Aptitude 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**

**Sur**

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**Vaibhav**

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

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**

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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**

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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**

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Give me some shortcut ?

**Sireesh**

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

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