A car has 4 tyres and 1 spare tyre. Each tyre can travel a maximum distance of 20000 miles before wearing off. What is the maximum distance the car can travel before you are forced to buy a new tyre? You are allowed to change tyres (using the spare tyre) unlimited number of times.

Answer: 25000

Divide the lifetime of spare tire into 4 equal part i.e., 5000 and swap it at each completion of 5000 miles distance.

Let four tyres be A, B, C and D and spare tyre be S.

5000 KMs: Replace A with S. Remaining distances (A, B, C, D, S) : 15000, 15000, 15000, 15000, 20000
10000 KMs: Put A back to its original position and replace B with S. Remaining distances (A, B, C, D, S) : 15000, 10000, 10000, 10000, 15000
15000 KMs: Put B back to its original position and replace C with S. Remaining distances (A, B, C, D, S) : 10000, 10000, 5000, 5000, 10000
20000 KMs: Put C back to its original position and replace D with S. Remaining distances (A, B, C, D, S) : 5000, 5000, 5000, 0, 5000
25000 KMs : Every tyre is now worn out completely.

All tyres are used upto their full strength.

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Related Amazon interview question

There are n pencils, each having l length. Each can write 4 kilometers. After writing 4 kilometers it has l/4 length. One can join 4 pencils which are having l/4 length and can make 1 pencil. One can’t make pencil of pieces if remaining pieces are 3 or 2 or 1 in number but one can include these remaining pieces whenever needed.
Write a relation independent of l, length of given pencil, for how much one can write from n pencils.

Examples:

Input : 4
Output : 20

Approach:
Suppose We use 3 pencils that will 12 and generate 3 used pencils, now if the remaining pencils are greater than zero at least 1 unused pencil can be used with those 3 unused to write 4 and that will generate 1 more unused pencil. This will keep repeating.

if(n-3 >= 1){
 f(n) = f(n-3) + 12 + 4
} else{
  // Used pencil that cannot be used
 f(n) = 4 + 4 
}

Above relation can be optimized in O(1)
// x is max no of time we can subtract 3 without n-3 <= 3 

n - 3 + x <= 3     
x  >  (n-3)/3
i.e. n/3 - 1 if it divides exactly else n/3

24

Recursively it can be written as:

int count(int n){
  if(n<4){
    return (n*4);
   } else{
     return (16+count(n-3));
   } 
}