16 MB of data is sent to the computer over a 32 Mbit / s data line. After receiving 4 Mbytes, the computer

16 MB of data is sent to the computer over a 32 Mbit / s data line. After receiving 4 Mbytes, the computer starts simultaneously transmitting this data over another communication line at a speed of 4 Mbit / s. How many seconds will elapse from the start of receiving data over a high-speed channel until it is fully transmitted over a low-speed channel?

Let’s convert 16 MB to Mbit.

Considering that 1 byte = 8 bits.

16 MB * 8 = 128 MBit.

4 MB * 8 = 32 MBit.

32 Mbps: 32 Mbps = 1 sec.

This means that 4 Mbytes are transferred over a line with a data transfer rate of 32 Mbit / s in 1 second.

128 Mbps: 4 Mbps = 32 sec.

This means that 16 Mbytes on a line with a data transfer rate of 4 Mbit / s are transferred in 32 seconds.

From here we get: 1 sec. + 32 sec. = 33 sec.

Answer: 33 seconds will pass from the start of data reception on the high-speed channel until the complete transmission of them on the low-speed channel.