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Problem 2
Solution (The typical method)$$ \frac{44}{11}+\frac{110}{44}+\frac{44}{1100} $$
We will simplify and get-
$$\frac {4} {1}+ \frac {10} {4}+ \frac {4} {100}$$
$$ = 4+ 2.5+0.04= 6.54 $$(C)
Problem 3
To ease the solution, I have given names to the points
Area of Grey Region= Area of HEFGJI+ Area of AKLMCB
Area of HEFGJI= \( 7^2 \) - \( 4^2 \)
= 49-16= 33
Area of AKLMCB
= \( 10^2 \) - \( 9^2 \)
= 100-81= 19
∴ 33+19=52(E)
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