Question

Agent Hunt transferred classified files from the CIA mainframe onto his flash drive. The drive had some files on it before the transfer, and the transfer happened at a rate of 4.4 megabytes per second. After 32 seconds, there were 384 megabytes on the drive. The drive had a maximum capacity of 1000 megabytes.
How full was the drive when the transfer began?
How long from the time that Agent Hunt started the transfer did it take the drive to be completely full?

Answer 1

Answer:

When the transfer began, the flash drive had 243.2 megabytes on it.

It took the flash drive 172 seconds for it to be completely full.

Step-by-step explanation:

Okay so we know that the files were transferred at a rate of 4.4 megabytes per second and with that we can say that 4.4T megabytes were transferred in T seconds.

The size of the files on the drive is comprised of the files that were on the drive before the transfer began and the files that were transferred. We can express this with the equation S=A+4.4T in which S represents the size of the files on said drive at a given amount of time, A represents the size of the files before the transfer began, and T represents the time (in seconds).

We want to find A, so let's solve the equation for A first which is:

S=A+4.4T now turns into  A=S-4.4T  

So now we know that after 32 seconds (T=32), there were 384 megabytes on the drive (S=384). Now let's plug these values into the equation to find the value of A.

A= 384-4.4*32= 243.2

So the answer is when the transfer began, the drive had 243.2 megabytes on it.

But to find how long it took the drive to be completely full, we need to plug in S=1000 into the equation and solve for T.

243.2=1000-4.4T

4.4T=756.8

T=172

Hopefully this helps!