Official Quant thread for CAT 2013

MBA CAT 2024
6781
@Brooklyn said:
sir can u explain a lil more???
Total no of ways in which a match can be held = C(6, 2)*C(6, 2)*2
(we can choose two men in C(6, 2) ways and two women in C(6, 2) ways. Now a match can be held in 2 ways for these 4 persons)

We need to remove those cases when a couple is in the same team
If both the teams have a couple, then C(6, 2) ways

If only one team has a couple, then we can choose that couple in C(6, 1) ways. For 2nd team, men can be chosen in C(5, 1) ways and women in C(4, 1) ways (as they shouldn't be a couple)

So, C(6, 2)*C(6, 2)*2 - C(6, 2) - C(6, 1)*C(5, 1)*C(4, 1) ways
6782
@chillfactor said:
C(6, 2)*C(6, 2)*2 - C(6, 2) - C(6, 1)*C(5, 1)*C(4, 1) = 450 - 15 - 120= 315@sujamait C(6, 2)*C(6, 2)*2 se zyaada kaise ho sakta hai answer
sirji explanation aise de rkhi hai, mera bhi galat hua

total matches possible 6c1.6c1.5c1.5c1 = 900
now number of matches with both teams having couples is 6C2=15
number of mathes when only one team is a couple and other is not is 5c1.4c1.6=120
MAX no of matches in touna...
= 900-15-120 = 765

any gadbad ?
6783
@catter2011 said:
3.xx ?
yeah
6784
@sujamait

yeh lo..

S = 1*997 + 2*996 + 3*995 +... + 996*2 + 997*1

if s = 997*499*A , then what is the sum of digits of A ?

8
9
10
14
6785
@catter2011 said:
3.xx ?
logic??
6786
@catter2011 said:
@sujamaityeh lo..S = 1*997 + 2*996 + 3*995 +... + 996*2 + 997*1if s = 997*499*A , then what is the sum of digits of A ?891014
9
6787
@catter2011 said:
@sujamaityeh lo..S = 1*997 + 2*996 + 3*995 +... + 996*2 + 997*1if s = 997*499*A , then what is the sum of digits of A ?891014
yeh concept toh pahle bhi hua hai.. sum is constant..998 of each term..waise yeh bhi karne ki jarurat nhn,,

sum of terms 9k hai..toh digit sum 9k..option B i guess..
6788
@sujamait said:
yeh concept toh pahle bhi hua hai.. sum is constant..998 of each term..waise yeh bhi karne ki jarurat nhn,,sum of terms 9k hai..toh digit sum 9k..option B i guess..
yes 9, please explain in detail..
6789
@19rsb yes, approach pls..
6790
@Brooklyn No, logic.. just proceed with the data..



26.325 gm => 45% M => M= 58.5

as 100 gm - 200cc => 58.5 gm - 117cc

=> 4/3 * 22/7 * r^3 = 117

=> r ^3 = 117*21/88 = 27.xxx


=> r = 3.xx
6791

In a certain country the average monthly income is calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 99 months per year. this leads people having an underestimation of their savings , since there would be an underestimation of the income and an overestimation of the expenditure per month.

then,

1. mr.jack comes back from ussr and convinces his community comprising 273 families to start calculating the average income on the basis of 12 months per year. now if it is know that the average estimated income in his community is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)
a. 251.60 rs b.282.75 rs c. 312.75 rs d . cannot be determined.

2. miss rose comes back from usa and convinces his community comprising 546 families to start calculating the average income and average expenditure on the basis of 12 months per year. now if it is know that the average estimated income on the island is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)
a.251.60 rs b. 565.5 rs c. 625.5 rs d . cannot be determined.


please tell.
6792
@catter2011 said:
@19rsb yes, approach pls..
nth term=n X (998-n)=998n-n^2
sum upto nth term= 998n(n+1)/2 - n(n+1)(2n+1)/6
for n=997 ......sigma=998 x 997 x499 - 997 x 1995 x 998/6
= 998 x 997 x499 -997 x665x 449=997 x499 x 333.......compairing it with given eqn A=333
sum =9
6793
@sujamait said:
sirji explanation aise de rkhi hai, mera bhi galat hua total matches possible 6c1.6c1.5c1.5c1 = 900now number of matches with both teams having couples is 6C2=15number of mathes when only one team is a couple and other is not is 5c1.4c1.6=120MAX no of matches in touna...= 900-15-120 = 765 any gadbad ?
Only the bold part is incorrect.

What he's done is, total ways = (Select 1 boy, 1 girl) * (Select 1 boy, 1 girl)
But the whole thing is to be divided by 2. Coz else, AB - XY is taken as distinct to XY - AB

So total matches = 450

6794
@catter2011 said:
@Brooklyn No, logic.. just proceed with the data..26.325 gm => 45% M => M= 58.5as 100 gm - 200cc => 58.5 gm - 117cc=> 4/3 * 22/7 * r^3 = 117=> r ^3 = 117*21/88 = 27.xxx=> r = 3.xx
oh !! :banghead:

i was confused as how 100gm maps to 58.5 gm :banghead:
6795
@Angadbir said:
Only the bold part is incorrect.What he's done is, total ways = (Select 1 boy, 1 girl) * (Select 1 boy, 1 girl)But the whole thing is to be divided by 2. Coz else, AB - XY is taken as distinct to XY - ABSo total matches = 450
Yeah seems right.may be solution provider considerd 2 sides of court ..but..that wud be crazy

6796
@IIM-A2013 said:
In a certain country the average monthly income is calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 99 months per year. this leads people having an underestimation of their savings , since there would be an underestimation of the income and an overestimation of the expenditure per month.then,1. mr.jack comes back from ussr and convinces his community comprising 273 families to start calculating the average income on the basis of 12 months per year. now if it is know that the average estimated income in his community is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a. 251.60 rs b.282.75 rs c. 312.75 rs d . cannot be determined.2. miss rose comes back from usa and convinces his community comprising 546 families to start calculating the average income and average expenditure on the basis of 12 months per year. now if it is know that the average estimated income on the island is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a.251.60 rs b. 565.5 rs c. 625.5 rs d . cannot be determined.please tell.
both cannot be determined ???
6797
@IIM-A2013 said:
In a certain country the average monthly income is calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 99 months per year. this leads people having an underestimation of their savings , since there would be an underestimation of the income and an overestimation of the expenditure per month.then,1. mr.jack comes back from ussr and convinces his community comprising 273 families to start calculating the average income on the basis of 12 months per year. now if it is know that the average estimated income in his community is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a. 251.60 rs b.282.75 rs c. 312.75 rs d . cannot be determined.2. miss rose comes back from usa and convinces his community comprising 546 families to start calculating the average income and average expenditure on the basis of 12 months per year. now if it is know that the average estimated income on the island is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a.251.60 rs b. 565.5 rs c. 625.5 rs d . cannot be determined.please tell.
1. b) 282.75 ?
6798
If f(x) + f(y) = f(xy) and f(3) = 1
find f(3) + f(3^1/3) + f(3^1/9).... infinity

3
4.5
9/8
1.5
6799
@catter2011 said:
If f(x) + f(y) = f(xy) and f(3) = 1find f(3) + f(3^1/3) + f(3^1/9).... infinity34.59/81.5
1.5?
6800
@catter2011 said:
If f(x) + f(y) = f(xy) and f(3) = 1find f(3) + f(3^1/3) + f(3^1/9).... infinity34.59/81.5
3/2 = 1.5

Let f(x) = k*log(x) (Observing from the equation f(x) + f(y) = f(xy) )
f(3) = 1 = k*log(3)
=> k= 1/log(3)
Hence f(3) + f(3^1/3) + f(3^1/9).... infinity = 1 + 1/3 + 1/3^2 +....
= 3/2 = 1.5