Quant by Arun Sharma

MBA CAT 2024
5926

chap 1(number systems), lod 3 , the maiden's hand question(in my book its q no 50 and 51).

196 satisfies all the conditions, yet the answer is can't be determined

5927

@leo12 said:

if u form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine,what can be maximum no. of elements in the subset.(include both 1 and 3000)

1)1668

2)1332

3)1333

4)1336

explain!!

is the answer "1668" ??
logic I used :
when we choose 1 for the subset, then we can't have 8,17,26,35....2996 (total =333) in the subset as these all will add up to either 9 or a multiple of 9 with 1.
for 2 to be in the subset we can't have 7,16,25,34....2995 (total = 333) in the subset for the same reason.
for 3, we can't have 6,15,24,33....2994 (total = 333)
for 4, we can't have 5,14,23,32....2993 (total = 333) in the subset
so out of 1-3000 we can't have 333*4=1332 numbers in the subset.
hence the answer is 3000-1332 =1668
and the subset can comprise of (1,2,3,4,9, 10,11,12,13,18, 19,20,21,22,23,.... 2997,2998,2999,3000)
hope that solves
5928

x+y+z=1
x,y,z are positive real.

maximum value of xy+yz+zx = ?
5929
@heycat1012 said:
x+y+z=1
x,y,z are positive real.
maximum value of xy+yz+zx = ?
1/3
5930
@heycat1012 said:
x+y+z=1
x,y,z are positive real.
maximum value of xy+yz+zx = ?
x=y=z=1/3 so maximum value of sum = 3/9 = 1/3
5931
@leo12 said:

if u form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine,what can be maximum no. of elements in the subset.(include both 1 and 3000)

1)1668

2)1332

3)1333

4)1336

explain!!

hey, I got that ans wrong previously
it should be 1333

along with the elimination of (5,6,7,8)'s series which I eliminated earlier, 9,18,27,36...2997 series should also be eliminated which I missed in my logic.
so 3000-333*5 = 1332
so we are left with (1,2,3,4, 10,11,12,13, 19,20,21,22, ..... 2998,2999,3000).
and any one of the numbers from 9,18,27,36....2997 can be included in this series as it will not sum up with any other number to give a multiple of 9.
so final ans = 1332+1 = 1333.
please let me know the answer
5932

Q. Find the number of consecutive zeroes at the end of 72!.


I solved it with following method -
72/5 + 72/(5^2) + ... = 16

But the answer is 18. Plz tell where is the mistake.
5933
@pandeyn said:
Q. Find the number of consecutive zeroes at the end of 72!.
I solved it with following method -
72/5 + 72/(5^2) + ... = 16
But the answer is 18. Plz tell where is the mistake.
from 1! to 4! = 0
5! to 9 = 1
10! to 14 = 2
15 to 19 = 3
20 to 24 = 4
now here is the catch according to you, there will be 5 0's in 25!
but it will have 6
this is because number of zero's is determined by 5 and 2
if you obeserve there is one 5 more in 25! since 25 = 5^2
similarly for 50 = 25*2 you will get one more so 16+2 = 18 😃
5934
@chandrakant.k said:
from 1! to 4! = 0
5! to 9 = 1
10! to 14 = 2
15 to 19 = 3
20 to 24 = 4
now here is the catch according to you, there will be 5 0's in 25!
but it will have 6
this is because number of zero's is determined by 5 and 2
if you obeserve there is one 5 more in 25! since 25 = 5^2
similarly for 50 = 25*2 you will get one more so 16+2 = 18
thanks a lot.. starting prep too late is making me commit mistakes
5935
@chandrakant.k said:
from 1! to 4! = 0
5! to 9 = 1
10! to 14 = 2
15 to 19 = 3
20 to 24 = 4
now here is the catch according to you, there will be 5 0's in 25!
but it will have 6
this is because number of zero's is determined by 5 and 2
if you obeserve there is one 5 more in 25! since 25 = 5^2
similarly for 50 = 25*2 you will get one more so 16+2 = 18
When is the formula not applicable in that case?
5936
@pandeyn said:
thanks a lot.. starting prep too late is making me commit mistakes
@Ibanez said:
When is the formula not applicable in that case?
hey sorry it has to be 16 only i guess... i just cross checked
ibanez sir : thankk you
5937
@chandrakant.k said:
hey sorry it has to be 16 only i guess... i just cross checked
ibanez sir : thankk you
yup it's 16
5938
@leo12 said:

K is a three digit nymber such that the ratio of the number to the sum of its digits is least.what is the difference between the hundreds and tens digit of K ?

1)9

2)8

3)7

4) none of these

plz explain also!!

is the answer 8 ??????????
gettig number 199
5939
@leo12 said:

if u form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine,what can be maximum no. of elements in the subset.(include both 1 and 3000)

1)1668

2)1332

3)1333

4)1336

explain!!

i am getting 1333 (ecluding 1 and 3000)
all elements of form 9k+1,9k+2,9k+3,9k+4
3000/9 =333 integers (9k form)
333*4 numbers =1332 (9k+1,9k+2,9k+3,9k+4 forms)
remove case where number =1
=> 1331
now 3000/9 leaves remainder 3
out of which only 2 are valid (3000 not counting)
so total 1333 numbers
5940

plz solve this

1.) 511 and 667 when divided by the same number,leave d same remainder. how many number can be used as the divisor in order to make this occur?
14
12
10
8
5941
@chandrakant.k said:
hey sorry it has to be 16 only i guess... i just cross checked
ibanez sir : thankk you
ok - so the answer given in the book is wrong... thanks
5942
@Revenous said:
plz solve this
1.) 511 and 667 when divided by the same number,leave d same remainder. how many number can be used as the divisor in order to make this occur?
14
12
10
8
12??/
667-511 = 156 = 2^2*3*13
the number of factors = 3*2*2 = 12
so these 12 divisors will give the same remainder
5943
72/5 + 72/(5^2) + ... = 18
18 will come bcoz
72/5=14+2=16
72/25=2
so total =16+2=18.
5944
@Revenous said:
72/5 + 72/(5^2) + ... = 18
18 will come bcoz
72/5=14+2=16
72/25=2
so total =16+2=18.
why did you take 75/5 = 14+2?? what is this 2??
5945
@techsurge yes thanks!!