Official Quant thread for CAT 2013

logrhythm · February 21, 2013, 12:32pm

@jain4444 said:
The total number of Integral solutions of u²v²w²x²y =648000

648000 = 2^3*5^3*2^3*3^4 = 2^6*3^4*5^3

u = 2^x1*3^y1*5^z1

v = 2^x2*3^y2*5^z2

w = 2^x3*3^y3*5^z3

x = 2^x4*3^y4*5^z4

y = 2^x5*3^y5*5^z5

2x1+2x2+2x3+2x4+x5 = 6

yaaaarrrrr aese ques kyu de rae ho...yahan pen paper nikaal kar solve nahi kar sakta...though ye hoga aese hi...

find the solution for these equations and in the end multiply all and then multiply that by 2^4 (as u,v,w and x can be -ve as well)...

srry for not being able to provide complete solutions...

nice problem though @jain4444

dexian · February 21, 2013, 12:36pm

@jain4444

arey ye thoda tough hai.....

vipul24 · February 21, 2013, 1:30pm

@jain4444: How you arrive to this?

when sum of even places = 17
9*5!*4!

krum · February 21, 2013, 1:33pm

@jain4444 said:
The total number of Integral solutions of u²v²w²x²y =648000

wont this be infinite?

subhashdec2 · February 21, 2013, 1:41pm

@Vipul24 said:
@jain4444: How you arrive to this?when sum of even places = 17 9*5!*4!

arrange kara h bhai odd and even places ko

vipul24 · February 21, 2013, 1:42pm

@Subhashdec2 said:

arrange kara h bhai odd and even places ko

thanks

krum · February 21, 2013, 1:42pm

@Vipul24 said:
@jain4444: How you arrive to this?when sum of even places = 17 9*5!*4!

sets with 4 elements are taking even positions and there are 9 of them

they can be arranged in 4! and rest in 5! so 9*4!*5!

abhishek.2011 · February 21, 2013, 1:48pm

@Logrhythm said:

648000 = 2^3*5^3*2^3*3^4 = 2^6*3^4*5^3

u = 2^x1*3^y1*5^z1

v = 2^x2*3^y2*5^z2

w = 2^x3*3^y3*5^z3

x = 2^x4*3^y4*5^z4

y = 2^x5*3^y5*5^z5

2x1+2x2+2x3+2x4+x5 = 6

yaaaarrrrr aese ques kyu de rae ho...yahan pen paper nikaal kar solve nahi kar sakta...though ye hoga aese hi...

find the solution for these equations and in the end multiply all and then multiply that by 2^4 (as u,v,w and x can be -ve as well)...

srry for not being able to provide complete solutions...

nice problem though @jain4444

lo brother m completing ur solution on ur behalf, hope its right :)

2x1+2x2+2x3+2x4+x5=6

here x5 is always of the form 2n

so 2x1+2x2+2x3+2x4+2n=6

x1+x2+x3+x4+n=3

number of solutions = 7c4=35

similarly for next case

2x1+2x2+2x3+2x4+x5=4

here also x5 is also even always

x1+x2+x3+x4+n=2

solutions 6c4 = 15

for next case

2x1+2x2+2x3+2x4+x5=3

here x5 shud always be odd and greater than 0

2x1+2x2+2x3+2x4+2n+1=3

x1+x2+x3+x4+x5=1

number of solutions 5c4=5

35*15*5=2625

abhishek.2011 · February 21, 2013, 1:49pm

@Logrhythm said:

648000 = 2^3*5^3*2^3*3^4 = 2^6*3^4*5^3

u = 2^x1*3^y1*5^z1

v = 2^x2*3^y2*5^z2

w = 2^x3*3^y3*5^z3

x = 2^x4*3^y4*5^z4

y = 2^x5*3^y5*5^z5

2x1+2x2+2x3+2x4+x5 = 6

yaaaarrrrr aese ques kyu de rae ho...yahan pen paper nikaal kar solve nahi kar sakta...though ye hoga aese hi...

find the solution for these equations and in the end multiply all and then multiply that by 2^4 (as u,v,w and x can be -ve as well)...

srry for not being able to provide complete solutions...

nice problem though @jain4444

lo brother m completing ur solution on ur behalf, hope its right :)

2x1+2x2+2x3+2x4+x5=6

here x5 is always of the form 2n

so 2x1+2x2+2x3+2x4+2n=6

x1+x2+x3+x4+n=3

number of solutions = 7c4=35

similarly for next case

2x1+2x2+2x3+2x4+x5=4

here also x5 is also even always

x1+x2+x3+x4+n=2

solutions 6c4 = 15

for next case

2x1+2x2+2x3+2x4+x5=3

here x5 shud always be odd and greater than 0

2x1+2x2+2x3+2x4+2n+1=3

x1+x2+x3+x4+x5=1

number of solutions 5c4=5

35*15*5=2625

hari_bang · February 21, 2013, 1:54pm

@saurabhlumarrai said:
Next number in the series should be 12....as it follows the series as:4,4*5(20),20+4(24),24/4(6),6-4(2),2*4(8),8+4(=12) [alternate +4 and -4 ]Correct me if I am wrong...

got it:)

dexian · February 21, 2013, 2:03pm

@abhishek.2011 said:

lo brother m completing ur solution on ur behalf, hope its right

2x1+2x2+2x3+2x4+x5=6

here x5 is always of the form 2n

so 2x1+2x2+2x3+2x4+2n=6

x1+x2+x3+x4+n=3

number of solutions = 7c4=35

similarly for next case

2x1+2x2+2x3+2x4+x5=4

here also x5 is also even always

x1+x2+x3+x4+n=2

solutions 6c4 = 15

for next case

2x1+2x2+2x3+2x4+x5=3

here x5 shud always be odd and greater than 0

2x1+2x2+2x3+2x4+2n+1=3

x1+x2+x3+x4+x5=1

number of solutions 5c4=5

35*15*5=2625

arey sahi hai.....
i was breaking my head over this for quite some time..
thanks..

soniaa · February 21, 2013, 2:51pm

Thank u guys... :)
here is another question Chapter1 LOD II Q 17 (ARUN SHARMA)
if 4^n+1 +x and 4^2n - x are divisible by 5, n being a positive integer, find the least vale of x.
a)1 b) 2 c) 3 d) 0 e) None of these

My answer is coming a) but in book the answer is b)... please help

mohitjain · February 21, 2013, 2:58pm

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.

logrhythm · February 21, 2013, 2:59pm

@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 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...

chirpibird · February 21, 2013, 3:04pm

@SoniaA said:
Thank u guys... here is another question Chapter1 LOD II Q 17 (ARUN SHARMA)if 4^n+1 +x and 4^2n - x are divisible by 5, n being a positive integer, find the least vale of x.a)1 b) 2 c) 3 d) 0 e) None of theseMy answer is coming a) but in book the answer is b)... please help

yar it has to be 1 only.

4 has cycle of 2. (last digit.)
where 4^1=4 and 4^2=6 and this continues.
so either we have to add 1 or subtract 1. hence least value of x is 1 only.

insane.vodka · February 21, 2013, 3:04pm

In a chess competition involving some boys and girls of a school, every student
had to play exactly one game with every other student. It was found that in 45 games
both the players were girls, and in 190 games both were boys. The number of games
in which one player was a boy and the other was a girl is?
a. 200
b. 216
c. 225
d. 212

soniaa · February 21, 2013, 3:06pm

Thankss :)

joyjitpal · February 21, 2013, 3:06pm

@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

GC2 equal to 45

and BC2 equal to 380

G and B are 10 and 20 respectively

s0 10*20 eual to 200

mailtoankit · February 21, 2013, 3:07pm

@SoniaA said:
Thank u guys... here is another question Chapter1 LOD II Q 17 (ARUN SHARMA)if 4^n+1 +x and 4^2n - x are divisible by 5, n being a positive integer, find the least vale of x.a)1 b) 2 c) 3 d) 0 e) None of theseMy answer is coming a) but in book the answer is b)... please help

its coming 1 only

put n = 2..x = 1

4^(2 + 1) + 1 = 65

4^(2*2) - 1 = 255

ilovetorres · February 21, 2013, 3:07pm

@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.

50?