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

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

Let length of train is is ‘L’ km and speed ‘S’ Km/hr
L/X-11 = 20/3600 = 1/180
L/X = 9/3600 = 1/400
X = 400 L
180 L = X – 11
180L = 400L-11
220L = 11
L = 11/220 = 1/20 KM = 1000/20 =50 meters

Each angle is 180(p-2)/p.
180-{360}/{p} = k
So 360/p has to be an integer.
360 = 2^3 * 3^2 * 5^1
So there are 4 * 3 * 2 = 24 possibilities, but we exclude 1 and 2, because p > = 3
So , 24 -2 = 22
Hence, choice (c) is the right answer

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