Question

Plz solve this asap ......

Answer 1

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$\Large\dagger\:\:\boxed{\bold{\red{(A) \:6 8 2 }}}$

$\sf\underline\pink{One \:Number \:is \:correct\: and \:well \:placed}$

Here, the number cannot be $\bold{6}$ as that will make Statement $\bold{B}$ wrong so number in code is $\bold{2}$ which is correctly placed also (on $\bold{3rd\: position}$)

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$\Large\dagger\:\:\boxed{\bold{\red{(B)\:6 1 4 }}}$

$\sf\underline\pink{One\: Number \:is \:correct\: but \:wrong \:placed}$

$\bold{6}$ cannot be that number as per $\bold{Statement\:A}$, It cannot be 1 also as if number is 1 it can be on either on $\bold{1st\: or \:3rd \: place}$ those are already taken by digits $\bold{0}$ and $\bold{2}$ so only number left is $\bold{4}$

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$\Large\dagger\:\:\boxed{\bold{\red{(C)\: 2 0 6 }}}$

$\sf\underline\pink{Two \:Numbers \:are\: correct \:but\: Wrong \:Places }$

One of digit which is correct is $\bold{0}$ as per Statement E but its place cannot be at $\bold{2nd}$ (as per statement C) and $\bold{3rd position}$ (as per Statement E) so it can be only one first place, other digit is $\bold{2}$ (as per statement A).

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$\Large\dagger\:\:\boxed{\bold{\red{(D)\: 7 3 8 }}}$

$\sf\underline\pink{Nothing \: is \: correct }$

We can rule out these $\bold{3\: digit}$ s from other statements

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$\Large\dagger\:\:\boxed{\bold{\red{(E)\: 8 7 0 }}}$

$\sf\underline\pink{One\: Number\: is\: correct\: but\: wrong\: place }$

We know $\bold{0}$ is one of digit in code but place for it would be not $\bold{3}$

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From statement $\sf\underline\red{A}$, $\sf\underline\red{B}$ and $\sf\underline\red{E}$ we got our numbers.

$\sf\underline\purple{Statement\:A}\:\longrightarrow\:$ X X 2

$\sf\underline\purple{Statement \:A,\:C \:and\: E}\:\longrightarrow\:$ 0 X 2

$\sf\underline\purple{Statement\:B}\:\longrightarrow\:$ Last digit 4 which will come in middle.

$\star\:\:\sf\underline\red{Hence\: \: the\:\: code\: \:is}\::$

$\huge\leadsto\:\:\mathfrak\pink{0\: 4 \:2}$

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