Latest Wipro Aptitude Question SOLUTION: min num of digits required to form every num greater than 900 and less than 9000. a.39 b.72 c.49 d.81

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(#M40139335) WIPRO Numerical Ability question aptitude Keep an EYE

min num of digits required to form every num greater than 900 and less than 9000.
a.39
b.72
c.49
d.81

Asked In Wipro Ritika (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

divyanshi  (9 years ago)

10+10+10+9=39

Is this solution Helpfull? Yes (2) | No (9)  

Ritika   (9 years ago)

please explain it clearly

Is this solution Helpfull? Yes (2) | No   

dipak kakadiya  (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

Is this solution Helpfull? Yes (2) | No   

vasuki  (9 years ago)

900

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kumari priyanka  (8 years ago)

it must be (a) i.e., 39.
10+10+10+9

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Ramadevi  (8 years ago)

10+10+10+9=39

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tani  (8 years ago)

plz explain properly

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sai  (8 years ago)

10+10+10+9=39 it is only for 1000 to 9000 na
what is about 900 to 1000

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Lakshmi  (8 years ago)

case 1 : 900-999
units place --- 9 digits [(1-9) since>900]
tens place --- 10 digits [0-9]
hundreds place - 1 digit [only 9 is possible]

sum=1+9+10=20

case --2 : 1000----9999

units place---10 digits [0-9]
tens place----10 digits [0-9]
hundreds place-- 10 digits [0-9]
thousands place ---8 digits [ 0 is prohobited because it becomes 3 digit number ex : 0345 and 9 is also prohobited since

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