Question 4

If $$A = \frac{0.216 + 0.008}{0.36 + 0.04 - 0.12}$$ and $$B = \frac{0.729 - 0.027}{0.81 + 0.09 + 0.27}$$, then what is the value of $$(A^2 + B^2)^2?$$

Solution

$$A = \frac{0.216 + 0.008}{0.36 + 0.04 - 0.12}$$

$$A = \frac{0.224}{0.28}$$

=0.8
$$B = \frac{0.729 - 0.027}{0.81 + 0.09 + 0.27}$$

$$B=\frac{0.702}{1.17}$$

=0.6
$$(A^2 + B^2)^2$$=0.36+0.64=1

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SSC CGL Tier-2 18-February-2018 Maths

If $$A = \frac{0.216 + 0.008}{0.36 + 0.04 - 0.12}$$ and $$B = \frac{0.729 - 0.027}{0.81 + 0.09 + 0.27}$$, then what is the value of $$(A^2 + B^2)^2?$$

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