Quant by Arun Sharma

MBA CAT 2024
2662
plz help me solve this

1. find the sum of all three digit numbers that give remaindeer of 4 when divided by 5

a. 98270
b. 99270
c. 1,02,090
d. 90270
e. none ofthese


since it gives 4 when divided by 5, they are of the form 104+109+114.........+999

so its a Ap, now a=104,d=5
so, 999=104+(n-1)5
n=180

=>s=(104+999)180/2
=>99270
2663
scan002 Says
104+109+......+999=99270



the sum of this AP comes to 2097984..how did u get 99270


ok got it thanx..was making mistake..
2664

is the question.

This can be expanded as ..





To find 21 + 22 + 23 + ... + 200:

n(n+1)/2 ==>
Using this in the orig prob

= 99270

2665
divyaakanksha Says
m not able to understand the solution..how did u get 5[(1+2+_ _ _+200)-(1+2__+20)]


First such number = 104
Last such number = 999
the Sum = 104+109+114+........+999
total no of terms = [ (999-104)/5 ] + 1 = 180

Therefore, Sum = n/2 * [ first term+last term] = (180/2) * [104+999] = 99270



PS:- how to attach fig as thumbnails ???


When you go for advanced reply there is an option given as Attachment (symboled with a safety-pin, just adjacent to the font options)....you can attach anything there (max size also mentioned there).
2666
divyaakanksha Says
m not able to understand the solution..how did u get 5[(1+2+_ _ _+200)-(1+2__+20)]


First such number = 104
Last such number = 999
the Sum = 104+109+114+........+999
total no of terms = [ (999-104)/5 ] + 1 = 180

Therefore, Sum = n/2 * [ first term+last term] = (180/2) * [104+999] = 99270



PS:- how to attach fig as thumbnails ???


When you go for advanced reply there is an option given as Attachment (symboled with a safety-pin, just adjacent to the font options)....you can attach anything there (max size also mentioned there).
2667

the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110

2668
the highest power on 990 that will exactly divide 1090! is
a.101
b.100
c.108
d.109
e.110

990=11x10x9..

ur job is to find which among the prime factors have least power dividing 1090!
obviously 11 it is
=> 1090/11+1090/121
=>99+9
=>108
2669

Hi fellow Puys,

Please help me solve this with a smart and better approach:

A cricketer scores following runs and averages:

Year Runs Avg
1919 891 34.27
1920 949 28.75
1921 1329 42.87
1922 1101 36.70

What was his avg for 4 years?

Thanks 😃

2670
chapter1
lod2
30)find d no. of zeros in d product 1^1 * 2^2 * 3^3 -- 100^100???
a 1200
b 1300
c1050
d 1225
e none of his


ans:1300

count the power of 5

(5+10+15+........+100)=1050

25 ,50 ,75 and 100 includes two 5s each

total no. of 5 will be 1050+2*(25+50+75+100)=1300

so no. of 0s= 1300
2671
The area of circle circumscribing 3 circles of unit radius touching each other is
a.(pie/3)(2+3^1/2)^2 b. 6*pie(2+3^1/2)^2 c. 3*pie(2+3^1/2)^2 d. pie/6(2+3^1/2)^2 e.pie/4(2+3^1/2)^2
ans=a
please give a detailed solution


center of the circumscribing circle = centroid of the equilateral triangle formed by joining the centers of three cicles

centroid of equilateral triangle = (2/3)*altitude of equilateral triangle= (2/3)*sqrt(3)/2*side of triangle= 2/3*sqrt(3)/2*2 =2/sqrt(3)

radius of circumscribing circle = (2/sqrt(3))+1 ( 1 is radius of a smaller circle)

now find the area and its option a...
2672
30)find d no. of zeros in d product 1^1 * 2^2 * 3^3 -- 100^100???
a 1200
b 1300
c1050
d 1225
e none of these
'm getting 1200...Any other ans...
2673
30)find d no. of zeros in d product 1^1 * 2^2 * 3^3 -- 100^100???
a 1200
b 1300
c1050
d 1225
e none of these
'm getting 1200...Any other ans...


complete solution..please
2674
kanak84 Says
complete solution..please

we have to calculate the number of 2's and 5's in the series...As we know that the number of 2's will be more than number of 5's,lets calculate 5 alone..
=(5+10+15+....100)
=5(1+2+3....20)==>1050
when we see the two number 50 and 100
50^50 can be written as (5*10)^50 which inturn can be written as (5*5*2)50
so its 1050+50=1100
100^100 can be written as (20*5)^100 which is written as (5*5*4)100
so its 1100+100=1200...
2675
we have to calculate the number of 2's and 5's in the series...As we know that the number of 2's will be more than number of 5's,lets calculate 5 alone..
=(5+10+15+....100)
=5(1+2+3....20)==>1050
when we see the two number 50 and 100
50^50 can be written as (5*10)^50 which inturn can be written as (5*5*2)50
so its 1050+50=1100
100^100 can be written as (20*5)^100 which is written as (5*5*4)100
so its 1100+100=1200...


then it shuould be 1300 as 25^25 is actually (5*5)^25 and 75^75 is (5*5*3)^75
2676
kanak84 Says
then it shuould be 1300 as 25^25 is actually (5*5)^25 and 75^75 is (5*5*3)^75

yep dude..Missed it out..
2677
30)find d no. of zeros in d product 1^1 * 2^2 * 3^3 -- 100^100???
a 1200
b 1300
c1050
d 1225
e none of these
'm getting 1200...Any other ans...

i think its 1300.
count total no. of 5.
2678

plz helpme solve


1. find gcd (2^100-1,2^120-1)
a. 2^20-1
b. 2^40-1
c. 2^60-1
d. 2^10-1


2. how many pair of natural numbers are there the difference of whose square is 45?

a. 1
b. 2
c. 3
d. 4
e. 5

2679
plz helpme solve


1. find gcd (2^100-1,2^120-1)
a. 2^20-1
b. 2^40-1
c. 2^60-1
d. 2^10-1


2. how many pair of natural numbers are there the difference of whose square is 45?

a. 1
b. 2
c. 3
d. 4
e. 5


1)2^20-1 as highest common factor to both are 2^20-1

2)I could only thk of 3...I am not aware of any method for this !!
2680
plz helpme solve


1. find gcd (2^100-1,2^120-1)
a. 2^20-1
b. 2^40-1
c. 2^60-1
d. 2^10-1




Solution :

Here in these types of questions you have to only find the GcD/HCF of the powers only !!

100= 2^2 X 5^5
120=2^3 X 5 X 3

hcf =20

Answer :2^20-1
2681
1)2^20-1 as highest common factor to both are 2^20-1

2)I could only thk of 3...I am not aware of any method for this !!



what r the three nos