Latest Capgemini Aptitude Question SOLUTION: sum of the digit of a five number is 41.find the probability that such a number is divisible by 11.

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sum of the digit of a five number is 41.find the probability that such a number is divisible by 11.

Asked In Capgemini Rita Shaw (13 years ago)

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SATHISH T.N  (11 years ago)

in order to get the sum as 41, the following 5 digit combination exist:
99995-> no.of 5 digits=5
99986-> no.of 5 digit=20
99977-> no.of 5 digit=10
99887-> no.of 5 digit=30
98888-> no.of 5 digit=5

now, 70 such number exists.,
now for a 5 digit number of form(pqrst) to be divisible by
(p,r,t)+(q+s)=41
(p,r,t)=(9,9,8) and (q,s)=(8,7)-----(1)
or (p,r,t)=(9,9,8) and (q,s)=(9,6)----(2)

using 1st eqn we can construct (3!/2!)*2! = 6 numbers
using 2nd eqn we can construct (3!/2!)*2! = 6 numbers

number of favble cases = 12.,

hence required probability = 12/70 = 6/35

Is this solution Helpfull? Yes (33) | No (12)  

krishna kumar gupta  (13 years ago)

since number divisible by 11 =sum of digit of number must be divisible by 11;

here 41 is not divisible by 11.
so probability of number divisible by 11 is 0;

Is this solution Helpfull? Yes (15) | No (76)  

aks  (8 years ago)

let the sum of digits at odd places be a and those at even places be b; such that a+b=41;a-b: divisible by 11; among all the possible combinations of (a,b) starting from (41,0) to (1,40) there lies only (26,15) and (37,4) only whose difference is divisible by 11 thus probability is (2/41)

Is this solution Helpfull? Yes (3) | No (5)  

Likhitha  (6 years ago)

In order to get the sum as 41, the following 5 digit combination exist :

99995--->(5!/4!)=5 numbers.

99986--->(5!/3!)=20 numbers.

99977--->5!/(3!2!)=10 numbers.

99887--->5!/(2!2!)=30numbers.

98888--->5!/4! =5 numbers.

Thus, 5 + 20 + 10 +30 +5 i.e 70 such number exists.

For a 5 digit number in the form (pqrst) to be divisible by 11, the condition is :-

=>(p+r+t)−(q+s)=11. ---------------(a)

Also, it is given :-

=>(p+r+t)+(q+s)=41. ---------------(b)

From (a) and (b), we get :-

p+r+t=26, q+s=15.

(p,r,t)=(9,9,8) and (q,s)=(8,7) ----------------(i)

or (p,r,t)=(9,9,8) and (q,s)=(9,6) ----------------(ii)

Using 1st equation, we have :-

=>(3!/2!)x2!= 6.

Using 2nd equation we have :-

=>(3!/2!)x2! =6.

Thus, the number of favourable cases = 12.

Hence, the required probability = 12/70 = 6/35.

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adisesharao  (8 years ago)

sorry to say this krishna kumar guptha 0 is wrong answer because the question is the probability for that 5 digit number not for the 41.

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