Quant by Arun Sharma

MBA CAT 2024
4886

3^32/50 => ((3^5)^6 * (3^2))/50

=> ((243)^6)/50 *( 9/50)

=> ((-7)^6)/50 * 9
=> (49^3)/50 * 9
=> (-1^3)/50 * 9
=> -1 * 9 = -9

now -9 can also be wriiten as 50-9=41
so answer is 41

4887

3^32/323 = 3^32/(17 * 19)
Now the eulers number of 17 when 3^32/17 is 16
so (3^16)/17 gives remainder 1
therefore 3^32 gives remainder 1

similary eulers number of 19 is 18 so we have
(3^1/19 * (3^14)/19 which gives 1 * (3^14)/19
so (243^2)/19 *(3^4)/19

=> (-4^2)/19 * 5
=> -3 * 5 = -15 or 4

Now we have to find the smallest number which gives remainder 1 when divided by 17 and rem 4 when
divided by 19

let the number be 17K+1
putting different values for k so that it gives remainder 4 or -15 with 19 ,we get value of k=8

so the number comes out to be 137
and that it is the answer( remainder is 137)

correct me If I am wrong

4888

41(3/17)% as fraction is equal to??

4889

41(3/17)% as fraction is equal to??
700/100*17 = 7/17

4890

7/17 should be the answer

4891
pungu Says
a team of miners planed to mine 1800 tons of ore durinng a ceratain number of days.due to technical problem in one third of the planned number of days,team was able to achieve an output of 20 tons of ore less than the planned output,to make for this,team over achieved for rest of days by 20 tons.result was that team completed the task one day ahead of time.how many tons of ore did the team intitaly plan to ore per day.


Let number of days be 3x
t be the decided output per day

so 3x*t = 1800

Now we also have

x(t-20)+(2x-1)(t+20) = 1800

Solve these two equation for t and get answer as 100tons
4892

Hi guys can some one plz help me out in this quest..

Q: find the remainder when 50^51^52 is divided by 11..

i hav tried to solve it using this way....plz correct me if i'm wrong somewhere..

since 50 and 11 are relatively co prime....so cyclicity of remainders for 11 is 10 using euler's theorem.

now 50^(51*52) % 11 = 50^(1*2) % 11

= 50^2 % 11
= 3

i'm getting 3 as answer but its given 6....can someone tell where i'm wrong...or is the answer given wrong..

4893
Hi guys can some one plz help me out in this quest..

Q: find the remainder when 50^51^52 is divided by 11..

i hav tried to solve it using this way....plz correct me if i'm wrong somewhere..


i think answer is correct..
51^52 will always give 6 as a last digit
then we need to find
50^6....solve this you will get R=6
4894
Hi guys can some one plz help me out in this quest..

Q: find the remainder when 50^51^52 is divided by 11..

i hav tried to solve it using this way....plz correct me if i'm wrong somewhere..

since 50 and 11 are relatively co prime....so cyclicity of remainders for 11 is 10 using euler's theorem.

now 50^(51*52) % 11 = 50^(1*2) % 11

= 50^2 % 11
= 3

i'm getting 3 as answer but its given 6....can someone tell where i'm wrong...or is the answer given wrong..


apply eulers
euler number for 11 is 10
so, 51^52/10 =1
therefore, 6^1%11=6

6 is the correct answer.
4895
apply eulers
euler number for 11 is 10
so, 51^52/10 =1
therefore, 6^1%11=6

6 is the correct answer.


Thanks man....but got others problems ryt except this one...

50^56^52 % 11 == ??

euler no. = 10
so, 56^52 % 10 = 6^52 = 6

therefore 50^6 % 11 = 6^6 = 6... but ans is given five....
4896

1) Let expense per student be 'x'
Hence initial expense = 42x
42x +31 = 55 (x-3)
--> x = 15 + 1/13
--> 42x = 630 + 42/13 = 633.23

4897
niveditasharma Says
for first the ans is 8 bt for 2nd also ans shud b 8 as we consider 7 ... why are we considering 3 as we have enough no. is 3's in this... limiting factor should b 7 ???? but it doesnt give the ans ...its the question from arun sharma n i also got struck here



Can somebody explain how is 3 deciding factor in second question !! m stuck here too !!
4898

Find the sum of divisors of 544 excluding 1 and 544

4899
chasingdreams Says
Find the sum of divisors of 544 excluding 1 and 544


my take is 589
4900
gudda1122 Says
my take is 589

Its correct . What is the approach bro ??
4901
chasingdreams Says
Find the sum of divisors of 544 excluding 1 and 544

approach is

chasingdreams Says
Its correct . What is the approach bro ??


sum of divisors/factors
544=(2)^5(17)
formula is
/a-1*/b-1...and so on
where a and b are prime bases...x and y are there powers
put the values and solve then in the end just remover=1+544=545..
that will be your answer
4902
approach is



sum of divisors/factors
544=(2)^5(17)
formula is
/a-1*/b-1...and so on
where a and b are prime bases...x and y are there powers
put the values and solve then in the end just remover=1+544=545..
that will be your answer


Thanks Bro !!
4903

Hi Puys!

Got doubt in the following question of Number Systems LOD II Question Number 62 Page 46.

M is a two digit number which has the property that: the product of factorials of its digits > sum of factorials of its digits. How many values of M exist?
I have got the answer to be 74 but I am not sure because no explanation for the solution has been given. and the answer according to the book is 63. Can anyone explain how?

4904

We will have to solve this problem thru general logic
Lets look at the numbers for which product of factorials 10 - 20 , 30,40,50,60,70,80,90 --> 11 + 7 = 18 values
21,31,41,51,61,71,81,91--> 8 vaues
22 is a 2 digit number in which 2! + 2! = 2! * 2! --> 1
So in total --> 18 +8 + 1 = 27
100 - 27 = 63.
Hope this helps 😃

4905

We will have to solve this problem thru general logic
Lets look at the numbers for which product of factorials 10 - 20 , 30,40,50,60,70,80,90 --> 11 + 7 = 18 values
21,31,41,51,61,71,81,91--> 8 vaues
22 is a 2 digit number in which 2! + 2! = 2! * 2! --> 1
So in total --> 18 +8 + 1 = 27
100 - 27 = 63.
Hope this helps 😃