for any weight we have 3 possibilities:-
Place it on the weight panPlace it on other pan, along with something whose weight we need to findWe don't use itSo, it is similar to base 3, hence by taking weight of form 3^n, we can ensure that we are using least number of weights and when we just have two possibilities. so 3^n >=300 so n=6

@Vikrant-Garg
Probability of catching the bus = Probability of getting faulty machines in the first 2 tests (this part is clear right?)
This is possible in 2 ways - if we get DD or if we get GG as the result of first 2 tests.
In your doubt, if we get DG or GD we will require to do one more test, which would make the mechanic miss the bus. So DG and GD are not favourable cases.
So total number of ways = 4!/(2! x 2!) = 6 ways.
Favourable cases = 2
Probability = 2/6 = 1/3
Is it clear ?

x^2 - 6x + 25 = (x-3)^2 + 16
so min of sqrt (x^2-6x+25) = 4
y^2 - 8y + 25 = (y-4)^2 + 9
so min of sqrt (y^2-8y+25) will be 3
so min sum will be 7
for x=3 and y=4

Sol: Remember that: If square of a number ends in same unit digit as that of number, then that unit digit can be 0, 1, 5 or 6 only. So possible values for G are 0, 1, 5 or 6. Now by checking you can easily negate that G can't be 0 as GOG will not remain a three digit number. It can't be 1 either as in this case the three digit number will be 1O1 and it'll be perfect square only when O is 2 i.e. T is 1 which is not possible as T and G are "DISTINCT" single digit positive integers. Similarly you can check for 5 also that no number of the form 5O5 is a perfect square.
Only possible case is 26² = 676 i.e. O = 7.