Official Quant thread for CAT 2013

MBA CAT 2024
23948
@Dexian said:
lekin bhai most of the word that u make .......... will also be not in dictionary... hai na...IMO Capitals shud be cosidered...
@chandrakant.k said:
Dont go on this notion. Question just asks the possibility. Whether that is used in reality or not in not a matter of our concern
But they have specified that the first letter should be a capital letter thats it.. and if you dont go by the general fact then the capital letter can also be a vowel or a consonant even that is not mentioned.. so arriving at the answer would be ambiguous

@Dexian ihave arranged jus the last three letters so the word will always start with a capital letter.. if i have done 7*(30+15+1)*4! then that would mean a capital letter can come anywhere in between
23949
@aditi88 said:
hi puys plz refer to the questions and quote me..thanks in advance
a)80
b)6
23950
@iLoveTorres said:
But they have specified that the first letter should be a capital letter thats it.. and if you dont go by the general fact then the capital letter can also be a vowel or a consonant even that is not mentioned.. so arriving at the answer would be ambiguous@Dexian ihave arranged jus the last three letters so the word will always start with a capital letter.. if i have done 7*(30+15+1)*4! then that would mean a capital letter can come anywhere in between
wahi to for 2nd 3rd and 4th place u r not selecting capitals......... which IMO shud be selected and then their permutations(3!)..........
23951
@aditi88 said:
hi puys plz refer to the questions and quote me..thanks in advance
A+B=8
no of sets=7(edit)
23952
There is a two-digit number, which is equal to the sum of the squares of its digits. What is the sum of the
digits of that number?
(1) 7 (2) 5 (3) 6
(4) 8 (5) Such a number is not possible
23953
@aditi88 said:
hi puys plz refer to the questions and quote me..thanks in advance
1. 6^60 - 6^(-60) = 8^20*3^60 - 8^(-20) * 3^(-60)
A+B = 80

2. 1+(n-1)d = 1000
=> (n-1)*d = 999 = 3^3 * 37

No of factors = 8
Case (n-1) = 1 is excluded

Hence total of 7 Ways?
23954
@hedonistajay said:
There is a two-digit number, which is equal to the sum of the squares of its digits. What is the sum of thedigits of that number?(1) 7 (2) 5 (3) 6(4) 8 (5) Such a number is not possible
No such number exists??
23955
@Logrhythm said:
No such number exists??
do not know the OA but I am also not able to find any number ...
23956
@iLoveTorres said:
a)80b)6 did not consider n=1, edited Ans=7
23957
A number n has 48 factors. Find the maximum number of ways n can be expressed as a product of two co-primes.

a) 12 b) 16 c) 20 d) 24 e) None of these
23958
@aditi88 said:
hi puys plz refer to the questions and quote me..thanks in advance
1) 6^60 - 6^-60 = 8^20*3^60 - 8^-20*3^-60
a+b=80...

2) a+(n-1)d = 1000
1+(n-1)d = 1000
(n-1)d = 999 = 3^3*37

so 4*2 = 8 sets, but removing the case where n=1

so total 7 sets...
23959
@hedonistajay said:
A number n has 48 factors. Find the maximum number of ways n can be expressed as a product of two co-primes.a) 12 b) 16 c) 20 d) 24 e) None of these
16 ways..

48 = 2^4*3

n = a*b*c*d*e^2

so, number of ways in which n can be expresses as a product of two co primes = 2^(5-1) = 16...
23960
@hedonistajay said:
A number n has 48 factors. Find the maximum number of ways n can be expressed as a product of two co-primes.a) 12 b) 16 c) 20 d) 24 e) None of these
2^(n-1) = 2^(5-1) = 2^4 = 16
23961
@hedonistajay said:
do not know the OA but I am also not able to find any number ...
see, first i tied doing it with the conventional way and then also with the options...

10a+b = a^2+b^2
a(10-a) = b(b-1)
a/b = (b-1)/(10-a)
tries various combinations and could not get the ratio equal, actually if u even look at it then u can see that it cannot be made equal...

now with the options..

7 --> (1,6);(2,5);(3,4) --> 37,29,25 (none equal)
5 --> (1,4);(2,3) --> 17,13 (none equal)
6 --> (1,5);(2,4);(3,3) --> 26,20,18 (none equal)
8 --> (1,7);(2,6);(3,5);(4,4) --> 50,40,34,32 (none equal)

so i take it that no such numbers exists...
23962
@Logrhythm said:
16 ways..48 = 2^4*3n = a*b*c*d*e^2so, number of ways in which n can be expresses as a product of two co primes = 2^(5-1) = 16...
@ScareCrow28 said:
2^(n-1) = 2^(5-1) = 2^4 = 16
right
23963
@hedonistajay said:
There is a two-digit number, which is equal to the sum of the squares of its digits. What is the sum of thedigits of that number?(1) 7 (2) 5 (3) 6(4) 8 (5) Such a number is not possible
a^2 + b^2 = 10a + b
(a-5)^2 + (b-1/2)^2 = 101/4
4 * [ (a-5)^2 + (b-1/2)^2 ] = 101

Now we can see that the inside-bracket part can be written as k^2 + l^2 / 4
Hence it becomes, 4*k^2 + l^2 = 101

Only possible solution of k and l is k=5 and l=1

But if k=5, then a = 10 ( which is not possible )

Hence No Solution
23964
@hedonistajay said:
A number n has 48 factors. Find the maximum number of ways n can be expressed as a product of two co-primes.a) 12 b) 16 c) 20 d) 24 e) None of these
16 ?

2^(n - 1) = 2^(5 - 1) = 2^4 = 16
23965
@hedonistajay said:
There is a two-digit number, which is equal to the sum of the squares of its digits. What is the sum of thedigits of that number?(1) 7 (2) 5 (3) 6(4) 8 (5) Such a number is not possible
equation = a^2 + b^2 = 10a + b
--> a(10-a) = b(b-1)
a
L.H.S can take values 9,16,21,24,25
R.H.S can take values 2,6,12,20,30 and so on...
so no value possible...
23966
@hedonistajay said:
A number n has 48 factors. Find the maximum number of ways n can be expressed as a product of two co-primes.a) 12 b) 16 c) 20 d) 24 e) None of these
should be when n is of the form = abcde^2
e^2 should be a part of one of the co-prime factor..
so in true sense it should be no. of the factors of abcd... which is 16...
i am not sure though...
23967

guys whats the remainder left when 5^41 is divided by 100?