Latest Wipro Aptitude Question SOLUTION: the minimum number of digits required to form every number which is greater tha 900 and less than 9000

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(#M40152469) WIPRO Numerical Ability question Keep an EYE

the minimum number of digits required to form every number which is greater tha 900 and less than 9000

Asked In Wipro farseena k u (8 years ago)

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kumar aryanand  (8 years ago)

The question is given less than 9000 so last number is 8999.So 1000ths digits can be 1-8.It is also given that greater than 900 so 1st no is 901.so unit plac range 1-9=9
Thus min no of digits =9+10+10+8=37

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vivek rawat  (8 years ago)

Answer - 39

for unit place numbers can range from 0-9 ie; total 10 digits
for tens(10th) place number can range from 0-9 ie; again 10 digits (0,1,2,3,4,5,6,7,8,9)
similarly for hundred(100th) place again 10 digits.
but for 1000th place it can only range from 1-9 ie; total 9. We cannot include 0 here as the number will then start from 0 means ex-0459 which only means 459.

thus minimum no of digits required are : 10+10+10+9 =39

39 is the right answer (100% sure)

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Nikhil   (8 years ago)

10 digits

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Aniruddha  (8 years ago)

Ans = 10+10+10+9=39

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Nikhil   (8 years ago)

1) for unit place we take 0 to 9 so =>10 digit
2)for 10nth place same as 0 to 9 so => 10 digit
3)for 100th place digit ==========>10digit
4)at last 1000nd digit (9000)====>9
so,10+10+10+9=39

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arijit biswas  (8 years ago)

The question is given less than 9000 so last number is 8999.So 1000ths digits can be 1-8.
Thus min no of digits =10+10+10+8=38

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shubham pal  (8 years ago)

39 is the correct......

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Abhishek Kumar  (8 years ago)

Answer is weird. Isn't it ? Only 10 digits are present in whole mathematics.

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