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Alex can download all his stored song from his hard drive to his ipod at speed of 30 000 kilobits per second (kbps)

 

Use the conversion 1GB aprox equals to 8.6x10^6kb to determine how long to the nearest min it will take Alex to download all his songs.

 

ANSWER -

time to download (seconds) = 4.24x8.6x10^6  / 300000

4= 1215.5  ---------> I don't get how this turns into 20 ?? SOME HELP!?!?

= 20 Mins

 Sep 9, 2014

Best Answer 

 #2
avatar+118608 
+5

I always use the units to help me do rates.

 

My method is quite unique I believe but when you get the hang of it it makes working out any rates related questions really easy - even really complicated looking ones.

 

I have explained this one in detail so it looks long but I normally would do it in 3 easy lines.  

 

I think it is worth puting the effort in and working out what I have done.     

 

$$\\\frac{30000kb}{1sec}\qquad \mbox{this can also be expressed as }\qquad \frac{1sec}{30000kb}\\\\
\mbox{Alex wants to download 4.24GB }\\
\mbox{(I have to assume this because it is not written into the question)}\\\\
\mbox{There are } \frac{8.6\times 10^6kb}{1GB}\quad \mbox {and Alex want to know seconds/download}\\\\\\
\mbox{Start with the ratio that has seconds on the top}\\\\
\frac{1sec}{30000kb}\\\\
\mbox{Now look for the one with kb on the top because we want to cancel those out}\\\\
\frac{1sec}{30000kb} \times \frac{8.6\times 10^6kb}{1GB}\\\\
\mbox{Now look for the one with GB on the top because we want to cancel those out}\\\\
\frac{1sec}{30000kb} \times \frac{8.6\times 10^6kb}{1GB}\times \frac{4.24GB}{1} \\\\
\mbox{So this will leave us with } \\\\$$

 

$$\\\frac{8.6\times10^6 \times 4.24 \;seconds}{30000} = 1215\;\; seconds\qquad \mbox{(All the other units cancelled out)}\\\\
\frac{1min}{60sec}\times \frac{1215 sec}{1} = \frac{1215 min}{60} = 20.25\;minutes\\\\
\mbox{Therefore download time is approximately 20 minutes}$$

 Sep 9, 2014
 #1
avatar+4473 
+5

AzizHusain Sep 9, 2014
 #2
avatar+118608 
+5
Best Answer

I always use the units to help me do rates.

 

My method is quite unique I believe but when you get the hang of it it makes working out any rates related questions really easy - even really complicated looking ones.

 

I have explained this one in detail so it looks long but I normally would do it in 3 easy lines.  

 

I think it is worth puting the effort in and working out what I have done.     

 

$$\\\frac{30000kb}{1sec}\qquad \mbox{this can also be expressed as }\qquad \frac{1sec}{30000kb}\\\\
\mbox{Alex wants to download 4.24GB }\\
\mbox{(I have to assume this because it is not written into the question)}\\\\
\mbox{There are } \frac{8.6\times 10^6kb}{1GB}\quad \mbox {and Alex want to know seconds/download}\\\\\\
\mbox{Start with the ratio that has seconds on the top}\\\\
\frac{1sec}{30000kb}\\\\
\mbox{Now look for the one with kb on the top because we want to cancel those out}\\\\
\frac{1sec}{30000kb} \times \frac{8.6\times 10^6kb}{1GB}\\\\
\mbox{Now look for the one with GB on the top because we want to cancel those out}\\\\
\frac{1sec}{30000kb} \times \frac{8.6\times 10^6kb}{1GB}\times \frac{4.24GB}{1} \\\\
\mbox{So this will leave us with } \\\\$$

 

$$\\\frac{8.6\times10^6 \times 4.24 \;seconds}{30000} = 1215\;\; seconds\qquad \mbox{(All the other units cancelled out)}\\\\
\frac{1min}{60sec}\times \frac{1215 sec}{1} = \frac{1215 min}{60} = 20.25\;minutes\\\\
\mbox{Therefore download time is approximately 20 minutes}$$

Melody Sep 9, 2014

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