[unique_1]

soltn to 2. -1<=x<=3
let a,ak,ak^2,ak^3 be the no:s. then we ve (1-k^3)/k(1-k) >=x or (1+k+k^2)/k >=x or K^2+(1-x)k+1>=0 then (1-x)^2-4<=0 or (x+1)(x-3)<=0 therefore -1<=x<=3.
nagi i guess u were almost rite but made a silly mistake in "lhs= p^3-1 / p2-p = (p-1)^3-3p(p-1)/(p(p-1))" it should be lhs= p^3-1 / p2-p = (p-1)^3+3p(p-1)/(p(p-1)).

[prade]

Originally Posted by prade

qn 3 answer option b rest will do tomorrow...

d/u-v = 84
d/u+v = d/u -9
let u =kv
d/v(k+1) = d/kv - 9
d/v(k-1) =84
84(k-1) ( 1/k - 1/(k+1)) = 9
9k^2 - 75k + 84 = 0
solving
k = 7, 4/3
for k = 7
d/v = 84*6
d/(v+u) = d/ v(k+1) = 84*6/8 = 63 hence answer option d

Q if a,b,c,d are in continued proportion then [(a-d)/(b-c)] >= x, then what is the calue of x?

a/b = b/c = c/d = k

a= dk^3
b=dk^2
c=dk

a-d/b-c =( k^3-1)/(k^2 -k) = (k-1)(k^2+k+1)/(k-1)k >= x

(k^2+k+1)/k >=x
k^2 + k(1-x) + 1 >= 0

for real K

(1-x)^2 - 4 >= 0
x<=-1 or x>=3

[clsuresh]

Folks, I am back again into my thread. I will be continously posting questions from now till the CAT exam( as done by me in the last year). Infact I'll be concentrating a lot on numbers, PG and pure maths part.

Here comes our first question.
1. Remainder when (56)^493 is divided by 126.

[nagi_nov3]

Originally Posted by clsuresh

Folks, I am back again into my thread. I will be continously posting questions from now till the CAT exam( as done by me in the last year). Infact I'll be concentrating a lot on numbers, PG and pure maths part.

Here comes our first question.
1. Remainder when (56)^493 is divided by 126.

here u go

56.56^492/126
= 4.56^492 mod 9 ( we have to multiply by 14 later)
=4. 2^492
= 4. 8^156
=4 . (-1)^156
=4 ( 14) we have divided intially
=56


56.56^492/126
= 4.56^492 mod 9 ( we have to multiply by 14 later)

euler number for 9 = 6 and 492 is divisible by 6 so
=4 .(14) = 56

[Superstar]

1. A stick of length 1 unit is cut into three parts. Probability that a triangle can be formed using these three sticks is?

Hi Suresh,

Great to see you back.

Could you please elaborate approach to above problem.

[nagi_nov3]

Originally Posted by Superstar

1. A stick of length 1 unit is cut into three parts. Probability that a triangle can be formed using these three sticks is?

lets take sides as a,b, 1-(a+b)

we know difference of 2 sides should be lessthan that of third side so

a-b < 1-(a+b) ==> a<1/2
b-a < 1 - (a+b) ==> b<1/2

so we just need a,b <1/2

we can say probailily to form a triangle with the given sticks is <1/4

it cant be even 1/4 it is slightly less than that.

[esjay]

Originally Posted by nagi_nov3

lets take sides as a,b, 1-(a+b)

we know difference of 2 sides should be lessthan that of third side so

a-b < 1-(a+b) ==> a<1/2
b-a < 1 - (a+b) ==> b<1/2

so we just need a,b <1/2

we can say probailily to form a triangle with the given sticks is <1/4

it cant be even 1/4 it is slightly less than that.

shudnt the condition be a,b,c<1/2.??: . i mean take a=b=0.1 which makes a+b<c and triangle cannot be formed
hence none of the sides can be of length 1/2 or more
so probability is slightly less the 1/8

[nagi_nov3]

Originally Posted by esjay

shudnt the condition be a,b,c<1/2.??: . i mean take a=b=0.1 which makes a+b<c and triangle cannot be formed
hence none of the sides can be of length 1/2 or more
so probability is slightly less the 1/8

u have a valid point

apart from a,b<1/2
we will have a+b > 1-(a+b) ( sum of 2 is more that 3rd one)
a+b >1/2
so third side 1-(a+b)/2 also <1/2.

so answer is less than 1/8.

Thanks buddy.

[esjay]

nagi but im not really sure about the probability thing. the ans "slightly less than 1/8" etc doesnt sound good to me. is that really how we find probabilities????
my probability sucks bigtime :(

[nagi_nov3]

Originally Posted by esjay

nagi but im not really sure about the probability thing. the ans "slightly less than 1/8" etc doesnt sound good to me. is that really how we find probabilities????
my probability sucks bigtime :(

normally for any probability qn we end up getting some fraction.
here we have considered all the possibilities so we may be right.

the traingle can only be formed when all sides satisifies triangle basic rules, so we are correct, unless some one come up with some other possibilities