Official Quant thread for CAT 2013

MBA CAT 2024
22887
@saurav205 said:
Question :ABCDEF is a 6-digit number with distinct digits. Further, the number is divisible by 11and the sum of its digits is 24. Further , A>C>E and B>D>FQ1: The sum A+C+E is equal to :1268Cannot be determinedQ2. A+B is always :1096Cannot be determined.P.S. : Source Nishit Sinha..Please dont pile on me if you find the question/options ambiguous
1) 12
2) CBD ??
22888
@bodhi_vriksha thanks bhai
22889
@saurav205 said:
Question :
ABCDEF is a 6-digit number with distinct digits. Further, the number is divisible by 11and the sum of its digits is 24. Further , A>C>E and B>D>F
Q1: The sum A+C+E is equal to :
12
6
8
Cannot be determined

Q2. A+B is always :
10
9
6
Cannot be determined.

P.S. : Source Nishit Sinha..Please dont pile on me if you find the question/options ambiguous
Q1 12
Q2 CBD
22890
@saurav205 said:
Question :ABCDEF is a 6-digit number with distinct digits. Further, the number is divisible by 11and the sum of its digits is 24. Further , A>C>E and B>D>FQ1: The sum A+C+E is equal to :1268Cannot be determinedQ2. A+B is always :1096Cannot be determined.P.S. : Source Nishit Sinha..Please dont pile on me if you find the question/options ambiguous
12,cbd ?
22891
@ScareCrow28 check the 2nd one again...
22892
@saurav205 said:
@ScareCrow28 check the 2nd one again...
Nos --> 685410, 674520..and others are also possible
A+B = 14 in one and 13 in other.. Hence CBD
22893
@saurav205 said:
Question :ABCDEF is a 6-digit number with distinct digits. Further, the number is divisible by 11and the sum of its digits is 24. Further , A>C>E and B>D>FQ1: The sum A+C+E is equal to :1268Cannot be determinedQ2. A+B is always :1096Cannot be determined.P.S. : Source Nishit Sinha..Please dont pile on me if you find the question/options ambiguous
OA :
12
6
I know the 2nd is the answer given is wrong. But since this was the OA I have posted it...
Everyone can put put in the values you were getting for A+B.

I know there are a lot of values poosible..so just a few of them :
18,16,15 ..etc..

22894
@saurav205 said:
Question :ABCDEF is a 6-digit number with distinct digits. Further, the number is divisible by 11and the sum of its digits is 24. Further , A>C>E and B>D>FQ1: The sum A+C+E is equal to :1268Cannot be determinedQ2. A+B is always :1096Cannot be determined.P.S. : Source Nishit Sinha..Please dont pile on me if you find the question/options ambiguous

A+B+C+D+E+F=24
A+C+E-B-D-F=11 OR 0
IF 11
A+C+E=17.5 NOT POSSIBLE

SO
A+C+E=12

B+D+F=12

A+B cannot be determined?
22895
@ScareCrow28 said:
Nos --> 685410, 674520..and others are also possibleA+B = 14 in one and 13 in other.. Hence CBD
I know yar...But since OA mein kuch aur tha so told you to check again..
even I was getting different possibilities.....
22896
@saurav205 said:
Question :ABCDEF is a 6-digit number with distinct digits. Further, the number is divisible by 11and the sum of its digits is 24. Further , A>C>E and B>D>FQ1: The sum A+C+E is equal to :1268Cannot be determinedQ2. A+B is always :1096Cannot be determined.P.S. : Source Nishit Sinha..Please dont pile on me if you find the question/options ambiguous
As number is divisible by 11, (B + D + F) - (A + C + E) = 11k

Also all digits are distinct, lowest sum in a bracket can be 0 + 1 + 2 = 3 and it is given that A + B + C + D + E + F = 24.

So (B + D + F) - (A + C + E) cannot be more than 21 - 3 = 18 and less than 3 - 21 = -18.

But it is multiple of 11, so it has to be -11, 0 or 11 only.

Further if sum of two numbers is even, then the two numbers will be both odd or both even. So their difference in either case has to be even.

Thus confirmed that (B + D + F) - (A + C + E) = 0

i.e. (B + D + F) = (A + C + E) = 12


Now A and B are the largest number in the two brackets.. But the two triplets can be formed in many ways e.g. (9, 2, 1) and (8, 4, 0) OR (9, 3, 0) and (7, 4, 1)

So A + B cannot be determined uniquely.

1. 12
2. CBD

Team BV
22897
@saurav205 said:
I know yar...But since OA mein kuch aur tha so told you to check again..even I was getting different possibilities.....
Ye Nishit Sinha hai kon??
22898
@ScareCrow28 said:
Ye Nishit Sinha hai kon??
bhai I do not know him personally..
Was in college when I bought this book...Abhi kuch aur nai hai so decided to use the book...questions are good, but there are a few mistakes..
22899
@ScareCrow28 @Subhashdec2 @bodhi_vriksha and others ..
Next question :
The last digit of the LCM of (3^2003 - 1) and (3^2003 + 1) is :
8
2
4
6
22900

Question :

a,b and c are positive integers such that a+b+c = 2003
Let E = (-1)^a + (-1)^b + (-1)^c. Find the number of possible values of E.
2004
3
1003
2
22901
@saurav205 said:
@ScareCrow28@Subhashdec2@bodhi_vriksha and others ..Next question :The last digit of the LCM of (3^2003 - 1) and (3^2003 + 1) is :8246
HCF is 2
HCF*LCM=P1*P2
LCM=(3^4006-1 )/ 2

(3^2-1)/2=4 or 9

option has 4 only
22902
@saurav205 said:
Question :a,b and c are positive integers such that a+b+c = 2003Let E = (-1)^a + (-1)^b + (-1)^c. Find the number of possible values of E.2004310032
a b c combinations possible
(odd,odd,odd)->-3
(even,even,odd)->1
2??
22903
@saurav205 said:
@ScareCrow28@Subhashdec2@bodhi_vriksha and others ..Next question :The last digit of the LCM of (3^2003 - 1) and (3^2003 + 1) is :8246
4 ??
22904
@saurav205 said:
@ScareCrow28 @Subhashdec2 @bodhi_vriksha and others ..
Next question :
The last digit of the LCM of (3^2003 - 1) and (3^2003 + 1) is :
8
2
4
6
4
16 and 18 :XX4
26 and 28 :XX4
22905
@saurav205 said:@ScareCrow28@Subhashdec2@bodhi_vriksha and others ..Next question

Since they are consecutive even integers, Gcd of (3^2003-1) and (3^2003+1) is 2

So, there LCM = (3^2003-1)(3^2003+1)/2 = (3^4006-1)/2

Last two digits of (3^4006-1) are given by (3^4006-1)mod100=28mod100

So, when divided by 2 , the last digit is 4

Team BV
22906
@saurav205 said:
@ScareCrow28@Subhashdec2@bodhi_vriksha and others ..Next question :The last digit of the LCM of (3^2003 - 1) and (3^2003 + 1) is :8246
OA : 4
both the numbers are even : hence HCF = 2
HCF*LCM = A*B
LCM = A*B/2
cycle of 3 : 3 9 7 3
hence unit digits will be 6 and 8
unit digit of (6*8/2) = 4