Quant by Arun Sharma

MBA CAT 2024
2258
200!=2^197*3^97*5^49*7^32*11^19*13^16*17^11*19^10*23^8*29^6*31^6*37^5*41^4*43^4*47^4*53^3*59^3*61^3*67*2......

70!=2^67*3^32*5^16*7^11*11^6*13^5*17^4*19^3*23^3....67^1

So,it is clear that the answer is 2.

Now,Arun Sharma has some misprints so you can never be sure if the answer is correct or not.Just the basics are important.If your approach is right,then answer should not be a problem :)


thanks a TON shashank bro n oder people as well..

.but bro how did u calculate the power of so many natural no. so smoothly???i mean did u use d normal approach,i.e d division method in which v go on dividing the no. by d divisor n square of d divisor etc,n add on...or dere is sum trick???plzzz lemme know bro....thanks 😃
2259

One litre of water is evaporated from 6 liters of a solution containing 5%of salt..Find the percentage of salt in the remaining solution???

2260
Ganu02 Says
One litre of water is evaporated from 6 liters of a solution containing 5%of salt..Find the percentage of salt in the remaining solution???

The answer should be 6%.

5% of 6L = 0.3L of salt
1L water evaporates -> 0.3L of salt in 5L solution = 6% salt
2261

Ams employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 .The probability that after 10 years atleast 6 of them still work in Ams is ???

a) 0.19 b) 1.22 c) 0.1 d) 0.16

2262
200!=2^197*3^97*5^49*7^32*11^19*13^16*17^11*19^10*23^8*29^6*31^6*37^5*41^4*43^4*47^4*53^3*59^3*61^3*67*2......

70!=2^67*3^32*5^16*7^11*11^6*13^5*17^4*19^3*23^3....67^1

So,it is clear that the answer is 2.

Now,Arun Sharma has some misprints so you can never be sure if the answer is correct or not.Just the basics are important.If your approach is right,then answer should not be a problem :)


Boss how did you expand 70! or say any factorial of a number??
2263
thanks a TON shashank bro n oder people as well..

.but bro how did u calculate the power of so many natural no. so smoothly???i mean did u use d normal approach,i.e d division method in which v go on dividing the no. by d divisor n square of d divisor etc,n add on...or dere is sum trick???plzzz lemme know bro....thanks :-)

s'squre Says
Boss how did you expand 70! or say any factorial of a number??


Err,just the age old trick of summing up the divisors while dividing continuously by a prime number.

Like say the power of 2 in 37! is

37/2=18
18/2=9
9/2=4
4/2=2
2/2=1

18+9+4+2+1=34

Hope that helps.
2264

one tenth percent of 36 is

2265
Ganu02 Says
one tenth percent of 36 is

What sort of a question is this? The answer would be 36/1000 = 0.036. Don't know how or why you couldn't solve this, or maybe you haven't posted the entire thing.
2266

hey sorry puys for spamming..I typed half the question...While copying from the source there was mistake..I ll soon post the full question..

2267

Hi,

I have few doubts in questions of Geometry and Mensuration section of Arun sharma Quantitative aptitude. I am mentioning questions below:

1 Find the area of the triangle inscribed in a circle circumscribed by a square made by a joining the mid points of the adjacent sides of a square of side a.
Options :
a 3a^2/16
b.3sqrt(3)a^2/16
c 3a^2(pie -12)/4
d 3sqrt(3)a^2/32
e none of these.

2 The section of a solid right circular cone by a plane containing vertex and perpendicular to base is an equilateral triangle of side 12 cm.Find the volume of the cone.
Options:
a.72cc
b.144cc
c.72sqrt(2) pie cc
d 72sqrt(3) pie cc
e 80Pie cc

Please explain approach of doing these type of questions.

Thanks in advance,

2268
Hi,

I have few doubts in questions of Geometry and Mensuration section of Arun sharma Quantitative aptitude. I am mentioning questions below:

1 Find the area of the triangle inscribed in a circle circumscribed by a square made by a joining the mid points of the adjacent sides of a square of side a.
Options :
a 3a^2/16
b.3sqrt(3)a^2/16
c 3a^2(pie -12)/4
d 3sqrt(3)a^2/32
e none of these.


I think some part is missing about the innermost triangle.It has to be an equilateral triangle for the options to fit.Else,we can draw a triangle of any area inside the circle :)

So,considering it to be an equilateral triangle,

The side of outermost square is a

Side of inner square is a/sqrt(2)

radius of circle is a/

Height of the equilateral triangle is 3a/ as the centroid divides the altitude in the ratio of 2:1.The centroid in this case is the center of the circumcircle.

So,side of the equilateral triangle is 2/sqrt(3)*3a/

So,area is 3*sqrt(3)*a^2/32

So,option (D)
2 The section of a solid right circular cone by a plane containing vertex and perpendicular to base is an equilateral triangle of side 12 cm.Find the volume of the cone.
Options:
a.72cc
b.144cc
c.72sqrt(2) pie cc
d 72sqrt(3) pie cc
e 80Pie cc



The diameter of the base is 12 cm and the height comes out to be sqrt(3)*0.5*12 which is the altitude of the triangle.

So,volume is

1/3*pi*r^2*h

1/3*pi*36*6*sqrt(3)

72*sqrt(3)*pi

option (D)
2269
I think some part is missing about the innermost triangle.It has to be an equilateral triangle for the options to fit.Else,we can draw a triangle of any area inside the circle :)

So,considering it to be an equilateral triangle,

The side of outermost square is a

Side of inner square is a/sqrt(2)

radius of circle is a/

Height of the equilateral triangle is 3a/ as the centroid divides the altitude in the ratio of 2:1.The centroid in this case is the center of the circumcircle.

So,side of the equilateral triangle is 2/sqrt(3)*3a/

So,area is 3*sqrt(3)*a^2/32

So,option (D)




The diameter of the base is 12 cm and the height comes out to be sqrt(3)*0.5*12 which is the altitude of the triangle.

So,volume is

1/3*pi*r^2*h

1/3*pi*36*6*sqrt(3)

72*sqrt(3)*pi

option (D)


Thanks for reply
2270

if the roots of the equation (a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0 are equal then which of the following is true?
a.ab=cd
b.ad=bc
c.ad=sqrtbc
d.ab=sqrtcd
abc=sqrtd
2.find the value of expression (sqrtx+(sqrtx+(sqrtx.....+)))
a.1/2[2sqrt(2x-1)+1
b.1/2[sqrt(4x+1)+1
c.1/2[2sqrt(2x-1)-1
d.1/2[sqrt(4x-1)-1
e.none of these

2271

the unit digit in the expression 36^234* 33^512* 39^180 - 54^29* 25^123* 31^512.
options:-
8
0
6
5
4

I keep getting the answer as 4. correct answer(bk) 6.
plz explain.
thanx in advance 😃

2272
the unit digit in the expression 36^234* 33^512* 39^180 - 54^29* 25^123* 31^512.
options:-
8
0
6
5
4

I keep getting the answer as 4. correct answer(bk) 6.
plz explain.
thanx in advance :)

Hey I got the answer for this problem 6

1st term= 6*1*1

2nd term Has 5 So If you multiply Unit digit will be 0 so 6-0 =6
2273
Hey I got the answer for this problem 6

1st term= 6*1*1

2nd term Has 5 So If you multiply Unit digit will be 0 so 6-0 =6

yes..the ans is 6..
(6*1*1-4*5*0)=6...:)
2274

oh ok. i got it. for some reason i kept assuming the first term as 6*1*9 (!!!)
thanx 😃

2275
yes..the ans is 6..
(6*1*1-4*5*0)=6...:)
How is it possible to get 0 there...It must be 1 only...Correct me if 'm wrong..
I thought 1 term is (6*1*1) and second term is (4*5*1)....
2276
the unit digit in the expression 36^234* 33^512* 39^180 - 54^29* 25^123* 31^512.
options:-
8
0
6
5
4

I keep getting the answer as 4. correct answer(bk) 6.
plz explain.
thanx in advance :)

the correct approach is
6*1*1-4*5*1=6-0
hence the answer is 6
u hav 2 consider d unit digit of the product which is 20
2277

yes u r rt but when u simplify dat u ll get as 6-0 u will get dis by considering the unit digit of 20