Official Quant thread for CAT 2013

MBA CAT 2024
17106
@TootaHuaDil said:
Find the least value of n such that n! has exactly 2394 zeroes.
9585/5 = 1917/5 = 383/5 = 76/5 = 15/5 = 3

1917+383+76+15+3 = 2394
@TootaHuaDil said:
Find the smallest positive integer n such that n! is a multiple of 10^2009
Do this like the 1st one...it just means that power of 5 should be 2009...
@TootaHuaDil said:
How many zeros are present at the end of 25! + 26! +27! +28! +30! ?
25!(1 + 26 + 27*26 + 28*27*26 + 30*29*28*27*26)
25! (17121105) = 5^6(x*5^1)
so seven 0's..
17107
@leonidas. said:
Suppose n is an integer such that the sum of the digits of n is 2, and10^10 a) 11b) 10c) 9d) 8
just find the numbers of the form 9k+2 between 10^10 and 10^11....
17108
@TootaHuaDil said:
How many zeros are present at the end of 25! + 26! +27! +28! +30! ?
6?
highest power of 5 in 25 = 6
17109
@pirateiim478 said:
In how many ways can 5 different balls can be arranged in 3 different boxes such that each box has atleast one ball?
150??

3 1 1 case ==> 3c1 * 5c3 * 2c1 = 60.

2 2 1 case ==> 3c1 * 5c1 *4c2 = 90.

total = 150 cases
17110
@pirateiim478 said:
In how many ways can 5 different balls can be arranged in 3 different boxes such that each box has atleast one ball?
A..B..C
1...1..3 -> 5c3*2c1*3!/2!
2...1...2 -> 5c2*3c1*3!/2!
@leonidas. said:
Ques:N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two €“minute song after the other. If the total time taken for singing is 28 minutes, what is N?a) 5b) 7 c) 9d) None of these
every person sings with n-2 ppl

try options now...
n-2c2 = 28
hence n = 9...
17111
@pirateiim478 said:
In how many ways can 5 different balls can be arranged in 3 different boxes such that each box has atleast one ball?
:banghead: confused
Let A B C be the balls
a b c d e be the boxes
Now cases possible are 311 and 221
case 311 : you need to select 3 balls out of 3 so 5C3
then you need to slect a box which can be done in 3 ways
so 5C3*3*2 = 10*3*3 = 60
Case 212 : you need to slect 2 out of 5 so 5C2.
then you need to slect 2out of 3 so 3C2
now again 3!/2!
so 5C2*3C2*3!/2! = 90
total = 150
One request to all please share your approach so that we all can learn easily :). We can avoid the posts everytime asking for solution 😃
17112
@pirateiim478 said:
In how many ways can 5 different balls can be arranged in 3 different boxes such that each box has atleast one ball?
150?

3 1 1 ----> 5c3*2c1*3!/2! = 60
2 2 1 ----> 5c2*3c2*3!/2! = 90

total = 150
17113
@Logrhythm said:
A..B..C1...1..3 -> 5c3*2c1*3!/2!2...1...2 -> 5c2*3c1*3!/2!

agar identical boxes ho to 3! se multiply nahi karnge na??

17114

I request the author of the question to post the correct answer as well.

17115

Q: kya ho rha hai bhai logon..

S=1/5+3/5^2+7/5^3+......+ 31/5^6 , find S

17116
@viewpt said:
Q: kya ho rha hai bhai logon..

S=1/5+3/5^2+7/5^3+......+ 31/5^6 , find S
Nth term is (2^n - 1)/5^n
S=(2/5)*(1 - (2/5)^6)/(3/5) -(1/5)*(1- (1/5)^6)/(4/5)
=(2/3)*(1-64/15625)-(1/4)*(1-1/15625)
=6468/15625
17117
Tilak went to a book shop. He bought a book worth Rs 125. He had notes in the denominations of Rs 5, Rs 10 and Rs 50 only. In how many ways can he pay the bill?
a) 25
b) 22
c) 23
d) 24
17118
@VISHAL.SHARMA20 said:
Tilak went to a book shop. He bought a book worth Rs 125. He had notes in the denominations of Rs 5, Rs 10 and Rs 50 only. In how many ways can he pay the bill?a) 25b) 22c) 23d) 24
24

5a + 10b + 50c = 125 . ==> here a cant be zero.

b=0,c=0 ==> 1 way.

only c=0 ==> b = 1 to 12 ==> 12 ways .

only b=0 ==> c= 1,2 ==> 2 ways.

c=1 ==> b = 1 to7 ==> 7 ways .

c= 2 ==> b = 1 , 2 ==> 2 ways.

total = 24 ways @VISHAL.SHARMA20
17119
@VISHAL.SHARMA20 said:
Tilak went to a book shop. He bought a book worth Rs 125. He had notes in the denominations of Rs 5, Rs 10 and Rs 50 only. In how many ways can he pay the bill?
a) 25
b) 22
c) 23
d) 24
d) 24
Please tag in OA.
17120

Yes its d) but how?

@hiteshgulati
@ravi.theja
17121
@VISHAL.SHARMA20 said:
Yes its d) but how?
@hiteshgulati
@ravi.theja
Make cases:
1) Lets select two INR 50 notes. The you can buy the book in 3 ways:
a)Five 5 Rs notes
b)Three 5 Rs notes and One 10Rs note
c)One 5 Rs note and Two 10 Rs notes
2) Lets select one INR 50 note. The you can buy the book in 8 ways:
a)Fifteen 5 Rs notes
b)Thirteen 5 Rs notes and One 10Rs note
Till.........One 5 Rs note and Seven 10 Rs notes
3) Now dont select any 50INR note. Then you will get 13 cases
Try working this out yourself
Total of 3 cases: 24
Let me know if you face any issues with this.
17122
@leonidas. said:
Suppose n is an integer such that the sum of the digits of n is 2, and
10^10

a) 11
b) 10
c) 9
d) 8
Ans is c) 9
Please tag in OA.
17123
@leonidas. said:
Suppose n is an integer such that the sum of the digits of n is 2, and
10^10

a) 11
b) 10
c) 9
d) 8
b)11
17124
@VISHAL.SHARMA20 said:
Tilak went to a book shop. He bought a book worth Rs 125. He had notes in the denominations of Rs 5, Rs 10 and Rs 50 only. In how many ways can he pay the bill?
a) 25
b) 22
c) 23
d) 24
Let 5 rupee notes used=a
Let 10 rupee notes used=b
Let 50 rupee notes used=c
Then 5a+10b+50c=125
a+2b+10c=25
since 25 is odd,a must be odd.
So let a=2a' +1
then, 2a'+2b+10c=24
a'+b+5c=12
c=0 => 13 ways
c=1 => 8 ways
c=2 => 3 ways
Sum=24
ANS=24
17125
@hiteshgulati got it