Quentin's Quiz

Let n be the number of points.

As number of points decreases,The change in percentage per point increases.

When percentage<51.5 it can be rounded to 51.

Assume n is odd.The 50% of mark will be at $\lfloor \frac{n}{2} \rfloor + 0.5$ ,where $\lfloor \frac{ n}{2} \rfloor$ is the integer part of the value of $\frac{n}{2}$ .

If n is the fewer number of questions,then $\lfloor \frac{n}{2}\rfloor+1$ must be less than 51.5%

That is 0.5<1.5% of total mark

$\frac{0.5\times100}{n}<1.5$

$\frac{50}{n}<1.5$

$\frac{50}{1.5}<n$

33.33<n

Therefore $\boxed{n=35}$ as n is odd.

Similarly we can calculate the value of n when n is even(68 which is greater than 35)

505/1000 < x/n < 515/1000 (where x is the number of points he got out of n points on the quiz).

101n < 200x < 103n (where both x and n must be integers).

The first case where 101n and 103n fall into different hundreds is where n=34, so just count up from there until they are separated by a hundred divisible by 200 (ex. 3400, 3600, 3800, 4000, etc).

You will find that at n=35...

3535 < 200x < 3605, and x=18.

So, the fewest number of points the quiz could be worth is 35.

Let Quentin scored x marks out of a total worth of n marks where n and x are positive integers with 0 <= x <= n. We have to find the smallest possible value for n such that 50.5 % < x/n < 51.5 % ==> 99/200 < x/n < 103/200.

99/200 < x/n < 103/200 => 200/99 > n/x > 200/103 => 2 + (2/99) > n/x > 1 + (97/103) Hence, either n = x + r or n = 2*x + r with 0 < r < x.

Case # (1): 2 > n/x > 1 + (97/103) n = x + r with 0 < r < x

1 > r/x > 97/103 => 1 < x/r < 103/97 => r < x < (103/97)*r Putting r = 1, 2, 3, ..., we find the lowest integral value for x that can satisfy the above inequality is when r = 17 making 17 < x < 18.05155 i.e. x = 18. n = x + r = 18 + 17 = 35.

In this case, Quentin scored 18/35 = 51.428571 %

Case # (2): n/x = 2 will not be considered as percentage scored in this case is always 1/2 = 50 %

Case # (3): 2 + (2/99) > n/x > 2 n = 2*x + r with 0 < r < x

2/99 > r/x > 0 => 99/2 < x/r => x > (99/2) r Putting r = 1, we get x > 49.5 i.e. x = 50. n = 2 x + r = 2*50 + 1 = 101.

In this case, Quentin scored 50/101 = 49.5049 %

Therefore, the fewest number of total points of the quiz = 35.

I used python to calculate it for me:

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for i in range (1,100):
   for j in range (1,i):
    j=float(j)
    if j/i < 0.55 and j/i > 0.45: print `i` + " " + `j` + " " + `j/i` 

(Code fixed by mod)

Let's try some numbers. We will start by trying 2/3, which is 67%. We move up by odd numbers, because even numbers will always be divisible by 2, therefore if we want 51% it will be 50% (half of the total number of questions) plus $x^{-1} \times 100$ , the value of one more question. We try 3/5, which is 60%. Obviously, it's going to be a bit larger than that, so let's try 21. 12/21 is 57%, still not good enough. Going to 17/31, we find 55%. Plugging in values, we find that 18/35 is approximately 51%.