hence from 1st statement we get X

sanyo Says

Part A) ans option- 1) Can be answered by using (i) alone.

explain the approach mate

Part A) ans option- 1) Can be answered by using (i) alone.

Part a-option it cannot be answered by the two statements....
let ar(apd) = A1 ;ar(drc) = A2; ar(PQM) = A3; ar(NQR)= A4; and let the required area be x;
then we know that
A1+x+A4=1\2(Area of rectangle)
A2+x+A3=1\2(area of rectanlgle);
Also,
A1+3+2+A4+20 = A1+A4+x
x= 25;
hence option 1;
Part b- option a;its easy to comprehend.....
@sanyo thanks for the help....

------------------------------------------------------
Quantitative Question # 136
------------------------------------------------------
Each question is followed by 2 statements, A and B. Anwer each question using the following instructions
Choose 1 if the question can be answered using A alone
Choose 2 if the question can be answered using B alone
Choose 3 if the question can be answered using either A or (exclusive) B
Choose 4 if the question can be answered using A and B together
Choose 5 if the question can be answered neither using A nor B
Part (B)
In a triangle ABC, D is a point on the side BC and X, Y are length of perpendicular dropped on line AD from the vertices B and C respectively. Is X > Y?
(A) BD < DC
(B) AB > AC

option 1.

let P and Q be the length of the perpendiculars from B and C respectively.
triangle BPD and trangle CDQ are similar. as

------------------------------------------------------
Quantitative Question # 136
------------------------------------------------------

Each question is followed by 2 statements, A and B. Anwer each question using the following instructions

Choose 1 if the question can be answered using A alone
Choose 2 if the question can be answered using B alone
Choose 3 if the question can be answered using either A or (exclusive) B
Choose 4 if the question can be answered using A and B together
Choose 5 if the question can be answered neither using A nor B

Part (A)

ABCD is a rectangle and M and N are the points on AB and BC respectively. AN and DM intersect at P, AN and CM at Q, and CM and DN at R. Area of triangle APM is 3, and area of triangle CRN is 2. What is the area of the region PQRD?
(A) Area of region MBNQ is 20
(B) Area of triangle APD+ Area of triangle DRC = 18
Part (B)

In a triangle ABC, D is a point on the side BC and X, Y are length of perpendicular dropped on line AD from the vertices B and C respectively. Is X > Y?

(A) BD < DC
(B) AB > AC

Thanks Aarav for telling us the source .. but I thought u came up with these questions urself i mean some of them ...and What Schooling in US !!!
Coming to today's QQ Part B
I got answer as 8 so none of these.. am i right?

Very seldom we present the question verbatim when we pick it up from some source as it has to molded to make CAT like and then options are to be set intelligently (if I'm allowed to use this word). Yes, we come up with original questions as well and we do it very often than we actually claim to do. Tomorrow's DS problems falls in that category only.

The answer to Part B is 14.

I got the answer as 12 for PART B. is it correct?

Thanks Aarav for telling us the source .. but I thought u came up with these questions urself i mean some of them ...and What Schooling in US !!!
Coming to today's QQ Part B
I got answer as 8 so none of these.. am i right?

Just in case if anyone is feeling that we went over the board to talk about bijections and binary tree in today's QQAD problems - these problems are picked from school level contests in US for 9/10/11/12th standard students. These are the same sources from where we have spotted some of the previous CAT quant problems as well.

We are not going to introduce here anything which is CAT unlike. Rest assured and be here with full energy.

We will see Data sufficiency problems in QQAD tomorrow. This is modelled on CAT quant paper of 3/4 years back where the quant section had 10% of these kind of questions.

Atticus Finch Says

Aarav and Sloth, u guys r superb...hadnt heard such beautiful conversation in years...thanks for posting great questions in qqad...we owe u a lot Aarav and QQAD team...I solved Part B this way:for first goal it will be always (1-0)For second goal, (2-0) and (2-1)For 3rd, (3-0), (3-1), (3-2)For 4th, (4-0), (4-1), (4-2), (4-3)For 5th, (5-0), (5-1), (5-2), (5-3), (5-4)...total 14 ways...

Can we think on htese lines?

Hey def it all started with 1-0 and ended up 5-4

5-1 = 4 and 4-0 = 4

In all it could have been possible in 5.5 ways( ) .....and subtracting not possible cases....can't we get?? ......Acgtually I dont' have enuf time that's why

Thanks Aarav, for sharing with us about your background

Hey ATTICUS FINCH, YOu got hte name from to kill a mocking bird ha ( I think I am not spamming :neutral:)

Ok perhaps that explains the brilliance we keep seeing from you now and then.

I completed Maths Hons from DU in 98. There we were taught on cyclic permutations in the course called Abstract Algebra. The concrete Mathematics (Tree and stuff) was part of our course in MCA days.

Aarav and Sloth, u guys r superb...hadnt heard such beautiful conversation in years...thanks for posting great questions in qqad...we owe u a lot Aarav and QQAD team...I solved Part B this way:for first goal it will be always (1-0)For second goal, (2-0) and (2-1)For 3rd, (3-0), (3-1), (3-2)For 4th, (4-0), (4-1), (4-2), (4-3)For 5th, (5-0), (5-1), (5-2), (5-3), (5-4)...total 14 ways...

In that case we are done for the day.

selective_sloth Says

Aarav, thanks for the admiration. But kindly don't reduce me to an IITian!

Apologies sir. Let's get back to problem solving now

Sloth, kekambus and other Maths lovers ... we hope you are aware of this site Art of Problem Solving Forum

The sky is the limit out here. But, if the focus is solely CAT then we recommend the intermediate section on this forum.

Aarav, thanks for the admiration. But kindly don't reduce me to an IITian!

Part A

selective_sloth , y shld one count identity map as only 1 , it is possible to have different functions for (1,1) (2,2) (3,3) (4,4)

so answer shld be 13

Part B

Answer is 15 option (2)

Crude approach

first two matches are definetely won by india
Further posiibilities

1. 2I,1P.....
2. 3I,1P.... or 3I,2P,....
3. 4I,1P... or 4I,2P... or 4I,3P...
4. 5I,4P => 1 way

CASE3
Now starting with the 3rd case 4I,3P... I can come only in 8th position
e.g. IIIIPPPIP
For 4I,2P ...I can come only in 7th position
For 4I,1P ...I can come only in 6th position
=> 3 ways
CASE2
3I,2P....Next value must be I which means the remaining I can come only in the 7th or 8th place => 2 ways
Similarily for 3I,1P.... I can come only in 6th or 7th or 8th place => 3 ways
=> overall 5 ways
CASE1
2I,1P..... Similarily N =1+ {2+3 } = 6

=> overall for the question a total of 15 ways

Case 1, case 2, case 3 are ok and gives all 14 ways but case 4 should not come as it will be covered in all of the cases in diff. manners.

Aarav a request, can we have a good link for more information on bijections.

Just see if this helps Bijection - Wikipedia, the free encyclopedia
Else, please put keywords in Google and look at the results.

kekambus Says

there is a MSc. Integrated in Mathematics at IIT Kanpur...but for Chemical Engineers like us we didn't get get to learn a lot of mathematics thr.

Ok ... quite few IITians out here. No wonder the level of QQAD is good mainly b'coz of its audience.

Not during my MBA, but yes some years back during my Maths Hons and later on in MCA course :-)

Where did you learn this stuff from? I'm not sure if Pure Mathematics is taught in IIT.

there is a MSc. Integrated in Mathematics at IIT Kanpur...but for Chemical Engineers like us we didn't get get to learn a lot of mathematics thr.

selective_sloth Says

Math Honours? That explains. From where? I was doing my M.Sc Math from I.I.T Bombay.

Ok perhaps that explains the brilliance we keep seeing from you now and then.

I completed Maths Hons from DU in 98. There we were taught on cyclic permutations in the course called Abstract Algebra. The concrete Mathematics (Tree and stuff) was part of our course in MCA days.

Koush where is my mistake ...would be helpful if u can point it out. Did it with the binary approach, a clean and simple way to get the answer.

Aarav a request, can we have a good link for more information on bijections.

****

Double Post.

selective_sloth Says

That is correct Aarav. Did you study Pure Mathematics?

Not during my MBA, but yes some years back during my Maths Hons and later on in MCA course :-)

Where did you learn this stuff from? I'm not sure if Pure Mathematics is taught in IIT.

That is correct Aarav. Did you study Pure Mathematics?

Part A

selective_sloth , y shld one count identity map as only 1 , it is possible to have different functions for (1,1) (2,2) (3,3) (4,4)

so answer shld be 13

Part B

Answer is 15 option (2)

Crude approach

first two matches are definetely won by india
Further posiibilities

1. 2I,1P.....
2. 3I,1P.... or 3I,2P,....
3. 4I,1P... or 4I,2P... or 4I,3P...
4. 5I,4P => 1 way

CASE3
Now starting with the 3rd case 4I,3P... I can come only in 8th position
e.g. IIIIPPPIP
For 4I,2P ...I can come only in 7th position
For 4I,1P ...I can come only in 6th position
=> 3 ways
CASE2
3I,2P....Next value must be I which means the remaining I can come only in the 7th or 8th place => 2 ways
Similarily for 3I,1P.... I can come only in 6th or 7th or 8th place => 3 ways
=> overall 5 ways
CASE1
2I,1P..... Similarily N =1+ {2+3 } = 6

=> overall for the question a total of 15 ways

See the solution of part B in the post above.
the answer is coming 14.

Part A

selective_sloth , y shld one count identity map as only 1 , it is possible to have different functions for (1,1) (2,2) (3,3) (4,4)

so answer shld be 13

There will be 4! one-one onto functions from X->X. Using sloth's notation ...

1- Identity will be (1)(2)(3)(4)
4C2 will be of types (12)(3)(4)
4C2/2 will be of types (12)(34)
(3-1)!*4 will be of types (123)(4)
(4-1)! will be of types (1234)

Only 1st three category among these 24 bijections fits our conditions.

selective_sloth , y shld one count identity map as only 1 , it is possible to have different functions for (1,1) (2,2) (3,3) (4,4)

?

Part A

selective_sloth , y shld one count identity map as only 1 , it is possible to have different functions for (1,1) (2,2) (3,3) (4,4)

so answer shld be 13

Part B

Answer is 15 option (2)

Crude approach

first two matches are definetely won by india
Further posiibilities

1. 2I,1P.....
2. 3I,1P.... or 3I,2P,....
3. 4I,1P... or 4I,2P... or 4I,3P...
4. 5I,4P => 1 way

CASE3
Now starting with the 3rd case 4I,3P... I can come only in 8th position
e.g. IIIIPPPIP
For 4I,2P ...I can come only in 7th position
For 4I,1P ...I can come only in 6th position
=> 3 ways
CASE2
3I,2P....Next value must be I which means the remaining I can come only in the 7th or 8th place => 2 ways
Similarily for 3I,1P.... I can come only in 6th or 7th or 8th place => 3 ways
=> overall 5 ways
CASE1
2I,1P..... Similarily N =1+ {2+3 } = 6

=> overall for the question a total of 15 ways

Part B) The answer is 14 ways.

Soln attached

selective_sloth Says

I am referring to the binary tree of possible score-lines. The root of this tree is 2 - 0 and the leaf nodes are all 5 - 4. At each node there are at most two branches. Hence, binary tree.

Superstar Says

I am getting the answer as 12 for PART B

but i am getting 16 as the answer ....:neutral:

first walla to sir ke upper se nikal gaya

selective_sloth Says

I am referring to the binary tree of possible score-lines. The root of this tree is 2 - 0 and the leaf nodes are all 5 - 4. At each node there are at most two branches. Hence, binary tree.

I am getting the answer as 12 for PART B

Answer for PART B)- 2) 15

Let me know if it is correct.

I am referring to the binary tree of possible score-lines. The root of this tree is 2 - 0 and the leaf nodes are all 5 - 4. At each node there are at most two branches. Hence, binary tree.

Part A -- Answer is (5)

If f:X -> X where X = {1,2,3,4} such that f.f (x) = x, then f is a bijection. There are 24 bijections from X to X. Of these only 10 are such that f.f (x) = x for all x in X.

In giving these maps explicitly I will use a cyclic notation for convenience.

eg, (1 2) stands for the map 1 -> 2, 2 -> 1, 3 -> 3, 4 -> 4

Note that in this cyclic notation 3 and 4 are not mentioned which means that they are mapped to themselves i.e. they are fixed.

The maps are :

(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)

and also

(1 2) (3 4)

(1 3) (2 4)

(1 4) (2 3)

Note that (1 2) (3 4) means 1 -> 2, 2 -> 1, 3 -> 4, 4 -> 3

And lastly the identity map which gives us a total of 10 maps satisfying the given condition.


Part B -- Answer is (3)

If one draws a binary tree to represent the problem, one is required to only count the number of leaf nodes of the tree. They turned out to be 14 in number.

Discrete Mathematics, Abstract Algebra. Wow, this looks more like real Mathematics!!

Particularly impressed with your use of notations in the first problem. Just elaborate the approach to Part B as well.

Part A -- Answer is (5)

If f:X -> X where X = {1,2,3,4} such that f.f (x) = x, then f is a bijection. There are 24 bijections from X to X. Of these only 10 are such that f.f (x) = x for all x in X.

In giving these maps explicitly I will use a cyclic notation for convenience.

eg, (1 2) stands for the map 1 -> 2, 2 -> 1, 3 -> 3, 4 -> 4

Note that in this cyclic notation 3 and 4 are not mentioned which means that they are mapped to themselves i.e. they are fixed.

The maps are :

(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)

and also

(1 2) (3 4)

(1 3) (2 4)

(1 4) (2 3)

Note that (1 2) (3 4) means 1 -> 2, 2 -> 1, 3 -> 4, 4 -> 3

And lastly the identity map which gives us a total of 10 maps satisfying the given condition.


Part B -- Answer is (3)

If one draws a binary tree to represent the problem, one is required to only count the number of leaf nodes of the tree. They turned out to be 14 in number.

------------------------------------------------------
Quantitative Question # 135 (The answer to both Part A and Part B are in the same set of options)
------------------------------------------------------
Part (A)

Let X = {1, 2, 3, 4}. How many functions f: X->X are there satisfying (f.f)(x) = x for all x in X?



Part (B)
In a hockey match India beat Pakistan 5-4. India scored first and kept the lead until the end. In how many different orders could the goals been scored?


(1) 13 (2) 15 (3) 14 (4) 16 (5) none of these
-------------------------------------------------------------------

Aarav I am writing the full solution to this problem as I took it...

Please let me know where I am wrong (if I am )

So, Finally

2x^4 - 4ax^3 + 6a^2.x^2 - 4a^3.x + (a^4-b) = 0

Now if assume a^4 > b then effectively there will be 4 sign changes. ( I included the sign for 2x^4 also ) ..........Am I correct
So, there are maximum 4 positive real roots to this equation.
For f (-x), I have no sign changes So there will be no negative roots to this equation.

So, all roots will have to be positive and they have to be 4 in number only as any number of roots less than 4 would introduce imaginary roots into the picture which is not possible.....So a^4 > b is a condition for roots to be real and satisy both equations

Aarav, Please finally enlighten.... So, that this freezes for everyone now.

Thanks for support..

Acchcha when to expect the next Practice test....:)

See, if you read the link pasted on Descrates rule of sign, you will see that the number of positive roots are either 4 or 2 or 0. Hence, you are not sure if this equation will have any real roots.
One example that explains this is (x-1)^4 = -1 as this has 0 real roots.

Next practice test is on 03/10.

Aarav I am writing the full solution to this problem as I took it...

Please let me know where I am wrong (if I am )

So, Finally

2x^4 - 4ax^3 + 6a^2.x^2 - 4a^3.x + (a^4-b) = 0

Now if assume a^4 > b then effectively there will be 4 sign changes. ( I included the sign for 2x^4 also ) ..........Am I correct
So, there are maximum 4 positive real roots to this equation.
For f (-x), I have no sign changes So there will be no negative roots to this equation.

So, all roots will have to be positive and they have to be 4 in number only as any number of roots less than 4 would introduce imaginary roots into the picture which is not possible.....So a^4 > b is a condition for roots to be real and satisy both equations

Aarav, Please finally enlighten.... So, that this freezes for everyone now.

Thanks for support..

Acchcha when to expect the next Practice test....:)

PART A

Option 5 none of these is the answer.

a^4 > b .

I did it by forming the equation which is of 4 degree which has a constant term as a^4 - b

Apply the logic ( There is some theorem) which helps to find out the maximum nember of positive and negative real roots. And One finds that the above condition has to hold true.

Aarav Bhaiyya , I hope the logic is fine

Descrates is sufficient in this case but not necessary. Sorry, for last minute hiccups ... just got terribly confused in the midst of work here.

Hi Aarav,

Even I applied the same logic as explained by SuperStar, Although i didnot posted the whole solution earlier.....Just wanted to know why we cant use the Descrat's theorem. Is there something which i am missing.
Thanks in advance

Perhaps, this can be answered using Descrates rule of signs had the question been just sufficient condition and not necessary.

For, x^4 + (x-a)^4 = b, we have 2x^4 - 4ax^3 + 6a^2.x^2 - 4a^3.x + (a^4-b) = 0 => b >= a^4. The change of signs is 3 in this case and hence we have atleast one positive real solution.

------------------------------------------------------
Quantitative Question # 134
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Part (A)

The necessary and sufficient condition for the equations x+y = a and x^4 + y^4 = b to have real roots is

(1) b >= a^4 (2) a >= 4b^4 (3) a >= b^4 (4) b >= 4a^4 (5) none of these

To solve take x=0,y=2 => options (2) ,(3) and (4) are eliminated.

Now take x=1,y=2 =>option (1) gone and hence the ans:)