Official Quant thread for CAT 2013

@ishu1991 said:
Four fair dice D1, D2, D3and D4 each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1 , D2and D3 is(A)91/216(B)108/216 (C)125/216(D)127/216
A hai kya...91/216..?

GUYS solve this too

@ishu1991 said:
Four fair dice D1, D2, D3and D4 each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1 , D2and D3 is(A)91/216(B)108/216 (C)125/216(D)127/216
(A) 91/216?

All 3 same numbers => 6
2 same numbers => 3 * 6 * 5 = 90
3 different numbers => 6 * 5 * 4 = 120
required probability = 6/216 * 1/6 + 90/216 * 2/6 + 120/216 * 3/6
= 91/216
@ishu1991 said:
Four fair dice D1, D2, D3and D4 each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1 , D2and D3 is(A)91/216(B)108/216 (C)125/216(D)127/216
a. 91/216

OA IS a) 91/216

@ishu1991 said:
Four fair dice D1, D2, D3and D4 each having six faces numbered 1,2,3,4,5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1 , D2and D3 is(A)91/216(B)108/216 (C)125/216(D)127/216
91/216
The sum of all two-digit numbers that are divisible by either 4 or 6 is
@Amrofa said:
The sum of all two-digit numbers that are divisible by either 4 or 6 is
1584?

6,12,18,24,30,..,96 => sum = 8 * (102) = 816
4,8,12,16,...,96 => sum = 12 * (100) = 1200
12,24,....,96 => sum = 4 * (108) = 432

Total sum = 816 + 1200 - 432 = 1584


@Amrofa
1684
@Amrofa
1584
@Amrofa said:
The sum of all two-digit numbers that are divisible by either 4 or 6 is
1584

What is the smallest integer N such that sqrt(N) + sqrt(N + 2005) is an integer?

@ishu1991 said:
@Amrofa1684
my bad its 1584 i counted 100 as well
@sunnychopra89

Is it 39204?? 198^2??
@sunnychopra89 said:
What is the smallest integer N such that sqrt(N) + sqrt(N + 2005) is an integer?
198^2?

sqrt(N) + sqrt(N + 2005) = k
sqrt(N + 2005) = k - sqrt(N)
N + 2005 = k^2 + N - 2ksqrt(N)
2005 = k (k - 2sqrt(N))
2005 = 5 * 401
k = 401
2sqrt(N) = 396
sqrt(N) = 198
N = 198^2
@sunnychopra89 said:
What is the smallest integer N such that sqrt(N) + sqrt(N + 2005) is an integer?
39204 hai kya....kuch jayada hi bada aa gaya hai mera no...
@grkkrg said:
198^2? sqrt(N) + sqrt(N + 2005) = ksqrt(N + 2005) = k - sqrt(N)N + 2005 = k^2 + N - 2ksqrt(N)2005 = k (k - 2sqrt(N))2005 = 5 * 401k = 4012sqrt(N) = 396sqrt(N) = 198N = 198^2
same approach mera bhi...
thank god..401 prime hai..nahin to punge padh jate isme...
@sunnychopra89 said:
What is the smallest integer N such that sqrt(N) + sqrt(N + 2005) is an integer?
198^2 ??

kahin silly mistake kiya hu, am sure!!
Ten families, each comprising 4 members attend a church for Christmas celebrations. If each person exchanges a greeting card with every other person of a different family exactly once, then find the number of greeting cards exchanged.

198^2