Official Quant thread for CAT 2013

MBA CAT 2024
6841
@maddy2807 said:
Find sum of the following series2/!1 + 3/2! + 6/3! + 11/4! + 18/5! + ...A) 3e-1B) 3(e-1)C) 3e+1D) 3(e+1)E) None of these
Edit: 3(e-1) ??

General term = (n^2 - 2n + 3)/n!
Tn = n^2/n! = (n^2-n+n)/n! = [n(n-1) + n]/n! = 1/(n-2)! + 1/(n-1)! = 2e
2n/n!= 2e
3/n! = 3e-3
Hence Ans: 3(e-1)

6842
@maddy2807 said:
Find sum of the following series2/!1 + 3/2! + 6/3! + 11/4! + 18/5! + ...A) 3e-1B) 3(e-1)C) 3e+1D) 3(e+1)E) None of these
ax^2 + bx + c
a+b+c=2
4a+2b+c=3
9a+3b+c=6

==>3a+b=1 and 5a+b=3 ==>a=1; b=-2; c=3
so num is like x^2 - 2x + 3
and den is x!

so gen term is (x^2 - 2x + 3)/x! = x/((x-1)!) - 2(1/(x-1)!) + 3/x!
term1:
1/(x-2)! + 1/(x-1)!
term 2:
2/(x-1)!

so 1/(x-2)! - 1/(x-1)! + 3/x!
=3(e -1)

so (b)....?

edited
6843
@19rsb d) none of these ?
6844
@nick_baba said:
You live 3km from the office, and you take an auto rickshaw to work every morning. Given how busy Mumbai Road is in Hyderabad, you have observed that 4 out of every 5 mornings, your auto rickshaw driver gets quite close to making an accident. If you move closer to the office, and now only live 1km away, how often do you expect to have a clear run with no close calls to the office?@anytomdickandhary @sujamait @Angadbir @Brooklyn @naga25french @krum @scrabbler @deedeedudu @techsurge@Estallar12 ..can u all please elaborate the concept behind it...i don't have an OA to it..
Suppose probability of having close call (when he lives 1 km away) = x
=> probability of having no close call (when he lives 1 km away) = 1 - x

Probability of having no close call (when he lives 3 km away) = (1 - x)^3
=> probability of having close call (when he lives 1 km away) = 1 - (1 - x)^3

So, 1 - (1 - x)^3 = 4/5
x = 1 - (1/5)^(1/3)
6845
@ScareCrow28 Dude I have two doubts , how to find the Nth term in this type of questions and what is "e" equal to ?
6846
@ashishgupta144 said:
@ScareCrow28 Dude I have two doubts , how to find the Nth term in this type of questions and what is "e" equal to ?
Just look at the series... Either you observe the numerator and can say that numerator is nothing but (2 + (n-1)^2) /n! = (n^2 -2n +3)/n!___(Just observation)

OR

You can do the conventional way

S= 2 + 3 + 6 +...........+ T(n-1) + T(n) (Just look at the numerator)
S= ___ 2 + 3 + 6....................T(n-1) + T(n)

Subtract
T(n) = 2 + 1 + 3 + 5 +...(2n-3)
Hence T(n) = (n-1)^2 + 2 = n^2- 2n +3

And e = 1/0! + 1/1! + 1/2! .....
I couldn't understand what you are asking
6847
@ScareCrow28 Your answer all my queries , i was not aware about :And e = 1/0! + 1/1! + 1/2! .....

Thanks for your detailed answer !!!
6848
@maddy2807 said:
Find sum of the following series2/!1 + 3/2! + 6/3! + 11/4! + 18/5! + ...A) 3e-1B) 3(e-1)C) 3e+1D) 3(e+1)E) None of these
3(e-1)???
6849

How many +ve integers can be formed using the digits 3,4,5,6,6,7 such that n>6000000?
320
360
540
720

6850
@sonamaries7 said:
How many +ve integers can be formed using the digits 3,4,5,6,6,7 such that n>6000000?320360540720
non of these bcoz there are only six digits
6851
@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.
@catter2011
I think in both the cases answer should be cannot be determined....because in both the cases the information is given about only certain number of families where as the question being asked is about change in the country's average savings. Until we know the total number of people in the country we cannot find the average of country is there are any changes in only for a certain number of families in the country.

ATDH.
6852
@nick_baba said:
with some fair discussions and arguments on previous page, we can assume the ans to be 11/15..but there are other possibilities also
chillfactor's post puts an end to all the arguments that we had. Please refer to his solution.

ATDH
6853
@sonamaries7 said:
How many +ve integers can be formed using the digits 3,4,5,6,6,7 such that n>6000000?320360540720
6!/2! = 360 :P

PS: Please check the question again. I don't think the question is correct. From the options I think the question should be 6999999>n>6000000. (I may be wrong though)

6854

Kunal has only 25 paise and 50 paise coins with him. The total amount in 50 paise denomination is Rs 4 more than the total amount in 25 paise denomination. The no. of 25 paise coins is 20 more than the number of 50 paise coins. What s the total amount with kunal?

6855
6856
@gupanki2 said:
Kunal has only 25 paise and 50 paise coins with him. The total amount in 50 paise denomination is Rs 4 more than the total amount in 25 paise denomination. The no. of 25 paise coins is 20 more than the number of 50 paise coins. What s the total amount with kunal?
Let x be the number of 50p coins and y be the number of 25p coins.

50x = 25y + 400
y = 20 + x

50x = 500 + 25x + 400
25x = 900
x = 36
y = 56

Total amount = 50x + 25y = 1800 + 1400 = 3200 paise = Rs.32
6857
@gupanki2 said:
Kunal has only 25 paise and 50 paise coins with him. The total amount in 50 paise denomination is Rs 4 more than the total amount in 25 paise denomination. The no. of 25 paise coins is 20 more than the number of 50 paise coins. What s the total amount with kunal?
32?
6858
@chillfactor said:
C)25 (edited)

Going north => a,b
Going east => x,y,z
Every consecutive N and E path can be replaced with an NE path to create a new path.
5!/(3!2!) = 120/(6 * 2) = 10
abxyz => 2
axbyz => 4
axybz => 4
axyzb => 2
xayzb => 2
xyazb => 2
xyzab => 1
xabyz => 2
xyabz => 2
xaybz => 4
Total = 25

6859
@chillfactor said:
25
6860
@chillfactor said:
25?