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(#M40074798) ELITMUS Numerical Ability question Quantitative Keep an EYE

What is probability (approximate) of a 4 digit cube root random number .
a .0004
b .0005
c .0011
d . 00331

Asked In Elitmus suresheca2012@gmail.com (10 years ago)

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Sunanda Anand  (10 years ago)

1000,1331,1728,2197 .......4000,9261,
(10^3,11^3,12^3............20^3,21^3)
total possibilities are 12
so probability is 12/9000(total number 4 digit nubers are 9000(9999-999)
answer is .00133

Is this solution Helpfull? Yes (44) | No   

SUNNY MAHATO  (10 years ago)

Please bro first write all option correctly. The right answer is- 12/9000= .00133..

Is this solution Helpfull? Yes (16) | No   

Surajit  (10 years ago)

1. first count how many 4digit numbers are threre?? (1000 to 9999)=9000(including 1000 and 9999)
2. now take cube root of 1000=10^3, likely as 11^3, 12^3, 13^3, 14^3 up to 9261=21^3(extreme 4 digit cube root)
3. so 10,11,12,13 up to 21 total numerics=12 (including 10 and 21).
4. now probability

P(N)=n(E) / n(S)
=12 /9000
= 0.0013333 (given Options are all wrong) -Thank you

Is this solution Helpfull? Yes (5) | No   

shekh firoz alam  (10 years ago)

I am not getting the question plz explain once

Is this solution Helpfull? Yes (2) | No (4)  

juhi  (9 years ago)

The answer might be 12/32...

Is this solution Helpfull? Yes | No (1)  

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