Official Quant thread for CAT 2013

MBA CAT 2024
10521
@Aman.Malhotra said:
In how many ways can the number 105 be written as a sum of two or more consecutive positive integers?If your answer is No. of odd factors -1 i.e 7 then kindly show that 7 No.'s in detail
105 = 3*5*7

Now, if odd number of consecutive integers are in a series, the mean is the middle number
and if even number of consecutive integers are in a series, then the mean is the average of the middle numbers

now, consider 3*( 35) => 34 +35 + 36 , basically n*(Mean) = Sum
Similarly, 5*(21) = 19 + 20 + 21 +22 + 23
7*(15) = 12 + 13 + 14 + 15 + 16 + 17 + 18

So, (Mean - (n-1)/2 ) >0 => the series will have positive integers

Now, also consider this : (6) *( 35/2) = 15 + 16 + 17 + 18 + 19 + 20
10*(21/2) = 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15
14 * (15/2) = 1 + 2 + 3 + 4 + 5 + 6 +7 + 8 +9 + 10 +11 + 12 + 13 + 14
2*(105/2) = 52+ 53

So, ([Mean] - n/2) >= 0 => the series will have positive integers

So, Total of 7 ways
10522
A spherical ball of radius 3cmis melted and recast into three spherical balls. the redii of two balls are 1.5 cm and 2 cm the redii of third is?
0.5cm
1cm
1.5cm
2.5cm
10523
@dream_IIMA said:
A spherical ball of radius 3cmis melted and recast into three spherical balls. the redii of two balls are 1.5 cm and 2 cm the redii of third is?0.5cm1cm1.5cm2.5cm
r^3=3^3-1.5^3-2^3=15.625
r=2.5
10524
@dream_IIMA said:
A spherical ball of radius 3cmis melted and recast into three spherical balls. the redii of two balls are 1.5 cm and 2 cm the redii of third is?0.5cm1cm1.5cm2.5cm
2.5??
10525
@krum said:
r^3=3^3-1.5^3-2^3=15.625r=2.5
thank u

10526

how many 4 digit no formed using 1,2,3,4 without repetition such tat 1 is not at ten's place, 2 is not at thousands place, 3 is not at 100's place & 4 is nt in unit's place?

8,9,11,13.
10527
@The_Loser said:
how many 4 digit no formed using 1,2,3,4 without repetition such tat 1 is not at ten's place, 2 is not at thousands place, 3 is not at 100's place & 4 is nt in unit's place?8,9,11,13.
Dearrangement of 4 . I.e 9 Ans
10528
@Aman.Malhotra said:
No is 1234N74N712314N723124N731234N7here 123 can be arranged in 3!=6 ways and possible value of n is 816*81*4=1944 ??@jain4444 bhai What is the OA ???
I missed tha part "between last 4 and 1st 7" if we consider this then we have only one 33 digit number

now try

A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number in between 1st digit and last digit of this 30 digit number and found the resulting number to be divisible by 11. Find the number of possible values of the 33 digit number.
10529
@The_Loser said:
how many 4 digit no formed using 1,2,3,4 without repetition such tat 1 is not at ten's place, 2 is not at thousands place, 3 is not at 100's place & 4 is nt in unit's place?8,9,11,13.
9?

1 at thousands place:
1243, 1432, 1423

3 at thousands place:
3142, 3241, 3421

4 at thousands place:
4123, 4132, 4231
10530
@Aman.Malhotra said:
Dearrangement of 4 . I.e 9 Ans
plz explain this derrangement method..
10531
@mailtoankit
@mailtoankit said:
plz explain this derrangement method..
http://www.pagalguy.com/news/permutations-combinations-cat-2011-distribute-box-chocolates-a-18150
10532
@jain4444 said:
I missed tha part "between last 4 and 1st 7" if we consider this then we have only one 33 digit numbernow try A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number in between 1st digit and last digit of this 30 digit number and found the resulting number to be divisible by 11. Find the number of possible values of the 33 digit number.
I think we can place that three digit no. anywhere in 30 digit no.

then No. is N12347 [where 12347 each is 6 times consecutive ]
1N2347
12N347
123N47
1234N7
>> 5!*4*81 ???
10533

In how many ways 4620 pizzas can be divided equallty among n people such that they cant furthur divide equally among n people.

10534

Each side of a cube is painted red or blue or green. How many distinct colour patterns possible?

10535
@jain4444

A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number in between 1st digit and last digit of this 30 digit number and found the resulting number to be divisible by 11. Find the number of possible values of the 33 digit number.

i would hv done it like this.
As we have a total of 81 nos which can be inserted.
so ans could be
= 81* 30C1 * 5!( no of ways of arranging 1,2,3,4,7 consecutive digits)

PS. Its a bizarre ans. so pardon me if incorrect. I m too skeptical about it.
10536
@jain4444 said:
I missed tha part "between last 4 and 1st 7" if we consider this then we have only one 33 digit numbernow try A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number in between 1st digit and last digit of this 30 digit number and found the resulting number to be divisible by 11. Find the number of possible values of the 33 digit number.
If the three digit number is inserted in one of the following gaps:-
11_11_11_22_22_22_33_33_33_44_44_44_77_77_77
then 14*81 possibilities

If in one of the following gaps
1_11_11_1222222333333444444777777
then 1abc1 should be divisible by 11
10001 + 10abc should be divisible by 11
2 - abc should be divisible by 2
abc should be of form 11k + 2
so, 3*82 possibilities

If in one of the following gaps
1111112_22_22_2333333444444777777
2abc2 should be divisible by 11
20002 + 10abc will be divisible by 11
4 - abc will be divisible by 11
abc should be of form 11k + 4
so, 82*3 possibilities

If in one of the following gaps
1111112222223_33_33_3444444777777
3abc3 should be divisible by 11
30003 + 10abc will be divisible by 11
6 - abc will be divisible by 11
abc should be of form 11k + 6
so, 82*3 possibilities

If in one of the following gaps
1111112222223333334_44_44_4777777
4abc4 should be divisible by 11
40004 + 10abc will be divisible by 11
8 - abc will be divisible by 11
abc should be of form 11k + 8
so, 82*3 possibilities

If in one of the following gaps
11111122222233333344447_77_77_7
7abc7 should be divisible by 11
70007 + 10abc will be divisible by 11
3 - abc will be divisible by 11
abc should be of form 11k + 3
so, 82*3 possibilities

So, total = 14*81 + 82*3*5 = 2364 possibilities
10537
@maddy2807 said:
10?
approach bhi share karlo ya. I know some symetry game is dre. OA dn know.
10538
@jaiswalicsi said:
In how many ways 4620 pizzas can be divided equallty among n people such that they cant furthur divide equally among n people.
4620 = 2^2 * 11 * 3 * 5 * 7

4620/n is an integer, Also, 4620/(n*n) is not an integer

n is a factor of 4620, but (n^2) is not a factor of 4620

=> Possible values of n = (all factors) - ( when n^2 is a factor n = 2,1) = 48 - 2 = 46 ?
10539

in a row of 20 seats , in hw many ways can 3 blocks of consecutive seats wid 5 seats in each block be selected??

Ans
10540
@The_Loser said:
approach bhi share karle ya. I know some symetry game is dre. OA dn know.
No gometry game. Only one formula. Burside's formula.
No of ways of painting a cube with n color
= (n^6+3*n^4+12*n^3+8*n^2)/24