# Difficult Probability Solved QuestionAptitude Discussion

 Q. Sum of digits of a 5 digit number is 41. Find the probability that such a number is divisible by 11?
 ✖ A. 2/15 ✖ B. 11/36 ✖ C. 3/35 ✔ D. 6/35

Solution:
Option(D) is correct

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

99995--->number of 5 digits =5
99986--->number of 5 digit =20
99977--->number of 5 digit =10
99887--->number of 5 digit =30
98888--->number of 5 digit =5

Now, 70 such number exists.

Now for a 5 digit number of form (pqrst) to be divisible by

$11(p+r+t)−(q+s)=11$, also $(p+r+t)+(q+s)=41$

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

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

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

Using $1^{st}$ equation we can construct

$=\dfrac{3!}{2!}\times 2!=6$ numbers.

Using $2^{nd}$ equation we can construct

$=\dfrac{3!}{2!}\times 2!=6$ numbers.

Number of favourable cases = 12.

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

## (7) Comment(s)

Sur
()

umm...the question says it must be divisible by 11 right...

Vaibhav
()

why in first equation its 11(p+r+t)-(q+s)?? why 11 as multiple factor with first term??

Yasaswani
()

it is not 11(p+q+r) actually the 11 in the first is from the above line:(pqrst) divisible by 11.

the actual equation is (p+q+r) - (q-s) =11.

Vishwanath
()

The sum of the digits of a four-digit number is 32. what is the probability that such a number is divisible by 11?

Vishwanath
()

The sum of the digits of a four-digit number is 32. what is the probability that such a number is divisible by 11?

Lakshmi
()

Give me some shortcut ?

Sireesh
()

Please post more questions in this level (with solutions)..