The product $(1.8)(40.3+.07)$ is closest to 
$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$
$(1.8)(40.37)\approx (1.8)(40)=72.$


Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$).
$74$ is the closest to $72$, so the answer is $\boxed{\text{C}}$.