Question

A computer has 3 ports to transfer data into flash-drives. The rate of data transfer varies as

the simultaneous transfers change as follows:

Single transfer means data transfer @ 3 MB per second

Two simultaneous transfers mean data transfer@ 2 MB per second for each transfer

Three simultaneous transfers mean data transfer @ 1.5 MB per second for each transfer

Rohit has to transfer 577 MB data into three flash-drives of data-holding capacity of 90 MB,

200MB and 300MB respectively. What is the shortest possible time to perform this transfer?

a.144
b.128
c.96
d.64

Answer 1

Since data is being transferred to three flash drives we will consider the rate of three simultaneous transfer.

In one second 1.5 MB is transferred to each of the three flash drives.

In one second the amount of data transferred equals :

3 × 1.5 = 4.5MB

In one second 4.5MB is transfered.

In how many seconds will 577MB be transferred?

In this case we use the concept of cross multiplication.

4.5 MB = 1 second

577MB =?

By cross multiplication :

(577 / 4.5) × 1 = 128.222

Which is approximately 128 rounded off to the nearest whole number.

Answer 2

Answer:

Step-by-step explanation:

the right answer is 144 seconds

in the first 60 seconds , each drive will fill up by 90 mb ( at 1.5 mb speed)

this means the one with the shortest capacity gets filled up.

in the next 55 seconds , the remaining two drives fill up by 110 mb ( at 2 mb per second speed )

this means that the drive with 200mb capacity gets filled up completely

now 577 - (90 *3 + 110 *2) = 87 mb of data is remaining to be transferred

and this gets transferred to the remaining one drive at 3 mb per second speed in 29 seconds .

so the shortest time needed is 60 + 55 + 29 = 144 seconds

hope this helps !