Official Quant thread for CAT 2013

MBA CAT 2024
21506
@joyjitpal said:
for GP i got 124 139 248and AP 123 135 147 159 234 246 258 345 357 369 456 468 567 579 678 789and 111 222 333 444 555 666 777 888 999getting 28 what ami i missing ?
@Exodia said:
How many three digit numbers have the property that their digits taken from left to right for an Airthmetic or Geometric Progression?1) 152) 363) 204) 42
Approach:

AP:
Numbers with common difference as 1 = 123,234,345,.....789 AND their reverse also. so total 7x2= 14
Numbers with cd as 2 =135,246,....579 and their reverse = 5x2 =10
Similarly numbers with cd as 3 = 6
numbers wtih cd as 4 = 2
Total = 14+10+6+2 = 32

Now 4 additional numbers are 210,420,630 and 840.

GP:
Numbers with common ratio as 2 are 124,248 and their reverse = 4
Numbers with common ratio as 3 = 139 and 931 =2

So total 32+4+6 = 42 such numbers.
21507
There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of one type. What is the minimum number of rows required for this to happen?
21508
@Ibanez : looks fine to me but is 469 included?
common ratio:1.5
21509
@CrookDinu said:
There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of one type. What is the minimum number of rows required for this to happen?
HCF ( 24, 36, 60) = 12
No of rows = 2 + 3 + 5 = 10
21510

HCF(24,36,60)=12
divide each by hcf
2+3+5=10

21511
@KhannaiiM said:
@Ibanez : looks fine to me but is 469 included?common ratio:1.5
If that is included then it becomes 43, not in the options :S So I belive the question is asking for only integral Common ratios in the GP
21512
@CrookDinu said:
There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of one type. What is the minimum number of rows required for this to happen?
hcf : 12..
2+3+5 .. 10?

btw i love the animated avatar! :)
21513
@CrookDinu 12*2=24
12*3=36
12*5=60
so minimum 10 rows
21514
@CrookDinu said:
There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of one type. What is the minimum number of rows required for this to happen?
HCF (24,36,60) = 12
min no. of rows = 24/12 + 36/12 + 60/12 = 2 + 3 + 5 = 10 ?
21515
@CrookDinu said:
There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of one type. What is the minimum number of rows required for this to happen?
20?
21516
Find the remainder (6^83+8^83) is divided by 49?
Options: 1 , 0 ,25, 35

PS : do not have OA so kindly give the approach
21517
@joyjitpal said:
200GC2 equal to 45and BC2 equal to 380G and B are 10 and 20 respectivelys0 10*20 eual to 200
@ChirpiBird yes OA : 200
21518
@hedonistajay said:
Find the remainder (6^83+8^83) is divided by 49?Options: 1 , 0 ,25, 35PS : do not have OA so kindly give the approach
N = 6^83 + 8^83 = (7-1)^83 + (7+1)^83
N mod 49 = 7*83 + 7*83 ( When expanding binomially other terms are multiple of 49)
Hence, N mod 49 = 35
21519
@hedonistajay said:
Find the remainder (6^83+8^83) is divided by 49?Options: 1 , 0 ,25, 35PS : do not have OA so kindly give the approach
(7-1)^83 + (7+1)^83
on expansion, we get all the terms more than 7^2 will be divisible by 49
we are left with 14/49
Remainder comes out to be 35... (solved same type of question long back, not sure of the answer)
21520
@abhishek.2011 said:
lo brother m completing ur solution on ur behalf, hope its right 2x1+2x2+2x3+2x4+x5=6here x5 is always of the form 2nso 2x1+2x2+2x3+2x4+2n=6x1+x2+x3+x4+n=3number of solutions = 7c4=35similarly for next case2x1+2x2+2x3+2x4+x5=4here also x5 is also even alwaysx1+x2+x3+x4+n=2solutions 6c4 = 15for next case2x1+2x2+2x3+2x4+x5=3here x5 shud always be odd and greater than 02x1+2x2+2x3+2x4+2n+1=3x1+x2+x3+x4+x5=1number of solutions 5c4=535*15*5=2625
u , v , w , x can be -ve also

so , answer would be 2^4*35*15*5 = 42000
21521
@junefever said:
(7-1)^83 + (7+1)^83on expansion, we get all the terms more than 7^2 will be divisible by 49we are left with 14/49Remainder comes out to be 35... (solved same type of question long back, not sure of the answer)
then ans shudn't be 14
21522
Find an 8-digit number using 4 2's and 4 1's that is divisible by 3 and 16.
21523
@jain4444 said:
Find an 8-digit number using 4 2's and 4 1's that is divisible by 3 and 16.
22112112
21524
@jain4444 said:
Find an 8-digit number using 4 2's and 4 1's that is divisible by 3 and 16.
xxxx2112 we can have remaining 2's and 1's in any order the number will be divisible by 3 and 16
21525
@jain4444 said:
Find an 8-digit number using 4 2's and 4 1's that is divisible by 3 and 16.
xxxx2112 ?.......xxxx containing 2's and 1's can be in any order ?