For the following questions answer them individually
For any $$a, b \epsilon R$$, define
a * b = max {a, b} and aob = min {a, b}. Then a * (boc) =
$$1 + \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + ...... $$ to infinity =
The greatest number which divides 121, 134 and 147 leaving 4 as the remainder in each case is
The correct order of $$a = \frac{3}{4}, b = \frac{4}{7}, c = \frac{11}{13}$$ and $$d = \frac{13}{15}$$ is
$$2\frac{1}{3} + 3\frac{3}{4} \div \frac{3}{4} - 2\frac{1}{3}\left(6\frac{1}{2} - 2\frac{1}{4}\right) \div 2\frac{1}{4} =$$
Two numbers x and y are respectively 20% and 25% more than the third number z. If x is a% of y, then a =