SSC GD 6th March 2019 Shift-1

Question 65

If $$\frac{a}{b}=\frac{2}{3}$$, then the value of $$(5a^3-2a^2b):(3ab^2-b^3)$$ is:

Solution

Let a = 2 and b = 3
Then, $$(5a^3-2a^2b):(3ab^2-b^3) = (5\times2^3 - 2\times2^2\times3) : (3\times2\times3^2 - 3^3)$$
$$= 5\times8 - 2\times4\times3 : 3\times2\times9 - 27$$
$$= 40-24 : 54-27 = 16 : 27$$

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