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

let total work=60 unit
3 M + 5 W per day work=20 unit
let M work done in one day=a
then work work done=(20-3a)/5
2 M work done in one day=2a
and 3 work done in one day=((60-9a)/5
now 60/2a = ((60/((60-9a)/5 -5
so 30/a +5 = 300/(60-9a)
find a then ratio

@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 ?

a^2 + ab + b^2 = n^2(a+b)^2 = n^2 + ab(a+b+n)(a+b-n) = ab
as 'a' and 'b' are prime numbersab has 4 factors: 1, a, b, abobviously, a + b + n > a + b - n
Case 1:a + b + n = aba + b - n = 1
Case 2:a + b + n = aa + b - n = b ... considering a>b
But, case 2 is not possible.b = -na = n,a = -b ... which is impossible
Case 1:adding the two equations2a + 2b = ab + 12a - ab + 2b = 1a(2-b) - 2(2-b) = -3(a-2)(b-2) = 3
only possibility for 'a' and 'b' being prime and satisfying the above condition is: 5 and 3
giving us he only possibility 49

Q30) There are 10 consecutive positive integers written on a blackboard. One number is erased. The sum of remaining nine integers is 2011. Which number was erased?