official quant thread for cat08

[selebratinglife]

Originally Posted by nikunj14_83

Hi all,
I newed ur help in this question....
In a triangle ABC,points D & E lie on BC and AC respectively .If AD & BE intersect at T so that AT/DT =3 & BT/ET =4 what is CD/BD??

From point T draw a line parallel to AC meeting BC at F,
In triangle BEC, BT:TE = 4:1, So BF:FC= 4:1(TF||EC)
In triangle ACD, AT:TD= 3:1 so DF:FC=1:3..(TF||AC)
So the ratio of BD: DC = (4-1/3): ((1/3)+1)=11:4

[shivam_01]

Originally Posted by saurabh1310

hii guys...m new to this site.......
hello to everyone.....
can anyone plzz solv this Q (may b an easy one 4 u ppl)...

Q: two cyclist are travelling at a speed of 3Km/hr and 10Km/hr around a circular track.What should be the ratio of number of distinct points that they would meet on that track if they travel in the same direction first and in the opposite next..??

using alligations we get the ratio as (3+10)/(10-3) answer sholud be 13/7 do u have the key

[saurabh1310]

Originally Posted by shivam_01

using alligations we get the ratio as (3+10)/(10-3) answer sholud be 13/7 do u have the key

answer given in the key is 7/13....
how did u approach by alligations

[implex]

New Problem 206)!!
FInd teh difference between teh largest and the smallest x which satisfies
x^2 + 7x + 6)^2 + 7(x^2 + 7x + 6)+ 6 = x

[Varun Khullar]

Originally Posted by implex

New Problem 206)!!
FInd teh difference between teh largest and the smallest x which satisfies
x^2 + 7x + 6)^2 + 7(x^2 + 7x + 6)+ 6 = x

i got

21^0.5

tried again ..

got root 3

[selebratinglife]

Originally Posted by implex

New Problem 206)!!
FInd teh difference between teh largest and the smallest x which satisfies
x^2 + 7x + 6)^2 + 7(x^2 + 7x + 6)+ 6 = x

Awesum problem..Really enjoyed doing it..i got the answer as 1+sqrt(3)-sqrt(2).
(t+1)(t+6)=x Where t=x^2 + 7x + 6
Now consider x1,x2 as max an min values..that satisfy the equation
So substituting x1,x2 an subtracting the equations, we get..
(x1-x2)(x1+x2+7)(x1^2 + x2^2 + 7x1 + 7x 2+ 19)=(x1-x2)..
Considering x1-x2<>0 an solving, we can get two values of x1-x2 since the above equation can be equated for 1*1 and also (-1)*(-1)..
We get (x1,x2)=(sqrt(3)-3,-sqrt(3)-3) or (sqrt(2)-4,-sqrt(2)-4)..The ones in bold are the highest and the least..So x1-x2 will b 1+sqrt(3)-sqrt(2)...

[anujsukhlecha]

Originally Posted by selebratinglife

Awesum problem..Really enjoyed doing it..i got the answer as 1+sqrt(3)-sqrt(2).
(t+1)(t+6)=x Where t=x^2 + 7x + 6
Now consider x1,x2 as max an min values..that satisfy the equation
So substituting x1,x2 an subtracting the equations, we get..
(x1-x2)(x1+x2+7)(x1^2 + x2^2 + 7x1 + 7x 2+ 19)=(x1-x2)..
Considering x1-x2<>0 an solving, we can get two values of x1-x2 since the above equation can be equated for 1*1 and also (-1)*(-1)..
We get (x1,x2)=(sqrt(3)-3,-sqrt(3)-3) or (sqrt(2)-4,-sqrt(2)-4)..The ones in bold are the highest and the least..So x1-x2 will b 1+sqrt(3)-sqrt(2)...

hey m getting a different answer...
take x^2 + 7x + 6 = x ; then we get the final equation as x^2 + 7x + 6 = x which is what we assumed. so the assumption is right and the difference in roots of the simplified equation will be 2*sqrt(3)!

[ankitthakur]

Originally Posted by vaibhav2cool

x=13y/y-13
y can be 14 and 26.
similarlily y=13x/x-13
x can be 14 and 26

Hi everybody, I am not satisfied with equation that u r giving and the solutions
as
1/x +1/y=1/13<=>x=13y(y-13)
and solution comes out to be are 4 (182,14),(14,182),26,26) and (39,16)

[implex]

Originally Posted by selebratinglife

Awesum problem..Really enjoyed doing it..i got the answer as 1+sqrt(3)-sqrt(2).
(t+1)(t+6)=x Where t=x^2 + 7x + 6
Now consider x1,x2 as max an min values..that satisfy the equation
So substituting x1,x2 an subtracting the equations, we get..
(x1-x2)(x1+x2+7)(x1^2 + x2^2 + 7x1 + 7x 2+ 19)=(x1-x2)..
Considering x1-x2<>0 an solving, we can get two values of x1-x2 since the above equation can be equated for 1*1 and also (-1)*(-1)..
We get (x1,x2)=(sqrt(3)-3,-sqrt(3)-3) or (sqrt(2)-4,-sqrt(2)-4)..The ones in bold are the highest and the least..So x1-x2 will b 1+sqrt(3)-sqrt(2)...

earlier u had done it incorrectly. now it is fine !

edit: there is a mistake it shoudld be 1+sqrt(3)+sqrt(2)

[themystique]

Originally Posted by implex

earlier u had done it incorrectly. now it is fine !

i lost it from here:

Considering x1-x2<>0 an solving, we can get two values of x1-x2 since the above equation can be equated for 1*1 and also (-1)*(-1)..
We get (x1,x2)=(sqrt(3)-3,-sqrt(3)-3) or (sqrt(2)-4,-sqrt(2)-4)..The ones in bold are the highest and the least..So x1-x2 will b 1+sqrt(3)-sqrt(2)...

could you please lead me from here,,
thanks
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