My first attempt at a lightning-fast calculation of this expression was as follows:
4.3 * 0.58 + 2.0E-5 - 0.9
Of the two numbers to multiply, $4.3$ is a few percent larger than $4$
while $0.58$ is a few percent smaller than $0.6$. Round $4.3$ to $4$
and $0.58$ to $0.6$ and hope that the errors cancel well enough within
the required precision of the problem. This produces
$$ 4 \times 0.6 + 2 \times 10^{-5} - 0.9 = 2.4 + 2 \times 10^{-5} - 0.9. $$
The term $2 \times 10^{-5}$ is so much smaller than the others that
I ignore it, so all that remains is
$$ 2.4 - 0.9 = 1.5. $$
But that's not accurate enough. My error was less than $6\%$ of the exact
result, but it was enough to change the second decimal digit after
rounding. The desired answer is $1.6$.
My mistake was in counting too heavily on the approximately
$3\%$ reduction from $0.6$ to $0.58$ to balance out the $7.5\%$ increase
from $4$ to $4.3$. If I had been more cautious and a little more practiced
at very fast estimation, I might have tried to actually estimate
how much error each rounding produced and how much of the errors canceled.
By increasing $0.58$ to $0.6$ I introduced an error of
$$ 0.02 \times 4.3 \approx 0.02 \times 4 \approx 0.1.$$
By decreasing $4.3$ to $4$ I introduced an error of
$$ -0.3 \times 0.58 \approx -0.3 \times 0.6 \approx -0.2.$$
So altogether the error was approximately $0.1 - 0.2 = -0.1$,
and to compensate for that I should have added $0.1$ to my estimate
of the product, so $4.3 \times 0.58 \approx 2.5$.
This takes much longer to explain than to do. Under time pressure I might
even look at the numbers, notice that I rounded one number down by a larger
percentage than I rounded the other up, and make a guess that this would be enough to cause an error in the second digit of my product, so the true product would be closer to $2.5$ than to $2.4$.
With practice, you may be able to recognize these patterns quickly enough.
Decades ago, when my mind was a little more nimble in this way
and gasoline pumps had mechanical (not electronic) displays showing only
the price per gallon and the number of gallons dispensed,
I developed a practice of predicting the cost of a tank of gasoline
in the few seconds between when the pump stopped and when the
attendant said how much money was due. This was usually a
product of a three-digit number (always ending in a $9$)
with a four-digit number,
with four digits of precision required in the output
(since the payment was rounded to the nearest penny).
The trick was to make an approximation based on round numbers,
then make successive corrections by estimating the round-off errors
until the remaining corrections were less than the required precision.
So at least some people should be able to do these tricks.
(To be honest, I only remember doing this trick when my father was
driving the car, and the main purpose may not have been to avoid being
cheated by the gas station, but rather to annoy my father.)
upto 2 digitsis notexact. For example, calculate4.3 * 0.6 = 2.58. Then consider that0.58is about 3% lower than0.6and 3% of 2.58 is around 0.08, so adjust2.58 - 0.08 = 2.5. Then ignore2.0E-5altogether since it's way below the required precision. Finally subtract2.5 - 0.9 = 1.6which matches the exact result1.59402to two significant digits. You can get more precision if you refine some of the steps, of course, but basic idea is know how to count what matters, and how to ignore the rest. – dxiv Jan 09 '16 at 04:48