From the given table find the values of a and b:

x	y
12	5
4	a
10	6
30	b

1. a = 8 and b = 25

2. a = 15 and b = 12

3. a = 12 and b = 90

4. a = 15 and b = 2

$∵ x$1y1 = 12 \times 5 = 60\\[1em] x3y3 = 10 \times 6 = 60\\[1em]

So, we can say that this is inverse variation.

$\Rightarrow a$1b1 = a2b2 \quad [\because \text{a varies inversely to b}] \\[1em] \Rightarrow 12 \times 5 = 4 \times a\\[1em] \Rightarrow 60 = 4 \times a\\[1em] \Rightarrow a = \dfrac{60}{4}\\[1em] \Rightarrow a = 15

And, similarly

$a$3b3 = a3b3 \quad [\because \text{a varies inversely to b}] \\[1em] \Rightarrow 10 \times 6 = 30 \times b\\[1em] \Rightarrow 60 = 30 \times b\\[1em] \Rightarrow b = \dfrac{60}{30}\\[1em] \Rightarrow b = 2

Hence, option 4 is the correct option.