Quant by Arun Sharma

MBA CAT 2024
5966

Hey can anybody please solve this problem (Arun Sharma quant - Page - 229 ,LOD 2 , Sum -22, Chapter - SI/CI)

A person lent out some money 6% per annum SI and after 18 months he again lent out the same money at 24% per annum. SI. In both cases he got 4704. Which of these could be the amt lent out in each case if interest is paid half yearly.

a) 4000 b)4200 c)4400 d)3600

5967
@MbaIIMA - language of the question looks confusing.

Solution - x + (x*2*6)/100 = 4704

Solve for x. The answer comes 4200.
5968

Q. There are 6 pups and 4 cats. In how many ways can they be seated in a row so that no cats sit together?


Explain plz
5969
@pandeyn is the ans. 35?
5970
@pandeyn the answer will be 6! x 35
firstly 6 pups can be arranged in 6! ways and now you can arrange 4 cats in 7p4 ways
P--P--P---P--P--P

5971
@pandeyn there is no contraint on pups hence it will be -p-p-p-p-p-p- now pups 6! ways and for cats 7C4( select any 4 places ) *4!
hence its 6! * 7P4
5972
ARUN SHARMA PNC LOD 3 Q 13. can someone please solve this
in an examination, the maximum marks for each of the three papers is 50 each. the maximum marks for the fourth paper is 100. find the number of ways with which a student can score 60% marks in aggregate.
a) 330850
b)233551
c)110551
d)none of these
5973
Can anyone please tell me the correct answer for the question"
find the total numbers between 122 and 442 that are divisible by 3 but not by 9
"?it is given 71 in the book but I am not getting the same.

5974

@nikita105

Total Marks in the exam = 50*3+100=250

60% of 250 = 0.6* 250=150

Now we can form the equation : A+B+C+D=150 , where A,B,C,D are the four subjects.

Now it boils down to dividing 150 similar things to 4 different groups.

Hence 153*152*151/6 =585276, the reqd no. of ways of getting an aggregate of 60% in the exam. The answer is : (D) None of these.

5975
@
@PKB22 said:

@nikita105

Total Marks in the exam = 50*3+100=250

60% of 250 = 0.6* 250=150

Now we can form the equation : A+B+C+D=150 , where A,B,C,D are the four subjects.

Now it boils down to dividing 150 similar things to 4 different groups.

Hence 153*152*151/6 =585276, the reqd no. of ways of getting an aggregate of 60% in the exam. The answer is : (D) None of these.

This solution is wrong. The mistake I did was in the equation A+B+C+D=150 , I considered that any of the subjects can have any marks between 0 and 150 which is not the case because in subjects A,B,C,D the maximum marks one can have either 50 or 100 only. There will be a condition on the equation : A+B+C+D=150, where 0
5976

hi guys .. please help me with a ques..
find the remainder when 50^51^52 is divided by 11 ???
(LOD-3 Arun Sharma .)

5977
@shishir91 said:
hi guys .. please help me with a ques..
find the remainder when 50^51^52 is divided by 11 ???
(LOD-3 Arun Sharma .)
As per Euler's theorem : 50^10 modulo 11 = 1
thus, express 51^52 in terms of 10K+r=> 1^52=1

So, 50^51^52 mod 11 reduces to 50^1 mod 11 = 6...


Let me know if you have any issues in the method.



5978
@shishir91 6?
5979
If (2 + 4 + 6 +... 50 terms)/(1 + 3 + 5 +... n terms) = 51/2
Find n ???
5980
@ziatron said:
If (2 + 4 + 6 +... 50 terms)/(1 + 3 + 5 +... n terms) = 51/2
Find n ???
10
5981
@deedeedudu said:
10
Pls explain how?
5982

There is a question in Geometry (Ques 43 Pg. 392, LOD 2, 5th edition)


Circles are drawn with four vertices as the center and radius equal to the side of the square. If the square is formed by joining the mid points of another square of side 2*(6^0.5), find the area common to all the four circles.

The answers are all in terms of pie and ^0.5 hence not typing the options....
5983
Hi. I am unable to solve this. Can anyone kindly help?
Q. What is the last digit of 62^43^54^65^76^87?
5984
@Nilanjan_007 said:
Hi. I am unable to solve this. Can anyone kindly help?
Q. What is the last digit of 62^43^54^65^76^87?
2
5985
@ChinmayaDave said:
Pls explain how?
Sum of numerator= 102 * 25
We want denominator to be 100
Sum of 10 terms will be 100