Official Quant thread for CAT 2013

MBA CAT 2024
29746
@Buck.up said:
The remainder of 11^11^11/9 ?P.S.Method other than Euler will be appreciated
5 I guess

11/9 = 2

So the problem is equivalent to 2^(11^11)/ 9

Now the sequence for powers of 2 with 9 is 2, 4, 8, 7, 5, 1 i.e. 2^6 leaves 1 (or we can see that 64 leaves 1...!)

Now 11^11 = (12-1)^11 = 12k-1 form on binomial expansion = 6m+5 form

So 2^(6m+5) = 2^6m *2^5 = 1 *5 = 5.

regards
scrabbler
29747
@djch1989 said:
there are 2 1s, 2 2s, 2 3s, in how many ways can these be arranged so that two same digits are not adjacent to each other?
Getting 30 by some awkward approach. Venn type logic...

Tried again, getting 30 still. Confused. OA?

regards
scrabbler
29748

@djch1989 said:
there are 2 1s, 2 2s, 2 3s, in how many ways can these be arranged so that two same digits are not adjacent to each other?
24 ways??
29749
@Buck.up said:
The remainder of 11^11^11/9 ?P.S.Method other than Euler will be appreciated
|11^11^11/9|=|2^11^11/9|...wid 2 9 shows a cycle of 6....[2,4,8,7,5,1]....
|11^11/9|=|11^11/9|=|2^11/9|=5....
..so |11^11^11/9|=|2^11^11/9|=|2^5/9|=|32/9|=5..
29750
@Buck.up said:
The remainder of 11^11^11/9 ?P.S.Method other than Euler will be appreciated
is it 5!!!! used euler only yaar
29751
@Buck.up said:
The remainder of 11^11^11/9 ?P.S.Method other than Euler will be appreciated
why not euler's ?? it is easier to use it here :D

11^11 % 6 = (-1)^odd % 6 = 5

11^5 % 9 = 2^5 % 9 = 5
29752
@Logrhythm nope, the answer is given as 30..i haven't found the solution..
29753
@djch1989 said:
@Logrhythm nope, the answer is given as 30..i haven't found the solution..
koi baat nahi... @scrabbler...explain karo pls :)
29754
@scrabbler pls explain..:)
29755
@djch1989 said:
there are 2 1s, 2 2s, 2 3s, in how many ways can these be arranged so that two same digits are not adjacent to each other?
Use principle of inclusion - exclusion (google it for more details)

6!/(2!2!2!) - 3*5!/(2!2!) + 3*4!/2! - 3! = 30 ways
29756
@djch1989 said:
@Logrhythm nope, the answer is given as 30..i haven't found the solution..
Oh! I thought for sure it would be wrong.

OK one approach I had was:

A = All possible arrangements = 6!/2!2!2!

B = Cases where 1 pair is together = 3C1 ( choosing which pair) * 5!/2!2!

C = Cases where two pairs are together = 3C2 (choosing the two pairs) * 4!/2!2!

D = Cases where all 3 are together = 3C3 * 3!

Now by Venn logic, the answer would be A - B + C - D = 90 - 90 + 36 - 6 = 30

Then I tried a different (shorter) approach, which also gave me 30 but was a little confusing (still not sure if it makes sense, but anyway fwiw....)

Consider 2, 2, 3, 3 ke possible arrangements

2323 & 3232 where no pair is together
For each of these, say 2323, we have x2x3x2x3x the two ones can come in any of the spots marked x so 5C2= 10 ways. So total 2 * 10 = 20 ways for these two cases.

2332 & 3223 where 1 pair is together
For each of these, say 2332, we have _2_3x_3_2_ One 1 has to come in the place marked x, the other can occupy any of the places marked _ so 4 ways. Hence total 4 * 2 = 8 ways for these two cases.

2233 & 3322 where both pairs are together.
For each of these, say 2233 we have 2x23x3 and the two 1s have to occupy the two places marked x. So only 1 way each leading to a total of 1*2 = 2 ways for these two cases.

So overall 20 + 8 + 2 = 30 cases.

regards
scrabbler
29757

a 3 digit no has its first 2 digits equal. Find the maximum possible ratio of the no and the sum of tis digit, if the no is not divisible by 10.

29758
@chillfactor said:
Use principle of inclusion - exclusion (google it for more details)6!/(2!2!2!) - 3*5!/(2!2!) + 3*4!/2! - 3! = 30 ways
sir i am little weak in PnC..

what i did was, we have 3 options for the first place, 2 for second, 2 for third, 2 for fourth and 1 each for 5th anc 6th.....what's wrong here?

Ok got my mistake
29759

the sum of the present ages of a mother and her daughter is 60 years When the mother attains her husband's present age, the ratio of her husband's age and her daughter's age would be 2:1 . FInd the present age of the daughter.

29760
@vbhvgupta said:
a 3 digit no has its first 2 digits equal. Find the maximum possible ratio of the no and the sum of tis digit, if the no is not divisible by 10.
991/19 ?
29761
@vbhvgupta said:
the sum of the present ages of a mother and her daughter is 60 years When the mother attains her husband's present age, the ratio of her husband's age and her daughter's age would be 2:1 . FInd the present age of the daughter.
Seems to be 20. But I am getting husband's age indeterminate :(

Edited to add equations:

Let husband be y and wife x and daughter 60 €“x
When wife reaches husband €™s age, y-x years have passed so their ages must be 2y-x, y and 60 +y -2x.
So 2y €“ x = 2[60+y-2x] => y cancels, giving x as 40. So daughter is 20.

regards
scrabbler
29762
@vbhvgupta said:
a 3 digit no has its first 2 digits equal. Find the maximum possible ratio of the no and the sum of tis digit, if the no is not divisible by 10.
aab is the number where b is not 0

R = (100a + 10a + b)/(a + a + b) = 55 - 54b/(2a + b)

R will be maximum when b/(2a + b) will be minimum
b/(2a + b) = 1/{2(a/b) + 1}
=> 2(a/b) should be maximum

=> a = 9 and b = 1

So maximum possible ratio = 991/19
29763
@vbhvgupta said:
a 3 digit no has its first 2 digits equal. Find the maximum possible ratio of the no and the sum of tis digit, if the no is not divisible by 10.
991/19 = 52.15 ??
29764
@vbhvgupta said:
the sum of the present ages of a mother and her daughter is 60 years When the mother attains her husband's present age, the ratio of her husband's age and her daughter's age would be 2:1 . FInd the present age of the daughter.
20 yrs?
29765
@vbhvgupta said:
the sum of the present ages of a mother and her daughter is 60 years When the mother attains her husband's present age, the ratio of her husband's age and her daughter's age would be 2:1 . FInd the present age of the daughter.
m+d =60
H+(H-M) / D+(H-M) = 2/1
SOLVE
2D-2M = -M
2D = M
put this in first equation
2D+ D = 60
D= 20
present age of daughter = 20 yrs