What is the probability that a two digit number selected at random will be a multiple of 4 and not a multiple of 7 ?

IBPS Mains Admit Card has been released on 29th September 2022. The candidates who are qualified in the Prelims are eligible to attend the Mains examination. The mains exam is going to be held on 8th October 2022. Earlier, the IBPS Clerk Prelims Result was released on 21st September 2022. The selection process for the IBPS Clerk exam consists of two stages - Prelims & Mains exam. A total number of 6035 vacancies are available for the IBPS Clerk recruitment. The finally appointed candidates will be entitled to a pay scale of INR 11,765 to INR 31,540.

"Either $d$ or $e$ has to be $6$", this is incorrect. In order for the number $de$ to be divisible by $3$, the sum of the two digits $(d+e)$ should be a multiple of $3$. Therefore, $(d+e)=(4+5)$ or $(5+7)$

Proof: Suppose you have a three-digit number $abc$ Then$$abc=100a+10b+c=99a+9b+0c+(a+b+c)$$ if $abc$ is divisible by $3$, then $99a+9b+0c+(a+b+c)$ must also be divisible by 3. Now because $99a,9b,0c$ can clearly be divided by $3$ (because of $99$), so $(a+b+c)$ must be a multiple by $3$.

For higher-digit numbers this proof also holds.

What is the probability that a two digit number selected at random will be a multiple of ' 3 ' and not a multiple of '5'?