To begin with this is not my idea, and has been borrowed from some other math site (i am not putting the name, will be against the rules of PG:), but i am sure our quant studs must be aware). So here are the rules.Ths basics: i start off with a pr…

To begin with this is not my idea, and has been borrowed from some other math site (i am not putting the name, will be against the rules of PG:), but i am sure our quant studs must be aware). So here are the rules.Ths basics: i start off with a problem, someone solves it(has to provide the full solution, and the solution can be discussed) and posts the new problem.

Few things worth noting:

1. Please keep the problems numbered.

2. The level of problems can be anyhting provided that it helps us in understanding the basics required to crack CAT(especially after trend in cat quant which focusses more on concepts rather than speed).

3. Please keep the thread moving by posting a new problem everytime u have solved a problem, if u have none with u..u can request somebody else to post a new problem.

So here we go:

PROB No.1: If 'a' is a real number and the equation (a-2)(x-)^2 +2(x-) +a^2 =0 (where denotes the greatest integer

a. (-1,2) b. (0,1) c. (-1,0) d. (2,3)

Happy solving..:)

Where are quant gods?...c,mmon folks at least give a try here...If nobody comes out with the solution i will have to post the solution which will be so boring ..:)...

To begin with this is not my idea, and has been borrowed from some other math site (i am not putting the name, will be against the rules of PG:), but i am sure our quant studs must be aware). So here are the rules.Ths basics: i start off with a problem, someone solves it(has to provide the full solution, and the solution can be discussed) and posts the new problem.

Few things worth noting:

1. Please keep the problems numbered.

2. The level of problems can be anyhting provided that it helps us in understanding the basics required to crack CAT(especially after trend in cat quant which focusses more on concepts rather than speed).

3. Please keep the thread moving by posting a new problem everytime u have solved a problem, if u have none with u..u can request somebody else to post a new problem.

So here we go:

PROB No.1: If 'a' is a real number and the equation (a-2)(x-)^2 +2(x-) +a^2 =0 (where denotes the greatest integer

a. (-1,2) b. (0,1) c. (-1,0) d. (2,3)

Happy solving..:)

artofproblemsolving.com

Intermediate section on that forum is something that is useful for CAT preparation.

That site runs marathon on many topics. I was interested in starting this, but decided against finally and instead went for "quant question a day" which gives you exactly what you need.

We are filtering out the best, and among many sources that site is one of our sources.

I see some competition to "quant question a day" from this thread.

And that's a good thing happening :-)

Happy solving.

f(2) = 4 + t

f(3) = 9 + 2t + t^2

where t = (x-)

=> f(2) 0 => t

(a-2)(x-2)^2 + 2(x-2) + a^2 = 0

=> 4 -4(a-2)a^2 >= 0

=> 1 - a^3 - 2a^2 >= 0

=> (1+a) (a^2 + a -1) => a lies in (-1, (5^1/2 -1)/2)

Hence, (a)

changing later...

F(0)F(1) exactly one.

I missed this basic point. Hence, (c) is the solution

**.**

PROB No.1: If 'a' is a real number and the equation (a-2)(x-)^2 +2(x-) +a^2 =0 (where denotes the greatest integer

a. (-1,2) b. (0,1) c. (-1,0) d. (2,3)

Happy solving..:)

the problem has only soln. in x belonging to (2,3) => (x-) belongs to (0,1) ....... (the key thing is the open interval and not the numbers 2 and 3)

let z = (x-)

=> (a-2)*z^2 + 2*z + a^2 = 0

=> z = -1 + sqrt(1 - (a-2)*a^2) .... (the other solution is

ignored as z cant be negative)

we have 0

=> 1

=> 1

=> 0

=> i) (a-2)*a^2

ii) (a-2)*a^2 > -3

i) => a

ii) => a^3 - 2*a^2 + 3 > 0

=> (a + 1)*(a^2 - 3*a + 3) > 0

=> (a + 1)*( (a - 1.5)^2 + 0.75 ) > 0

=> a > -1

i), ii) => -1

artofproblemsolving.com

Intermediate section on that forum is something that is useful for CAT preparation.

That site runs marathon on many topics. I was interested in starting this, but decided against finally and instead went for "quant question a day" which gives you exactly what you need.

We are filtering out the best, and among many sources that site is one of our sources.

I see some competition to "quant question a day" from this thread.

And that's a good thing happening :-)

Happy solving.

f(2) = 4 + t

f(3) = 9 + 2t + t^2

where t = (x-)

=> f(2) 0 => t

(a-2)(x-2)^2 + 2(x-2) + a^2 = 0

=> 4 -4(a-2)a^2 >= 0

=> 1 - a^3 - 2a^2 >= 0

=> (1+a) (a^2 + a -1) => a lies in (-1, (5^1/2 -1)/2)

Hence, (a)

The interval in option a) includes the interval in option c) also...How do we check for that..

I think there is more to it. We can proceed by taking into account the fact that if 2

**anyways where is next problem, Aarav bhai and Chautauqa bhai. Any one of you please post the new problem**

Since, the norm is to give a question after solving it, here is one from me.

I solved this, but looking for an alternate approach.

**Problem #2**

Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

The interval in option a) includes the interval in option c) also...How do we check for that..

I think there is more to it. We can proceed by taking into account the fact that if 2

Amazing solution. I'm impressed.

And I also realise my error.

Please give no to the problem also that you have posted...It keeps things easier ...I hope you don't mind doing that :). Regarding the options, i think i have tried to explain the previous post

For problem no.2, ie m/n + (n+1)/m=4..I am just giving a try...

We can write the given equation as m/n+n/m+1/m=4

Using A.M>=G.M , we can write 4>= 2+1/m

ie 1/m

I dont know wht to do next:)

For problem no.2, ie m/n + (n+1)/m=4..I am just giving a try...

We can write the given equation as m/n+n/m+1/m=4

Using A.M>=G.M , we can write 4>= 2+1/m

ie 1/m

I dont know wht to do next:)

AM-GM doesn't apply for 2 numbers when m/n

Requires a bit of number theory unless you come up with some other approach 😃

Since, the norm is to give a question after solving it, here is one from me.

I solved this, but looking for an alternate approach.

Problem #2

Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

lhs= m/n+n/m+1/m=4

= (root(m/n)-root(n/m))^2+1/m=2

clearly , m, n have to be greater than zero , which limits the set of integers to 01 then either of root(m/n) or root(n/m) will be irrational and square of irrational = irrational. to that add 1/m , answer again irrational , while2 is rational.

so lhs= irrational and rhs = rational , hence the proof.

i realise this isnt a gr8 solution ... so shud I still post a problem or not??

anyways here isproblem #3

This one a bit easier than above 2:

At what time are the hands of clock together between 2 and 3?

1. 10 10/11 mins past 2

2. 10 10/11 mins to 3

3. 9 1/11 mins past 2

4. 11 1/11 mins past 2

Since, the norm is to give a question after solving it, here is one from me.

I solved this, but looking for an alternate approach.

Problem #2

Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

lhs= m/n+n/m+1/m=4

= (root(m/n)-root(n/m))^2+1/m=2

clearly , m, n have to be greater than zero , which limits the set of integers to 01 then either of root(m/n) or root(n/m) will be irrational and square of irrational = irrational. to that add 1/m , answer again irrational , while2 is rational.

so lhs= irrational and rhs = rational , hence the proof.

i realise this isnt a gr8 solution ... so shud I still post a problem or not??

anyways here isproblem #3

This one a bit easier than above 2:

At what time are the hands of clock together between 2 and 3?

1. 10 10/11 mins past 2

2. 10 10/11 mins to 3

3. 9 1/11 mins past 2

4. 11 1/11 mins past 2

I cud not understand the part of ur solution which assumes/concludes m, and n to be positive integers. Could u elaborate a little please..:).. i think we should post the new problem only after the solution to the previous problem has been discussed and clarified..:)...that wud be better ...wht say..

problem #3

This one a bit easier than above 2:

At what time are the hands of clock together between 2 and 3?

1. 10 10/11 mins past 2

2. 10 10/11 mins to 3

3. 9 1/11 mins past 2

4. 11 1/11 mins past 2

Hmm an easy one Option 2 is the answer.

Hmm at exactly 2pm or a.m angle between hours nd minutes hand is 60 deg.

So say in P mts they are together..

then we have 60 + 0.5P = 6P

per minute - Hours hand covers 1/2 a degree nd minutes hand covers 6 deg.

So we have P = 120/11 = 10 10/11 mins past 2

**Prob 4**: DS qstion basically Hope DS qstion is ok in this thread.:)

Is a+b

I . a,b,c,d are multiples of 20

II. a+c

Am not able to solve this fully..??:??:

Hi i am posting prblm#4

If theres a square of each side 'a'..den if we draw a quadrant of a circle wd one of th vertices as a center and th side as the radius, inside th square,i.e. th arc is drawn with in the square and so if we draw 4 such arcs or quadrants with in the square wd all th 4 vertices as centres...den find th area of th portion common to all th 4 quadrants i.e th area where all of dem meet..the figure will b like two leaves intersecting one another with in the square..i hope i hav elucidated th prblm..

**Varun please edit your post, the 4th problem has already been posted and yet to be solved:)**. Solve the 4th problem and then post the new problem. I am not sure about the 2nd question,

**AARAV bhai problem 2 ka kuch karo**:)

And yeah, Varun you post a new problem only when you have solved the previous problem...I hope that is clear..It keeps things simple and in order..:)

Vijay please edit your post, the 4th problem has already been posted and yet to be solved:). Solve the 4th problem and then post the new problem. I am not sure about the 2nd question,AARAV bhai problem 2 ka kuch karo:)

And yeah, Vijay you post a new problem only when you have solved the previous problem...I hope that is clear..It keeps things simple and in order..:)

Hey I think I solved the 3rd problem nd then only posted the 4th one..and that too I posted it before the second 4th problem is posted.SO who needs to edit whose posts amar??:grab:

**I am sorry vijay, actually the message was meant for Varun..:)...I have edited my post**

You know what you did here, created a square root and then had put the restrictions.

It's like saying -3 = x = (root(x))^2 = -3 has no solution as the square is always positive.

Problem #2

Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

lhs= m/n+n/m+1/m=4

= (root(m/n)-root(n/m))^2+1/m=2

clearly , m, n have to be greater than zero , which limits the set of integers to 0

Hi

M soo sorry to post it..

Regards

Varun

Varun please edit your post, the 4th problem has already been posted and yet to be solved:). Solve the 4th problem and then post the new problem. I am not sure about the 2nd question,AARAV bhai problem 2 ka kuch karo:)

And yeah, Varun you post a new problem only when you have solved the previous problem...I hope that is clear..It keeps things simple and in order..:)

For** problem 2**. i give one more try:::

m/n+(n+1)/m=4

m^2-4n(m)+n^2+n=0, for n, m not equal to 0

The above equation is quadratic in 'm'.

Solving for m, we have m= 2n+-sqrt(n.(3n-1)),**Assuming that 'n' is an integer, if i can prove that n(3n-1) is never a square, we arrive at the result that 'm' is not an integer...which i think will complete the proof.****Aarav, need ur help in proving this ...n(3n-1) is never a square for any integer value of 'n'....:)**