Official Quant thread for CAT 2013

@mohitjain said:
Two series A (a1, a2, a3, ....an) and B (b1, b2, b3, ....bn) are in A.P. such that an – bn = n – 1, where an stands for the nth term of the series A, and bn for the nth term of the series B. It is also known that a6 = b8. Find the value of a101 – b121.
let B series be.. b1=b,b2=b+d, b3=b+2d ... and so on

it is given a6=b8
also, an-bn=n-1

a6-b6=5
b8-b6=5
or b+7d- b-5d=5
d=5/2

now a101-b101=100
or a101= 100+b101

substitute this in what we have to find..
100+b101-b121
100+ b + 100*5/2 - b -120*5/2
350-300=50
@Logrhythm said:
bhai i am not sure if the bold part is correct....how about the case 2x1+2x2+x5 = 6
here x5 can be odd as well...

PS - take this with a pinch of salt as i am not saying this based on a lot of thought, office mein hun abhi tak...
bhai 2x1 is even 2x2 is even if x3 is odd
then how can even+even+odd gives u even ???????
@iLoveTorres said:
50?
@ChirpiBird said:
let B series be.. b1=b,b2=b+d, b3=b+2d ... and so onit is given a6=b8also, an-bn=n-1a6-b6=5b8-b6=5or b+7d- b-5d=5d=5/2now a101-b101=100or a101= 100+b101substitute this in what we have to find..100+b101-b121100+ b + 100*5/2 - b -120*5/2350-300=50
@Logrhythm said:
bhai i am not sure if the bold part is correct....how about the case 2x1+2x2+x5 = 6
here x5 can be odd as well...

PS - take this with a pinch of salt as i am not saying this based on a lot of thought, office mein hun abhi tak...
bhai how can even+even+odd gives u even??
think ??????
@insane.vodka said:
In a chess competition involving some boys and girls of a school, every studenthad to play exactly one game with every other student. It was found that in 45 gamesboth the players were girls, and in 190 games both were boys. The number of gamesin which one player was a boy and the other was a girl is?a. 200b. 216c. 225d. 212
200..

bC2=190 (2 boys played against each other)... we get b=20
gC2 =45 ... we get g=10

10C1 *20C1 = 10*20 = 200matches where one player was girl and other was boy.
@abhishek.2011 said:
bhai 2x1 is even 2x2 is even if x3 is oddthen how can even+even+odd gives u odd ???????
ahh...i forgot it was equal to 6...meine 5 le liya...

mea culpa...told u was in office.. :)
@abhishek.2011 said:
lo brother m completing ur solution on ur behalf, hope its right 2x1+2x2+2x3+2x4+x5=6here x5 is always of the form 2nso 2x1+2x2+2x3+2x4+2n=6x1+x2+x3+x4+n=3number of solutions = 7c4=35similarly for next case2x1+2x2+2x3+2x4+x5=4here also x5 is also even alwaysx1+x2+x3+x4+n=2solutions 6c4 = 15for next case2x1+2x2+2x3+2x4+x5=3here x5 shud always be odd and greater than 02x1+2x2+2x3+2x4+2n+1=3x1+x2+x3+x4+x5=1number of solutions 5c4=535*15*5=2625
bhai 2^4 (as u,v,w and x can be -ve as well) se multiply nahi karne ka kya?
@joyjitpal said:
bhai 2^4 (as u,v,w and x can be -ve as well) se multiply nahi karne ka kya?
ye nahi kara @abhishek.2011 bhai?? 2^4 se toh multiply hoga pakka...
@Logrhythm said:
ahh...i forgot it was equal to 6...meine 5 le liya...mea culpa...told u was in office..
bro even if the power of 2 is even still there is no compulsion on taking x5 as even? jus want to kno as i am not so good with numbers
@joyjitpal said:
bhai 2^4 (as u,v,w and x can be -ve as well) se multiply nahi karne ka kya?
ya if negative integers are allowed
i solved taking only positive integers....
@Logrhythm said:
ye nahi kara @abhishek.2011 bhai?? 2^4 se toh multiply hoga pakka...
pakka hoga mere ko lagta hai waisa :)
@iLoveTorres said:
bro even the power of 2 is even still there is no compulsion on taking x5 as even? jus want to kno as i am not so good with numbers
bhai dint get u..

power of 2 even hai tabhi toh even+even+odd cannot be even...hence x5 has to be even..
@insane.vodka said:
In a chess competition involving some boys and girls of a school, every studenthad to play exactly one game with every other student. It was found that in 45 gamesboth the players were girls, and in 190 games both were boys. The number of gamesin which one player was a boy and the other was a girl is?a. 200b. 216c. 225d. 212
let no. of girls be m...and no. of boys be n

mc2 = 45----> m(m - 1) = 90---> m = 10
nc2 = 190----> n(n - 1) = 380 ---> n = 20

20c1*10c1 = 200 ?

@Logrhythm said:
bhai dint get u..power of 2 even hai tabhi toh even+even+odd cannot be even...hence x5 has to be even..
so x5 should be taken as 2n?
@iLoveTorres said:
so x5 should be taken as 2n?
yes bhai....actually the question was set this way (may be) so that it becomes easy for us... :)

nahi toh cases lene padte hum loagon ko...which would have been a little arduous
@insane.vodka said:
In a chess competition involving some boys and girls of a school, every studenthad to play exactly one game with every other student. It was found that in 45 gamesboth the players were girls, and in 190 games both were boys. The number of gamesin which one player was a boy and the other was a girl is?a. 200b. 216c. 225d. 212
200!!!

200 it is
20 boys
10 girls

How many three digit numbers have the property that their digits taken from left to right for an Airthmetic or Geometric Progression?
1) 15
2) 36
3) 20
4) 42

@Exodia said:
How many three digit numbers have the property that their digits taken from left to right for an Airthmetic or Geometric Progression?1) 152) 363) 204) 42
for GP i got 124 139 248

and AP 123 135 147 159 234 246 258 345 357 369 456 468 567 579 678 789

and 111 222 333 444 555 666 777 888 999

getting 28 what ami i missing ?
@Exodia said:
How many three digit numbers have the property that their digits taken from left to right for an Airthmetic or Geometric Progression?1) 152) 363) 204) 42
42