Quant by Arun Sharma

MBA CAT 2024
3314
hi Puys ,
Q67.K is a three digit number such that the ratio of the number to the sum of its digit is least?What is the Difference between the hundreds and the tens digits of K?

a)9 b)8 c)7 d)None of these.

Please provide the detailed soln and also advice how to approch such problems ?

Is the answer 0 ?
I mean option d ?


This problem was discussed some time ago in Quant thread.
Here, we need to mimize number and maximize sum of digits.
Least number = 100 => But it gives least sum of digits as well
Higest sum means number to be 999. But it maximizes number and ratio becomes larger.

So, we need a balance.
First digit should be 1 and last digit should be 9. So, we get both lesser number and higher sum.
So, we have 1_9. Now, we just need try two values for middle digit 0 and 9.
So, 109 / 10 or 199 / 19.
We can easily say, 199/19 is less. So, number is 199.
Difference between the hundreds and the tens = 8
3315
My take:
Answer: (d) 1336
Explanation:
Consider numbers from 1 to 9.
If you take 1, 8 is out.
If you take 2, 7 is out.
If you take 3, 6 is out.
If you take 4, 5 is out.

So, out of 9, you can take 4 numbers i.e. numbers of form 9k+1, 9k+2, 9k+3, 9k+4.
So, out of 2997 you can take 4*(2997/9)=1332 numbers
You can also take 2998, 2999, 3000.

Additionaly, you can pick one number divisible by 9.
So, totally you can pick 1332+3+1 = 1336 numbers.

thanks for the soln buddy.
3316
Chill bhai, as per your method, number we get is 199, so, as per the question, the difference would be 9 - 1 = 8.

But, I did not get how you solved it. Could you pls elaborate.


Ah..thanks for the correction.

Actually to get the least value of (100a + 10b + c)/(a + b + c), a should be minimum, i.e, 1 and denominator should be maximum, i.e, b = c = 9
3317
This problem was discussed some time ago in Quant thread.
Here, we need to mimize number and maximize sum of digits.
Least number = 100 => But it gives least sum of digits as well
Higest sum means number to be 999. But it maximizes number and ratio becomes larger.

So, we need a balance.
First digit should be 1 and last digit should be 9. So, we get both lesser number and higher sum.
So, we have 1_9. Now, we just need try two values for middle digit 0 and 9.
So, 109 / 10 or 199 / 19.
We can easily say, 199/19 is less. So, number is 199.
Difference between the hundreds and the tens = 8

@ ThinkAce

the ratio will be highest when Number is greatest and sum of the digits is least.
so we have 100/1 , 200/2 and so on as such numbers.

So the max ratio can be 100 .
Please let me know if I am wrong ..
and thanks for the detailed solution.
3318
@ ThinkAce

the ratio will be highest when Number is greatest and sum of the digits is least.
so we have 100/1 , 200/2 and so on as such numbers.

So the max ratio can be 100 .
Please let me know if I am wrong ..
and thanks for the detailed solution.


You are right!! Maximum ratio will be 100.
In this case, we find a balance between maximizing number and minimizing sum of digits.
3319

Please solve the following question : LOD 2 ,Number Systems

Q 79.A set S is formed by including some of the First one thousand naural numbers.S contains the max no. of numbers such that they satisfy the following questions.
1.No number of the set S is prime.
2.When the numbers of the set S are selected two at a time , we always see co prime numbers.

what is the number of elements in the set S?
a)11 b)12 c)13 d)7

3320

Hi all, I've a few questions I could not find solution.

Kindly post the approach/.

1) Find the least no which 2800 should be multiplied so that the product is a perfect square.

Ans : 2,
7,
14,
N-O-These



2)Least multiple of 7 which leave a remainder of 4 when * by 6,9,15,18

a)94
b)364
c)184
d)74

3)A number is that when divided by 4,5,6,7 leaves remainder 2,3,4,5 respec.
Find largest no below 4000 satisfying this.

a)3358
b)78
c)3898
d)2938

5) No's below 200-300 not divisible by 2,3,4,5

a) 26
b)27
c)25
d)28

6)The mean of 1,2,2^2,2^3,.................2^31lies between

a)2^24 -2^25
b)2^26-2^27
c)2^29-2^30
3321

Hi all, I've a few questions I could not find solution.

Kindly post the approach/.

1) Find the least no which 2800 should be multiplied so that the product is a perfect square.

Ans : 2,
7,
14,
N-O-These



2)Least multiple of 7 which leave a remainder of 4 when * by 6,9,15,18

a)94
b)364
c)184
d)74

3)A number is that when divided by 4,5,6,7 leaves remainder 2,3,4,5 respec.
Find largest no below 4000 satisfying this.

a)3358
b)78
c)3898
d)2938

5) No's below 200-300 not divisible by 2,3,4,5

a) 26
b)27
c)25
d)28

6)The mean of 1,2,2^2,2^3,.................2^31lies between

a)2^24 -2^25
b)2^26-2^27
c)2^29-2^30


1. 2800 = 2*2*7*2*5*2*5 = 2^4 * 5^2 * 7^1.
So, the number to be multiplied to be a perfect square is 7.

2. LCM of (6,9,15,1 = 90, and it leaves a remainder of 4.
So, it is of the form 90k + 4.
For, k = 1, number = 94, not divisible by 7.
k = 2, number = 184, not divisible by 7.
k = 3, number = 274, not divisible by 7.
k = 1, number = 364, divisible by 7.

So, the number is 364.

3. LCM of (4,5,6,7) = 420.
And the difference between any of the number and its remainder as per the question is (4-2) or (5-3) or (6-4) or (7-5) = 2.
So, the number is of the form 420K - 2.

Now, by the options, 3358 is the number satisfying the above condition.

5. Question is not clear.
6. Don't have pen/paper to solve this .. 😛 So, didn't try ..
3322

Hi all, I've a few questions I could not find solution.

Kindly post the approach/.

1) Find the least no which 2800 should be multiplied so that the product is a perfect square.

Ans : 2,
7,
14,
N-O-These



2)Least multiple of 7 which leave a remainder of 4 when * by 6,9,15,18

a)94
b)364
c)184
d)74

3)A number is that when divided by 4,5,6,7 leaves remainder 2,3,4,5 respec.
Find largest no below 4000 satisfying this.

a)3358
b)78
c)3898
d)2938

5) No's below 200-300 not divisible by 2,3,4,5

a) 26
b)27
c)25
d)28

6)The mean of 1,2,2^2,2^3,.................2^31lies between

a)2^24 -2^25
b)2^26-2^27
c)2^29-2^30



Q1 to Q4 -- soln already posted so not posting it again.
Q6.the series in ques can be written as

2^0 + 2^1 +2^2 + .......+ 2^31
to take the mean of the above sum divide it by 32 i.e the no. of terms .
(2^0 + 2^1 +2^2 + .......+ 2^31) /32.
taking the sum of the series in numerator :
sum of this series is ( 2^32 - 1 )/(2-1) = (2^32 -1 ).

so the expression becomes (2^32 -1 )/2^5 . note :(32 = 2^5).
dividing we get 2^27 - (1/2^5 ).
as the expression is less than 2^27 and greater than 2^26.
so the answer is ( b).
3323

5) No's below 200-300 not divisible by 2,3,4,5

a) 26
b)27
c)25
d)28


if we replace the word below with between them (Both included ) ...

No's between 200-300 not divisible by 2,3,4,5

Soln:
nos. divisible by 2 :((300-200) /2 )+1 = 51.
so numbers not divisible by 2 = 101 - 51 = 50.
now numbers divisible by 3 but not dicisible by 2.
first number = 201 , then 207 ,then 213 ..and so on ..
a series with c.d = 6 . ( 3*2 = 6).
total such numbers = (297 - 201 )/6 + 1 = 16 + 1 = 17.
numbera left = 50 - 17 = 33.
now we will not check for numbers divisible by 4 as the numbers divisible by 4 are even numbers which are removed already.
checking for odd numbers divisible by 5.
1st num :205 , then 215 and so on c.d = 10 . (5 * 2).
(295 -205 ) /10 + 1 = 9 + 1 = 10.

but here we have double counted those numbers which are divisible by both 3 and 5 both.

such no. are 225 , 255 , 285 . (c.d = 30 ) i.e 3 numbers.
so we need to remove this 3 numbers from the 10 counted above as they are already removed from the original. so 10 - 3 = 7.

so total no. remaining = 33 - 7 = 26.

So 26 is the answer.
so try to solve your problem similarly.
u will get the ans.

3324
5) No's below 200-300 not divisible by 2,3,4,5

a) 26
b)27
c)25
d)28


if we replace the word below with between them (Both included ) ...

No's between 200-300 not divisible by 2,3,4,5

Soln:
nos. divisible by 2 :((300-200) /2 )+1 = 51.
so numbers not divisible by 2 = 101 - 51 = 50.
now numbers divisible by 3 but not dicisible by 2.
first number = 201 , then 207 ,then 213 ..and so on ..
a series with c.d = 6 . ( 3*2 = 6).
total such numbers = (297 - 201 )/6 + 1 = 16 + 1 = 17.
numbera left = 50 - 17 = 37.
now we will not check for numbers divisible by 4 as the numbers divisible by 2 are also divisible by 4 .
checking for numbers divisible by 5 but not by 2.
1st num :205 , then 215 and so on c.d = 10 . (5 * 2).
(295 -205 ) /10 + 1 = 9 + 1 = 10.

so numbers left = 37 - 10 = 27.

but here we have conted those numbers which are divisible by both 3 and 5 but not by 2.

such no. are 225 , 255 , 285 . (c.d = 30 )

so total no. remaining = 27 - 3 = 24.

so try to solve your problem similarly.
u will get the ans.


I get the point that the numbers which are divisible by 2 are removed should again not be counted when checking for divisibility by 3. But, in that case, how do u arrive at 6 ? I mean, why do u check the numbers divisible by 6. Similarly for the numbers that you check for divisibility by 10.
3325
pkaman Says
I get the point that the numbers which are divisible by 2 are removed should again not be counted when checking for divisibility by 3. But, in that case, how do u arrive at 6 ? I mean, why do u check the numbers divisible by 6. Similarly for the numbers that you check for divisibility by 10.

we are not checking for divisibility by 6 or 10.
As I have given above , the first no. divisible by 3 but not by 2 i.e the odd numbers which are divisible by 3.

1st num : 201.
then the next number divisible by 3 is 204 but it is even as we are looking for only odd numbers which are divisible by 3 we will discard it.
the next number is 207, then 213 ,219 and so on...

so the common Diff C.d is 6 here as u can see.
Similarly 10 is the C.d there .
3326
Need help with this one from Progressions LOD3:
If x, y, z are in GP and ax, by, cz are equal then a, b, c are in
1. AP
2. GP
3. HP
4. None of these
5. Cannot be determined



a,b,c, are in GP

ax*cz=acxz=(b^2)*(y^2)
but y^2=xz
hence b^2=ac
3327
Numbers of form 9k + 1, 9k + 2, 9k + 3, 9k + 4 can be part of the set and one of number of form 9k.
=> Total such numbers = 334 + 334 + 334 + 333 + 1 = 1336

Final result will be (x + 1)(y + 1)(z + 1) ..... (a + 1) - 1
= (1 + 1)(1 + 1/2)(1 + 1/3)......(1 + 1/1972) - 1
= 2(3/2)(4/3)(5/4)...(1973/1972) - 1
= 1973 - 1
= 1972

abc/(a + b + c) = 100 - (90b + 99c)/(a + b + c)

(90b + 99c)/(a + b + c) is maximum for a = 1 and b = c = 9
=> difference is 0


Hi there,

I did not understand the explanation for the first one:
why do we take 9k +1 ...upto 9k+4 and not till 9k+ 8.. ?
3328
Q1 to Q4 -- soln already posted so not posting it again.
Q6.the series in ques can be written as

2^0 + 2^1 +2^2 + .......+ 2^31
to take the mean of the above sum divide it by 32 i.e the no. of terms .
(2^0 + 2^1 +2^2 + .......+ 2^31) /32.
taking the sum of the series in numerator :
sum of this series is ( 2^32 - 1 )/(2-1) = (2^32 -1 ).

so the expression becomes (2^32 -1 )/2^5 . note :(32 = 2^5).
dividing we get 2^27 - (1/2^5 ).
as the expression is less than 2^27 and greater than 2^26.
so the answer is ( b).


Hi there,
HOw did you find sum of the series in numerator ???
3329
Hi there,
HOw did you find sum of the series in numerator ???


the given sequence is in g.p so find the sum of the series of the given g.p wid r =2.
3330

Could anyone help me with this. Its from Arun Sharma's Probability LOD 3:banghead:

Probabilities that Rajesh passes in Maths, Physics and Chemistry are m, p and c respectively. Of theses subjects, Rajesh has a 75% chance of passing , in at least two and 40% chance of passing in exactly two. Find which of the following is true?

a) p+m+c =19/20
b)p+m+c = 27/20
c)pmc = 1/20
d)pmc =1/8

I did not get even using the hint

The answer is "b"
I dont understand what "p+m+c" exactly represents:splat:

thanx in advance

3331

puys, how do u solve question where u have to find out the last 2 digits of set of multiples.

EX: 102*106*109*197*199*196.. find the last 2 digits of this multiple.

3332
puys, how do u solve question where u have to find out the last 2 digits of set of multiples.

EX: 102*106*109*197*199*196.. find the last 2 digits of this multiple.

Fot finding the last 2 digits you have to divide by 100.
So virtually think of the expression as :

(102*106*109*197*199*196)/100

now divide each term in the numerator by 100.
and replace them by the remainders.

= R{ 2*6*9*(-3)*(-1)*(-4)} /100 note :Use Negative Remainder Theorem
= R{108 * (-12)}/100
= R{8 * (-12)}/100
= R {-96}

As remainder is always +ve.
100 - 96 = 04.

So the last 2 digits are 04.
Try to solve such Questions with this approach.
3333
puys, how do u solve question where u have to find out the last 2 digits of set of multiples.

EX: 102*106*109*197*199*196.. find the last 2 digits of this multiple.


Refer to the below link :

http://www.pagalguy.com/forum/quanti...as-tips-5.html (concepts ,fundas and tips to crack quants section)

And for the above question last two digits would be 04.