You can do it without a calculator, by approximating.
I know 56,700bps to be 7KB/s (tops), see previous pages
Given 9600bps, that is rougly 1/6th of 56K
therefore the bandwidth at any given time is given by ~1 to 1.1KB/s
In a 4KB file, it would not exceed 4 seconds, transferring at 1KB/s (at a minimum).
The possible answers were:
a) 3
b) 0.4
c) 6
d) 4.3
A was the correct answer.
You could also do it mathematically (as you've quoted), but again, it will be an approximation and making use of guesswork.
You could also do it such that:
9600 bps remains constant (leaving it in bits per second)
4KB approximates to 4 x 1KB
1KB approximates to 1000 bytes (or 1024 for the pedantic)
therefore it is 4000 bytes,
multiply 4000 by 8 (8 bits per byte)
gives 32000 (you can do this using multiplication, no calculator)
now, given that 4KB = 4000 bytes or 32000 bits
knowing that the transfer rate is 9600 bits per second
32000 / 96000
crossing out the three zero's
32 / 9
= 3.5 seconds (as 3*9 = 27, remainder of 5/9th's of a second)
3.5 is closer to 3, than it is to 4.3
Leaving 3 seconds (A) to be the answer, discounting 0.4 seconds (B), 6 seconds (C), and 4.3 seconds (D)
Hope that helps you
