Official Quant thread for CAT 2013

@gs4890 said:
@amresh_maverick kya haal chal boss ? CAT 13 ?-----------------------------------------------------------Itz not a spam Q. Find the number of pairs possible. x^2-y^2 = 98
Infinite.

Oh, you meant integers? Well, why didn't you say so? Zero, that I can spot.

regards
scrabbler




@gs4890 said:
@amresh_maverick kya haal chal boss ? CAT 13 ?-----------------------------------------------------------Itz not a spam Q. Find the number of pairs possible. x^2-y^2 = 98
0 pairs possible.
(x-y)(x+y) = 98

Both x-y and x+y can be either odd or even.
Case 1. Both are Odd,
Product of two odd numbers can't be even. So not possible

Case2. Both are even
98 cannot be expressed as a product of two even numbers.
Hence no case possible.

@jain4444 said:
hmm did mistake so , answer should be (21C4 + 21C3*3 + 21C2)*4!*20! right ?
Not sure....was not so comfortable with your route so didn't follow to logical conclusion....am not even sure of mine btw :P

regards
scrabbler

@gs4890 said:
@amresh_maverick kya haal chal boss ? CAT 13 ?-----------------------------------------------------------Itz not a spam Q. Find the number of pairs possible. x^2-y^2 = 98
zero , done a similar one

x^2 -y^2 =875622
no of +ve integral solns


@Joey_Sharma said:
how did u conclude that third smallest sum will be b+c only and not a+d ? i didn't get this.
see it like this...
a=0,b=1,c=2,d=4
a+d=4
and b+c=3

or

a=1,b=2,c=3,d=4
b+c=5
and a+d=5

so it would either be same or be less...but never more than a+d

it wld always happen in cases where a
u can try with random numbers and see for yourself...
@amresh_maverick said:
zero , done a similar onex^2 -y^2 =875622 no of +ve integral solns
875622 is 4k+2 form...hence no solution??
@Logrhythm said:
see it like this...a=0,b=1,c=2,d=4a+d=4and b+c=3ora=1,b=2,c=3,d=4b+c=5and a+d=5so it would either be same or be less...but never more than a+dit wld always happen in cases where au can try with random numbers and see for yourself...
1, 3, 4, 5

regards
scrabbler

In the given figure, there are two concentric circles with radii 10 cm and 20 cm. Also, AB
and AC are tangents to the circle with radius 10 cm at P and Q respectively, PQ = 19 cm
and BC is a tangent to the circle with radius 20 cm. Find the area of Δ ABC in cm2.

(Take 3 = 1.732 and 39 = 6.25).

1] 885.5 cm2
2] 306.5 cm2
3] 353.5 cm2
4] 994.5 cm2


P.S : Refer to the figure attached, kindly provide the approach !
@scrabbler said:
1, 3, 4, 5regardsscrabbler
oh...i was taking consecutive numbers...

that question had a condition on sum (dnt remember exactly atm), wo satisfy bhi toh karni thi.. :D
@Logrhythm said:
oh...i was taking consecutive numbers... that question had a condition on sum (dnt remember exactly atm), wo satisfy bhi toh karni thi..
Take -0.5, 1.5, 2.5, 3.5 satisfy ho jayega.

So d is either 3.5 or 4, total 15/2...

regards
scrabbler

@scrabbler said:
Take -0.5, 1.5, 2.5, 3.5 satisfy ho jayega.regardsscrabbler
bohot sahi...and there was an option of 15/2...wohi answer hona chahiye fir toh...

ye wala case kaise nikaala?? matlab how did u come to it??
@Logrhythm said:
see it like this...a=0,b=1,c=2,d=4a+d=4and b+c=3ora=1,b=2,c=3,d=4b+c=5and a+d=5so it would either be same or be less...but never more than a+dit wld always happen in cases where au can try with random numbers and see for yourself...
is it true for this question only? because cases like a=0,b=3,c=4 and d=5 don't satisfy this...
@tani90 said:
In the given figure, there are two concentric circles with radii 10 cm and 20 cm. Also, ABand AC are tangents to the circle with radius 10 cm at P and Q respectively, PQ = 19 cmand BC is a tangent to the circle with radius 20 cm. Find the area of Δ ABC in cm2.(Take 3 = 1.732 and 39 = 6.25).1] 885.5 cm22] 306.5 cm2 3] 353.5 cm2 4] 994.5 cm2P.S : Refer to the figure attached, kindly provide the approach !
two options eliminated it has to be 1 or 4, Taking a guess here option 4> 994.5 cm^2

could be option 1 as well , let me solve


@Logrhythm said:
bohot sahi...and there was an option of 15/2...wohi answer hona chahiye fir toh...ye wala case kaise nikaala?? matlab how did u come to it??
Consider a, b, c, d in order

Now min cases are (a+b), (a+c) and max are (c+d), (b+d) only the middle two are uncertain (a+c) and (b+d). So solve two sets of eqns - one which you did and the second taking (a+d) = 3 and (b+c) = 4, which yield the other solution.

regards
scrabbler

@amresh_maverick said:
A lot of repetitionsABC from state X ,MNO from state Y , PQR from state Z Suppose u selected 1 from each state : A , M and Pand to random, let us say : Q, RYour group : A,M ,P,Q,RNow suppose u chose : A, M, R (1 from each state) + P, Q (2 from each state)Your group : A,M ,P,Q,R ------a repeatu get the drift ?
yeah..got it. thanks man!!
@Joey_Sharma said:
is it true for this question only? because cases like a=0,b=3,c=4 and d=5 don't satisfy this...
nah..i was wrong...check @scrabbler's post on this...
@amresh_maverick said:
two options eliminated , Taking a guess here option A 885.5 cm^2
Don't know the actual answer, I also zeroed it to 1 & 4 options... Because area of the trapesium is itself close to 1... 😃

If x+y+z=0 then value of

([x/(z+y) + y/(z+x) +z/(x+y)]/ [(x+y)-(y+z)-(z+x)] ) * 2(x+y)

@amresh_maverick said:
If x+y+z=0 then value of([x/(z+y) + y/(z+x) +z/(x+y)]/ [(x+y)-(y+z)-(z+x)] ) * 2(x+y)
-3? undefined?
Hii,This is my first post
78. The following is known about three real numbers x,y and z.-4