Quant by Arun Sharma

MBA CAT 2024
985
1) Four Bells ring at the intervals of 6,8,12,&18 seconds.They started ringing together at 12'0 Clock.After how many seconds they will ring together again??

a) 72 b) 84 c) 60 d) 48


2)For the above problem, How many times they eill ring together durnig the next 12 mins.

a) 9 b) 10 c) 11 d) none


3)The Product of digits of two digit number is twice as large as the sum of its digits.If we subtract "27" from the number, we get a number consisting of same digits written in reverse order.Find the No.??

4)Find no. of Zero's in the Product 1 X 2^2 X 3^3 X 4^4 X 5^5...............100^100

1) Take Lcm(6,8,12,18 ) = 72

2)(12*60)/Lcm(6,8,12,18 ) = 720/72 = 10

3)ab = 2(a+b), 10a+b - 27 = 10b+a

9a-9b = 27, a-b = 3

easily we can see the number is 63

4)calculate number of 5's

5 + 15 + 50 + 35 + 45 + 55 + 65 + 150 + 85 + 95 + 700 = 1300
986

HI,
Pls answer these following qsns......

1) Three numbers are such that second is as much lesser than the third as the first is lesser than the second.If the product of two smaller numbers is 85 and product of two larger numbers is 115.Find the middle no
a) 9 b) 8 c) 10 d) 30


2) AB+XY=2XP where A!=0 all the letters signify different digits from 0 to 9 , then the value of "A" is
a) 6 b) 7 c)9 d)any number more than "6"


3) X-4 + Y-4 = 4 , then how many Integer values can (X,Y) have??

4) A Two digit number is thrice as large as the sum of its digits and square of that sum is equal to trebled of required number,then find the no.??

solve and explain....
Thanks,

987
HI,
Pls answer these following qsns......

1) Three numbers are such that second is as much lesser than the third as the first is lesser than the second.If the product of two smaller numbers is 85 and product of two larger numbers is 115.Find the middle no
a) 9 b) 8 c) 10 d) 30

2) AB+XY=2XP where A!=0 all the letters signify different digits from 0 to 9 , then the value of "A" is
a) 6 b) 7 c)9 d)any number more than "6"

3) X-4 + Y-4 = 4 , then how many Integer values can (X,Y) have??

4) A Two digit number is thrice as large as the sum of its digits and square of that sum is equal to trebled of required number,then find the no.??

solve and explain....
Thanks,

1)a,b,c

b-a = c-b {Implies they are in AP}

ab = 85, bc = 115

c/a = 115/85 = 23/17{ since they are prime, they can be equated to c & a}

17, b, 23

so,b = 20

2)Its not possible!

two digit numbers when added can't giv a number more than 200,{Max is 99 + 99 = 198}

If the question is AB +XY = 1XP then out of options A can take 9 since A + x = x, so A can be 9 or 0.

3)If we draw a graph, we get a right angle triangle with vertices (0,0),(0,4),(4,0)

Now the integer points which lies with in hte area of triangle are
(1,3)(3,1)
(2,2)
(0,4)(4,0)

4)(10a + b) = 3*(a+b) ..... i

(a+b)^2 = = 3*(10a+b) ... ii

solve we get a+b = 9

so number is 27
988

hello puy.........my doubt is

The numbers 2,4,6,8...98,100 are multiplied together. The number of zeros at the end of the product must be :
13
12
11
10
9

989
hello puy.........my doubt is

The numbers 2,4,6,8...98,100 are multiplied together. The number of zeros at the end of the product must be : 13
12
11
10
9

Calculate the no of 5 in that
10,20,30,40,60,70,80,90 - one 5's each
50,100 - two 5's each
So total = 8+4 =12 zeros
990
Calculate the no of 5 in that
10,20,30,40,60,70,80,90 - one 5's each
50,100 - two 5's each
So total = 10 zeros


from 10 to 90 total 8 zero

and for 50 & 100 total 4 zero

so ans =12
991
from 10 to 90 total 8 zero

and for 50 & 100 total 4 zero

so ans =12

but the ans given is 11
992
shona_ne Says
but the ans given is 11


It is not given factorial to calculate number of 5's

Its just 2*4*6*8*10*.....*100

zeros are contributed by 10,20,30,...100

so will be having 11 zeros at the end
993
It is not given factorial to calculate number of 5's

Its just 2*4*6*8*10*.....*100

zeros are contributed by 10,20,30,...100

so will be having 11 zeros at the end


but menacebhai 50 and 100 both have 2 times 5
994
jigar_p_civil Says
but menacebhai 50 and 100 both have 2 times 5


multiply them individually

50*100 = 5000

we are not going to consider number of 5's

unless it is

50! ,here number of zeros wld be + = 10 + 2 = 12
995
multiply them individually

50*100 = 5000

we are not going to consider number of 5's

unless it is

50! ,here number of zeros wld be + = 10 + 2 = 12


ok i got it

but if question like 2,4,6,.........1000

then how can we judge?
996
multiply them individually

50*100 = 5000 * some even no = 10000

we are not going to consider number of 5's

unless it is

50! ,here number of zeros wld be + = 10 + 2 = 12


But we have other even nos which on mulltiplication gives one more zero...see it in bold
997
sumitbce Says
But we have other even nos which on mulltiplication gives one more zero...see it in bold


Maafi 😃

we get 12 zeros at the end

coz, extra zero is contributed by 50
998
ok i got it

but if question like 2,4,6,.........1000

then how can we judge?


single zeros, 10,20,...,90,110120,...190,..like wise till 990 we get 99 zeros

Double zeros contributed by 100,200 ...till 900 we get 9 more

so, 99+9 = 108

3 zeros contibuted by 1000 so,108+3= 111

and there are few more zeros contributed by

50,150,250,350...950 so 10 more

total will be 111 + 10 = 121 zeros in all
999
single zeros, 10,20,...,90,110120,...190,..like wise till 990 we get 99 zeros

Double zeros contributed by 100,200 ...till 900 we get 9 more

so, 99+9 = 108

3 zeros contibuted by 1000 so,108+3= 111

and there are few more zeros contributed by

50,150,250,350...950 so 10 more

total will be 111 + 10 = 121 zeros in all



Double zeros contributed by 100,200 ...till 900 we get 9 more

its 18 zero not 9
1000
Double zeros contributed by 100,200 ...till 900 we get 9 more

its 18 zero not 9


No, we have already considered 1 zero, while counting 10,20...90,100,..200..


so counting only single zero from 100, 200, 300 ..an so on
1001
ok i got it

but if question like 2,4,6,.........1000

then how can we judge?


My method was more like a trial and error, errors wld creep in easily

Else we can convert it in to factorials, tht wld be easy

2^500 * 500!

now, + + = 100 + 20 + 4 = 124 zeros

2*4*6*..*100

2^50*50!

+ = 10 + 2 = 12 zeros
1002

Hi ,
I tried to solve this problem in Time and work but I couldn't.Could someone help me out

It takes six days for 3 women and 2 men working together to complete a work.Three men would do the same work 5 days sooner than nine women.How many times does the output of the man exceed that of a women?

a) 3 times b)4 times c)5 times d)6 times

I need explanation too not answer alone:

1003
Hi ,
I tried to solve this problem in Time and work but I couldn't.Could someone help me out

It takes six days for 3 women and 2 men working together to complete a work.Three men would do the same work 5 days sooner than nine women.How many times does the output of the man exceed that of a women?

a) 3 times b)4 times c)5 times d)6 times

I need explanation too not answer alone:


Need to form three equation

3W + 2M = 6 DAYS---(1)

9W = X DAYS
1W= 9X DAYS----(2)
1W per day = 1/9X
3W per day = 3/9X

3M = X-5 Days
1M = 3(X-5) Days --(3)
1M per day= 1/3(X-5)
2M per day = 2/3(X-5)

3/9X + 2/3(X-5)= 1/6 -- From (1)
solving this u will get X=10 days

from 2 -- 1W=90 Days
from 3 --- 1M = 3(10-5) = 15 days

1M=6W
1004
Hi ,
I tried to solve this problem in Time and work but I couldn't.Could someone help me out

It takes six days for 3 women and 2 men working together to complete a work.Three men would do the same work 5 days sooner than nine women.How many times does the output of the man exceed that of a women?

a) 3 times b)4 times c)5 times d)6 times

I need explanation too not answer alone:

My take is 6 times
(2m+3w)*6 = work ....(1)
where m is the amount of work done by one man in one day and w is the amount of work done by one women in one day.

Also Let t be the the time taken by 9 women to complete the task so 3 man will take t-5 days to complete the same task

since 3 man will take t-5 days to complete 1 work
so one man will take 3(t-5) days to complete the task
similarly one women will take 9t days to complete the task

Hence the amount of work by one man in one day = work/3(t-5) = m
amount of work by one women in one day = work/9t = w

Putting it in equation 1 we will get t = 10( 1 discarded)

So m/w = 90/15 = 6