Official Quant thread for CAT 2013

MBA CAT 2024
6741
@maddy2807 said:
210a min of two colors wud be needed.choose two colors. 3C2=3Now choose the color for 3 strips= 2C1=2total no of wys= 2C1*3C2* 7!/4!*3!= 210correct me if wrng?
i will not correct u u r already correct
6742
@rubikmath said:
i will not correct u u r already correct
6743
@maddy2807 said:
1/17?
@mailtoankit said:
1/17
@soumitrabengeri said:
1/17
Explain please.
6744
@sauravd2001 said:
@soumitrabengeri bhai good one i was doing something out of this owrld for this thinking of 7 strips so 4,3 ka pair and 4 strips in 4! ways and 3 in 3! ways so 144 ways with one strip and pata ni kahan pahunch agay answeer but ur approach is ryt
Happens..prob is sometimes damn confusing..
6745

aaj kal Quant form par logo ko jagana padta hai...

6746
@seetharam7 (13*12)/52*51
one honour ka meaning same type ke jaise dil ke pan ke aise toh
first time 13 cards then 1 more taken out without replacement means that 12 cards one is already out so
we get 1/17
6747
@seetharam7 said:
Explain please.
favourable case: 52C1 * 3C1
sample space: 52C1 * 51C1
probability = 3/51 = 1/17
6748
A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?

(A) 79
(B) 83
(C) 85
(D) 87
(E) 88
6749
@soumitrabengeri did i tell u what a simple question i screwed in my cat exam it was not more than 10 second question of pnc 😞 i text u in inbox because they some nda or something
6750
@seetharam7 said:
Explain please.
total way of selecting two cards= 52C2
There are 4 honors, selecting 2= 4C2= 6
there are 13 no.

so
6*13/52C2= 1/17
6751
Help me in calculation:
A. 54129/29648
B. 10518/5577

Option A>B, B>A

With approach
6752
@pankaj1988 said:
Help me in calculation: A. 54129/29648 begin_of_the_skype_highlighting54129/29648end_of_the_skype_highlightingB. 10518/5577Option A>B, B>AWith approach
B>A
use percentage method.
i guess that will be helpful
6753
@rubikmath said:
John has to paint a wall with seven horizontal stripes. He only has enough paint for four red stripes, four blue stripes, and four yellow stripes. If he can use at most two colors, how many different ways can he paint the wall?(A) 420 (B) 210 (C) 84 (D) 35 (E) 7no OA
210
6754
@rubikmath bhai 79 ??
6755
@sauravd2001 nope
6756
@rubikmath said:
A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?(A) 79 (B) 83 (C) 85 (D) 87 (E) 88
83??
6757
@kingsleyx
@kingsleyx said:
83??
yup
6758
@rubikmath said:
A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?(A) 79 (B) 83 (C) 85 (D) 87 (E) 88
division 1=9x2-1=17
division2=10x2-1=19
division3=11x2-1=21
division4=12x2-1=23
for over all =4-1=3
tot=83?
6759

the sum of 1st 7 terms of

1+4+18+96+600 +..
6760
@19rsb bhai sahi hai m so tard n stupid i canttell u i took it out right upto 80 match in first round later i didnot count the last 3 matches ohh man katta ho gaya itni mehnat ke baad
:/