CRACK THE CODE
CODE
A NUMERIC LOCK HAS A 3 DIGIT KEY
HINT
482
One number is correctand well placed
416
one number is correct but wrong placed
204
two numbersare correct but wrong placed
738
nothing is correct
780
one number is correct but wrong placed
Answers
Answer:
ANSWER: 0 4 2
The code is 0 4 2
A. 6 8 2 – One Number is correct and well placed – Number cannot be 6 as that will make Statement B wrong so number in code is 2 which is correctly placed also(on Position 3)
B. 6 1 4 – One Number is correct but wrong place – 6 cannot be that number as per Statement A, It cannot be 1 also as if number is 1 it can be on either on 1st or 3rd place those are already taken by digits 0 and 2 so only number left is 4
C. 2 0 6 – Two Numbers are correct but Wrong Places – One of digit which is correct is 0 as per Statement E but its place cannot be at second(as per statement C) and third position( as per Statement E) so it can be only one first place, other digit is 2 as per statement A.
D. 7 3 8 – Nothing is correct – We can rule out these 3 digits from other statements
E. 8 7 0 – One Number is correct but wrong place – We know 0 is one of digit in code but place for it would be not 3
From statement A, B and E we got our numbers. Statement A gave position for Number 2 . Statement C and E gave position for other 2 numbers.
Statement A -> X X 2
Statement A, C and E-> 0 X 2
Statement B -> last digit 4 which will come in middle so code would be 0 4 2
Hence the code is 0 4 2
Step-by-step explanation:
Therefore, the co-de of the given numerical lock is 042.
Decoding the clues given:
On observing the rules given, from the fourth rule, we get that the lock contains no numbers from the list [7, 3, 8]. So, from the first rule 6, 2 remains.
From the second rule, we get that one number among 6, 1, 4 is correct but wrongly placed. But the number 6 is in the same position in rules 1 and 2. So, there is no chance of 6 being among the three numbers of the lock.
So, from the first rule, we get that the number at the last position is 2 as it says that the number is well placed.
_ _ 2
In the third rule, if we eliminate 6, the remaining numbers are 2, 0. They are correct but wrongly placed. As 2 is fixed with the last position, 0 goes for the first position as per the rule.
0 _ 2
Now, from rule 2, we get that one number is correct among 1, 4 but is wrongly placed. If 1 is correct, it would be well placed and that contradicts the rule which says that the number is wrongly placed. So, the correct number that should be in the second position is 4
0 4 2
Therefore, the co-de of the given numerical lock is 042.
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