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Official Quant Thread for CAT 2011 [Part 8] Quantitative

Please continue here with all the quant queries and their discussions.
link to previous thread :-
http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html
:cheerio: :cheerio: :cheerio: KEEP LEARNING !!! :cheerio: :cheerio: :ch...
28 28
it is 7 koi condition hi ni hai jo option mein minimum hai wo ans,..
15x+10y=380
y=8
x=20
ho gya satisfy...
hTc
Its 1180. check again bro
1 1
what about x,y,z..are they real no's or integers?
Reminder is 3..
Substitute the values...when z=7,15(20)+10(+7(7)=429
z=7 , x=20, y=8
remainder = 3
2011 = 6K + 1 ;(E (7)= 6)
3 it should be.
Euler (7) = 6
3^6K = 1 Mod(7)
3^2010 = 1 Mod(7)
The remainder is 3
15x + 10y + 7z = 429
What is the minimum value z can have?
a)7
b)12
c)17
d)22
varun bhai plz ise explain kar do samajh ni aaya ye
is it 1080rs..with 1 guava of cost 80...
1 1
The cost of 6 mangoes, 4 Guavas and 10 Water melons = 750*10 - 1580*4 = 1180
For no green balls = 1 way
For 1 green ball = 6 ways
For 2 green balls = 4+3+2+1 = 10 ways
For 3 green balls = 4 ways
Total = 21 ways
1 1
six boxes numbered 1 to 6 are arranged in a row. each is to be filled by either a blue or a green coloured ball in such a way that no two adjacent boxes contain green coloured balls.in how many ways can the boxes be filled with the balls?
1. 15
2. 16
3. 20
4. 21
hTc
Hi Puys,
Q- 3 mangoes, 4 guavas and 5 watermelons cost rs 750. 10 watermelons, 6 mangoes and 9 guavas cost 1580. what is the cost of 6 mangoes, 10 watermelons and 4 guavas ?
If we take p=1, r=0, q can be anything from 1 to 9.
say q=6
=> pq = 16, pr = 10, qp = 61, pp = 11
pqr = pq*pr = 16*10 = 160
prrp = 1001
(qp prrp)/pp = 61*1001/11 = 0 remainder
bhaiyo
i use to have have problems in identical ball and box concept in combinatories/permutation
how to tackle those problems
the discussed question related to filling of boxes from 1 to 6 so that no two adjacent boxes contain green coloured box for case iii)when there are two ball...
Taking a sample AP 1/2,1...where q is 2 and p is 1...
so sum to 2 terms is 3/2 ...
now using options.... (pq+1)/2 satisfies
Option 3 it would be.
Integral solutions of the equation:
a+b+c+d+.....upto r terms = n
is given by C(n+r-1,r-1)...In this case n = 12 and r = 3
C(14,2) = 91
@Koshti bhai: Welcome back bhai!!!! How was your CAT ??