Latest Elitmus Aptitude Question SOLUTION: If all the odd number from 0 to 10000 were written.How many digits would be written down?

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(#M40151296) ELITMUS Numerical Ability question Maths Keep an EYE

If all the odd number from 0 to 10000 were written.How many digits would be written down?

Asked In Elitmus sunil vj (9 years ago)

Unsolved Read Solution (13)

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Read Solution (13) :

RAKESH  (9 years ago)

1 digit ======> 5

2 digits =====> 45*2 = 90 [11,13,.......,99]

3 digits =====> 450*3 = 1350 [101,103,........,999]

4 digits =====> 4500*4= 18000 [1001,1003,......,9999]

total = 5 + 90 + 1350 + 18000 = 19445

Is this solution Helpfull? Yes (85) | No (5)  

kumar avinash  (9 years ago)

To count number of terms, (l-a)/d +1

so, 1 digit no (1,3,..,9) =>> (9-1)/2 +1 =5
Consequently no. of digits= 5*1= 5

2 digit no (11,13,....,99) =>> (99-1)/2 +1 =45
Consequently no. of digits= 45*2= 90


3 digit no (101,103,....,999) =>> (999-101)/2 +1 =450
Consequently no. of digits= 450*3= 1350

4 digit no (1001,1003,....,9999) =>> (9999-1001)/2 +1 =4500
Consequently no. of digits= 4500*4=18000

So , total no f digits are = 5+90+1350+18000=19445

Is this solution Helpfull? Yes (29) | No   

padam  (9 years ago)

5000 odd numbers

Is this solution Helpfull? Yes (3) | No (2)  

Narottam  (9 years ago)

it asked total digits not total no.
so p&c used

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Pravin Kumar  (9 years ago)

19945 same as mentioned

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pratham patil  (9 years ago)

5000
in 10000 every alternate number is odd number so 5000 odd numbers

Is this solution Helpfull? Yes (1) | No (10)  

anand kumar kushwaha  (9 years ago)

from
0 to 10 there will be =5 digits
11 to 99= 2*45=90
100 to 999=450*3=1350
1000 to 9999=4500*4=18000
total =5+90+1350+18000=19445

Is this solution Helpfull? Yes | No (1)  

soni kumari  (9 years ago)

19445 digits are possible if digits can be repeated

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rohit  (8 years ago)

49999
all the alternate number is odd except 0 and 1

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Vikash kumar  (8 years ago)

correct ans is 5000
explanation:
odd no between:
0 to 10000
1,3,5,7,9,.......................9999
a=1,d=2, tn=9999
we know that
tn =a+(n-1)d;
9999=1+(n-1)2
9998/2=n-1
4999=n-1
n=5000

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jhuma kumari  (7 years ago)

last term=9999
a=1,d=2
l=a+(n-1)d
9999=1+(n-1)2
9999-1=2n-2
9998+2=2n
n=10000/2=5000

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saurabh kumar  (7 years ago)

alternate numberi:e5000

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bradraj kumar  (7 years ago)

i think that answer would be 10 because digits from (0-9) is used to form any odd number from 1 to 10000,according to me combination of digits make number.
for ex.- 121 is a number with 3- digits.
please review if it seems to be relevant.

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