Question

# One byte is a unit of memory size used with computers. One megabyte contains approximately $10^6$ bytes and Paul’s computer hard-drive accesses the information at a rate of 150 megabytes per second. If Paul wants to work with 5 files on his computer that are $2,340 \times 10^6$ bytes each, approximately how long will it take his computer to access them?A) 15.6 secondsB) 9 secondsC) 1.3 minutesD) 78 minutes

Verified
129k+ views
Hint:The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.
While using the unitary method, we have to find the amount needed for one unit of the substance and then we multiply by the quantity provided.
For example, if 3 boxes can be filled by 12 chocolates, 1 box will be filled by 4 chocolates and hence 5 jugs will be filled by 20 chocolates. It is simply done by calculating multiplication $5 \times 4$.

Complete step by step solution:
Step 1:
1 megabytes $= \mathop {{\text{10}}}\nolimits^{\text{6}} {\text{ }}$bytes
Step 2:
Also, $\mathop {{\text{10}}}\nolimits^{\text{6}} {\text{ }}$bytes = 1 megabytes
Hence, ${\text{1 byte=}}\dfrac{{\text{1}}}{{\mathop {{\text{10}}}\nolimits^{\text{6}} }}{\text{ megabytes}}$
$2340 \times 10^6 \text{bytes} = 2340 \times 10^6 \times \dfrac{1}{10^6} \text{megabytes}$
= 2,340 megabytes …… (1)
Step 3:
The rate at which the information is accessed on the computer = 150 megabytes per second
($\because$given)
$\Rightarrow$In one second, 150 megabytes of information is accessed on the computer.
Or, 150 megabytes of information is accessed in = 1 sec
1 megabyte of information is accessed in $= \dfrac{1}{{150}}$sec
2,340 megabytes of information is accessed in $= 2,340 \times \dfrac{1}{{150}}{\text{ sec}}$
${\text{ =2,34{0}}} \times \dfrac{1}{{15{0}}}{\text{ sec}} \\ {\text{ =}}\dfrac{{234}}{{15}}{\text{ sec}} \\$
${\text{ =15}}{\text{.6 sec}}$…… (2)
Step 4:
Size of one file$= 2,340 \times \mathop {10}\nolimits^6$bytes($\because$given)
= 2,340 megabytes(from (1))
Step 5:
Time to access one file on computer = 15.6 sec(from (2))
Time to access five files on computer $= 5 \times 15.6{\text{ sec}}$
= 78 sec
Step 6
1 minutes = 60 seconds
Or, 60 seconds = 1 minutes
${\text{1 seconds=}}\dfrac{{\text{1}}}{{{\text{60}}}}{\text{ minutes}}$
$\therefore {\text{78 seconds=}}\dfrac{{{\text{78}}}}{{{\text{60}}}}{\text{ minutes}}$
= 1.3 minutes
Step 7
Time to access five files on computer = 78 sec = 1.3 minutes.

$\therefore$ Time taken to access five files on the computer is 1.3 minutes. Hence, the correct answer is option (C).

$7$ minutes = $7 \times 60$ seconds