Latest Elitmus Aptitude Question SOLUTION: A set M contains element all even number between 1 and 23 and all odd numbers 24 and 100. if all the elements of the set multiplied than how many trailing 0 , resulting number

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(#M40157641) ELITMUS Numerical Ability question Quant Keep an EYE

A set M contains element all even number between 1 and 23 and all odd numbers 24 and 100. if all the elements of the set multiplied than how many trailing 0 , resulting number will contains?
a-10
b-12
c-9
d- not remember?

Asked In Elitmus amit maurya (7 years ago)

Solved sachin dubey Read Solution (12)

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Read Solution (12) :

priyadarshan kumar  (7 years ago)

( Answer will be 12 )

because the result trailing 0 occurs when one 2 and one 5 multiplied.

here we see the set M:

( 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 35, 45, 55, 65, 75, 85, 95 )

after 22 I have taken only the term which contains factor of 5

now we find prime factor

2 = 2
4 = 2 * 2
6 = 2 * 3
8 = 2 * 2 * 2
10 = 2 * 5
12 = 2 * 2 * 3
14 = 2 * 7
16 = 2 * 2 * 2 * 2
18 = 2 * 3 * 3
20 = 2 * 2 * 5
22 = 2 * 11
25 = 5 * 5
35 = 5 * 7
45 = 5 * 9
55 = 5 * 11
65 = 5 * 13
75 = 5 * 5 * 3
85 = 5 * 17
95 = 5 * 19

here we find
total number of 5 is 12
and
total number of 2 is 19

and
we get total number of 5 * 2 pair is 12
so number of trailing zero will be 12

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sachin dubey  (7 years ago) Best Solution

12 because m={10,20 } &{25,35,45,55,65,75,85,95}
in set m number of 5 is 12.

Is this solution Helpfull? Yes (23) | No (8)  

Vikash kumar  (7 years ago)

correct answer is 12
explanation:
m=(2,4,6,8,10,12,14,16,18,20,22,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,91,93,95,97,99)
question asked how many trailing 0
there are two condition
1st if we multiply 2*5=10
end with =0
if we count 5 in set m we get 12
if we count 2 in set m we get more than 12
so total no end with 0 equal to 12

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andru venu gopal  (7 years ago)

10
because there are two 0's and eight 5's
So total 10 zeros will come

Is this solution Helpfull? Yes (1) | No (3)  

sachin pratap singh  (7 years ago)

12 is the answer..bcoz (5*2)=10 , 5 and 2 makes one zero, and there are 12 '5'. and no of 2's will be more than 12 ....so 12 '5' will make 12 zeros...

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lekhraj jeengar  (7 years ago)

find no. of 5 =no. 0's

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neelaksh  (7 years ago)

last option was 11

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somanita  (7 years ago)

any one know when 2nd oct result will be announce?

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bharat gabrani  (7 years ago)

9 because 2,4,8,10,12,14,18,20,22 will be multiplied with odd no 25,35,45,55,65,75.85.95

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Bharadwaj  (7 years ago)

10
I will tell you why
2*25
4*35
5*45
6*55
8*65
10
12*75
14*85
16*95
20

all even numbers multiplied by 5 multiples give leading zeros and all ten multiples five leading zero ! Thus 10 !

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Debarati Gupta  (7 years ago)

25-100=there are 10 fives
1-23=there are 19 two's
subtracting 10 two's=19-10=9
we excluded 10 two's beacuse we are multiplying 5*2 ten times.
5x2=10 and 4x2=8
we exclude 4 two's because we are not getting reaminder as zero.

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Deepak  (6 years ago)

The answer will be 12....
Because 10 and 20 gives 2 zeros
(25,35,45,55,65,75,85,95)* multiply with even digits gives 8 more
2 even digits left. multiplied with upper product they will give 2 more zeros
so total= 2+8+2= 12 zeros

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